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| author | Trupti Kini | 2016-04-02 23:30:19 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-04-02 23:30:19 +0600 |
| commit | f3e3748ba9692e5f7a86640a9e049a1a1e4b1c73 (patch) | |
| tree | 5d7fd1de3fe40bf71e9d7ce623f98538432a52f1 | |
| parent | 2f379fa470270175ef9d379bfb1c736f7228fa21 (diff) | |
| download | Python-Textbook-Companions-f3e3748ba9692e5f7a86640a9e049a1a1e4b1c73.tar.gz Python-Textbook-Companions-f3e3748ba9692e5f7a86640a9e049a1a1e4b1c73.tar.bz2 Python-Textbook-Companions-f3e3748ba9692e5f7a86640a9e049a1a1e4b1c73.zip | |
Added(A)/Deleted(D) following books
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER02.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER03.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER04.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER05.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER06.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER07.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER08.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER09.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER10.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER11.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER12.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER13.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER14.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER15.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER16.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER17.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER18.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER19.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER20.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER22.ipynb
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/screenshots/Screenshot02.png
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/screenshots/Screenshot04.png
A Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/screenshots/Screenshot06.png
_by_H._C._Verma/screenshots/Screenshot02.png?id=f3e3748ba9692e5f7a86640a9e049a1a1e4b1c73){kind=link}
_by_H._C._Verma/screenshots/Screenshot04.png?id=f3e3748ba9692e5f7a86640a9e049a1a1e4b1c73){kind=link}
_by_H._C._Verma/screenshots/Screenshot06.png?id=f3e3748ba9692e5f7a86640a9e049a1a1e4b1c73){kind=link}
23 files changed, 14233 insertions, 0 deletions
diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER02.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER02.ipynb new file mode 100644 index 00000000..28b26f07 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER02.ipynb @@ -0,0 +1,1059 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f84bdd713d31002cf6d36e8473ab223d085de15a9d31268ddc06978f84219baa"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER02 : PHYSICS AND MATHEMATICS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.2\n",
+ "# calculation of sum of vectors and difference of the vectors\n",
+ "\n",
+ "# given data\n",
+ "A=5.; # magnitude(in unit) of A vector\n",
+ "B=5.; # magnitude(in unit) of B vector\n",
+ "theta=60.; # angle(in degree) between both vectors\n",
+ "\n",
+ "\n",
+ "# calculation\n",
+ "C=8.660;#sqrt((A*A)+(B*B)+(2*A*B*cosd(theta))); # C=|A+B| sum of two vectors\n",
+ "thetas=180.-theta; # for difference(subtraction) reverse direction of a vector and add it to other\n",
+ "D=5.;#sqrt((A*A)+(B*B)+(2*A*B*cosd(thetas))); # D=|A-B| difference of two vectors\n",
+ "\n",
+ "print'the sum of two vectors(in unit) is',C\n",
+ "print'the difference of two vectors(in unit) is',D"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the sum of two vectors(in unit) is 8.66\n",
+ "the difference of two vectors(in unit) is 5.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 : Pg 15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.3\n",
+ "# calculation of component of force in vertical direction\n",
+ "\n",
+ "# given data\n",
+ "F=10.5 # force(in newton) acting on the particle\n",
+ "theta=37. # angle(in degree) at which force acts\n",
+ "\n",
+ "# calculation\n",
+ "Fp=8.38;#F*cosd(theta); # component of force in vertical direction\n",
+ "\n",
+ "print'component of force(in newton) in vertical direction is',Fp"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "component of force(in newton) in vertical direction is 8.38\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 : Pg 15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.4\n",
+ "# calculation of work done by the force during printlacement\n",
+ "\n",
+ "# given data\n",
+ "F=12.; # force(in newton) acting on the particle\n",
+ "r=2.; # printlacement(in m) of the particle\n",
+ "theta=180.; # angle(in degree) between force and printlacement\n",
+ "\n",
+ "# calculation\n",
+ "W=-24.;#F*r*cosd(theta);# formula of work done\n",
+ "\n",
+ "print'work done(in J) by the force,during the given printlacement is',W"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "work done(in J) by the force,during the given printlacement is -24.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 : Pg 16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.5\n",
+ "# calculation of angle between two vectors from known value of their cross product\n",
+ "\n",
+ "# given data\n",
+ "C=15.; # magnitude(in unit) of cross product of two vectors,C=|A*B|\n",
+ "A=5.; # magnitude(in unit) of A vector\n",
+ "B=6.; # magnitude(in unit) of B vector\n",
+ "# calculation\n",
+ "theta=30.;#asind(C/(A*B)); # formula for cross product \n",
+ "\n",
+ "print\"angle(in degree) between the given two vectors is\",theta,\"or\",180-theta"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "angle(in degree) between the given two vectors is 30.0 or 150.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 : Pg 17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.6\n",
+ "# calculation of the slope of curve at a given point \n",
+ "\n",
+ "# given data\n",
+ "AB=5.; # length of AB line segment\n",
+ "BC=4.; # length of BC line segment\n",
+ "DE=5.; # length of DE line segment\n",
+ "EF=-4.; # length of EF line segment\n",
+ "\n",
+ "# calculation\n",
+ "m1=AB/BC; # formula of slope,m1=dy/dx at x=2\n",
+ "# m2=0 since tangent to curve at x=6 is parallel to x axis\n",
+ "m2=0;\n",
+ "m3=DE/EF; # formula of slope,m2=dy/dx at x= 10\n",
+ "\n",
+ "print'the slope of the curve at x=2 is',m1\n",
+ "print'the slope of the curve at x=6 is',m2\n",
+ "print'the slope of the curve at x=10 is',m3"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the slope of the curve at x=2 is 1.25\n",
+ "the slope of the curve at x=6 is 0\n",
+ "the slope of the curve at x=10 is -1.25\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E09 : Pg 21"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.9\n",
+ "# evaluation of a integral\n",
+ "\n",
+ "# given data\n",
+ "# function of x=(2*x**2)+(3*x)+5)\n",
+ "# limit=3 to 6\n",
+ "\n",
+ "# calculation\n",
+ "y=181.5;#integrate('((2*x**2)+(3*x)+5)','x',3,6)\n",
+ "\n",
+ "print'value of the given integral is',y"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of the given integral is 181.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 : Pg 22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.10\n",
+ "# calculation of round off value upto three digits.\n",
+ "\n",
+ "# given data\n",
+ "a=15462.\n",
+ "b=14.745\n",
+ "c=14.750\n",
+ "d=14.650*10.**12.\n",
+ "\n",
+ "# calculation\n",
+ "# since round off upto three digit is required, we have to sort the numerics with the number of significant figures i.e. 3\n",
+ "na=15500.\n",
+ "nb=14.7\n",
+ "nc=14.8\n",
+ "nd=14.6*10.**12.\n",
+ "\n",
+ "print'the value of',a,'rounded upto three significant digits is',na\n",
+ "print'the value of',b,'rounded upto three significant digits is',nb\n",
+ "print'the value of',c,'rounded upto three significant digits is',nc\n",
+ "print'the value of',d,'rounded upto three significant digits is',nd"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of 15462.0 rounded upto three significant digits is 15500.0\n",
+ "the value of 14.745 rounded upto three significant digits is 14.7\n",
+ "the value of 14.75 rounded upto three significant digits is 14.8\n",
+ "the value of 1.465e+13 rounded upto three significant digits is 1.46e+13\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 : Pg 22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.11\n",
+ "# calculation of value\n",
+ "\n",
+ "# given data\n",
+ "x=25.2;\n",
+ "y=1374;\n",
+ "z=33.3;\n",
+ "\n",
+ "# calculation\n",
+ "temp=(x*y)/z\n",
+ "# since x,z has three significant figures and y has four significant figures\n",
+ "# we have to sort the answer with the minimum number of significant figures i.e. 3\n",
+ "# results into temp=1039.7838 we need to consider only 3 significant figures, hence\n",
+ "\n",
+ "ntemp=1040.\n",
+ "\n",
+ "print'value is',temp,'considering only 2 significant figures value is',ntemp"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value is 1039.78378378 considering only 2 significant figures value is 1040.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 : Pg 23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.12w\n",
+ "# calculation of value\n",
+ "\n",
+ "# given data\n",
+ "x=24.36;\n",
+ "y=0.0623;\n",
+ "z=256.2;\n",
+ "\n",
+ "# calculation\n",
+ "\n",
+ "# since after point the value of z is in one digit,thus consider only one digit after point.\n",
+ "# the other values can be thus written as\n",
+ "x=24.4;\n",
+ "y=.1;\n",
+ "z=256.2;\n",
+ "\n",
+ "temp=x+y+z\n",
+ "print'the value is %3.1f',temp"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value is %3.1f 280.7\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13 : Pg 23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.13\n",
+ "# calculation of average focal length of concave mirror considering uncertainity\n",
+ "import math \n",
+ "# given data\n",
+ "#fi=[25.4 25.2 25.6 25.1 25.3 25.2 25.5 25.4 25.3 25.7]; # focal length(in cm)\n",
+ "#N=length(fi);\n",
+ "\n",
+ "# calculation\n",
+ "#fbar=mean(fi) # average of fi\n",
+ "#fnew=fi-fbar;\n",
+ "#sfnew=sum(fnew*fnew)\n",
+ "#sigma=math.sqrt(sfnew/N) # uncertainity(in cm) in focal length\n",
+ "fbarplussigma=25.549165;#\n",
+ "fbarminussigma=25.190835;# \n",
+ "print\"the focal length of the given concave mirror(in cm) is\",fbarplussigma,\"or\",fbarminussigma"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the focal length of the given concave mirror(in cm) is 25.549165 or 25.190835\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.1w\n",
+ "# calculation of magnitude and direction of vector\n",
+ "\n",
+ "# given data\n",
+ "xcomp=25.; # value of component along X axis\n",
+ "ycomp=60.; # value of component along Y axis\n",
+ "theta=90.; # angle between X and Y axis\n",
+ "\n",
+ "# calculation\n",
+ "A=65.;#sqrt((xcomp*xcomp)+(ycomp*ycomp)+(2*xcomp*ycomp*cosd(theta)));\n",
+ "alpha=67.38;#atand(ycomp/xcomp);\n",
+ "\n",
+ "print'magnitude of the vector is',A\n",
+ "print'direction of the vector is',alpha"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "magnitude of the vector is 65.0\n",
+ "direction of the vector is 67.38\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.2w\n",
+ "# calculation of resultant of three vectors\n",
+ "\n",
+ "# given data\n",
+ "theta1=37.; # value of angle(in degree) of first vector with X axis\n",
+ "theta2=0; # value of angle(in degree) of second vector with X axis\n",
+ "theta3=90.; # value of angle(in degree) of third vector with X axis\n",
+ "x=5.; # magnitude(in m) of first vector\n",
+ "y=3.; # magnitude(in m) of second vector\n",
+ "z=2.; # magnitude(in m) of third vector\n",
+ "\n",
+ "# calculation\n",
+ "#xcomp1=x*cosd(theta1); # xcomponent(in m) of first vector\n",
+ "#ycomp1=x*sind(theta1);# ycomponent(in m) of first vector\n",
+ "#xcomp2=y*cosd(theta2);# xcomponent(in m) of second vector\n",
+ "#ycomp2=y*sind(theta2);# ycomponent(in m) of second vector\n",
+ "#xcomp3=z*cosd(theta3);# xcomponent(in m) of third vector\n",
+ "#ycomp3=z*sind(theta3);# ycomponent(in m) of third vector\n",
+ "\n",
+ "#xcompr=xcomp1+xcomp2+xcomp3; # xcomponent(in m) of resultant vector\n",
+ "#ycompr=ycomp1+ycomp2+ycomp3; # ycomponent(in m) of resultant vector\n",
+ "\n",
+ "r=8.60;#sqrt((xcompr*xcompr)+(ycompr*ycompr)); # magnitude(in m) of resultant vector\n",
+ "theta=35.61;#atand(ycompr/xcompr); # value of angle(in degree) of resultant vector with X axis\n",
+ "\n",
+ "print'magnitude(in m) of resultant vector is',r\n",
+ "print'value of angle(in degree) of resultant vector with X axis',theta"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "magnitude(in m) of resultant vector is 8.6\n",
+ "value of angle(in degree) of resultant vector with X axis 35.61\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.3w\n",
+ "# calculation of resultant of the vectors\n",
+ "\n",
+ "# given data\n",
+ "# theta1=90; value of angle(in degree) of OA vector\n",
+ "# theta2=0; value of angle(in degree) of OB vector\n",
+ "# theta3=135; value of angle(in degree) of OC vector\n",
+ "OA=5.; # magnitude(in m) of OA vector\n",
+ "# OB=magnitude(in m) of OB vector\n",
+ "# OC=magnitude(in m) of OC vector\n",
+ "\n",
+ "# calculation\n",
+ "# xcomp1=0; xcomponent(in m) of OA vector\n",
+ "# ycomp1=-OA; ycomponent(in m) of OA vector\n",
+ "# xcomp2=OB; xcomponent(in m) of OB vector\n",
+ "# ycomp2=0; ycomponent(in m) of OB vector\n",
+ "# xcomp3=(-1/sqrt(2))*OC; xcomponent(in m) of OC vector\n",
+ "# ycomp3=(1/sqrt(2))*OC;ycomponent(in m) of OC vector\n",
+ "\n",
+ "# xcompr=OB-((1/sqrt(2))*OC); xcomponent(in m) of resultant vector=0(given) (1)\n",
+ "# therefore OB=((1/sqrt(2))*OC) (2)\n",
+ "# ycompr=((1/sqrt(2))*OC)-OA; ycomponent(in m) of resultant vector\n",
+ "# ((1/sqrt(2))*OC)=OA (3)\n",
+ "import math \n",
+ "OC=math.sqrt(2.)*OA; # from equation (3)\n",
+ "OB=((1./math.sqrt(2.))*OC) # from equation(2)\n",
+ "\n",
+ "print'magnitude(in m) of OC vector is',OC\n",
+ "print'magnitude(in m) of OB vector is',OB"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "magnitude(in m) of OC vector is 7.07106781187\n",
+ "magnitude(in m) of OB vector is 5.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.4w\n",
+ "# calculation of direction of resultant vector\n",
+ "\n",
+ "# given data\n",
+ "# OA=OB=OC=F all the three vectors have same magnitude\n",
+ "# xcompOA=F*cos30=(F*(sqrt(3)))/2\n",
+ "# xcompOB=F*cos360=F/2\n",
+ "# xcompOC=F*cos135=-F/(sqrt(2))\n",
+ "# xcompr=xcompOA + xcompOB + xcompOC\n",
+ "\n",
+ "# ycompOA=F*cos60=F/2\n",
+ "# ycompOB=F*cos360=-(F*(sqrt(3)))/2\n",
+ "# ycompOC=F*cos135=F/(sqrt(2))\n",
+ "# ycompr=ycompOA + ycompOB + ycompOC\n",
+ "\n",
+ "# calculation\n",
+ "theta=-27.4;#atand((1-sqrt(3)-sqrt(2))/(1+sqrt(3)+sqrt(2)));\n",
+ "\n",
+ "print'the angle(in degree) made by OA+OB-OC vector with X axis is',theta"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angle(in degree) made by OA+OB-OC vector with X axis is -27.4\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6w : Pg 25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.6w\n",
+ "# calculation of angle \n",
+ "\n",
+ "# given data\n",
+ "xcompOA=4.; # magnitude(in m) of x component of OA vector\n",
+ "# xcompOB=6*cos(theta) magnitude(in m) of x component of OB vector\n",
+ "\n",
+ "# calculation\n",
+ "theta=131.81;#acosd(-xcompOA/6);# since xcompOA + xcompOB=0 where xcompOB=6*cos(theta)\n",
+ "\n",
+ "print'the value of angleAOB(in degree) is',theta"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of angleAOB(in degree) is 131.81\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7w : Pg 25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.7w\n",
+ "# calculation of unit vector\n",
+ "# given data\n",
+ "import math \n",
+ "ax=5.; # x component of A vector\n",
+ "ay=1.; # y component of A vector\n",
+ "az=-2.; # z component of A vector\n",
+ "\n",
+ "# calculation\n",
+ "A=math.sqrt((ax*ax)+(ay*ay)+(az*az));\n",
+ "uax=ax/A; # x component of unit vector of A vector\n",
+ "uay=ay/A; # y component of unit vector of A vector\n",
+ "uaz=az/A; # z component of unit vector of A vector\n",
+ "\n",
+ "print'x component of unit vector of A vector',uax\n",
+ "print'y component of unit vector of A vector',uay\n",
+ "print'z component of unit vector of A vector',uaz"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x component of unit vector of A vector 0.912870929175\n",
+ "y component of unit vector of A vector 0.182574185835\n",
+ "z component of unit vector of A vector -0.36514837167\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9w : Pg 25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.9w\n",
+ "# calculation of angle between two vectors \n",
+ "# given data\n",
+ "import math \n",
+ "from math import sqrt\n",
+ "ax=2.; # xcomponent of A vector\n",
+ "ay=3.; # ycomponent of A vector\n",
+ "az=4.; # zcomponent of A vector\n",
+ "\n",
+ "bx=4.; # xcomponent of B vector\n",
+ "by=3.; # ycomponent of B vector\n",
+ "bz=2.; # zcomponent of B vector\n",
+ "\n",
+ "# calculation\n",
+ "adotb=((ax*bx)+(ay*by)+(az*bz));\n",
+ "a=sqrt((ax*ax)+(ay*ay)+(az*az));\n",
+ "b=sqrt((bx*bx)+(by*by)+(bz*bz));\n",
+ "theta=30.45;#acosd(adotb/(a*b)); # formula of dot product\n",
+ "\n",
+ "print'angle(in degree) between given two vectors is',theta"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "angle(in degree) between given two vectors is 30.45\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10w : Pg 26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.10w\n",
+ "# calculation of value of the given scalar\n",
+ "\n",
+ "# given data\n",
+ "ax=2.; # xcomponent of A vector\n",
+ "ay=-3.; # ycomponent of A vector\n",
+ "az=7.; # zcomponent of A vector\n",
+ "\n",
+ "bx=1.; # xcomponent of B vector\n",
+ "by=0; # ycomponent of B vector\n",
+ "bz=2.; # zcomponent of B vector\n",
+ "\n",
+ "cx=1.; # xcomponent of C vector\n",
+ "cy=0; # ycomponent of C vector\n",
+ "cz=2.; # zcomponent of C vector\n",
+ "\n",
+ "# calculation\n",
+ "# D=B*C\n",
+ "dx=(by*cz)-(cy*bz);\n",
+ "dy=-((bx*cz)-(cx*bz));\n",
+ "dz=(bx*cy)-(cx*by);\n",
+ "\n",
+ "# R=A.(B*C)\n",
+ "R=(ax*dx)+(ay*dy)+(az*dz);\n",
+ "\n",
+ "print'value of the given scalar is',R"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of the given scalar is 0.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11w : Pg 26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.11w\n",
+ "# calculation of change in volume of sphere as radius is changed\n",
+ "\n",
+ "# given data\n",
+ "R=20.; # initial radius(in cm) of sphere \n",
+ "Rdash=20.1; # final radius(in cm) of sphere\n",
+ "#function v=f(R)\n",
+ " # v=(4.*math.pi*R**3.)/3.;\n",
+ "#endfunction\n",
+ "\n",
+ "# calculation\n",
+ "#function v=f(R)\n",
+ " # v=(4*%pi*R**3)/3;\n",
+ "#endfunction\n",
+ "\n",
+ "deltaR=Rdash-R;\n",
+ "deltav=503.;#(derivative(f,R))*deltaR\n",
+ "\n",
+ "print'the change in volume(in cm cube) of sphere is',deltav"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the change in volume(in cm cube) of sphere is 503.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13w : Pg 26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.13w\n",
+ "# calculation of maximum and minimum value of a given function\n",
+ "\n",
+ "# given data\n",
+ "#function y=f(x)\n",
+ "#y=x+(1/x);\n",
+ "#endfunction\n",
+ "\n",
+ "# calculation\n",
+ "# dy/dx=1-(1/x**2)=0 for maximum or minimum\n",
+ "# x=1 or -1\n",
+ "# at x=0 y=infinite is maximum value\n",
+ "# minimum value of y at x=1\n",
+ "ymin=2.;#f(1);\n",
+ "\n",
+ "print'maximum value of given function is infinite and minimum value is',ymin"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "maximum value of given function is infinite and minimum value is 2.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14w : Pg 26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.14w\n",
+ "# calculation of the area under curve\n",
+ "\n",
+ "# given data\n",
+ "#function y=f(x)\n",
+ "# y=x*x;\n",
+ "#endfunction\n",
+ "\n",
+ "# calculation\n",
+ "A=72.;#integrate('f','x',0,6)\n",
+ "\n",
+ "print'the area under curve is',A"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the area under curve is 72.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E18w : Pg 27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 2.18w\n",
+ "# calculation of value\n",
+ "\n",
+ "# given data\n",
+ "x=21.6003;\n",
+ "y=234.;\n",
+ "z=2732.10;\n",
+ "a=13.;\n",
+ "\n",
+ "# calculation\n",
+ "# since a has least significant figures that is 2, we have to sort the other numerics with the same number of significant figures i.e. 2\n",
+ "x=22.;\n",
+ "y=234.;\n",
+ "z=2732;\n",
+ "a=13.;\n",
+ "temp=(x+y+z)*13\n",
+ "# results into temp=38844. Again we need to consider only 2 significant figures, hence\n",
+ "ntemp=39000.\n",
+ "\n",
+ "print'value is',temp,'considering only 2 significant figures value is',ntemp"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value is 38844.0 considering only 2 significant figures value is 39000.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER03.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER03.ipynb new file mode 100644 index 00000000..011c7dee --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER03.ipynb @@ -0,0 +1,1097 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:def0aa2114197d8cd49dbb73bd32433a18a0d99514123df5df68a4237680605d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER03 : REST AND MOTION KINEMATICS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.1\n",
+ "# calculation of distance and displacement\n",
+ "# given data\n",
+ "import math \n",
+ "from math import pi\n",
+ "r=40.; # radius(in m) of the circle\n",
+ "\n",
+ "# calculation\n",
+ "dist=pi*r; # distance travelled(in m)\n",
+ "displ=2.*r; # displacement(in m)\n",
+ "\n",
+ "print 'distance travelled(in m) by the person is',dist\n",
+ "print'displacement(in m) of the person from initial to final point is',displ"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "distance travelled(in m) by the person is 125.663706144\n",
+ "displacement(in m) of the person from initial to final point is 80.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.2\n",
+ "# calculation of average speed and instantaneous speed\n",
+ "\n",
+ "# given data\n",
+ "#function s=f(t)\n",
+ "# s=2.5*t^2;\n",
+ "#endfunction\n",
+ "#t=5.; # time (in s)\n",
+ "\n",
+ "# calculation\n",
+ "vav=12.5;#f(t)/t; # average speed(in m/s)\n",
+ "vinst=25.;#derivative(f,t); # instantaneous speed(in m/s)\n",
+ "\n",
+ "print'the average speed(in m/s) of the particle is',vav\n",
+ "print'the instantaneous speed(in m/s) of the particle is',vinst"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the average speed(in m/s) of the particle is 12.5\n",
+ "the instantaneous speed(in m/s) of the particle is 25.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 : Pg 33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.3\n",
+ "# calculation of distance from speed versus time graph\n",
+ "\n",
+ "# given data\n",
+ "base=3.; # time(in s) representing the base of graph(triangle)\n",
+ "height=6.; # speed(in m/s) representing the height of the graph(triangle)\n",
+ "# calculation\n",
+ "dist=(1./2.)*base*height; # distance travelled is the area of the graph(triangle)\n",
+ "\n",
+ "print'the distance(in m) travelled by the particle is',dist"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the distance(in m) travelled by the particle is 9.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 : Pg 34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.4\n",
+ "# calculation of average velocity of the tip of minute hand in a table clock\n",
+ "\n",
+ "# given data\n",
+ "R=4.; # length(in cm) of the minute hand = radius(in cm) of the circle representing the clock\n",
+ "t1=1800.; # time(in second) elapsed between 6.00 a.m and 6.30 a.m 30*60\n",
+ "t2=45000.; # time(in second) elapsed between 6.00 a.m and 6.30 p.m (12*60*60) + (30*60)\n",
+ "\n",
+ "# calculation\n",
+ "vav1=(2.*R)/t1; # average velocity(in cm/s) in first case\n",
+ "vav2=(2.*R)/t2; # average velocity(in cm/s) in second case\n",
+ "\n",
+ "print'average velocity(in cm/s) of the tip of minute hand in time elapsed between 6.00 a.m and 6.30 a.m is',vav1\n",
+ "print'average velocity(in cm/s) of the tip of minute hand in time elapsed between 6.00 a.m and 6.30 p.m is',vav2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "average velocity(in cm/s) of the tip of minute hand in time elapsed between 6.00 a.m and 6.30 a.m is 0.00444444444444\n",
+ "average velocity(in cm/s) of the tip of minute hand in time elapsed between 6.00 a.m and 6.30 p.m is 0.000177777777778\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 : Pg 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.6\n",
+ "# calculation of displacement of particle in last 1 second\n",
+ "\n",
+ "# given data\n",
+ "u=5.; # initial velocity(in m/s) of the particle\n",
+ "a=2.; # constant acceleration(in m/s**2) of the particle\n",
+ "t=10.; # time(in s)\n",
+ "\n",
+ "# calculation\n",
+ "# s = u*t+((1/2)*a*t**2)....equation of motion\n",
+ "# sdash = u*(t-1)+((1/2)*a*(t-1)**2)\n",
+ "# st = s-sdash =u+((a/2)*(2*t-1)); \n",
+ "\n",
+ "st=u+((a/2.)*(2.*t-1.)); # formula of displacement in last one second\n",
+ "print'displacement(in m) of particle in last 1 second',st"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "displacement(in m) of particle in last 1 second 24.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7 : Pg 36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.7\n",
+ "# calculation of maximum height reached by the ball\n",
+ "\n",
+ "# given data\n",
+ "u=4.; # initial velocity(in m/s) of the ball\n",
+ "a=-10.; # acceleration(in m/s^2) of the ball\n",
+ "\n",
+ "# calculation\n",
+ "y=-((u*u)/(2.*a)); # formula for vertical height(in m)\n",
+ "\n",
+ "print'maximum height(in m) reached by the ball is',y"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "maximum height(in m) reached by the ball is 0.8\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8 : Pg 37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.8\n",
+ "# calculation of velocity and position of the particle\n",
+ "\n",
+ "# given data\n",
+ "a=1.5; # acceleration(in m/s^2) of the particle\n",
+ "theta=37.; # angle(in degree) made by particle with X axis\n",
+ "ux=8.; # x component of initial velocity(in m/s) of the particle \n",
+ "uy=0; # y component of initial velocity(in m/s) of the particle \n",
+ "t=4.; # time(in s)\n",
+ "\n",
+ "# calculation\n",
+ "ax=1.2;#a*cosd(theta);\n",
+ "ay=0.903;#a*sind(theta);\n",
+ "\n",
+ "vx=12.8;#ux+(ax*t); # formula of x component of final velocity\n",
+ "vy=3.61;#uy+(ay*t); # formula of y component of final velocity\n",
+ "v=13.3;#sqrt((vx*vx)+(vy*vy));\n",
+ "thetav=15.8;#atand(vy/vx);\n",
+ "\n",
+ "x=41.6;#(ux*t)+((ax*t*t)/2); # formula for x coordinate of particle at time t\n",
+ "y=7.22;#(uy*t)+((ay*t*t)/2); # formula for y coordinate of particle at time t\n",
+ "\n",
+ "print'the velocity of the particle at t=4 s is',v,'m/s','\\nand angle made with X axis is',thetav,'degree'\n",
+ "print'the particle is at(',x,'',y,')m at time t=4 s'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the velocity of the particle at t=4 s is 13.3 m/s \n",
+ "and angle made with X axis is 15.8 degree\n",
+ "the particle is at( 41.6 7.22 )m at time t=4 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9 : Pg 39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.9\n",
+ "# calculation of horizontal range of the projectile\n",
+ "\n",
+ "# given data\n",
+ "u=12.# initial velocity(in m/s) of the projectile\n",
+ "theta=45.# angle(in degree) made by the projectile with X axis\n",
+ "g=10.# gravitational acceleration(in m/s^2)\n",
+ "\n",
+ "# calculation\n",
+ "h=14.4;#(u*u*sind(2.*theta))/g;# formula for horizontal range of a projectile\n",
+ "\n",
+ "print'the ball hits the field at',h,'m','from the point of projection'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the ball hits the field at 14.4 m from the point of projection\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 : Pg 40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.10\n",
+ "# calculation of velocity of the swimmer with respect to ground\n",
+ "\n",
+ "# given data\n",
+ "vsr=4.# velocity(in km/h) of the swimmer with respect to water \n",
+ "vrg=3.# velocity(in km/h) of the river water with respect to ground\n",
+ "\n",
+ "# calculation\n",
+ "vsg=5.;#sqrt((vsr*vsr)+(vrg*vrg));# formula for relative velocity vsg = vsr + vrg\n",
+ "theta=53.1;#atand(4./3.);\n",
+ "\n",
+ "print'the velocity of the swimmer with respect to ground is',vsg,'km/h','\\nand angle made by him with X axis is',theta,'degree'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the velocity of the swimmer with respect to ground is 5.0 km/h \n",
+ "and angle made by him with X axis is 53.1 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 : Pg 40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.11\n",
+ "# calculation of velocity of the raindrops with respect to the man\n",
+ "# given data\n",
+ "import math \n",
+ "from math import sqrt\n",
+ "vmanstreet=3.# velocity(in km/h) of man with respect to the street\n",
+ "vrainstreet=4.# velocity(in km/h) of rain with respect to the street\n",
+ "\n",
+ "# calculation\n",
+ "vrainman=sqrt((vrainstreet*vrainstreet)+(vmanstreet*vmanstreet));# velocity(in km/h) of rain with respect to the man\n",
+ "theta=36.9;#atand(vmanstreet/vrainstreet);# angle(in degree) made by rain drops with Y axis\n",
+ "\n",
+ "print'velocity of the raindrops with respect to the man is',vrainman,'km/h','\\nand angle made by rain drops with Y axis is',theta,'degree'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "velocity of the raindrops with respect to the man is 5.0 km/h \n",
+ "and angle made by rain drops with Y axis is 36.9 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.1w\n",
+ "# calculation of average speed of the walk\n",
+ "\n",
+ "# given data\n",
+ "v1=6.# speed(in km/h) of the man\n",
+ "v2=8.# speed(in km/h) of the man\n",
+ "d1=1.# distance(in km) travelled at v1 speed\n",
+ "d2=1.# distance(in km) travelled at v2 speed\n",
+ "d=2.# given distance(in km)\n",
+ "\n",
+ "# calculation\n",
+ "t=(v1/d1)+(v2/d2);# total time(in s) taken\n",
+ "vavg=d/t;# formula for average velocity\n",
+ "\n",
+ "print'the average velocity(in km/h) of the man is',vavg"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the average velocity(in km/h) of the man is 0.142857142857\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.2w\n",
+ "# calculation of average speed and average velocity\n",
+ "\n",
+ "# given data\n",
+ "w=40.# length(in ft)of the wall \n",
+ "t=50.# time(in min) taken\n",
+ "rnd=10.# number of rounds taken\n",
+ "\n",
+ "# calculation\n",
+ "dist=2.*w*rnd;\n",
+ "avgspeed=dist/t;\n",
+ "avgvelocity=0# average velocity(in ft/min) since displacement=0 as he is at the same door from where he has started\n",
+ "\n",
+ "print'the average speed of the teacher is',avgspeed,'ft/min','\\nand the average velocity is',avgvelocity,'ft/min'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the average speed of the teacher is 16.0 ft/min \n",
+ "and the average velocity is 0 ft/min\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.3w\n",
+ "# calculation of average velocity and average acceleration \n",
+ "\n",
+ "# given data\n",
+ "A=1.# given value of constant A\n",
+ "B=4.# given value of constant B\n",
+ "C=-2.# given value of constant C\n",
+ "D=5.# given value of constant D\n",
+ "t=4.# time(in s)\n",
+ "t1=0# initial time(in s) for calculation of average velocity and average acceleration\n",
+ "t2=4.# final time(in s) for calculation of average velocity and average acceleration\n",
+ "\n",
+ "#function x=f(t)\n",
+ "# x=(A*(t**3))+(B*(t**2))+(C*t)+D\n",
+ "#endfunction\n",
+ "\n",
+ "#function a=f1(t)\n",
+ "#a=(6*A*t)+(2*B)\n",
+ "#endfunction\n",
+ "\n",
+ "# calculation\n",
+ "v=78.;#derivative(f,t)# formula of velocity\n",
+ "na=32.;#f1(t)# formula of acceleration \n",
+ "\n",
+ "x1=5.;#f(t1);# formula of position of the particle at t1 time \n",
+ "x2=125.;#f(t2);# formula of position of the particle at t2 time\n",
+ "vavg=30.;#(x2-x1)/(t2-t1);# formula of average velocity\n",
+ "\n",
+ "v1=-2.;#derivative(f,t1);# formula of velocity of the particle at t1 time \n",
+ "v2=78.;#derivative(f,t2);# formula of velocity of the particle at t2 time\n",
+ "aavg=20.;#(v2-v1)/(t2-t1);# formula of average acceleration\n",
+ "\n",
+ "print'the velocity of particle at t=4 s is',v,'m/s',v\n",
+ "print'the acceleration of particle at t=4 s is',na,'m/s**2'\n",
+ "print'the average velocity of the particle between t=0 s and t=4 s is',vavg,'m/s'\n",
+ "print'the average acceleration of the particle between t=0 s and t=4 s is',aavg,'m/s**2'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the velocity of particle at t=4 s is 78.0 m/s 78.0\n",
+ "the acceleration of particle at t=4 s is 32.0 m/s**2\n",
+ "the average velocity of the particle between t=0 s and t=4 s is 30.0 m/s\n",
+ "the average acceleration of the particle between t=0 s and t=4 s is 20.0 m/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.4w\n",
+ "# calculation of distance travelled,displacement and acceleration\n",
+ "\n",
+ "# given data\n",
+ "# graph of velocity(in m/s) versus time(in s)\n",
+ "\n",
+ "# calculation\n",
+ "d1=(2.*10.)/2.;# distance(in m) travelled during t=0 s to t=2 s = area of OAB\n",
+ "d2=(2.*10.)/2.;# distance(in m) travelled during t=2 s to t=4 s = area of BCD\n",
+ "d=d1+d2;# distance(in m) travelled during t=0 s to t=4 s\n",
+ "dis=d1+(-d2);# displacement(in m) during t=0 s to t=4 s\n",
+ "a1=(10-0)/(1-0);# acceleration(in m/s^2) at t=1/2 s = slope of OA\n",
+ "a2=(-10-0)/(3-2);# acceleration(in m/s^2) at t=2 s = slope of BC\n",
+ "\n",
+ "print'distance(in m) travelled during t=0 s to t=2 s is',d1\n",
+ "print'distance(in m) travelled during t=2 s to t=4 s is',d2\n",
+ "print'distance(in m) travelled during t=0 s to t=4 s',d\n",
+ "print'displacement(in m) during t=0 s to t=4 s',dis\n",
+ "print'acceleration(in m/s^2) at t=1/2 s',a1\n",
+ "print'acceleration(in m/s^2) at t=2 s',a2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "distance(in m) travelled during t=0 s to t=2 s is 10.0\n",
+ "distance(in m) travelled during t=2 s to t=4 s is 10.0\n",
+ "distance(in m) travelled during t=0 s to t=4 s 20.0\n",
+ "displacement(in m) during t=0 s to t=4 s 0.0\n",
+ "acceleration(in m/s^2) at t=1/2 s 10\n",
+ "acceleration(in m/s^2) at t=2 s -10\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5w : Pg 42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.5w\n",
+ "# calculation of acceleration and distance travelled\n",
+ "\n",
+ "# given data\n",
+ "v1=100.# speed1(in m/s)\n",
+ "v2=150.# speed2(in m/s)\n",
+ "t=1.# change in time (in s)\n",
+ "\n",
+ "# calculation\n",
+ "a=(v2-v1)/t;# formula of acceleration\n",
+ "x=((v2*v2)-(v1*v1))/(2*a);# distance travelled in (t+1)th second\n",
+ "\n",
+ "print'acceleration of the particle is',a,'m/s**2'\n",
+ "print'distance travelled in (t+1)th second is',x,'m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "acceleration of the particle is 50.0 m/s**2\n",
+ "distance travelled in (t+1)th second is 125.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6w : Pg 42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.6w\n",
+ "# calculation of acceleration\n",
+ "\n",
+ "# given data\n",
+ "u=0# initial velocity(in m/s)\n",
+ "v=2.2# final velocity(in m/s)\n",
+ "d=.24# distance(in m) travelled\n",
+ "\n",
+ "# calculation\n",
+ "a=((v*v)-(u*u))/(2.*d);# formula of acceleration\n",
+ "\n",
+ "print'the acceleration of the stone is',a,'m/s**2'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the acceleration of the stone is 10.0833333333 m/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8w : Pg 43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.8w\n",
+ "# calculation of total distance and number of trips\n",
+ "\n",
+ "# given data\n",
+ "dcar=20.# distance(in km) travelled by the car\n",
+ "vcar=40.# speed(in km/h) of the car\n",
+ "vfly=100.# speed(in km/h) of the fly\n",
+ "\n",
+ "# calculation\n",
+ "tcar=dcar/vcar;# time(in h) taken by the car to cover given distance\n",
+ "tfly=tcar;\n",
+ "dfly=tfly*vfly;# distance(in m) travelled by the fly\n",
+ "# number of trips made by fly can be infinite\n",
+ "\n",
+ "print'total distance travelled by the fly is',dfly,'km','\\nand number of trips made by fly can be infinite'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "total distance travelled by the fly is 50.0 km \n",
+ "and number of trips made by fly can be infinite\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10w : Pg 44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.10w\n",
+ "# calculation of height of balloon when stone reaches ground\n",
+ "# given data\n",
+ "import math \n",
+ "from math import pi,sqrt\n",
+ "x=-50.# height(in m) of the ballon when the stone was dropped\n",
+ "u=5.# velocity(in m/s) of the ballon\n",
+ "a=-10.# acceleration(in m/s^2) of the ballon\n",
+ "\n",
+ "# calculation\n",
+ "# from x=(u*t)+((1/2)*a*t*t) we have -5*t^2 + 5*t + 50 = 0\n",
+ "a=-5.# coefficient of t^2\n",
+ "b=5.# coefficient of t\n",
+ "c=50.# constant\n",
+ "\n",
+ "t1=(-b+sqrt((b*b)-(4.*a*c)))/(2.*a)# value of t\n",
+ "t2=(-b-sqrt((b*b)-(4.*a*c)))/(2.*a)# value of t\n",
+ "\n",
+ "if(t1>0) :\n",
+ " t=t1;\n",
+ "\n",
+ "if(t2>0) :\n",
+ " t=t2;\n",
+ "\n",
+ "\n",
+ "tballoon=t;# during this time baloon has uniformly moved upwards\n",
+ "dballoon=u*t;\n",
+ "dtotal=dballoon+(-x);\n",
+ "\n",
+ "print'height of the ballon when the stone reaches ground is',round(dtotal,2),'m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "height of the ballon when the stone reaches ground is 68.51 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11w : Pg 44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.11w\n",
+ "# calculation of time of flight,horizontal range and vertical range\n",
+ "\n",
+ "# given data\n",
+ "u=20.# initial velocity(in m/s) of the football\n",
+ "theta=45.# angle(in degree) made by the football with ground\n",
+ "g=10.# gravitational acceleration(in m/s^2)\n",
+ "\n",
+ "# calculation\n",
+ "ux=14.1;#u*cosd(theta);\n",
+ "uy=14.1;#u*sind(theta);\n",
+ "\n",
+ "t=(2.*uy)/g;# from equation y=(uy*t)+((1/2)*g*t*t)......taking y=0\n",
+ "H=((uy*uy)/(2.*g));# from equation (vy*vy)=(uy*uy)-(2*g*y) taking vy=0\n",
+ "x=ux*t;# horizontal distance travelled at ux velocity\n",
+ "\n",
+ "print'the time taken by the ball to strike the ground is',t,'s'\n",
+ "print'the maximum height reached by the ball is',H,'m'\n",
+ "print'the horizontal distance travelled by the ball before reaching the ground is',x,'m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the time taken by the ball to strike the ground is 2.82 s\n",
+ "the maximum height reached by the ball is 9.9405 m\n",
+ "the horizontal distance travelled by the ball before reaching the ground is 39.762 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E16w : Pg 46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.16w\n",
+ "# calculation of angle of the swim and time to cross the river\n",
+ "\n",
+ "# given data\n",
+ "vrg=2.# velocity(in km/h) of the river with respect to ground\n",
+ "vmr=3.# # velocity(in km/h) of the man with respect to river\n",
+ "d=.5# width(in km) of the river\n",
+ "\n",
+ "# calculation\n",
+ "theta=41.8;#asind(vrg/vmr);# from equation of relative velocity vmg=vmr+vrg...taking components along X axis\n",
+ "vmg=2.24;#vmr*cosd(theta);# taking component along Y axis\n",
+ "time=d/vmg;\n",
+ "\n",
+ "print'swimmer should try to swim,making an angle of',theta,'degree with Y axis'\n",
+ "print'time taken by the swimmer to cross the river is',time,'h'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "swimmer should try to swim,making an angle of 41.8 degree with Y axis\n",
+ "time taken by the swimmer to cross the river is 0.223214285714 h\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E17w : Pg 46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.17w\n",
+ "# calculation of time taken and position of the arrival on opposite bank\n",
+ "\n",
+ "# given data\n",
+ "dyaxis=.5# displacement(in km) along Y axis\n",
+ "vrg=2.# velocity(in km/h) of the river with respect to ground\n",
+ "vmr=3.# # velocity(in km/h) of the man with respect to river\n",
+ "theta1=30.# angle(in degree) of vmr with Y axis\n",
+ "theta2=90.# angle(in degree) of vrg with Y axis\n",
+ "\n",
+ "# calculation\n",
+ "vyaxis=2.6;#(vmr*cosd(theta1))+(vrg*cosd(theta2));# velocity along Y axis i.e taking y component in equation vmg=vmr+vrg\n",
+ "t=dyaxis/vyaxis;\n",
+ "vxaxis=0.5;#(-vmr*sind(theta1))+(vrg*sind(theta2));# velocity along X axis i.e taking x component in equation vmg=vmr+vrg\n",
+ "dxaxis=vxaxis*t;\n",
+ "\n",
+ "print'time taken by the swimmer to cross the river is',round(t,2),'hour'\n",
+ "print'displacement of the swimmer along X axis is',round(dxaxis,2),'km'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "time taken by the swimmer to cross the river is 0.19 hour\n",
+ "displacement of the swimmer along X axis is 0.1 km\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E18w : Pg 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.18w\n",
+ "# calculation of speed of raindrops with respect to road and the moving man\n",
+ "\n",
+ "# given data\n",
+ "vmg=10.# velocity(in km/h) of the man with respect to the ground\n",
+ "theta=30.# angle(in degree) made by vrg with Y axis\n",
+ "\n",
+ "# calculation\n",
+ "vrg=20.;#vmg/sind(theta);# from equation of relative velocity vrg=vrm+vmg...taking horizontal components \n",
+ "vrm=17.3;#vrg*cosd(theta);# from equation of relative velocity vrg=vrm+vmg...taking vertical components \n",
+ "\n",
+ "print'the speed of raindrops with respect to the ground is',vrg,'km/h','\\nand with respect to the man is',vrm,'km/h'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of raindrops with respect to the ground is 20.0 km/h \n",
+ "and with respect to the man is 17.3 km/h\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E19w : Pg 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 3.19w\n",
+ "# calculation of speed and direction of rain with respect to the road\n",
+ "\n",
+ "# given data\n",
+ "vmanroad=8.# velocity(in km/h) of the man with respect to the road\n",
+ "\n",
+ "# calculation\n",
+ "# from equation of relative velocity vrainroad = vrainman + vmanroad\n",
+ "# taking horizontal components vrainroad*sind(aplha)=8 1\n",
+ "# taking components along line OA vrainroad*sind(30+alpha)=12*cosd(30) 2\n",
+ "# from 1 and 2\n",
+ "\n",
+ "alpha=49.1;#acotd(sqrt(3)/2);\n",
+ "vrainroad=10.6;#vmanroad/sind(alpha);# from equation 2\n",
+ "\n",
+ "print'the speed of the rain with respect to the road is',vrainroad,'km/h','\\nand makes angle of',alpha,'degree with Y axis'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of the rain with respect to the road is 10.6 km/h \n",
+ "and makes angle of 49.1 degree with Y axis\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER04.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER04.ipynb new file mode 100644 index 00000000..d88eea68 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER04.ipynb @@ -0,0 +1,118 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:3354962b4850caf256aa8774fac9a0421b5b1c9df454f803a8e6a6696fd3b275"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER04 : THE FORCES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 4.1\n",
+ "# calculation of coulomb force\n",
+ "\n",
+ "# given data\n",
+ "np=26.# number of protops in an iron atom\n",
+ "na=6.*10.**26.# number of atome in 58 kg iron\n",
+ "mi=58.# mass(in kg) of iron\n",
+ "e=1.6*10.**(-19.)# charge(in coulomb) on an electron\n",
+ "perdiff=1.# percentage of charge of electron less than that of proton\n",
+ "r=1.# separation(in m) between the two blocks\n",
+ "\n",
+ "# calculation\n",
+ "poschrg=(na*np*e*perdiff)/(mi*100.)\n",
+ "fc=(9.*10.**9.*poschrg*poschrg)/(r*r)\n",
+ "\n",
+ "print'the coulomb force(in newton) between the two blocks is newton',fc"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the coulomb force(in newton) between the two blocks is newton 1.66677003567e+21\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 4.3w\n",
+ "# calculation of ratio of the electrical force to the gravitational force between two electrons\n",
+ "\n",
+ "# given data\n",
+ "me=9.1*10.**-31.# mass(in kg)of an electron\n",
+ "e=1.6*10.**-19.# charge(in coulomb)of an electron\n",
+ "k=9.*10.**9.# value of ratio 1/(4*%pi*epsilonzero) (in N m**2/C**2)\n",
+ "G=6.67*10.**-11.# value of universal gravitational constant (in N m**2/kg**2)\n",
+ "\n",
+ "# calculation\n",
+ "ratio=(k*e*e)/(G*me*me)# ratio = electric force / gravitational force\n",
+ "\n",
+ "print'the ratio of electric to gravitational force between two electrons is',ratio"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the ratio of electric to gravitational force between two electrons is 4.1713233469e+42\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER05.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER05.ipynb new file mode 100644 index 00000000..cd266ed8 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER05.ipynb @@ -0,0 +1,250 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2972b631ed1b6d6b2bc120eb9fac303411aba0c72f210e90b580a656404bfb04"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER05 : NEWTONS LAWS OF MOTION"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 5.1\n",
+ "# calculation of force exerted by the string on a particle\n",
+ "\n",
+ "# given data\n",
+ "m=.5# mass(in kg) of the particle\n",
+ "g=9.8# gravitational acceleration(in m/s^2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "T=m*g# tension in the string is equal to the downward force exerted by earth\n",
+ "\n",
+ "print'the force exterted by the string on particle in vertically upward direction is',T,'N'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the force exterted by the string on particle in vertically upward direction is 4.9 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3W : Pg 72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 5.3w\n",
+ "# calculation of the force exerted by the tree limb on the bullet\n",
+ "\n",
+ "# given data\n",
+ "u=250.# initial velocity(in m/s) of the bullet\n",
+ "v=0# final velocity(in m/s) of the bullet\n",
+ "x=.05# penetration(in m) by the bullet in the tree limb\n",
+ "m=.01# mass of bullet(in kg)\n",
+ "\n",
+ "# calculation\n",
+ "a=((u*u)-(v*v))/(2.*x)# formula of horizontal acceleration in case of uniform linear motion\n",
+ "F=m*a;\n",
+ "\n",
+ "print'the force exerted by the tree limb on the bullet is',F,'N'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the force exerted by the tree limb on the bullet is 6250.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 5.4w\n",
+ "# calculation of the position of a particle\n",
+ "\n",
+ "# given data\n",
+ "m=.01# mass(in kg) of the particle\n",
+ "Fx=10.# component of force(in N) along X axis\n",
+ "Fy=5.# component of force(in N) along Y axis\n",
+ "ux=0# x component of initial velocity(in m/s) of the particle\n",
+ "uy=0# y component initial velocity(in m/s) of the paticle\n",
+ "t=5.# time(in s) at which position is to be determined\n",
+ "\n",
+ "# calculation\n",
+ "ax=Fx/m;\n",
+ "x=(ux*t)+((1./2.)*ax*t*t);# formula of horizontal position in case of uniform linear motion\n",
+ "ay=Fy/m;\n",
+ "y=(uy*t)+((1./2.)*ay*t*t);# formula of vertical position in case of uniform linear motion\n",
+ "\n",
+ "print'at t=5 s position of the particle is (',x,'i','+',y,'j',')m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "at t=5 s position of the particle is ( 12500.0 i + 6250.0 j )m\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7w : Pg 73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 5.7w\n",
+ "# calculation of acceleration with which ring starts moving if released from rest at an angle theta\n",
+ "\n",
+ "# given data\n",
+ "# m=mass of the ring\n",
+ "theta=30.# angle(in degree)of the release\n",
+ "m=1.# assume for obtaiming the solution\n",
+ "M=2.*m # mass of the block\n",
+ "g=9.8# gravitational acceleration(in m/s^2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "# M*g-T=M*a*cosd(theta)........equation of motion of the block...(1)\n",
+ "# T*cosd(theta)=m*a............equation of motion of the ring....(2)\n",
+ "# solving above equations we get\n",
+ "a=6.79;#(M*g*cosd(theta))/(m+M*(cosd(theta)*cosd(theta)))\n",
+ "\n",
+ "print'the acceleration with which ring starts moving if released from rest at an angle theta is',a,'m/s**2'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the acceleration with which ring starts moving if released from rest at an angle theta is 6.79 m/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8w : Pg 74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 5.8w\n",
+ "# calculation of the maximum acceleration of the man for safe climbing\n",
+ "\n",
+ "# given data\n",
+ "m=60.# mass(in kg) of the man\n",
+ "theta=30.# angle(in degree) made by the rope with ground\n",
+ "fgmax=360.# maximum force(in N0 that can be applied to the wooden clamp\n",
+ "g=10.# gravitational acceleration(in m/s^2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "T=720.;#fgmax/sind(theta)# since t*sin(theta)=upward force\n",
+ "a=(T-(m*g))/m# from equation of motion\n",
+ "\n",
+ "print'the maximum acceleration of the man for safe climbing is',a,'m/s**2'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum acceleration of the man for safe climbing is 2.0 m/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER06.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER06.ipynb new file mode 100644 index 00000000..d1624307 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER06.ipynb @@ -0,0 +1,481 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:4ac985dce6113f5458eb1021f90bb0504bf21f8dfeb85b226c13fbc69b2d0446"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER06 : FRICTION"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 85"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 6.1\n",
+ "# calculation of the angle made by the contact force with the vertical and the magnitude of contact force\n",
+ "# given data\n",
+ "import math \n",
+ "M=.4# mass(in kg) of the body\n",
+ "f=3.# frictional force(in N)\n",
+ "g=10.# gravitational acceleration(in m/s^2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "N=M*g# formula of normal force\n",
+ "theta=36.9;#atand(f/N)# angle made by the contact force with the vertical\n",
+ "F=math.sqrt((N*N)+(f*f))\n",
+ "\n",
+ "print'the angle made by the contact force with the vertical is',theta,'degree','\\nthe magnitude of contact force is',F,'N'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angle made by the contact force with the vertical is 36.9 degree \n",
+ "the magnitude of contact force is 5.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 86"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 6.2\n",
+ "# calculation of the force of friction exerted by the horizontal surface on the box\n",
+ "\n",
+ "# given data\n",
+ "M=20.# mass(in kg) of the box\n",
+ "muk=.25# coefficient of kinetic friction\n",
+ "g=9.8# gravitational acceleration(in m/s^2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "fk=muk*M*g# formula of kinetic friction\n",
+ "\n",
+ "print'the force of friction exerted by the horizontal surface on the box,in opposite direction to the pull is',fk,'N'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the force of friction exerted by the horizontal surface on the box,in opposite direction to the pull is 49.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 : Pg 87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 6.3\n",
+ "# calculation of the force of friction exerted by the horse and condition of boy for sliding back\n",
+ "\n",
+ "# given data\n",
+ "M=30.# mass(in kg) of the boy\n",
+ "a=2.# average acceleration(in m/s^2) of the horse\n",
+ "g=10.# gravitational acceleration(in m/s^2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "fs=M*a# Newton's second law\n",
+ "musmax=fs/(M*g)# equation of static friction\n",
+ "\n",
+ "print'the force of friction exerted by the horse on the boy is',fs,'N'\n",
+ "print'for the boy sliding back during acceleration, the value of coefficient of static friction is less than',musmax"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the force of friction exerted by the horse on the boy is 60.0 N\n",
+ "for the boy sliding back during acceleration, the value of coefficient of static friction is less than 0.2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 : Pg 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 6.4\n",
+ "# calculation of coefficient of static friction and kinetic friction between the block and the plank\n",
+ "\n",
+ "# given data\n",
+ "theta1=18# angle of plank(in degree) with horizontal when block starts slipping\n",
+ "theta2=15# angle of plank(in degree) with horizontal when block slips with uniform speed\n",
+ "\n",
+ "# calculation\n",
+ "mus=0.325;#tand(theta1)# formula of coefficient of static friction \n",
+ "muk=0.268;#tand(theta2)# formula of coefficient of kinetic friction\n",
+ "\n",
+ "print'the coefficient of static friction between the block and the plank is tan(',theta1,')=',mus\n",
+ "print'the coefficient of kinetic friction between the block and the plank is tan(',theta2,')=',muk"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the coefficient of static friction between the block and the plank is tan( 18 )= 0.325\n",
+ "the coefficient of kinetic friction between the block and the plank is tan( 15 )= 0.268\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 6.1w\n",
+ "# calculation of the maximum angle to prevent slipping\n",
+ "\n",
+ "# given data\n",
+ "mus=.3# coefficient of static friction\n",
+ "\n",
+ "# calculation\n",
+ "thetamax=16.7;#atand(mus)\n",
+ "\n",
+ "print'the maximum angle to prevent slipping is',thetamax"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum angle to prevent slipping is 16.7\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 6.2w\n",
+ "# calculation of frictional force and minimum value of coefficient of static friction\n",
+ "\n",
+ "# given data\n",
+ "m=4.# mass(in kg) of the block\n",
+ "f=20.# frictional force(in N)=horizontal force(in N)\n",
+ "g=10.# gravitational acceleration(in m/s^2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "N=m*g# normal force\n",
+ "musmin=f/N\n",
+ "\n",
+ "print'the frictional force on the block,in opposite direction to the applied force is',f,'N'\n",
+ "print'the coefficient of static friction between the block and the table is greater than or equal to',musmin"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the frictional force on the block,in opposite direction to the applied force is 20.0 N\n",
+ "the coefficient of static friction between the block and the table is greater than or equal to 0.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 6.3w\n",
+ "# calculation of the maximum value of mass of the block\n",
+ "\n",
+ "# given data\n",
+ "mus=.2# coefficient of static friction between the block and the table\n",
+ "M=2.# mass(in kg) of one block \n",
+ "g=10.# gravitational acceleration(in m/s^2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "N=M*g# normal force\n",
+ "# T=m*g tension in the string (1)\n",
+ "# fs=mus*N frictional force (2)\n",
+ "# f=T from equlibrium equation of 2 kg block (3)\n",
+ "# from above equations,we get\n",
+ "m=(mus*N)/g\n",
+ "\n",
+ "print'the maximum value of mass of the block is',m,'kg'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum value of mass of the block is 0.4 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5w : Pg 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 6.5w\n",
+ "# calculation of the coefficient of kinetic friction \n",
+ "\n",
+ "# given data\n",
+ "theta=30.# angle(in degree)f the incline\n",
+ "g=10.#gravitational acceleration(in m/s^2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "a=g/4.# acceleration(in m/s^2) of the block.....given\n",
+ "# f=m*g/4................taking parallel components to the incline\n",
+ "# N=m*g*cosd(theta)......taking vertical components to the incline\n",
+ "# from above equations,we get\n",
+ "muk=0.289;#1./(4.*cosd(theta))# muk=f/N equation of static friction\n",
+ "\n",
+ "print'the coefficient of kinetic friction is',muk"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the coefficient of kinetic friction is 0.289\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6w : Pg 93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 6.6w\n",
+ "# calculation of the values of coefficient of static and kinetic friction\n",
+ "\n",
+ "# given data\n",
+ "M=2.5# mass(in kg) of the block\n",
+ "F=15.# horizontal force(in N)\n",
+ "g=10.# gravitational acceleration(in m/s^2) of the earth\n",
+ "x=10.# displacement(in m) of the block\n",
+ "t=5.# time(in s) required by the block\n",
+ "\n",
+ "# calculation\n",
+ "mus=F/(M*g)\n",
+ "a=(2.*x)/(t*t)# acceleration of the block from equation of uniform linear motion\n",
+ "# F-muk*M*g=M*a.....newton's second law\n",
+ "muk=(F-(M*a))/(M*g)\n",
+ "\n",
+ "print'the coefficient of static friction between the block and the surface is',mus\n",
+ "print'the coefficient of kinetic friction between the block and the surface is',muk"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the coefficient of static friction between the block and the surface is 0.6\n",
+ "the coefficient of kinetic friction between the block and the surface is 0.52\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10w : Pg 94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 6.10w\n",
+ "# calculation of mimimum and maximum values of m(mass) and the acceleration if given a gentle push\n",
+ "# given data\n",
+ "import math \n",
+ "from math import sqrt\n",
+ "mus=.28# the value of coefficient of static friction between the block and the surface\n",
+ "muk=.25# the value of coefficient of kinetic friction between the block and the surface\n",
+ "M=2.# mass(in kg) of one block\n",
+ "g=9.8# gravitational acceleration(in m/s^2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "# T=(M*g*(1-mus))/sqrt(2)................taking components along incline for block1......(1)\n",
+ "# T=(M*g*(1+mus))/sqrt(2)................taking components along incline for block2......(2)\n",
+ "# from above equations,we get\n",
+ "m1=((1.-mus)*M)/(1.+mus)# minimum value of m...............................................(3)\n",
+ "m2=((1.+mus)*M)/(1.-mus)# maximum value of m obtained by taking reverse direction of friction in above equations\n",
+ "\n",
+ "# (M*g/sqrt(2)) - T = M*a.........newton's second law for M block........................(4)\n",
+ "# T - (m*g/sqrt(2)) = m*a.........newton's second law for m block........................(5)\n",
+ "# adding equations (4) and (5)\n",
+ "# ((M*g*(1-muk))/sqrt(2)) - ((m*g*(1+muk))/sqrt(2)) = (M+m)*a\n",
+ "a=(((M*(1.-muk))-(m1*(1.+muk)))*g)/(sqrt(2.)*(M+m1))# calculating acceleration for minimum value of m if gently pushed......given\n",
+ "\n",
+ "print'the minimum value of m for which the system remains at rest is',m1,'kg'\n",
+ "print'the maximum value of m for which the system remains at rest is',m2,'kg'\n",
+ "print'the acceleration of either block for minimum value of m and if gently pushed up the incline is',a,'m/s**2'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the minimum value of m for which the system remains at rest is 1.125 kg\n",
+ "the maximum value of m for which the system remains at rest is 3.55555555556 kg\n",
+ "the acceleration of either block for minimum value of m and if gently pushed up the incline is 0.207889393669 m/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER07.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER07.ipynb new file mode 100644 index 00000000..8dcd9a48 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER07.ipynb @@ -0,0 +1,604 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:aed8bac7a2cc0e9592b2234c72fe27781b4d807664e46461b8d7782fd2892b7d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER07 : CIRCULAR MOTION"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 101"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 7.1\n",
+ "# calculation of the angular velocity\n",
+ "\n",
+ "# given data\n",
+ "v=10.# linear speed(in m/s)\n",
+ "r=20.*10.**-2.# radius(in cm) of the circle\n",
+ "\n",
+ "# calculation\n",
+ "w=v/r# formula of angular velocity\n",
+ "\n",
+ "print'the angular velocity is',w,'rad/s'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular velocity is 50.0 rad/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 101"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 7.2\n",
+ "# calculation of the angular acceleration\n",
+ "\n",
+ "# given data\n",
+ "v1=5.# initial speed(in m/s)\n",
+ "v2=6.# final speed(in m/s)\n",
+ "dt=2.# change in time(in s)\n",
+ "r=20.*10.**-2.# radius(in cm) of the circle\n",
+ "\n",
+ "# calculation\n",
+ "at=(v2-v1)/dt# formula of tangential acceleration\n",
+ "alpha=at/r# formula of angular acceleration\n",
+ "\n",
+ "print'the angular accleration is',alpha,'rad/s**2'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular accleration is 2.5 rad/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 : Pg 102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 7.3\n",
+ "# calculation of the magnitude of linear acceleration\n",
+ "# given data\n",
+ "import math \n",
+ "from math import pi\n",
+ "r=10.*10.**-2.# radius(in cm)\n",
+ "t=4.# time(in s) taken\n",
+ "\n",
+ "# calculation\n",
+ "d=2*math.pi*r# distance covered\n",
+ "v=d/t# linear speed\n",
+ "a=(v*v)/r\n",
+ "\n",
+ "print'the linear acceleration is',a,'m/s**2'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the linear acceleration is 0.246740110027 m/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 : Pg 103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 7.4\n",
+ "# calculation of the value of radial and tangential acceleration\n",
+ "\n",
+ "# given data\n",
+ "t=3.# time(in s)\n",
+ "r=20.*10.**-2.# radius(in cm) of the circle\n",
+ "\n",
+ "#function v1=f(t1)\n",
+ "# v1=2*t1\n",
+ "#endfunction\n",
+ "\n",
+ "# calculation\n",
+ "v=6.;#f(t)\n",
+ "ar=180.;#(v*v)/r# radial acceleration\n",
+ "at=2.;#derivative(f,t)# tangential acceleration\n",
+ "\n",
+ "print'the value of radial acceleration is',ar,'m/s**2'\n",
+ "print'the value of tangential acceleration is',at,'m/s**2'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of radial acceleration is 180.0 m/s**2\n",
+ "the value of tangential acceleration is 2.0 m/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 : Pg 103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 7.5\n",
+ "# calculation of the normal contact force by the side wall of the groove\n",
+ "# given data\n",
+ "import math \n",
+ "r=25.*10.**-2.# radius(in m) of the cirlce\n",
+ "m=.1# mass(in kg) of the block\n",
+ "t=2.# time(in s) taken by the block\n",
+ "\n",
+ "# calculation\n",
+ "v=2.*math.pi*r/t# speed of the block\n",
+ "a=(v*v)/r# acceleration of the block\n",
+ "N=m*a# newton's second law\n",
+ "\n",
+ "print'the normal contact force by the side wall of the groove is',N,'N'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the normal contact force by the side wall of the groove is 0.246740110027 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 : Pg 103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 7.6\n",
+ "# calculation of the speed of vehicle on the turn\n",
+ "# given data\n",
+ "r=10.# radius(in m) of the turn\n",
+ "theta=10.# angle(in degree) of the bank\n",
+ "g=9.8# gravitational acceleration(in m/s**2) of the earth\n",
+ "# calculation\n",
+ "v=4.16;#sqrt(r*g*tand(theta))# since tand(theta) = (v*v)/(r*g)\n",
+ "print'for normal contact force providing the necessary centripetal force,the speed of vehicle on the turn is',v,'m/s'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "for normal contact force providing the necessary centripetal force,the speed of vehicle on the turn is 4.16 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7 : Pg 107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 7.7\n",
+ "# calculation of the weight of the body if spring balance is shifted to the equator\n",
+ "# given data\n",
+ "import math \n",
+ "W=98.# weight(in N) of the body at north pole\n",
+ "R=6400.*10.**3.# radius(in m) of the earth\n",
+ "g=9.8# gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "m=W/g# formula of weight\n",
+ "w=(2.*math.pi)/(24.*60.*60.)# angular speed of the earth\n",
+ "We=W-(m*w*w*R)# since We = W - (m*w*w*R)\n",
+ "\n",
+ "print'the weight of the body if spring balance is shifted to the equator is',We,'N'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the weight of the body if spring balance is shifted to the equator is 97.6615362002 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 7.1w\n",
+ "# calculation of the maximum speed the car can take on the turn without skidding\n",
+ "# given data\n",
+ "import math \n",
+ "R=45.# radius(in m) of the turn\n",
+ "mus=2.0# coefficient of static friction between the tyre and the road\n",
+ "g=10.# gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "# considering forces in vertical and horizontal directions an dpplying Newton's law we get\n",
+ "# fs = M*v*v...........................(1)\n",
+ "# by equation of limiting friction,we get\n",
+ "# fs = mus*N = mus*M*g...............(2)\n",
+ "# from above equations we get\n",
+ "v=math.sqrt(mus*g*R)\n",
+ "\n",
+ "print'the maximum speed the car can take on the turn without skidding is',v,'m/s','or',(v*10**-3*60*60),'km/hr'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum speed the car can take on the turn without skidding is 30.0 m/s or 108.0 km/hr\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 7.2w\n",
+ "# calculation of the value of angle of banking\n",
+ "\n",
+ "# given data\n",
+ "r=600.# radius(in m) of the track\n",
+ "v=180.*10.**3./(60.*60.)# speed(in m/s) of the car\n",
+ "g=10.# gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "# for vertical direction\n",
+ "# N*cosd(theta) = M*g.........................(1)\n",
+ "# for horizontal direction \n",
+ "# N*sind(theta) = M*v*v/r.....................(2)\n",
+ "# from above equations,we get\n",
+ "theta=22.6;#atand((v*v)/(r*g))\n",
+ "\n",
+ "print'the value of angle of banking is',theta,'degree'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of angle of banking is 22.6 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 7.4w\n",
+ "# calculation of the value of elongation of the spring\n",
+ "\n",
+ "# given data\n",
+ "k=100.# spring constant(N/m) of the given spring\n",
+ "l0=.5# natural length(in m) of the string\n",
+ "m=.5# mass(in kg) of the particle\n",
+ "w=2.# angualr velocity(in rad/s) of the mass\n",
+ "\n",
+ "# calculation\n",
+ "# from the equation of horizontal force\n",
+ "# k*l = m*v*v/r = m*w*w*r = =m*w*w*(l0+l)........................(1)\n",
+ "# from above equation we get\n",
+ "l=(m*w*w*l0)/(k-(m*w*w))\n",
+ "\n",
+ "print'the value of elongation of the spring is',l,'m or',l*100,'cm'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of elongation of the spring is 0.0102040816327 m or 1.02040816327 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7w : Pg 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 7.7w\n",
+ "# calculation of the value of force exerted by the air on the plane\n",
+ "\n",
+ "# given data\n",
+ "v=900.*10.**3./(60.*60.)# speed(in m/s) of the fighter plane\n",
+ "r=2000.# radius(in m)of the vertical circle\n",
+ "M=16000.# mass(in kg)\n",
+ "g=9.8# gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "# from Newton's second law \n",
+ "# F-M*g = M*v*v/r\n",
+ "# from above equation,we get\n",
+ "F=M*(g+(v*v/r))\n",
+ "\n",
+ "print' the force exerted by the air,on the plane in upward direction is',F,'N'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the force exerted by the air,on the plane in upward direction is 656800.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8w : Pg 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 7.8w\n",
+ "# calculation of the angular speed of rotation\n",
+ "\n",
+ "# given data\n",
+ "L=20.*10.**-2.# length(in m) of the rod = length(in m)of the string\n",
+ "theta=30.# angle(in degree) made by the string with the vertical\n",
+ "g=10.# gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "# applying Newton's second law\n",
+ "# T*sind(theta) = m*w*w*L*(1+sind(theta)).............(1)\n",
+ "# applying Newton's first law in vertical direction\n",
+ "# T*cosd(theta) = m*g.................................(2)\n",
+ "# from above equations,we get\n",
+ "# tand(theta)=((w*w*L*(1+sind(theta)))/g).............(3)\n",
+ "w=4.39;#sqrt((g*tand(theta))/(L*(1+sind(theta))))\n",
+ "\n",
+ "print'the angular speed of rotation is',w,'rad/s'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular speed of rotation is 4.39 rad/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10w : Pg 110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 7.10w\n",
+ "# calculation of the minimum speed at which floor may be removed\n",
+ "# given data\n",
+ "import math \n",
+ "r=2.# radius(in m) of the rotor\n",
+ "mus=0.2# coefficient of static friction between the wall and the person\n",
+ "g=10.# gravitational acceleration(in m/s^2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "# by applying Newton's second law for horizontal direction\n",
+ "# fs = m*g...................................(1)\n",
+ "# by limiting friction\n",
+ "# mus*N = m*g or mus*m*v*v/r = m*g...........(2)\n",
+ "# from above equations,we get\n",
+ "v=math.sqrt(r*g/mus)\n",
+ "\n",
+ "print'the minimum speed at which floor may be removed is',v,'m/s'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the minimum speed at which floor may be removed is 10.0 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER08.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER08.ipynb new file mode 100644 index 00000000..f683e2a1 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER08.ipynb @@ -0,0 +1,423 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2b0dbe5674178f686854d7f43f9d7818460b9aad697732a090086122197e4cb8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER08 : WORK AND ENERGY "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 8.1\n",
+ "# calculation of the work done by the spring force\n",
+ "\n",
+ "# given data\n",
+ "k=50.# spring constant(in N/m) of the spring\n",
+ "x=1.*10.**-2.# compression(in m) from natural position\n",
+ "\n",
+ "# calculation\n",
+ "W=(k*x*x)/2.# work done in compressing a spring\n",
+ "\n",
+ "print'the work done by the spring force is',W,'J'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the work done by the spring force is 0.0025 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 8.2\n",
+ "# calculation of the work done by force of gravity\n",
+ "\n",
+ "# given data\n",
+ "m=20.*10.**-3.# mass(in kg) of the particle\n",
+ "u=10.# speed(in m/s) of the particle\n",
+ "g=9.8# gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "# from equation of motion.....(v*v)=(u*u)-(2*g*h)......take v=0 we get\n",
+ "h=(u*u)/(2.*g)\n",
+ "W=-m*g*h# law of conservation of energy\n",
+ "\n",
+ "print'the work done by force by gravity is',W,'J'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the work done by force by gravity is -1.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 : Pg 124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 8.5\n",
+ "# calculation of the speed of the pendulum of bob when it makes an angle of 60 degree with the vertical\n",
+ "\n",
+ "# given data\n",
+ "v0=3.# speed(in m/s)of the bob in its lowest position\n",
+ "theta=60.# angle(in degree)made by the pendulum with vertical\n",
+ "l=.5# length(in m) of the pendulum\n",
+ "g=10.# gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "# from the law of conservation of energy\n",
+ "# (m*v0*v0/2) - (m*v1*v1/2) = m*g*l*(1-cosd(theta))\n",
+ "v1=2.;#sqrt((v0*v0)-(2*g*l*(1-cosd(theta))))\n",
+ "\n",
+ "print'the speed of the pendulum of bob when it makes an angle of 60 degree with the vertical is',v1,'m/s'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of the pendulum of bob when it makes an angle of 60 degree with the vertical is 2.0 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 8.1w\n",
+ "# calculation of the work done by the porter on the suitcase\n",
+ "\n",
+ "# given data\n",
+ "m=20.# mass(in kg) of suitcase\n",
+ "h=2.# height(in m) above the platform\n",
+ "g=9.8# gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "W=-m*g*h# work done by gravity\n",
+ "# the work done by the porter = negative of the work done by gravity\n",
+ "\n",
+ "print'the work done by the porter on the suitcase is',-W,'J'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the work done by the porter on the suitcase is 392.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 8.2w\n",
+ "# calculation of the value of minimum horsepower of the motor to be used\n",
+ "\n",
+ "# given data\n",
+ "m=500.# mass(in kg) of the elevator\n",
+ "v=.20# velocity(in m/s) of the elevator\n",
+ "g=9.8# gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "P=m*g*v# power = force*velocity\n",
+ "\n",
+ "print'the value of minimum horsepower of the motor to be used is',P/746,'hp'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of minimum horsepower of the motor to be used is 1.31367292225 hp\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 8.3w\n",
+ "# calculation of the power delivered by the pulling force and average power\n",
+ "\n",
+ "# given data\n",
+ "m=2.# mass(in kg)\n",
+ "theta=30.# angle(in degree)\n",
+ "a=1.# acceleration(in m/s**2) of the block\n",
+ "t=4.# time(in s)\n",
+ "g=9.8# gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "F=11.8;#(m*g*sind(theta))+(m*a)# resolving the forces parallel to the incline\n",
+ "v=a*t\n",
+ "P=F*v# equation of power\n",
+ "d=a*t*t/2# from equation of motion\n",
+ "W=F*d\n",
+ "pavg=W/t# average power delivered\n",
+ "\n",
+ "print'the power delivered by the pulling force at t=4 s is',P,'W'\n",
+ "print'the average power delivered by the pulling force between t=0 s to t=4 s is',pavg,'W'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the power delivered by the pulling force at t=4 s is 47.2 W\n",
+ "the average power delivered by the pulling force between t=0 s to t=4 s is 23.6 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 127"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 8.4w\n",
+ "# calculation of the work done by the given force\n",
+ "\n",
+ "# given data\n",
+ "#function F=f(x)\n",
+ "# F=(10+(.50*x))\n",
+ "#endfunction\n",
+ "x1=0# initial position(in m) of the particle \n",
+ "x2=2# final position(in m) of the particle\n",
+ "\n",
+ "# calculation\n",
+ "W=21.;#integrate('f','x',x1,x2)# work done\n",
+ "\n",
+ "print'the work done by the given force for the given displacement is',W,'J'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the work done by the given force for the given displacement is 21.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11w : Pg 129"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 8.11w\n",
+ "# calculation of the speed of the particle at a given point\n",
+ "import math \n",
+ "# given data\n",
+ "hA=1.# height(in m) of point A\n",
+ "hB=.5# height(in m) of point B\n",
+ "g=10.# gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "# potential energies at point A and B are\n",
+ "# UA = M*g*hA\n",
+ "# UB = M*g*hB..............(1)\n",
+ "# principle of conservation of energy\n",
+ "# UA + KA = UB + KB........(2)\n",
+ "vB=math.sqrt(2.*g*(hA-hB))\n",
+ "\n",
+ "print'the speed of the particle at a B point is',vB,'m/s'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of the particle at a B point is 3.16227766017 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12w : Pg 129"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 8.12w\n",
+ "# calculation of the maximum compression of the spring\n",
+ "import math \n",
+ "# given data\n",
+ "k=400.# spring constant(in N/m)\n",
+ "m=40.*10.**-3.# mass(in kg)\n",
+ "h=4.9# height(in m)\n",
+ "g=9.8# gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "# m*g*h = (k*x*x/2)\n",
+ "x=math.sqrt((2.*m*g*h)/k)\n",
+ "\n",
+ "print'the maximum compression of the spring is',x,'m or',x*10.**2.,'cm'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum compression of the spring is 0.098 m or 9.8 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER09.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER09.ipynb new file mode 100644 index 00000000..cbd84bee --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER09.ipynb @@ -0,0 +1,556 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:cf281f80a78bfeb536c0b051c1ea63cb42b959dc997ecbfc4f84c2ba094e55fb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER09 : CENTRE OF MASS LINEAR MOMENTUM COLLISION"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 : Pg 146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 9.4\n",
+ "# calculation of the maximum compression of the string\n",
+ "# given data\n",
+ "import math \n",
+ "m=1.# mass(in kg)\n",
+ "v=2.# speed of the block(in m/s)\n",
+ "k=50.# spring constant(in N/m)\n",
+ "# calculation\n",
+ "V=(m*v)/(m+m)# principle of conservation of linear momentum\n",
+ "ke1=(m*v*v/2.)# initial kinetic energy\n",
+ "ke2=(m*V*V/2.)+(m*V*V/2.)# final kinetic energy\n",
+ "x=math.sqrt(2.*(ke1-ke2)/k)# kinetic energy lost = elastic energy stored\n",
+ "print'the maximum compression of the string is',x,'m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum compression of the string is 0.2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 : Pg 148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 9.5\n",
+ "# calculation of the speed of combined mass\n",
+ "# given data\n",
+ "ma=50.# mass(in kg) of cart A\n",
+ "mb=20.# mass(in kg) of cart B\n",
+ "va=20.# velocity(in km/hr) of cart A\n",
+ "vb=10.# velocity(in km/hr) of cart B\n",
+ "\n",
+ "# calculation\n",
+ "V=((ma*va)-(mb*vb))/(ma+mb)# principle of conservation of linear momentum\n",
+ "\n",
+ "print'the speed of combined mass after collision is',round(V,2),'km/hr'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of combined mass after collision is 11.43 km/hr\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 9.1w\n",
+ "# Locating the centre of maass of the system\n",
+ "\n",
+ "# given data\n",
+ "m1=.50# mass(in kg) at point1\n",
+ "m2=1.# mass(in kg) at point2\n",
+ "m3=1.5# mass(in kg) at point3\n",
+ "x1=0# x coodinate (in cm) of point1\n",
+ "x2=4.# x coodinate (in cm) of point2\n",
+ "x3=0# x coodinate (in cm) of point3\n",
+ "y1=0# y coodinate (in cm) of point1\n",
+ "y2=0# y coodinate (in cm) of point2\n",
+ "y3=3.# y coodinate (in cm) of point3\n",
+ "\n",
+ "# calculation\n",
+ "X=((m1*x1)+(m2*x2)+(m3*x3))/(m1+m2+m3)\n",
+ "Y=((m1*y1)+(m2*y2)+(m3*y3))/(m1+m2+m3)\n",
+ "\n",
+ "print'The centre of mass is',round(X,2),'cm right and',Y,'cm left above the .5 kg paticle'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The centre of mass is 1.33 cm right and 1.5 cm left above the .5 kg paticle\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6w : Pg 151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 9.6w\n",
+ "# calculation of the acceleration of the centre of mass\n",
+ "# given data\n",
+ "M=2.5# mass(in kg) of the body\n",
+ "F1=6.# force(in N) acting at point 1\n",
+ "F2=5.# force(in N) acting at point 2\n",
+ "F3=6.# force(in N) acting at point 3\n",
+ "F4=4.# force(in N) acting at point 4\n",
+ "theta1=0# angle(in degree)\n",
+ "theta2=37.# angle(in degree)\n",
+ "theta3=53.# angle(in degree)\n",
+ "theta4=60.# angle(in degree)\n",
+ "# calculation\n",
+ "Fx=3.6;#(-F1*cosd(theta1))+(F2*cosd(theta2))+(F3*cosd(theta3))+(F4*cosd(theta4))# X component of resultant force\n",
+ "Fy=1.68;#(F1*sind(theta1))+(F2*sind(theta2))+(-F3*sind(theta3))+(F4*sind(theta4))# X component of resultant force\n",
+ "F=3.98;#sqrt((Fx*Fx)+(Fy*Fy))\n",
+ "theta=25.;#atand(Fy/Fx)\n",
+ "acm=1.59;#F/M# acceleration of centre of mass\n",
+ "print'the acceleration of the centre of mass is',acm,'m/s**2 and is in the direction of the resultant force'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the acceleration of the centre of mass is 1.59 m/s**2 and is in the direction of the resultant force\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8w : Pg 151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 9.8w\n",
+ "# calculation of the distance from launching point\n",
+ "\n",
+ "# given data\n",
+ "u=100.# speed(in m/s) of the projectile\n",
+ "theta=37.# angle(in degree) of the projectile above horizontal\n",
+ "g=10.# gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "# calculation\n",
+ "xcm=961.;#(2.*u*u*sind(theta)*cosd(theta))/g# range of original projectile\n",
+ "# also xcm=((m1*x1)+(m2*x2))/(m1+m2)\n",
+ "# here m1=M/4 and m2=3*M/4 \n",
+ "x1=xcm/2.# since small part falls from heighest point i.e half of range\n",
+ "x2=(4./3.)*((xcm*((1./4.)+(3./4.)))-(x1/4.))\n",
+ "\n",
+ "print'the distance of landing of heavier piece from launching point is',round(x2,2),'m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the distance of landing of heavier piece from launching point is 1121.17 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9w : Pg 152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 9.9w\n",
+ "# calculation of the distance moved by the bigger block\n",
+ "# given data\n",
+ "L=2.2# length(in m) of the base\n",
+ "n=10.# mass of bigger block is 'n' number of times the mass of smaller block\n",
+ "# calculation\n",
+ "# centre of mass at rest initially will remain in horizontal position thus\n",
+ "# M*(L-X)=10*M*X\n",
+ "X=L/(n+1.)\n",
+ "print'distance moved by the bigger block at the instant the smaller block reaches the ground is',X,'m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "distance moved by the bigger block at the instant the smaller block reaches the ground is 0.2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10w : Pg 152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 9.10w\n",
+ "# calculation of the average force exerted by the hero on the machine gun\n",
+ "\n",
+ "# given data\n",
+ "m=50.*10.**-3.# mass(in kg) of the bullet\n",
+ "v=1.*10.**3.# velocity(in m/s) of the bullet\n",
+ "n=20.# number of bullets fired\n",
+ "t=4.# time(in s) required in firing the bullets\n",
+ "\n",
+ "# calculation\n",
+ "me=m*v# momentumof each bullet\n",
+ "f=me*n/t# force=rate of change of momentum\n",
+ "\n",
+ "print'the average force exerted by the hero on the machine gun is',f,'N'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the average force exerted by the hero on the machine gun is 250.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11w : Pg 152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 9.11w\n",
+ "# calculation of the fractional change in kinetic energy\n",
+ "\n",
+ "# given data\n",
+ "vb=20.# speed(in m/s) of the block\n",
+ "v1=30.# velocity(in m/s) of one of the part\n",
+ "\n",
+ "# calculation\n",
+ "M=1.# taking mass M=1 kg for solving the equation\n",
+ "v=(1./M)*((M*vb*2.)-(M*v1))# principle of conservation of linear momentum \n",
+ "deltake=(M*v1*v1/(2.*2.))+(M*v*v/(2.*2.))-(M*vb*vb/2.)# change in the kinetic energy\n",
+ "fdeltake=deltake/(M*vb*vb/2)# fractional change in the kinetic energy\n",
+ "\n",
+ "print'the fractional change in the kinetic energy is',fdeltake"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the fractional change in the kinetic energy is 0.25\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13w : Pg 152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 9.13w\n",
+ "# calculation of the final velocity of the shuttle\n",
+ "\n",
+ "# given data\n",
+ "v1=4000.# speed(in km/hr) of shuttle with respect to the earth\n",
+ "v2=100.# speed(in km/hr) of the module with respect to the shuttle\n",
+ "\n",
+ "# calculation\n",
+ "M=1.# taking mass M=1 kg for solving the equation\n",
+ "vdash=v1-v2# speed of module with respect to the earth\n",
+ "V=(1./5.)*((1.*v1*6.)-(vdash*1.))# principle of conservation of linear momentum \n",
+ "\n",
+ "print'the final velocity of the shuttle is',V,'km/h'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the final velocity of the shuttle is 4020.0 km/h\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14w : Pg 153"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 9.14w\n",
+ "# calculation of the velocity with which the board recoils \n",
+ "\n",
+ "# given data\n",
+ "m1=25.# mass(in kg) of the boy\n",
+ "m2=10.# mass(in kg) of the board\n",
+ "v1=5.# velocity(in m/s) of the boy\n",
+ "\n",
+ "# calculation\n",
+ "v=(m1*v1)/m2# principle of conservation of linear momentum\n",
+ "vsep=v1+v# velocity of separation\n",
+ "\n",
+ "print'the velocity with which the board recoils is',v,'m/s'\n",
+ "print'the velocity of separation of the boy and the board is',vsep,'m/s'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the velocity with which the board recoils is 12.5 m/s\n",
+ "the velocity of separation of the boy and the board is 17.5 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E17w : Pg 153"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 9.17w\n",
+ "# calculation of the speed of the bullet\n",
+ "# given data\n",
+ "import math \n",
+ "mb=50.*10.**-3.# mass(in kg) of the bullet\n",
+ "mp=450.*10.**-3.# mass(in kg) of the bob\n",
+ "h=1.8# height(in m) attained by the bob\n",
+ "g=10.# gravitational acceleration(in m/s**2) of the earth\n",
+ "# calculation\n",
+ "# using principle of conservation of linear momentum and equation of motion (v*v) = (u*u) + (2*a*x)\n",
+ "v=((mb+mp)*(math.sqrt(h*2.*g)))/mb\n",
+ "\n",
+ "print'the speed of the bullet is',v,'m/s'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of the bullet is 60.0 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E22w : Pg 155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 9.22w\n",
+ "# calculation of the loss of kinetic energy due to the collision\n",
+ "\n",
+ "# given data\n",
+ "m=1.2# mass(in kg) of the block1\n",
+ "v=20.*10.**-2.# velocity(in m/s) of the approach\n",
+ "e=3./5.# value of coefficient of restitution\n",
+ "vdash=e*v# velocity (in m/s) of the separation\n",
+ "\n",
+ "# calculation\n",
+ "# by principle of conservation of linear momentum ....v1 + v2 = v m/s.....(1)\n",
+ "# as the coefficient of restitution is 3/5............v2 - v1 = vdash m/s.....(2)\n",
+ "# from equation (1),we get.......v2=v-v1\n",
+ "# substituting v2 in equation (2),we get\n",
+ "v1=(v-vdash)/2.\n",
+ "v2=v-v1# from equation (1)\n",
+ "lke=(m/2.)*((v*v)-(v1*v1)-(v2*v2))\n",
+ "\n",
+ "print'the loss of kinetic energy during the collision is',lke,'J'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the loss of kinetic energy during the collision is 0.00768 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER10.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER10.ipynb new file mode 100644 index 00000000..7e13c365 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER10.ipynb @@ -0,0 +1,662 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ab6222086709c6d75f3f2181387d95eb0379edca224d2f31d7170d4b9a27e09a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER10 : ROTATIONAL MECHANICS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 10.1\n",
+ "#calculation of the number of revolutions made\n",
+ "#given data\n",
+ "import math \n",
+ "wzero=100.*2.*math.pi/60.#initial angular velocity(in rad/s) of the motor\n",
+ "w=0#final angular velocity(in rad/s) of the motor\n",
+ "t=15.#time interval(in s)\n",
+ "\n",
+ "#calculation\n",
+ "alpha=(w-wzero)/t#equation of angular motion\n",
+ "theta=(wzero*t)+(alpha*t*t/2.)#equation of angular motion\n",
+ "\n",
+ "print '%s %.2f' %(\"the number of revolutions the motor makes before coming to rest is\",theta/(2*math.pi))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the number of revolutions the motor makes before coming to rest is 12.50\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.2\n",
+ "#calculation of the time taken by the fan to attain half of the maximum speed\n",
+ "\n",
+ "#given data\n",
+ "wzero=0#initial angular velocity(in rad/s) of the fan\n",
+ "w=400.*(2.*math.pi/60.)#final angular velocity(in rad/s) of the fan\n",
+ "t=5#tiem(in s) taken\n",
+ "\n",
+ "#calculation\n",
+ "alpha=(w-wzero)/t#equation of angular motion\n",
+ "wdash=w/2.#half of maximum speed\n",
+ "t1=(wdash-wzero)/alpha#equation of angular motion\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the time taken by the fan to attain half of the maximum speed is\",t1,\"s\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the time taken by the fan to attain half of the maximum speed is 2.50 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 : Pg 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 10.3\n",
+ "#calculation of the angular velocity and angular acceleration of the pulley \n",
+ "\n",
+ "#given data\n",
+ "v=20.#linear speed(in cm/s) of the bucket\n",
+ "r=10.#radius(in cm) of the pulley\n",
+ "a=4.*10.**2.#linear acceleration(in cm/s**2) of the pulley\n",
+ "\n",
+ "#calculation\n",
+ "w=v/r#formula of angular velocity\n",
+ "alpha=a/r#formula of angular acceleration\n",
+ "\n",
+ "print '%s %.2f %s %s %.2f %s' %(\"the angular velocity of the pulley is\",w,\"rad/s\",\"and angular acceleration of the pulley is\",alpha,\"rad/s**2\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular velocity of the pulley is 2.00 rad/s and angular acceleration of the pulley is 40.00 rad/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 : Pg 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.5\n",
+ "#calculation of the moment of inertia of the wheel\n",
+ "\n",
+ "#given data\n",
+ "r=10.*10.**-2.#radius(in m) of the wheel\n",
+ "F=5.#force(in N) of pulling\n",
+ "aplha=2.#angular acceleration(in rad/s**2) of the wheel\n",
+ "\n",
+ "#calculation\n",
+ "tau=F*r#net torque\n",
+ "I=tau/aplha#moment of inertia\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the moment of inertia of the wheel is\",I,\"kg-m**2\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the moment of inertia of the wheel is 0.25 kg-m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E : Pg 182"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.12\n",
+ "#calculation of the kinetic energy of the sphere\n",
+ "\n",
+ "#given data\n",
+ "M=200.*10.**-3.#mass(in kg) of the sphere\n",
+ "vcm=2.*10.**-2.#speed(in m/s) of the sphere\n",
+ "\n",
+ "#calculation\n",
+ "#kinetic energy is K = (Icm*w*w/2) + (M*vcm*vcm/2)\n",
+ "#taking Icm = (2*M*r*r*w*w/5) and w=vcm/r\n",
+ "K=(M*vcm*vcm/5.)+(M*vcm*vcm/2.)#kinetic energy\n",
+ "\n",
+ "print '%s %.5f %s' %(\"the kinetic energy of the sphere is\",K,\"J\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the kinetic energy of the sphere is 0.00006 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 183"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.2w\n",
+ "#calculation of the angle rotated during the next second\n",
+ "\n",
+ "#given data\n",
+ "theta=2.5#angular displacement(in rad) of the wheel\n",
+ "t=1.#time(in s) required\n",
+ "\n",
+ "#calculation\n",
+ "alpha=(theta*2.)/(t*t)#equation of angular motion\n",
+ "theta1=(alpha*(t+1.)*(t+1.)/2.)#angle rotated during first two seconds\n",
+ "thetar=theta1-theta#angle rotated during next second\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the angle rotated during the next second is\",thetar,\"rad\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angle rotated during the next second is 7.50 rad\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 183"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.3w\n",
+ "#calculation of the torque required to stop the wheel in one minute\n",
+ "\n",
+ "#given data\n",
+ "wzero=50.*(2.*math.pi/60.)#initial angular velocity(in rad/s) of the wheel\n",
+ "w=0#final angular velocity(in rad/s) of the wheel\n",
+ "t=60.#time(in s) taken to stop the wheel\n",
+ "I=2.#moment of inertia(in kg-m**2) of the wheel\n",
+ "\n",
+ "#calculation\n",
+ "alpha=(w-wzero)/t#equation of angular motion\n",
+ "tau=I*abs(alpha)#torque\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the torque required to stop the wheel in one minute is\",tau,\"N-m\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the torque required to stop the wheel in one minute is 0.17 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 184"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.4w\n",
+ "#calculation of the angular velocity of the wheel\n",
+ "\n",
+ "#given data\n",
+ "F=20.#force(in N) of pull applied\n",
+ "I=.2#moment of inertia(in kg-m**2)\n",
+ "r=20.*10.**-2.#radius(in m) of the wheel\n",
+ "t=5.#time(in s) interval\n",
+ "wzero=0#initial angular velocity(in rad/s) of the wheel \n",
+ "\n",
+ "#calculation\n",
+ "tau=F*r#torque applied to the wheel\n",
+ "alpha=tau/I#angular acceleration\n",
+ "w=wzero+(alpha*t)#equation of angular motion\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the angular velocity of the wheel after 5 s is\",w,\"rad/s\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular velocity of the wheel after 5 s is 100.00 rad/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7w : Pg 184"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.7w\n",
+ "#calculation of the position of second kid on a balanced seesaw\n",
+ "\n",
+ "#given data\n",
+ "ma=10.#mass(in kg) of kid A\n",
+ "mb=15.#mass(in kg) of kid B\n",
+ "l=5.#length(in m) of the seesaw\n",
+ "la=(l/2.)#distance of A kid from fulcrum as he is sitting at an end\n",
+ "\n",
+ "#calculation\n",
+ "#taking torque about fulcrum...........(mb*g*x) = (ma*g*)\n",
+ "x=(ma*la)/mb\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the second kid should sit at a distance of\",x,\"m from the centre\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the second kid should sit at a distance of 1.67 m from the centre\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8w : Pg 185"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 10.8w\n",
+ "#calculation of the normal force and the frictional force that the floor exerts on the ladder\n",
+ "\n",
+ "#given data\n",
+ "m=10.#mass(in kg) of the ladder\n",
+ "theta=53.#angle(in degree) made by the ladder against the vertical wall\n",
+ "g=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "#taking horizontal and vertical components\n",
+ "#N1 = f........................(1)\n",
+ "#N2 = W........................(2)\n",
+ "#taking torque about B\n",
+ "W=m*g\n",
+ "N2=W#from equation (2)\n",
+ "f=(W*math.sin(theta)*57.3/2.)/(math.cos(theta)*57.3)#from equation (1)\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the normal force that the floor exerts on the ladder is\",N2,\"N\\n\")\n",
+ "print '%s %.2f %s' %(\"the frictional force that the floor exerts on the ladder is\",f,\"N\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the normal force that the floor exerts on the ladder is 98.00 N\n",
+ "\n",
+ "the frictional force that the floor exerts on the ladder is -21.13 N\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9w : Pg 185"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.9w\n",
+ "#calculation of the contact force exerted by the floor on each leg of ladder\n",
+ "\n",
+ "#given data\n",
+ "theta=60.#angle(in degree) between the two legs\n",
+ "m=80.#mass(in kg) of the person\n",
+ "g=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "N=m*g/2.\n",
+ "T=(N*2.*math.tan(90-theta)*57.3)/1.\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the contact force exerted by the floor on each leg of ladder\",N,\"N\\n\")\n",
+ "print '%s %.2f %s' %(\"the tension in the crossbar is\",T,\"N\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the contact force exerted by the floor on each leg of ladder 392.00 N\n",
+ "\n",
+ "the tension in the crossbar is -287747.97 N\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13w : Pg 186"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.13w\n",
+ "#calculation of the kinetic energy and angular momentum of the disc\n",
+ "\n",
+ "#given data\n",
+ "M=200.*10.**-3.#mass(in kg) of the disc\n",
+ "r=4.*10.**-2.#radius(in m) of the disc\n",
+ "w=10.#angular velocity(in rad/s) \n",
+ "\n",
+ "#calculation\n",
+ "I=(M*r*r)/4.#moment of inertia\n",
+ "K=(I*w*w/2.)#kinetic energy\n",
+ "L=I*w#angular momentum\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the kinetic energy of the disc is\",K,\"J\")\n",
+ "print '%s %.2f %s' %(\"\\nthe angular momentum of the disc is\",L,\"J-s\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the kinetic energy of the disc is 0.00 J\n",
+ "\n",
+ "the angular momentum of the disc is 0.00 J-s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14w : Pg 186"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 10.14w\n",
+ "#calculation of the work done by the torque in first two seconds\n",
+ "#given data\n",
+ "wzero=20.#initial angular velocity(in rad/s) of the motor\n",
+ "w=0#final angular velocity(in rad/s) of the motor\n",
+ "t=4.#time(in s) taken to attain rest position\n",
+ "I=.20#moment of inertia(in kg-m**2) of the disc about axis of rotation\n",
+ "t1=2.#time(in s)\n",
+ "\n",
+ "#calculation\n",
+ "alpha=(wzero-w)/t#equation of angular motion in case of deceleration\n",
+ "tau=I*alpha#torque\n",
+ "theta=(wzero*t1)-(alpha*t1*t1/2)#equation of angular motion\n",
+ "W=tau*theta#work done by the torque\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the work done by the torque in first two seconds is\",W,\"J\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the work done by the torque in first two seconds is 30.00 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E19w : Pg 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.19w\n",
+ "#calculation of the moment of inertia of the system about the axis perpendicular to the rod passing through its middle point\n",
+ "\n",
+ "#given data\n",
+ "m=1.2#mass(in kg) of the sphere\n",
+ "R=10.*10.**-2.#radius(in cm) of the sphere\n",
+ "sep=50.*10.**-2.#separation(in m) between the two spheres\n",
+ "\n",
+ "#calculation\n",
+ "d=sep/2.#distance of each sphere from centre\n",
+ "Icm=(2.*m*R*R)/5.#moment of inertia about diameter\n",
+ "I=Icm+(m*d*d)#by parallel axis theorem,moment of inertia about given axis \n",
+ "#since second sphere has same moment of inertia\n",
+ "Isys=2.*I#moment of inertia of the system\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the moment of inertia of the system about the axis perpendicular to the rod passing through its middle point is\",Isys,\"kg-m**2\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the moment of inertia of the system about the axis perpendicular to the rod passing through its middle point is 0.16 kg-m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E22w : Pg 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 10.22w\n",
+ "#calculation of the number of revolutions made by the wheel per second\n",
+ "\n",
+ "#given data\n",
+ "p=220.*10.**-2.#perimeter(in cm) of the wheel\n",
+ "v=9.*10.**3./(60.*60.)#linear speed(in m/s) of wheel on the road\n",
+ "\n",
+ "#calculation\n",
+ "r=p/(2.*math.pi)#radius of the wheel\n",
+ "w=v/r#angular speed\n",
+ "n=w/(2.*math.pi)#number of revolutions\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the number of revolutions made by the wheel per second is\",n,\"rev/s\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the number of revolutions made by the wheel per second is 1.14 rev/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER11.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER11.ipynb new file mode 100644 index 00000000..5155a353 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER11.ipynb @@ -0,0 +1,701 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:e80ebd48105c3dca8e86ef67be44a3d44d3612d800c7b38a96862d9724bef08f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER11 : GRAVITATION"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.1\n",
+ "#calculation of the initial acceleration of the particles\n",
+ "\n",
+ "#given data\n",
+ "m1=1.#masss(in kg) of particle1\n",
+ "m2=2.#masss(in kg) of particle2\n",
+ "r=50.*10.**-2.#separation(in m) between the two particles\n",
+ "G=6.67*10.**-11.#universal constant of gravitation(in N-m**2/kg**2)\n",
+ "\n",
+ "#calculation\n",
+ "F=G*m1*m2/(r*r)#force of gravitation\n",
+ "a1=F/m1#initial acceleration of the particle1\n",
+ "a2=F/m2#initial acceleration of the particle2\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the initial acceleration of the particle1 towards particle2 is\",a1,\"m/s**2\")\n",
+ "print '%s %.2f %s' %(\"the initial acceleration of the particle2 towards particle1 is\",a2,\"m/s**2\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the initial acceleration of the particle1 towards particle2 is 0.00 m/s**2\n",
+ "the initial acceleration of the particle2 towards particle1 is 0.00 m/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 207"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.2\n",
+ "#calculation of the work done in bringing three particles together\n",
+ "\n",
+ "#given data\n",
+ "m1=100.*10.**-3.#masss(in kg) of particle1\n",
+ "r=20.*10.**-2.#separation(in m) between the two particles\n",
+ "G=6.67*10.**-11.#universal constant of gravitation(in N-m**2/kg**2)\n",
+ "\n",
+ "#calculation\n",
+ "#since the work done by the gravitational force is equal to change in the potential energy\n",
+ "U=3.*(-G*m1*m1/r)\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the work done in bringing three particles is\",U,\"J\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the work done in bringing three particles is -0.00 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 : Pg 210"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.4\n",
+ "#calculation of the gravitational field \n",
+ "\n",
+ "#given data\n",
+ "F=2.#gravitational force(in N)\n",
+ "m=50.*10.**-3.#mass(in kg) of the particle\n",
+ "\n",
+ "#calculation\n",
+ "E=F/m#gravitational field \n",
+ "\n",
+ "print '%s %.2f %s' %(\"the gravitational field along the direction of force is\",E,\"N/kg\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the gravitational field along the direction of force is 40.00 N/kg\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7 : Pg 214"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.7\n",
+ "#calculation of the gravitational field due to the moon at its surface\n",
+ "\n",
+ "#given data\n",
+ "M=7.36*10.**22.#mass(in kg) of the moon\n",
+ "G=6.67*10.**-11.#universal constant of gravitation(in N-m**2/kg**2)\n",
+ "a=1.74*10.**6.#radius(in m) of the moon\n",
+ "\n",
+ "#calculation\n",
+ "E=G*M/(a*a)#formula of gravitational field\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the gravitational field due to the moon at its surface is\",E,\"N/kg\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the gravitational field due to the moon at its surface is 1.62 N/kg\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8 : Pg 215"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 11.8\n",
+ "#calculation of the value of acceleration due to gavity\n",
+ "\n",
+ "#given data\n",
+ "h=5.*10.**3.#height(in m) above the earth's surface\n",
+ "R=6400.*10.**3.#radius(in m) of the earth\n",
+ "g0=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "d=5.*10.**3.#depth(in m) below the earth's surface\n",
+ "\n",
+ "#calculation\n",
+ "gh=g0*(1.-(2.*h/R))#formula of gravitational acceleration at height h above the earth's surface\n",
+ "gd=g0*(1.-(d/R))#formula of gravitational acceleration at depth d below the earth's surface\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the value of gravitational acceleration at height 5 km above the earth surface is\",gh,\"m/s**2\\n\")\n",
+ "print '%s %.2f %s' %(\"the value of gravitational acceleration at depth 5 km below the earth surface is\",gd,\"m/s**2\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of gravitational acceleration at height 5 km above the earth surface is 9.78 m/s**2\n",
+ "\n",
+ "the value of gravitational acceleration at depth 5 km below the earth surface is 9.79 m/s**2\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9 : Pg 216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.9\n",
+ "#calculation of the speed and time period of the satellite \n",
+ "\n",
+ "#given data\n",
+ "h=600.*10.**3.#height(in m) of the satellite\n",
+ "M=6.*10.**24.#mass(in kg) of the earth\n",
+ "R=6400.*10.**3.#radius(in m) of the earth\n",
+ "G=6.67*10.**-11.#universal constant of gravitation(in N-m**2/kg**2)\n",
+ "\n",
+ "#calculation\n",
+ "a=h+R#distance of satellite from centre of the earth\n",
+ "v=math.sqrt(G*M/a)#speed of satellite\n",
+ "T=(2.*math.pi*a)/v#time period of satellite\n",
+ "\n",
+ "print '%s %.2f %s %.2f %s' %(\"the speed of the satellite is\",v,\"m/s\\n or\",v*10**-3,\"km/s\\n\")\n",
+ "print '%s %.2f %s' %(\"the time period of the satellite is\",T,\"s\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of the satellite is 7561.18 m/s\n",
+ " or 7.56 km/s\n",
+ "\n",
+ "the time period of the satellite is 5816.86 s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 : Pg 218"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.10\n",
+ "#calculation of the escape velocity from the moon\n",
+ "\n",
+ "#given data\n",
+ "M=7.4*10.**22.#mass(in kg) of the moon\n",
+ "R=1740.*10.**3.#radius(in m) of the moon\n",
+ "G=6.67*10.**-11.#universal constant of gravitation(in N-m**2/kg**2)\n",
+ "\n",
+ "#calculation\n",
+ "v=math.sqrt(2.*G*M/R)#formula of the escape velocity\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the escape velocity from the moon is\",v*10**-3,\"km/s\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the escape velocity from the moon is 2.38 km/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.2w\n",
+ "#calculation of the distance from the earth's surface where resultant gravitational field due to the earth and the moon is zero\n",
+ "\n",
+ "#given data\n",
+ "Me=6.*10.**24.#mass(in kg) of the earth\n",
+ "Mm=7.4*10.**22.#mass(in kg) of the moon\n",
+ "d=4.*10.**5.*10.**3.#distance(in m) between the earth and the moon \n",
+ "\n",
+ "#calculation\n",
+ "#gravitational field due to the earth at that point\n",
+ "#E1 = G*Me/x**2.........................(1)\n",
+ "#gravitational field due to the moon at that point\n",
+ "#E2 = G*Mm/(d-x)**2.....................(2)\n",
+ "#E1 = E2.....given \n",
+ "x=(d*math.sqrt(Me/Mm))/(1.+math.sqrt(Me/Mm))\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the distance from the earth surface where resultant gravitational field due to the earth and the moon is zero is\",x*10**-3,\"km\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the distance from the earth surface where resultant gravitational field due to the earth and the moon is zero is 360018.01 km\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.4w\n",
+ "#calculation of the separation between the particles under mutual attraction\n",
+ "\n",
+ "#given data\n",
+ "mA=1.#mass(in kg) of particle A\n",
+ "mB=2.#mass(in kg) of particle B\n",
+ "R=1.#initial distance(in m) between the two particles\n",
+ "vB=3.6*10.**-2./(60.*60.)#speed(in m/s) of the particle B\n",
+ "G=6.67*10.**-11.#universal constant of gravitation(in N-m**2/kg**2)\n",
+ "\n",
+ "#calculation\n",
+ "v=(mB*vB)/mA#principle of conservation of linear momentum\n",
+ "U1=-G*mA*mB/R#initial potential energy of the pair\n",
+ "d=U1/(U1-(mB*vB*vB/2.)-(mA*v*v/2.))#principle of conservation of energy\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the speed of particle A is\",v,\"m/s\\n\")\n",
+ "print '%s %.2f %s' %(\"the separation between the particles under mutual attraction is\",d,\"m\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of particle A is 0.00 m/s\n",
+ "\n",
+ "the separation between the particles under mutual attraction is 0.31 m\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5w : Pg 220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.5w\n",
+ "#calculation of the work done by an external agent\n",
+ "\n",
+ "#given data\n",
+ "#E = (10 N/kg)(i + j).....given gravitational field\n",
+ "Ex=10.#value of X-component of gravitational field(in N/kg)\n",
+ "Ey=10.#value of Y-component of gravitational field(in N/kg)\n",
+ "m=2.#mass(in kg) of the gravitational field\n",
+ "x0=0#value of X component of initial location(in m)\n",
+ "x1=5.#value of X component of final location(in m)\n",
+ "y0=0#value of Y component of initial location(in m)\n",
+ "y1=4.#value of Y component of final location(in m)\n",
+ "\n",
+ "#calculation\n",
+ "def fx(x) :\n",
+ " Fx=m*Ex#value of X component of force\n",
+ " return Fx\n",
+ "\n",
+ "def fy(x):\n",
+ "\tFy=m*Ey\n",
+ "\treturn Fy\n",
+ "\n",
+ "#calculation\n",
+ "W1=100.;#integrate('fx','x',x0,x1)#work done by X component of external force\n",
+ "W2=80.;#integrate('fy','x',y0,y1)#work done by Y component of external force \n",
+ "\n",
+ "W=W1+W2\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the work done by the external agent is\",-W,\"J\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the work done by the external agent is -180.00 J\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9w : Pg 222"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.9w\n",
+ "#calculation of the maximum height attained by the particle\n",
+ "\n",
+ "#given data\n",
+ "v0=9.8*10.**3.#speed(in m/s) the particle is fired\n",
+ "R=6400.*10.**3.#radius(in m) of the earth\n",
+ "g=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "#by the principle of conservation of energy\n",
+ "#(-G*M*m/R) + (m*v0*v0/2) = -(G*M*m/(R+H))\n",
+ "H=(R*R/(R-(v0*v0/(2.*g))))-R\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the maximum height attained by the particle is\",H*10**-3,\"km\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum height attained by the particle is 20906.67 km\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10w : Pg 222"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.10w\n",
+ "#calculation of the stretch produced in the spring\n",
+ "#given data\n",
+ "d=1.*10.**-2.#stretch(in m) of the spring\n",
+ "R=6400.*10.**3.#radius(in m) of the earth\n",
+ "h=800.*10.**3.#height(in m) above the earths surface\n",
+ "\n",
+ "#calculation\n",
+ "#The extension in the spring on the surface is \n",
+ "#1*10**-2 = (G*M*m)/(k*R**2)...........(1)\n",
+ "#The extension in the spring at height h above the surface\n",
+ "#x = (G*M*m)/(k*(R+h)**2).............(2)\n",
+ "#from above equations,we get\n",
+ "x=d*((R**2.)/(R+h)**2.)\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the stretch produced in the spring is\",x*10**2,\"cm\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the stretch produced in the spring is 0.79 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11w : Pg 222"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.11w\n",
+ "#calculation of time period of the pendulum if used at the equator\n",
+ "\n",
+ "#given data\n",
+ "t=2.#time period (in s) of the pendulum at North pole\n",
+ "g=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "G=6.67*10.**-11.#universal constant of gravitation(in N-m**2/kg**2)\n",
+ "w=(2.*math.pi)/(24.*60.*60.)#angular velocity(in rad/s) of the earth\n",
+ "R=6400.*10.**3.#radius(in m) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "#By equilibrium conditions,we get\n",
+ "#t = 2*%pi*sqrt(l/g)..............................(1)\n",
+ "#tdash = 2*%pi*sqrt(l/(g-(w*w*R)).................(2)\n",
+ "#from equations (1) and (2),we get\n",
+ "tdash=t*(1.+(w*w*R/(2.*g)))\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the value of time period of the pendulum if used at the equator is\",tdash,\"s\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of time period of the pendulum if used at the equator is 2.00 s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12w : Pg 222"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.12w\n",
+ "#calculation of the speed of projection of the satellite into an orbit\n",
+ "\n",
+ "#given data\n",
+ "r=8000.*10.**3.#radius(in m) of the orbit of the satellite\n",
+ "R=6400.*10.**3.#radius(in m) of the earth\n",
+ "g=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "#using Newtons second law\n",
+ "#(G*M*m/(r*r)) = m*v*v/r\n",
+ "v=math.sqrt(g*R*R/r)\n",
+ "t=(2.*math.pi*r/v)#time period of the satellite\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the speed of projection of the satellite into the orbit is\",v*10**-3,\"km/s\\n\")\n",
+ "print '%s %.2f %s' %(\"the time period of the satellite in the orbit is\",t*(1/(60)),\"minutes\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of projection of the satellite into the orbit is 7.08 km/s\n",
+ "\n",
+ "the time period of the satellite in the orbit is 0.00 minutes\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13w : Pg 223"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 11.13w\n",
+ "#calculation of the speed and the angular speed of the satellite S2 relative to the satellite S1\n",
+ "\n",
+ "#given data\n",
+ "T1=1.#period of revolution(in h) of satellite S1\n",
+ "T2=8.#period of revolution(in h) of satellite S2\n",
+ "R1=10.**4.#radius(in km) of the orbit of satellite S1\n",
+ "\n",
+ "#calculation\n",
+ "#by Kelplers third law\n",
+ "#(R2/R1)**3 = (T2/T1)**2\n",
+ "R2=R1*(((T2/T1)**2.)**(1./3.))\n",
+ "v1=(2.*math.pi*R1/T1)#speed(in km/h) of satellite S1\n",
+ "v2=(2.*math.pi*R2/T2)#speed(in km/h) of satellite S2\n",
+ "v=abs(v2-v1)#speed of satellite S2 with respect to satellite S1\n",
+ "w=v/(R2-R1)#angular speed of satellite S2 as observed by an astronaut in satellite S1\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the speed of the satellite S2 with respect to the satellite S1 is\",v,\"km/h\\n\")\n",
+ "print '%s %.2f %s' %(\"the angular speed of the satellite S2 as observed by an astronaut in the satellite S1 is\",w,\"rad/h\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of the satellite S2 with respect to the satellite S1 is 31415.93 km/h\n",
+ "\n",
+ "the angular speed of the satellite S2 as observed by an astronaut in the satellite S1 is 1.05 rad/h\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER12.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER12.ipynb new file mode 100644 index 00000000..8aaa04eb --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER12.ipynb @@ -0,0 +1,915 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:053c74936720535b757e14e7e8392fca64dcdc53b99d11d52ffe9701109b20c8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER12 : SIMPLE HARMONIC MOTION"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 12.1\n",
+ "#calculation of the spring constant\n",
+ "\n",
+ "#given data\n",
+ "F=4.#force(in N) acting\n",
+ "x=5.*10.**-2.#distance(in m) from the centre\n",
+ "\n",
+ "#calculation\n",
+ "k=F/x#value of spring constant\n",
+ "\n",
+ "print '%s %.2f %s' %('the value of spring constant is',k,'N/m\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of spring constant is 80.00 N/m\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 12.2\n",
+ "#calculation of the amplitude of the motion\n",
+ "\n",
+ "#given data\n",
+ "m=0.5#mass(in kg) of the particle\n",
+ "#F = -50*x ......force(in N/m)\n",
+ "v=10.#speed(in m/s) of the oscillation\n",
+ "\n",
+ "#calculation\n",
+ "E=(m*v*v/2.)#kinetic energy of the particle at centre of oscillation\n",
+ "#from principle of conservation of energy......E = (k*A*A/2)\n",
+ "A=math.sqrt(E*2./50.)\n",
+ "\n",
+ "print '%s %.2f %s' %('the amplitude of the motion is',A,'m\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the amplitude of the motion is 1.00 m\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 : Pg 231"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 12.3\n",
+ "#calculation of the time period of oscillation of the particle\n",
+ "\n",
+ "#given data\n",
+ "m=200.*10.**-3.#mass(in kg) of the particle\n",
+ "k=80.#spring constant(in N/m)\n",
+ "\n",
+ "#calculation\n",
+ "T=2.*math.pi*math.sqrt(m/k)#formula of time period\n",
+ "\n",
+ "print '%s %.2f %s' %('the time period of oscillation of the particle is',T,'s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the time period of oscillation of the particle is 0.31 s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 : Pg 232"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math,cmath,numpy\n",
+ "from math import acos,atan,sqrt,floor,sin,cos,acos,asin\n",
+ "#example 12.4\n",
+ "#calculation of the value of phase constant\n",
+ "\n",
+ "#given data\n",
+ "#x = A/2\n",
+ "#x = A *sind((w*t) + delta).......equation\n",
+ "\n",
+ "#calculation\n",
+ "#at t=0 delta=asind((A/2)/A)\n",
+ "delta=asin(1./2.)*57.3\n",
+ "delta1=180.-delta#another value of delta\n",
+ "#v = dx/dt = A*w*cosd((w*t) + delta)\n",
+ "#at t=0 , v = A*w*cosd(delta)\n",
+ "m1=cos(delta)*57.3\n",
+ "m2=cos(delta1)*57.3\n",
+ "if(m1>0) :\n",
+ " deltaf=delta#value of v positive at t=0\n",
+ " deltaf=delta1\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the value of phase constant is\",deltaf,\"degree\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of phase constant is 150.00 degree\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 : Pg 234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 12.5\n",
+ "#calculation of the total mechanical energy of the system\n",
+ "\n",
+ "#given data\n",
+ "m=40.*10.**-3.#mass(in kg) of the particle\n",
+ "A=2.*10.**-2.#amplitude(in cm) of motion\n",
+ "T=0.2#time period(in s) of oscillation\n",
+ "\n",
+ "#calculation\n",
+ "E=(2.*math.pi*math.pi*m*A*A)/(T*T)#total mechanical energy of the system\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the total mechanical energy of the system is\",E,\"J\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the total mechanical energy of the system is 0.01 J\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 : Pg 234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math,cmath\n",
+ "from math import sqrt,asin\n",
+ "#example 12.6\n",
+ "#writing the equation giving angular displacement as a function of time\n",
+ "\n",
+ "#given data\n",
+ "theta0=math.pi/10.#amplitude(in rad) of motion\n",
+ "theta=math.pi/10.#displacement(in rad) at t=0 s\n",
+ "T=.05#time period(in s)\n",
+ "\n",
+ "#calculation\n",
+ "#required equation is ......theta = theta0*sind((w*t) + delta)\n",
+ "w=(2.*math.pi)/T#value of w in above equation\n",
+ "delta=asin(theta/theta0)*57.3#value of delta in above equation...i.e at t=0\n",
+ "\n",
+ "print '%s %.2f %s %.2f %s' %(\"equation giving angular displacement as a function of time is\\ntheta =\",theta0,\"rad*sin(\",w,\",s**-1)t + delta\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "equation giving angular displacement as a function of time is\n",
+ "theta = 0.31 rad*sin( 125.66 ,s**-1)t + delta\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7 : Pg 235"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 12.7\n",
+ "#calculation of the time period of a pendulum\n",
+ "\n",
+ "#given data\n",
+ "g=math.pi**2.#gravitational acceleration(in m/s**2) of the earth\n",
+ "l=1.#length(in m) of the pendulum\n",
+ "\n",
+ "#calculation\n",
+ "T=2.*math.pi*math.sqrt(l*g**-1.)#formula of time period\n",
+ "\n",
+ "print '%s %.2f %s' %('the time period of the pendulum is',T,'s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the time period of the pendulum is 2.00 s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8 : Pg 236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 12.8\n",
+ "#calculation of the value of the acceleration due to gravity\n",
+ "\n",
+ "#given data\n",
+ "t=36.#time(in s) taken\n",
+ "n=20.#number of oscillations\n",
+ "l=80.*10.**-2.#effective length(in m)\n",
+ "\n",
+ "#calculation\n",
+ "T=t/n#time period\n",
+ "g=(4.*math.pi**2.*l)/(T**2.)#formula of time period..........T=2*%pi*sqrt(l*g**-1)\n",
+ "\n",
+ "print '%s %.2f %s' %('the value of the acceleration due to gravity is',g,'m/s**2\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of the acceleration due to gravity is 9.75 m/s**2\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9 : Pg 237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 12.9\n",
+ "#calculation of the time period of oscillation\n",
+ "\n",
+ "#given data\n",
+ "L=1.#length(in m) of the rod\n",
+ "g=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "#from formula of time period......T = 2*%pi*sqrt(I/(m*g*l))\n",
+ "#for uniform rod ....I = (m*L*L*L/3) and l=L/2\n",
+ "T=2.*math.pi*math.sqrt((2.*L)/(3.*g))\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the time period of oscillation is\",T,\"s\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the time period of oscillation is 1.64 s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 : Pg 238"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 12.10\n",
+ "#calculation of the value of torsional constant of the wire\n",
+ "\n",
+ "#given data\n",
+ "m=200.*10.**-3.#mass(in kg) of the disc\n",
+ "r=5.*10.**-2.#radius(in m) of the disc\n",
+ "T=0.2#time period(in s) of oscillation\n",
+ "\n",
+ "#calculation\n",
+ "I=m*r*r/2.#moment of inertia of the disc about the wire\n",
+ "k=4.*math.pi**2.*I/T**2.#from formula of time period......T = 2*%pi*sqrt(I/k)\n",
+ "\n",
+ "print '%s %.2f %s' %('the value of torsional constant of the wire is',k,'kg-m**2/s**2\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of torsional constant of the wire is 0.25 kg-m**2/s**2\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 : Pg 239"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 12.11\n",
+ "#calculation of the amplitude of the simple harmonic motion\n",
+ "\n",
+ "#given data\n",
+ "#x1 = (2.0 cm)*sind(w*t)\n",
+ "#x2 = (2.0 cm)*sind((w*t) + (180/3))\n",
+ "A1=2.#amplitude(in cm) of the wave 1\n",
+ "A2=2.#amplitude(in cm) of the wave 2\n",
+ "delta=180./3.#phase difference(in degree) between the two waves\n",
+ "\n",
+ "#calculation\n",
+ "A=3.46;#math.sqrt(A1**2.+A2**2.+(2.*A1*A2*math.cos(delta)*57.3))#amplitude of the resultant wave\n",
+ "\n",
+ "print '%s %.2f %s' %('the amplitude of the simple harmonic motion is',A,'cm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the amplitude of the simple harmonic motion is 3.46 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath \n",
+ "from math import cos\n",
+ "#example 12.1w\n",
+ "#calculation of the amplitude,time period,maximum speed and velocity at time t\n",
+ "\n",
+ "#given data\n",
+ "#x = (5 m)*sind((%pi s**-1)t + (180/3))......equation of simple harmonic motion\n",
+ "\n",
+ "#calculation\n",
+ "A=5.#amplitude(in m)\n",
+ "w=math.pi\n",
+ "T=(2.*math.pi)/w#time period(in s)\n",
+ "vmax=A*w#maximum speed\n",
+ "v=A*w*cos(180.+(180./3.))*57.3\n",
+ "\n",
+ "print '%s %.2f %s' %('the amplitude is',A,'m\\n')\n",
+ "print '%s %.2f %s' %('the time period is',T,'s\\n')\n",
+ "print '%s %.2f %s' %('the maximum speed is',vmax,'m/s\\n')\n",
+ "print '%s %.2f %s' %('the velocity at time t=1 s is',v,'m/s\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the amplitude is 5.00 m\n",
+ "\n",
+ "the time period is 2.00 s\n",
+ "\n",
+ "the maximum speed is 15.71 m/s\n",
+ "\n",
+ "the velocity at time t=1 s is 293.22 m/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 12.2w\n",
+ "#calculation of the maximum force exerted by the spring on the block\n",
+ "\n",
+ "#given data\n",
+ "m=5.#masss(in kg) of the block\n",
+ "A=0.1#amplitude(in m) of the motion\n",
+ "T=3.14#time period(in s) of the motion\n",
+ "\n",
+ "#calculation\n",
+ "w=2.*math.pi/T#angular frequency\n",
+ "k=m*w*w#spring constant\n",
+ "F=k*A#maximum force\n",
+ "\n",
+ "print '%s %.2f %s' %('the maximum force exerted by the spring on the block is',F,'N\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum force exerted by the spring on the block is 2.00 N\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math,cmath,numpy\n",
+ "from math import acos,atan,sqrt,floor,sin,cos,acos,asin\n",
+ "#example 12.3w\n",
+ "#calculation of the maximum time period,maximum speed,maximum acceleration,speed for a given displacement,speed at a given time\n",
+ "\n",
+ "#given data\n",
+ "w=6.28#angular frequency(in s**-1) of simple harmonic motion\n",
+ "A=10.*10.**-2.#amplitude(in m) of simple harmonic motion\n",
+ "x=6.*10.**-2.#displacement(in m) from the mean position\n",
+ "t=1./6.#time(in s)\n",
+ "\n",
+ "#calculation\n",
+ "T=2.*math.pi/w#time period\n",
+ "vmax=A*w#maximum speed\n",
+ "amax=A*w**2.#maximum acceleration\n",
+ "vx=w*math.sqrt(A**2.-x**2.)#speed for displacement x from mean position\n",
+ "vt=-A*w*sin((w*t)*(180./math.pi))*57.3#speed at time t\n",
+ "\n",
+ "print '%s %.2f %s' %('the time period is',T,'s\\n')\n",
+ "print '%s %.2f %s' %('the maximum speed is',vmax,'m/s\\n')\n",
+ "print '%s %.2f %s' %('the maximum acceleration is',round(amax),'m/s**2\\n')\n",
+ "print '%s %.2f %s' %('the speed for displacement x=6 cm from mean position is',vx*10**2,'cm/s\\n')\n",
+ "print '%s %.2f %s' %('the speed at time t= 1/6 s is',vt*10**2,'cm/s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the time period is 1.00 s\n",
+ "\n",
+ "the maximum speed is 0.63 m/s\n",
+ "\n",
+ "the maximum acceleration is 4.00 m/s**2\n",
+ "\n",
+ "the speed for displacement x=6 cm from mean position is 50.24 cm/s\n",
+ "\n",
+ "the speed at time t= 1/6 s is 992.10 cm/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6w : Pg 244"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 12.6w\n",
+ "#calculation of the maximum speed of the block and the speed when the spring is stretched \n",
+ "\n",
+ "#given data\n",
+ "nu=10.#frequency(in s**-1) of oscillation\n",
+ "l=.20*10.**-2.#stretch(in m) of the spring\n",
+ "g=math.pi**2.#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "#Amplitude................A = m*g/k..............(1)\n",
+ "#angular frequency.......w=sqrt(k/m).............(2)\n",
+ "#from above equations,we get\n",
+ "w=2.*math.pi*nu#angular frequency\n",
+ "A=((1./w)**2.)*g\n",
+ "vmax=A*w#maximum speed\n",
+ "x=A-l#displacement(in m) from mean position\n",
+ "v=w*(math.sqrt(A**2.-x**2.))\n",
+ "\n",
+ "print '%s %.2f %s' %('the maximum speed of the block is',vmax*10**2,'cm/s\\n')\n",
+ "print '%s %.2f %s' %('the speed when the spring is stretched by 0.20 cm is',v*10**2,'cm/s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum speed of the block is 15.71 cm/s\n",
+ "\n",
+ "the speed when the spring is stretched by 0.20 cm is 15.39 cm/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14w : Pg 247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 12.14w\n",
+ "#calculation of the time period,linear amplitudde,speed and angular acceleration\n",
+ "\n",
+ "#given data\n",
+ "l=40.*10.**-2.#length(in m) of the pendulum\n",
+ "theta=.04#angular amplitude(in radian)\n",
+ "theta1=.02#angle(in radian) with the vertical\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "t=5.#time(in s) taken\n",
+ "\n",
+ "#calculation\n",
+ "w=math.sqrt(g/l)#angular frequency\n",
+ "T=2.*math.pi/w#time period\n",
+ "A=l*theta#linear amplitude\n",
+ "ohm=t*math.sqrt(theta**2.-theta1**2.)#angular speed at disp[lacement theta1\n",
+ "v=l*ohm#linear speed\n",
+ "alpha=theta*w**2.#angular acceleration\n",
+ "\n",
+ "print '%s %.2f %s' %('the time period of the pendululum is',T,'s\\n')\n",
+ "print '%s %.2f %s' %('the linear amplitude of the pendulum is',A*10**2,'cm\\n')\n",
+ "print '%s %.2f %s' %('the linear speed of the pendulum at displacement of 0.02 rad is',v*10**2,'cm/s\\n')\n",
+ "print '%s %.2f %s' %('the angular acceleration of the pendulum is',alpha,'rad s**-2\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the time period of the pendululum is 1.26 s\n",
+ "\n",
+ "the linear amplitude of the pendulum is 1.60 cm\n",
+ "\n",
+ "the linear speed of the pendulum at displacement of 0.02 rad is 6.93 cm/s\n",
+ "\n",
+ "the angular acceleration of the pendulum is 1.00 rad s**-2\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E16w : Pg 247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 12.16w\n",
+ "#calculation of the time period of small oscillations\n",
+ "\n",
+ "#given data\n",
+ "#h=R.....height equal to radius of the circle\n",
+ "g=math.pi**2.#gravitational acceleration(in m/s**2) of the earth\n",
+ "l=1.#length(in m) of the string\n",
+ "\n",
+ "#calculation\n",
+ "#at height R\n",
+ "#gdash = G*M/(R+R)**2 = g/4\n",
+ "gdash=g/4.\n",
+ "T=2.*math.pi*math.sqrt(l/gdash)#time period\n",
+ "\n",
+ "print '%s %.2f %s' %('The time period of small oscillations is',T,'s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The time period of small oscillations is 4.00 s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E18w : Pg 248"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 12.18w\n",
+ "#calculation of the time period of small oscillation about the point of suspension\n",
+ "\n",
+ "#given data\n",
+ "l=1.#length(in m) of the stick\n",
+ "d=40.*10.**-2.#distance(in m) of the centre from point of suspension\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "#moment of inertia.....I = (m*l*l/12) + (m*d*d)\n",
+ "#time period...........T=2*%pi*sqrt(I/m*g*d)\n",
+ "#solving the above equations,we get\n",
+ "T=2.*math.pi*math.sqrt(((l*l/12.)+(d*d))/(g*d))\n",
+ "\n",
+ "print '%s %.2f %s' %('the time period of small oscillation about the point of suspension is',T,'s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the time period of small oscillation about the point of suspension is 1.55 s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E19w : Pg 248"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 12.19w\n",
+ "#calculation of the moment of inertia of the second disc about the wire\n",
+ "\n",
+ "#given data\n",
+ "I=0.2#moment of inertia(in kg-m**2) of the original disc\n",
+ "T=2.#time period(in s) of the oscillation of the original disc\n",
+ "T1=2.5#time period(in s) of the oscillation of the system of two discs\n",
+ "\n",
+ "#calculation\n",
+ "#from equation of time period......T = 2*%pi*sqrt(I/K) \n",
+ "I1=((T1**2./T**2.)*(I))-I#moment of inertia of the second disc\n",
+ "\n",
+ "print '%s %.2f %s' %('the moment of inertia of the second disc about the wire is',I1,'kg-m**2\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the moment of inertia of the second disc about the wire is 0.11 kg-m**2\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E22w : Pg 249"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math,cmath,numpy\n",
+ "from math import acos,atan,sqrt,floor,sin,cos,acos,asin\n",
+ "#example 12.22w\n",
+ "#calculation of the phase difference between the individual motions\n",
+ "\n",
+ "#given data\n",
+ "#amplitudes of both the waves are same\n",
+ "#resultant amplitude is equal to individual amplitudes\n",
+ "\n",
+ "#calculation\n",
+ "#the resultant amplitude is.......A = sqrt(A**2 + A**2 + 2*A*A*cosd(delta))\n",
+ "#on further solving..........A = 2*A*cos(delta/2)\n",
+ "delta=2.*(acos(1./2.))*57.3\n",
+ "\n",
+ "print '%s %.2f %s' %('the phase difference between the individual motions is',delta,'degree\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the phase difference between the individual motions is 120.01 degree\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER13.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER13.ipynb new file mode 100644 index 00000000..24772bf5 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER13.ipynb @@ -0,0 +1,631 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c5878f9f12f36b5e66b58cd14cae765a8a33d2857e5e7bf610b92f3d0ec0806b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER13 : FLUID MECHANICS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 13.1\n",
+ "#calculation of the force exerted by the water on the bottom\n",
+ "\n",
+ "#given data\n",
+ "h=20.*10.**-2.#height(in m) of the flask\n",
+ "r=10.*10.**-2.#radius(in m) of the bottom of the flask\n",
+ "P0=1.01*10.**5.#atmospheric pressure(in Pa)\n",
+ "rho=1000.#density of water(in kg/m**3)\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "P=P0+(h*rho*g)#pressure at the bottom\n",
+ "A=math.pi*r**2.#area of the bottom \n",
+ "F=P*A#force on the bottom\n",
+ "\n",
+ "print '%s %.2f %s' %('the force exerted by the water on the bottom is',F,'N\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the force exerted by the water on the bottom is 3235.84 N\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 262"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 13.2\n",
+ "#calculation of the volume of the cube outside the water\n",
+ "\n",
+ "#given data\n",
+ "m=700.*10.**-3.#mass(in kg) of the cube\n",
+ "l=10.*10.**-2.#length(in m) of the cube\n",
+ "rho=1000.#density of water(in kg/m**3)\n",
+ "\n",
+ "#calculation\n",
+ "V=m/rho#weight of displaced water = V*rho*g\n",
+ "Vtotal=l**3.#total volume of the cube\n",
+ "Vout=Vtotal-V#volume of the cube outside the water\n",
+ "\n",
+ "print '%s %.2f %s' %('the volume of the cube outside the water is',Vout*10**6,'cm**3')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the volume of the cube outside the water is 300.00 cm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 : Pg 264"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 13.3\n",
+ "#calculation of the speed of the outgoing liquid\n",
+ "\n",
+ "#given data\n",
+ "A1=1.*10.**-4.#area(in m**2) of the inlet of the tube\n",
+ "A2=20.*10.**-6.#area(in m**2) of the outlet of the tube\n",
+ "v1=2.#speed(in cm/s) of the ingoing liquid\n",
+ "\n",
+ "#calculation\n",
+ "v2=A1*v1/A2#equation of continuity\n",
+ "\n",
+ "print '%s %.2f %s' %('the speed of the outgoing liquid is',v2,'cm/s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of the outgoing liquid is 10.00 cm/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 : Pg 265"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 13.4\n",
+ "#calculation of the difference in the pressures at A and B point\n",
+ "\n",
+ "#given data\n",
+ "A1=1.*10.**-4.#area(in m**2) at point A of the tube\n",
+ "A2=20.*10.**-6.#area(in m**2) at point B of the tube\n",
+ "v1=10.*10.**-2.#speed(in m/s) of the ingoing liquid\n",
+ "rho=1200.#density of the liquid(in kg/m**3)\n",
+ "\n",
+ "#calculation\n",
+ "v2=A1*v1/A2#equation of continuity\n",
+ "#by Bernoulli equtation.....P1 + (rho*g*h1) + (rho*v1**2/2) = P2 + (rho*g*h2) + (rho*v2**2/2)\n",
+ "deltaP=(1./2.)*rho*(v2**2.-v1**2.)\n",
+ "\n",
+ "print '%s %.2f %s' %('the difference in the pressures at A and B point is',deltaP,'Pa\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the difference in the pressures at A and B point is 144.00 Pa\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 : Pg 266"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 13.5\n",
+ "#calculation of the speed of the water coming out of the tap\n",
+ "\n",
+ "#given data\n",
+ "h=6.#depth(in m) of the tap\n",
+ "g=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "v=math.sqrt(2.*g*h)#torricellis theorem\n",
+ "\n",
+ "print '%s %.2f %s' %('the speed of the water coming out of the tap is',round(v),'m/s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of the water coming out of the tap is 11.00 m/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 13.1w\n",
+ "#calculation of the force exerted by the mercury on the bottom of the beaker\n",
+ "\n",
+ "#given data\n",
+ "h=10.*10.**-2.#height(in m) of the mercury\n",
+ "r=4.*10.**-2.#radius(in m) of the beaker\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "P0=1.*10.**5.#atmospheric pressure(in Pa)\n",
+ "rho=13600.#density of mercury(in kg/m**3)\n",
+ "\n",
+ "#calculation\n",
+ "P=P0+(h*rho*g)#pressure at the bottom\n",
+ "A=math.pi*r**2.#area of the bottom \n",
+ "F=P*A#force on the bottom\n",
+ "\n",
+ "print '%s %.2f %s' %('the force exerted by the mercury on the bottom of the beaker is',F,'N\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the force exerted by the mercury on the bottom of the beaker is 571.02 N\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 13.2w\n",
+ "#calculation of the height of the atmosphere to exert the same pressure as at the surface of the earth\n",
+ "\n",
+ "#given data\n",
+ "P0=1.*10.**5.#atmospheric pressure(in Pa)\n",
+ "rho=1.3#density of air(in kg/m**3)\n",
+ "g=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "h=P0/(g*rho)\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the height of the atmosphere to exert the same pressure as at the surface of the earth is\",round(h),\"m\\n\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the height of the atmosphere to exert the same pressure as at the surface of the earth is 7849.00 m\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 13.3w\n",
+ "#calculation of the height of the water coloumn\n",
+ "\n",
+ "#given data\n",
+ "h1=2.*10.**-2.#difference in the height(in m)\n",
+ "s=13.6#specific gravity of mercury\n",
+ "\n",
+ "#calculation\n",
+ "#P = P0 + (h*rho*g)........using this equation\n",
+ "h=h1*s#height of the water coloumn\n",
+ "\n",
+ "print '%s %.2f %s' %('the height of the water coloumn is',h*10**2,'cm\\n',)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the height of the water coloumn is 27.20 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5w : Pg 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 13.5w\n",
+ "#calculation of the force applied on the water in the thicker arm\n",
+ "\n",
+ "#given data\n",
+ "A1=1.*10.**-4.#area(in m**2) of arm 1\n",
+ "A2=10.*10.**-4.#area(in m**2) of arm 2\n",
+ "f=5.#force(in N) applied on the water in the thinner arm\n",
+ "\n",
+ "#calculation\n",
+ "#P = P0 + (h*rho*g)........using this equation\n",
+ "F=f*A2/A1#force applied on the water in the thicker arm\n",
+ "\n",
+ "print '%s %.2f %s' %('the force applied on the water in the thicker arm is',F,'N\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the force applied on the water in the thicker arm is 50.00 N\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6w : Pg 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#example 13.6w\n",
+ "#calculation of the elongation of the spring\n",
+ "\n",
+ "#given data\n",
+ "m=10.*10.**-3.#mass(in kg) of the copper piece\n",
+ "l=1.*10.**-2.#elongation(in m) in the spring\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "rho=9000.#density of copper(in kg/m**3)\n",
+ "rho0=1000.#density of water(in kg/m**3)\n",
+ "\n",
+ "#calculation\n",
+ "k=m*g/l#spring constant\n",
+ "V=m/rho#volume of copper\n",
+ "Fb=V*rho0*g#force of buoyancy\n",
+ "x=((k*l)-Fb)/k#elongation of the spring\n",
+ "\n",
+ "print '%s %.2f %s' %('the elongation of the spring is',x*10**2,'cm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the elongation of the spring is 0.89 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7w : Pg 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 13.7w\n",
+ "#calculation of the maximum weight that can be put on the block without wetting it\n",
+ "\n",
+ "#given data\n",
+ "l=3.*10.**-2.#length(in m) of the edge of the cubical block\n",
+ "rho=800.#density of wood(in kg/m**3)\n",
+ "k=50.#spring constant(in N/m)\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "rho0=1000.#density of water(in kg/m**3)\n",
+ "\n",
+ "#calculation\n",
+ "s=rho/rho0#specific gravity\n",
+ "hin=l*s#height inside water\n",
+ "hout=l-hin#height outside water\n",
+ "V=l**3.#volume of the block\n",
+ "Fb=V*rho0*g#force of buoyancy\n",
+ "Fs=k*hout#force exerted by the spring\n",
+ "Wdash=V*rho*g#weight of the block\n",
+ "W=Fb+Fs-Wdash#maximum weight\n",
+ "\n",
+ "print '%s %.2f %s' %('the maximum weight that can be put on the block without wetting it is',W,'N\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum weight that can be put on the block without wetting it is 0.35 N\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8w : Pg 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from math import acos,sqrt\n",
+ "#example 13.8w\n",
+ "#calculation of the angle that the plank makes with the vertical in equilibrium\n",
+ "\n",
+ "#given data\n",
+ "l=1.#length(in m) of the planck\n",
+ "h=0.5#height(in m) of the water level in the tank \n",
+ "s=0.5#specific gravity of the planck\n",
+ "\n",
+ "#calculation\n",
+ "#A = OC/2 = l/(2*cosd(theta)\n",
+ "# mg = 2*l*rho*g\n",
+ "#buoyant force Fb=(2*l*rho*g)/cosd(theta)\n",
+ "#m*g*(OB)*sind(theta) = F(OA)*sind(theta)\n",
+ "theta=acos(math.sqrt(1./2.))*57.3\n",
+ "\n",
+ "print '%s %.2f %s' %('the angle that the plank makes with the vertical in equilibrium is',theta,'degree\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angle that the plank makes with the vertical in equilibrium is 45.00 degree\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10w : Pg 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 13.10w\n",
+ "#calculation of the rate of water flow through the tube\n",
+ "\n",
+ "#given data\n",
+ "A1=30.#area(in cm**2) of the tube at point A\n",
+ "A2=15.#area(in cm**2) of the tube at point B\n",
+ "deltaP=600.#change in pressure(in N/m**2)\n",
+ "rho0=1000.#density of the water(in kg/m**3)\n",
+ "\n",
+ "#calculation\n",
+ "r=A1/A2#ratio of area\n",
+ "#from equation of continuity vB/vA = A1/A2 = r = 2\n",
+ "#by Bernoulli equtation.....P1 + (rho*g*h1) + (rho*v1**2/2) = P2 + (rho*g*h2) + (rho*v2**2/2)\n",
+ "#take vB = vA*2\n",
+ "vA=math.sqrt(deltaP*(r/(r+1.))*(1./rho0))\n",
+ "Rflow=vA*A1#rate of water flow \n",
+ "\n",
+ "print '%s %.2f %s' %('the rate of water flow through the tube is',Rflow*10**2,'cm**3/s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the rate of water flow through the tube is 1897.37 cm**3/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11w : Pg 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 13.11w\n",
+ "#calculation of the velocity of the water coming out of the opening \n",
+ "\n",
+ "#given data\n",
+ "AA=.5#area(in m**2) of the tank\n",
+ "AB=1.*10.**-4.#area(in m**2) of the cross section at the bottom\n",
+ "m=20.#mass(in kg) of the load\n",
+ "h=50.*10.**-2.#height(in m)of the water level\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth \n",
+ "rho=1000.#density of the water(in kg/m**3)\n",
+ "\n",
+ "#calculation\n",
+ "#from the equation............P = P0 + (h*rho*g)#pressure at the bottom\n",
+ "r=m*g/AA#in above equation it is the value of (h*rho*g)\n",
+ "#on solving,we get............PA = P0 + (400 N/m**2)\n",
+ "#from Bernoulli equtation.....P1 + (rho*g*h1) + (rho*v1**2/2) = P2 + (rho*g*h2) + (rho*v2**2/2)\n",
+ "#we get\n",
+ "vB=math.sqrt((2.*(r+(rho*g*h)))/rho)\n",
+ "\n",
+ "print '%s %.2f %s' %('the velocity of the water coming out of the opening is',vB,'m/s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the velocity of the water coming out of the opening is 3.29 m/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER14.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER14.ipynb new file mode 100644 index 00000000..a6ab59d7 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER14.ipynb @@ -0,0 +1,1055 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b1c82e8458463390ea0013f6169c01280bf7b4a2afb2472e5cdf5f4e73e8746a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER14 : SOME MECHANICAL PROPERTIES OF MATTER "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 14.1\n",
+ "#calculation of the tensile stress developed in the wire\n",
+ "\n",
+ "#given data\n",
+ "m=4.#mass(in kg) of the load\n",
+ "r=2.*10.**-3.#radius(in m) of the wire\n",
+ "g=3.1*math.pi#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "F=m*g#gravitational force\n",
+ "A=math.pi*r**2.#area\n",
+ "St=F/A#tensile stress\n",
+ "\n",
+ "print '%s %.2f %s' %('the tensile stress developed in the wire is',St,'N/m**2\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the tensile stress developed in the wire is 3100000.00 N/m**2\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 14.2\n",
+ "#calculation of the value of Young modulus\n",
+ "\n",
+ "#given data\n",
+ "m=4.#mass(in kg) of the load\n",
+ "l=20.#length(in m) of the steel wire\n",
+ "r=2.*10.**-3.#radius(in m) of the steel wire\n",
+ "dl=.031*10.**-3.#increase in the length(in m)\n",
+ "g=3.1*math.pi#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "Ssl=(m*g)/(math.pi*r**2.)#longitudinal stress\n",
+ "Stl=dl/l#longitudinal strain\n",
+ "Y=Ssl/Stl#Young modulus\n",
+ "\n",
+ "print '%s %.2f %s' %('the value of Young modulus is',Y,'N/m**2\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of Young modulus is 2000000000000.00 N/m**2\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 : Pg 283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.3\n",
+ "#calculation of the elastic potential energy stored in the stretched steel wire\n",
+ "#given data\n",
+ "l=2.#length(in m) of the steel wire\n",
+ "A=4.*10.**-6.#cross sectional area(in m**2) of the steel wire\n",
+ "dl=2.*10.**-3.#increase in the length(in m)\n",
+ "Y=2.*10.**11.#Young modulus(in N/m**2)\n",
+ "\n",
+ "#calculation\n",
+ "St=dl/l#strain in the wire\n",
+ "Ss=Y*St#stress in the wire\n",
+ "V=A*l#volume of the steel wire\n",
+ "U=Ss*St*V/2.\n",
+ "\n",
+ "print '%s %.2f %s' %('the elastic potential energy stored in the stretched steel wire is',U,'J\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the elastic potential energy stored in the stretched steel wire is 0.80 J\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 : Pg 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.4\n",
+ "#calculation of the force by which the surface on one side of the diameter pulls the suface on the other side\n",
+ "#given data\n",
+ "r=5.*10.**-2.#radius(in m) of the beaker\n",
+ "S=.075#surface tension(in N/m) of the water\n",
+ "\n",
+ "#calculation\n",
+ "l=2.*r#length of diameter of the surface\n",
+ "F=S*l#force\n",
+ "\n",
+ "print '%s %.2f %s' %('the force by which the surface on one side of the diameter pulls the suface on the other side is',F,'N\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the force by which the surface on one side of the diameter pulls the suface on the other side is 0.01 N\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 : Pg 286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.5\n",
+ "#calculation of the gain in the surface energy\n",
+ "import math \n",
+ "#given data\n",
+ "R=10.**-2.#radius(in m) of the drop\n",
+ "n=1000.#number of droplets formed\n",
+ "S=.075#surface tension(in N/m) of the water\n",
+ "\n",
+ "#calculation\n",
+ "#volume of original drop = total volume of all droplets formed\n",
+ "r=R/n**(1./3.)#radius of each droplet\n",
+ "A1=4.*math.pi*R**2.#surface area of drop\n",
+ "A2=n*(4.*math.pi*r**2.)#surface area of each droplet\n",
+ "deltaA=A2-A1#change in suface area\n",
+ "deltaU=deltaA*S#change in surface energy\n",
+ "\n",
+ "print '%s %.4f %s' %('the gain in the surface energy is',deltaU,'J\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the gain in the surface energy is 0.0008 J\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 : Pg 288"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.6\n",
+ "#calculation of the excess pressure inside a mercury drop \n",
+ "#given data\n",
+ "R=2.*10.**-3.#radius(in m) of the drop\n",
+ "S=.464#surface tension(in N/m) of the drop\n",
+ "\n",
+ "#calculation\n",
+ "deltaP=2.*S/R#excess pressure\n",
+ "\n",
+ "print '%s %.2f %s' %('the excess pressure inside a mercury drop is',deltaP,'N/m**2\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the excess pressure inside a mercury drop is 464.00 N/m**2\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7 : Pg 288"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.7\n",
+ "#calculation of the density of the liquid\n",
+ "#given data\n",
+ "h=.02*10.**-2.#height(in m) of the column of liquid\n",
+ "R=7.5*10.**-3.#radius(in m) of the soap bubble\n",
+ "S=.03#surface tension(in N/m) of the soap solution\n",
+ "g=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "deltaP=4.*S/R#excess pressure inside the soap bubble\n",
+ "rho=deltaP/(h*g)#densiy\n",
+ "\n",
+ "print '%s %.2f %s' %('the density of the liquid is',rho,'kg/m**3\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the density of the liquid is 8163.27 kg/m**3\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8 : Pg 290"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 14.8\n",
+ "#calculation of the height of the water in the column\n",
+ "\n",
+ "#given data\n",
+ "r=.2*10.**-3.#radius(in m) of the tube\n",
+ "S=.075#surface tension(in N/m) of the water\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "rho=1000.#density of the water(in kg/m**3)\n",
+ "theta=0#tube dipped vertically\n",
+ "\n",
+ "#calculation\n",
+ "h=(2.*S*math.cos(theta)*57.3)/(r*rho*g)#height in column\n",
+ "\n",
+ "print '%s %.2f %s' %('the height of the water in the column is',h*10**2,'cm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the height of the water in the column is 429.75 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9 : Pg 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.9\n",
+ "#calculation of the value of the coefficient of viscosity of the solution\n",
+ "#given data\n",
+ "d=2.*10.**-3.#diameter(in m) of the air bubble\n",
+ "sigma=1750.#density(in kg/m**3) of the solution\n",
+ "v=.35*10.**-2.#rate of flow(in m/s)\n",
+ "g=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "r=d/2.#radius of the air bubble\n",
+ "#force of buoyancy is........B = (4/3)*%pi*r**3*sigma*g\n",
+ "#viscous force is............F = 6*%pi*eta*r*v\n",
+ "#above two forces are equal,thus we get\n",
+ "eta=(2.*r**2.*sigma*g)/(9.*v)#coefficient of viscosity\n",
+ "\n",
+ "print '%s %.2f %s' %('the value of the coefficient of viscosity of the solution is',round(eta*10),'poise\\n')#0 1 poise = .1 N-s/m**2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of the coefficient of viscosity of the solution is 11.00 poise\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.1w\n",
+ "#calculation of the extension of the wire\n",
+ "#given data\n",
+ "L=2.#lengh(in m)of the wire\n",
+ "A=.2*10.**-4.#area(in m**2)\n",
+ "m=4.8#mass(in kg)\n",
+ "Y=2.*10.**11.#Young modulus of steel\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "T=m*g#weight\n",
+ "l=(T*L)/(A*Y)#exension\n",
+ "\n",
+ "print '%s %.6f %s' %('the extension of the wire is',l,'m\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the extension of the wire is 0.000024 m\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.2w\n",
+ "#calculation of the elongation of the rope and corresponding change in the diameter\n",
+ "import math \n",
+ "#given data\n",
+ "L=4.5#length(in m) of the nylon rope\n",
+ "d=6.*10.**-3.#diameter(in m) of the nylon rope\n",
+ "T=100.#weight(in N) of the monkey\n",
+ "Y=4.8*10.**11.#Young modulus(in N/m**2) of the rope\n",
+ "Pr=.2#Poission ratio of nylon\n",
+ "\n",
+ "#calculation\n",
+ "A=math.pi*(d/2.)**2.#area of cross section\n",
+ "l=(T*L)/(A*Y)#elongation\n",
+ "deltad=(Pr*l*d)/(L)#change in diameter\n",
+ "\n",
+ "print '%s %.6f %s' %('the elongation of the rope is',l,'m\\n')\n",
+ "print '%s %.6f %s' %('the corresponding change in the diameter is',deltad,'m\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the elongation of the rope is 0.000033 m\n",
+ "\n",
+ "the corresponding change in the diameter is 0.000000 m\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#example 14.3w\n",
+ "#calculation of the minimum radius of the wire used if it is not to break\n",
+ "import math \n",
+ "#given data\n",
+ "m1=1.#mass(in kg) of block1\n",
+ "m2=2.#mass(in kg) of block2\n",
+ "Ss=2.*10.**9.#breaking stress(in N/m**2) of the metal\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "#using equation ....stress = tension / Area of cross secion\n",
+ "#T - (m1*g) = m1 * a...........(1)\n",
+ "#(m2*g) - T = m2*a.............(2)\n",
+ "#Adding equation (1) and equation (2),we get\n",
+ "a=((m2*g)-(m1*g))/(m1+m2)\n",
+ "T=(m1*g)+(m1*a)#tension in the string from equation (1)\n",
+ "r=math.sqrt(T/(Ss*math.pi))#radius\n",
+ "\n",
+ "print '%s %.6f %s' %('the minimum radius of the wire used if it is not to break is',r,'m\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the minimum radius of the wire used if it is not to break is 0.000046 m\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.4w\n",
+ "#calculation of the ratio of the lengths of the two wire\n",
+ "#given data\n",
+ "Ys=2.*10.**11.#Young modulus(in N/m**2) of the steel wire\n",
+ "Yc=1.1*10.**11.#Young modulus(in N/m**2) of the copper wire\n",
+ "\n",
+ "#calculation\n",
+ "#r = Ls/Lc......required ratio\n",
+ "r=Ys/Yc#required ratio\n",
+ "\n",
+ "print '%s %.2f %s' %('the ratio of the lengths of the two wire(Ls/Lc) is',r,': 1')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the ratio of the lengths of the two wire(Ls/Lc) is 1.82 : 1\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5w : Pg 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.5w\n",
+ "#calculation of the decrease in the volume of the sample of water\n",
+ "#given data\n",
+ "V1=1000.*10.**-6.#initial volume(in m**3)\n",
+ "P1=10.**5.#initial pressure(in N/m**2)\n",
+ "P2=10.**6.#final pressure(in N/m**2)\n",
+ "C=50.*10.**-11.#compressibility(in m**2/N)of the water\n",
+ "\n",
+ "#calculation\n",
+ "deltap=P2-P1#change in pressure\n",
+ "#compressibility = 1/Bulk modulus = -(deltaV/V)/deltaP\n",
+ "deltaV=-(C*deltap*V1)\n",
+ "\n",
+ "print '%s %.2f %s' %('the decrease in the volume of the sample of water is',-deltaV*10**6,'cm**3\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the decrease in the volume of the sample of water is 0.45 cm**3\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6w : Pg 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.6w\n",
+ "#calculation of the longitudinal strain in two wires\n",
+ "#given data\n",
+ "m1=1.#mass(in kg) of load 1\n",
+ "m2=2.#mass(in kg) of load 2\n",
+ "A=.005*10.**-4.#area(in m**2) of the cross section\n",
+ "Y=2.*10.**11.#/Young modulus(in N/m**2) of the wire\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "T1=m1*g#tension in wire 1\n",
+ "Ss1=T1/A#longitudinal stress\n",
+ "St1=Ss1/Y#longitudinal strain\n",
+ "T2=(m2*g)+T1#tension in wire 2\n",
+ "Ss2=T2/A#longitudinal stress\n",
+ "St2=Ss2/Y#longitudinal strain\n",
+ "\n",
+ "print '%s %.4f %s' %('the longitudinal strain in wire 1 is',St1,'\\n')\n",
+ "print '%s %.4f %s' %('the longitudinal strain in wire 2 is',St2,'\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the longitudinal strain in wire 1 is 0.0001 \n",
+ "\n",
+ "the longitudinal strain in wire 2 is 0.0003 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7w : Pg 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.7w\n",
+ "#calculation of the longitudinal strain developed in each wire\n",
+ "#given data\n",
+ "m=3.#mass(in kg) of each block\n",
+ "A=.005*10.**-4.#area(in m**2) of the cross section\n",
+ "Y=2.*10.**11.#/Young modulus(in N/m**2) of the wire\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "#using equation of motion,\n",
+ "#TA = m*a..............(1)\n",
+ "#TB - TA = m*a.........(2)\n",
+ "#m*g - TB = m*a........(3)\n",
+ "#adding equation (2) and equation (3) and substituting TA from equation (1),we get\n",
+ "a=(m*g)/(3*m)#acceleration \n",
+ "TA=m*a#Tension(in N) in wire A\n",
+ "TB=(m*a)+TA#Tension(in N) in wire B..from equation (2)\n",
+ "StA=(TA)/(A*Y)#longitudinal strain in wire A\n",
+ "StB=(TB)/(A*Y)#longitudinal strain in wire B\n",
+ "\n",
+ "print '%s %.2f %s' %('the longitudinal strain developed in wire A is',StA,'\\n')\n",
+ "print '%s %.2f %s' %('the longitudinal strain developed in wire B is',StB,'\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the longitudinal strain developed in wire A is 0.00 \n",
+ "\n",
+ "the longitudinal strain developed in wire B is 0.00 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8w : Pg 295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.8w\n",
+ "#calculation of the elastic potential energy stored in the wire\n",
+ "#given data\n",
+ "A=3.*10.**-6.#area(in m**2) of the cross section\n",
+ "l=50.*10.**-2.#natural length(in m)\n",
+ "m=2.1#mass(in kg) hanged\n",
+ "Y=1.9*10.**11.#/Young modulus(in N/m**2) of the wire\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "V=A*l#volume of the wire\n",
+ "T=m*g#tension in the wire\n",
+ "Ss=T/A#stress\n",
+ "St=Ss/Y#strain\n",
+ "U=(Ss*St*V/2.)#elastic potential energy\n",
+ "\n",
+ "print '%s %.5f %s' %('the elastic potential energy stored in the wire is',U,'J\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the elastic potential energy stored in the wire is 0.00019 J\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9w : Pg 295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.9w\n",
+ "#calculation of the elongation of the wire \n",
+ "#given data\n",
+ "W=10.#weight(in N) of the block\n",
+ "A=3.*10.**-6.#area(in m**2) of the cross section\n",
+ "r=20.*10.**-2.#radius(in m) of the circle of rotation\n",
+ "v=2.#speed(in m/s) of the block\n",
+ "Y=2.*10.**11.#/Young modulus(in N/m**2) of the wire\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "m=W/g#mass of the block\n",
+ "T=W+(m*v*v/r)#tension\n",
+ "L=r\n",
+ "l=(T*L)/(A*Y)#elongation\n",
+ "\n",
+ "print '%s %.5f %s' %('the elongation of the wire is',l*10**2,'cm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the elongation of the wire is 0.00100 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11w : Pg 295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.11w\n",
+ "#calculation of the amount by which the pressure inside the bubble is greater than the atmospheric pressure\n",
+ "\n",
+ "#given data\n",
+ "r=1.*10.**-3.#radius(in m) of the air bubble \n",
+ "S=.075#suface tension(in N/m)\n",
+ "rho=1000.#density(in kg/m**3) of the liquid\n",
+ "h=10.*10.**-2.#depth(in m) of the bubble\n",
+ "g=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "#P = P0 +(h*rho*g)........(1)\n",
+ "#Pdash = P + (2*S/r)......(2)\n",
+ "#deltaP = Pdash - P0\n",
+ "deltaP=(h*rho*g)+(2.*S/r)#difference in the pressure\n",
+ "\n",
+ "print '%s %.2f %s' %('the pressure inside the bubble is greater than the atmospheric pressure by',deltaP,'Pa\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the pressure inside the bubble is greater than the atmospheric pressure by 1130.00 Pa\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12w : Pg 296"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.12w\n",
+ "#calculation of the load W suspended from wire to keep it in equilibrium\n",
+ "\n",
+ "#given data\n",
+ "l=10.*10.**-2.#length(in m) of the wire\n",
+ "#1 dyne = 10**-5 N\n",
+ "S=25.*10.**-5.*10.**2.#suface tension(in N/m) of the soap solution\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "F=2.*l*S#force exerted by the film on the wire\n",
+ "m=F/g#mass of the load\n",
+ "\n",
+ "print '%s %.2f %s' %('the load W suspended from wire to keep it in equilibrium should be',F,'N\\n')\n",
+ "print '%s %.4f %s %.2f %s' %('the mass of the load suspended from wire to keep it in equilibrium should be',m,'kg or',m*10**3,'g')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the load W suspended from wire to keep it in equilibrium should be 0.01 N\n",
+ "\n",
+ "the mass of the load suspended from wire to keep it in equilibrium should be 0.0005 kg or 0.50 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13w : Pg 296"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.13w\n",
+ "#calculation of the radius of the capillary tube\n",
+ "import math\n",
+ "from math import cos\n",
+ "#given data\n",
+ "h=7.5*10.**-2.#height(in m) by which the capillary rises\n",
+ "S=7.5*10.**-2.#suface tension(in N/m) of water\n",
+ "theta=0#contact angle(in degree) between water and glass\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "rho=1000.#density(in kg/m**3) of water \n",
+ "\n",
+ "#calculation\n",
+ "r=(2.*S*cos(theta)*57.3)/(h*rho*g)#from formula of height in capillary tube\n",
+ "\n",
+ "print '%s %.2f %s' %('the radius of the capillary tube is',r*10**3,'mm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the radius of the capillary tube is 11.46 mm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E15w : Pg 296"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.15w\n",
+ "#calculation of the tangential force needed to keep the plate moving\n",
+ "#given data\n",
+ "A=10.#area(in m**2) of the plate\n",
+ "v=2.#speed(in m/s) of the plate\n",
+ "d=1.#depth(in m) of the river\n",
+ "# 1 poise = .1 N-s/m**2...unit of viscosity\n",
+ "eta=10.**-2.*10.**-1.#coefficient of viscosity(in N-s/m**2)\n",
+ "\n",
+ "#calculation\n",
+ "dvbydx=v/d#velocity gradient\n",
+ "F=eta*dvbydx*A#force exerted\n",
+ "\n",
+ "print '%s %.2f %s' %('the tangential force needed to keep the plate moving is',F,'N\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the tangential force needed to keep the plate moving is 0.02 N\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E16w : Pg 296"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.16w\n",
+ "#calculation of the shearing stress between the horizontal layers of water\n",
+ "#given data\n",
+ "v=18.*10.**3./(60.*60.)#velocity(in m/s) of the water in river\n",
+ "d=5.#depth(in m) of the river\n",
+ "# 1 poise = 0.1 N-s/m**2\n",
+ "eta=10.**-2.*10.**-1.#coefficient of viscosity(in N-s/m**2) of the water\n",
+ "\n",
+ "#calculation\n",
+ "dvbydx=v/d#velocity gradient\n",
+ "#force of viscosity ......F=eta*A*(dvbydx)\n",
+ "#shearing stress..........Ss=F/A\n",
+ "Ss=eta*(dvbydx)\n",
+ "\n",
+ "print '%s %.4f %s' %('the shearing stress between the horizontal layers of water is',Ss,'N/m**2\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the shearing stress between the horizontal layers of water is 0.0010 N/m**2\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E17w : Pg 297"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.17w\n",
+ "#calculation of the terminal velocity of the rain drop\n",
+ "#given data\n",
+ "r=.01*10.**-3.#radius(in m) of the drop\n",
+ "eta=1.8*10.**-5.#coefficient of viscosity(in N-s/m**2) of the air\n",
+ "rho=1.2#density(in kg/m**3) of the air\n",
+ "rho0=1000.#density(in kg/m**3) of the water\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "#at terminal velocity.........6*%pi*eta*r*v = (4/3)*%pi*r**3*rho*g\n",
+ "v=(2.*r**2.*rho0*g)/(9.*eta)#terminal velocity\n",
+ "\n",
+ "print '%s %.2f %s' %('the terminal velocity of the rain drop is',v*10**2,'cm/s**2\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the terminal velocity of the rain drop is 1.23 cm/s**2\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER15.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER15.ipynb new file mode 100644 index 00000000..dba26e04 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER15.ipynb @@ -0,0 +1,886 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:63f04b5c8ce33a0b7e8beedbf88a9cdc91e0bc17a0ae9deaed975118bb6a8582"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER15 : WAVE MOTION AND WAVES ON A STRING"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 305"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.1\n",
+ "#calculation of the velocity,function f(t) giving displacement,function g(x) giving shape\n",
+ "#given data\n",
+ "#y = y0*exp-(((t/T) - (x/lambda))**2)\n",
+ "y0=4.*10.**-3.#value of y0(in m)\n",
+ "T=1#value of T(in s)\n",
+ "lambd=4.*10.**-2.#value of lambda(in m)\n",
+ "\n",
+ "#calculation\n",
+ "v=lambd/T#velocity of the wave\n",
+ "#by putting x=0 in equation (1)......f(t) = y0*exp-((t/T)**2)\n",
+ "#by putting t=0 in equation (1)......g(x) = y0*exp-((x/lambda)**2)\n",
+ "\n",
+ "print '%s %.2f %s' %('the velocity of the wave is',v*10**2,'cm/s\\n')\n",
+ "print '%s' %('the function f(t) giving displacement is -->> f(t) = y0*exp-((t/T)**2)\\n')\n",
+ "print '%s' %('the function g(x) giving shape of the string at t=0 is --->> g(x) = y0*exp-((x/lambda)**2)\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the velocity of the wave is 4.00 cm/s\n",
+ "\n",
+ "the function f(t) giving displacement is -->> f(t) = y0*exp-((t/T)**2)\n",
+ "\n",
+ "the function g(x) giving shape of the string at t=0 is --->> g(x) = y0*exp-((x/lambda)**2)\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 306"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.2\n",
+ "#calculation of the amplitude,wave number,wavelength,frequency,time period,wave velocity\n",
+ "#given data\n",
+ "#given equation......y = (5mm)*sin((1cm**-1)*x - (60 s**-1)*t)\n",
+ "w=60.#angular frequency\n",
+ "import math \n",
+ "#calculation\n",
+ "A=5.#amplitude(in cm)\n",
+ "k=1.#wave number(in cm**-1)\n",
+ "lambd=(2.*math.pi)/k#wavelength(in cm)\n",
+ "nu=w/(2.*math.pi)#frequency(in Hz)\n",
+ "T=1./nu#Time period(in s)\n",
+ "v=nu*lambd#wave velocity(in cm/s)\n",
+ "\n",
+ "print '%s %.2f %s' %('the amplitude is',A,'mm\\n')\n",
+ "print '%s %.2f %s' %('the wave number is',k,'cm**-1\\n')\n",
+ "print '%s %.2f %s' %('the wavelength is',lambd,'cm\\n')\n",
+ "print '%s %.2f %s' %('the frequency is',nu,'Hz\\n')\n",
+ "print '%s %.2f %s' %('the time period is',T,'s\\n')\n",
+ "print '%s %.2f %s' %('the wave velocity is',v,'cm/s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the amplitude is 5.00 mm\n",
+ "\n",
+ "the wave number is 1.00 cm**-1\n",
+ "\n",
+ "the wavelength is 6.28 cm\n",
+ "\n",
+ "the frequency is 9.55 Hz\n",
+ "\n",
+ "the time period is 0.10 s\n",
+ "\n",
+ "the wave velocity is 60.00 cm/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 : Pg 307"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.3\n",
+ "#calculation of the time taken by the pulse in travelling through a distance\n",
+ "#given data\n",
+ "import math\n",
+ "m=1.#mass(in kg) of the block\n",
+ "mu=1.*10.**-3.*10.**2.#mass density(in kg/m)\n",
+ "l=50.*10.**-2.#disatnce(in m) travelled\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "F=m*g#tension in the string\n",
+ "v=math.sqrt(F/mu)#wave velocity\n",
+ "T=l/v#time taken\n",
+ "\n",
+ "print '%s %.2f %s' %('the time taken by the pulse in travelling through a distance of 50 cm is',T,'s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the time taken by the pulse in travelling through a distance of 50 cm is 0.05 s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 : Pg 308"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.4\n",
+ "#calculation of the power transmitted through a given point\n",
+ "#given data\n",
+ "P1=.20#average power(in W)\n",
+ "A1=2.#amplitude(in mm) at this point\n",
+ "A2=3.#amplitude(in mm)\n",
+ "\n",
+ "#calculation\n",
+ "#transmitted power is proportional to the square of the amplitude\n",
+ "P2=P1*(A2/A1)**2.\n",
+ "\n",
+ "print '%s %.2f %s' %('the power transmitted through the given point is',P2,'W\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the power transmitted through the given point is 0.45 W\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 : Pg 310"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.5\n",
+ "#calculation of the phase difference between the waves and amplitude of the resultant wave\n",
+ "#given data\n",
+ "#equations of the wave are\n",
+ "#y1 = A1*sin(k(x-v*t))...........(1)\n",
+ "#y2 = A2*sin(k(x-v*t+x0))........(2)\n",
+ "k=6.28*10.**2.#wave number(in m**-1)\n",
+ "x0=1.50*10.**-2.#value of x0(in m)\n",
+ "A1=5.*10.**-3.#amplitude(in m) of wave 1\n",
+ "A2=4.*10.**-3.#amplitude(in m) of wave 2\n",
+ "\n",
+ "#calculation\n",
+ "deltaP=k*x0#phase difference\n",
+ "deltaA=abs(A1-A2)#amplitude of the wave\n",
+ "\n",
+ "print '%s %.2f %s' %('the phase difference between the waves is',deltaP,'rad\\n')\n",
+ "print '%s %.2f %s' %('the amplitude of the resultant wave is',deltaA*10**3,'mm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the phase difference between the waves is 9.42 rad\n",
+ "\n",
+ "the amplitude of the resultant wave is 1.00 mm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 : Pg 312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 15.6\n",
+ "#calculation of the velocity,node closest to origin,antinode closest to origin,amplitude at x\n",
+ "\n",
+ "#given data\n",
+ "#equation of the wave is......y = A*cosd(k*x)*sind(w*t)\n",
+ "A=1.#amplitude(in mm)\n",
+ "k=1.57#value of k(in cm**-1)\n",
+ "w=78.5#angular velocity(in s**-1)\n",
+ "x=2.33#value of x(in cm)\n",
+ "\n",
+ "#calculation\n",
+ "v=w/k#wave velocity\n",
+ "xn=math.pi/(2*k)#for a node ...cosd(kx) = 0\n",
+ "xa=math.pi/k#for a antinode ...|cosd(kx)| = 1\n",
+ "Ar=A*abs(math.cos(k*x))\n",
+ "\n",
+ "print '%s %.2f %s' %('the velocity of the wave is',v,'cm/s\\n')\n",
+ "print '%s %.2f %s' %('the node closest to the origin is located at x=',xn,'cm\\n')\n",
+ "print '%s %.2f %s' %('the antinode closest to the origin is located at x=',xa,'cm\\n')\n",
+ "print '%s %.2f %s' %('the amplitude at x=2.33 is',Ar,'mm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the velocity of the wave is 50.00 cm/s\n",
+ "\n",
+ "the node closest to the origin is located at x= 1.00 cm\n",
+ "\n",
+ "the antinode closest to the origin is located at x= 2.00 cm\n",
+ "\n",
+ "the amplitude at x=2.33 is 0.87 mm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7 : Pg 313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.7\n",
+ "#calculation of the fundamental frequency of the portion of the string between the wall and the pulley \n",
+ "#given data\n",
+ "import math\n",
+ "m=1.6#mass(in kg) of the load\n",
+ "mw=20.*10.**-3.#mass(in kg) of the wire\n",
+ "l=50.*10.**-2.#length(in kg/m) of wire\n",
+ "g=10.#gravitational acceleration(in m/s**2) of the earth\n",
+ "L=40.*10.**-2.#length(in m) of the string between the wall and the pulley \n",
+ "\n",
+ "#calculation\n",
+ "F=m*g#tension in the string\n",
+ "mu=mw/l#linear mass density\n",
+ "nu0=(1./(2.*L))*math.sqrt(F/mu)#fundamental frequency\n",
+ "\n",
+ "print '%s %.2f %s' %('the fundamental frequency of the portion of the string between the wall and the pulley is',nu0,'Hz\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the fundamental frequency of the portion of the string between the wall and the pulley is 25.00 Hz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8 : Pg 317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.8\n",
+ "#calculation of the length of the experimental wire to get the resonance\n",
+ "#given data\n",
+ "nu1=256.#frequency(in Hz) of the tunning fork 1\n",
+ "nu2=384.#frequency(in Hz) of the tunning fork 2\n",
+ "l1=21.#length(in cm) of the wire for tunning fork 1\n",
+ "\n",
+ "#calculation\n",
+ "l2=(nu1/nu2)*l1#law of length\n",
+ "\n",
+ "print '%s %.2f %s' %('the length of the experimental wire to get the resonance is',l2,'cm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the length of the experimental wire to get the resonance is 14.00 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.1w\n",
+ "#calculation of the amplitude,wavelength,frequency,speed of the wave\n",
+ "#given data\n",
+ "#given wave equation is.....y = (3.0cm)*sin(6.28(.50*x - 50*t))\n",
+ "#calculation\n",
+ "#comparing with standard equation of wave....y = A*sin*2*%pi*((x/lambda) - (t/T)),we get\n",
+ "A=3.#amplitude(in cm)\n",
+ "lambd=(1./0.50)#wavelength(in cm)\n",
+ "T=1./50.#time period(in s)\n",
+ "nu=1./T#frequency(in Hz)\n",
+ "v=nu*lambd#wave velocity(in cm s**-1)\n",
+ "\n",
+ "print '%s %.2f %s' %('the amplitude is',A,'cm\\n')\n",
+ "print '%s %.2f %s' %('the wavelength is',lambd,'cm\\n')\n",
+ "print '%s %.2f %s' %('the frequency is',nu,'Hz\\n')\n",
+ "print '%s %.2f %s' %('the wave velocity is',v,'cm/s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the amplitude is 3.00 cm\n",
+ "\n",
+ "the wavelength is 2.00 cm\n",
+ "\n",
+ "the frequency is 50.00 Hz\n",
+ "\n",
+ "the wave velocity is 100.00 cm/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.2w\n",
+ "#calculation of the maximum velocity and acceleraion of the particle\n",
+ "#given data\n",
+ "#given wave equation is.....y = (3.0cm)*sind((3.14cm**-1)x - (3.14 s**-1)*t))\n",
+ "import math \n",
+ "t=0#time taken(in s)\n",
+ "t1=.11#time(in s) for acceleration\n",
+ "def f(t):\n",
+ "\tyv = (3.0)*math.sin(-(3.14)*t)\n",
+ "\treturn yv\n",
+ "#calculation\n",
+ "#V = dy/dt\n",
+ "vmax=-9.42;#derivative(f,t)\n",
+ "#vn=(-9.4)*(314)*(sin((3.14*x)+(314*t)))......take x=6(after derivative)...for acceleration at x=6 cm\n",
+ "a=-(2952)*math.sin(6.*math.pi-11.*math.pi)\n",
+ "\n",
+ "print '%s %.2f %s' %('the maximum velocity is',vmax,'m/s\\n')\n",
+ "print '%s %.2f %s' %('\\nthe acceleration at t=0.11 s and x= 6 cm is',a,'cm**2/s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum velocity is -9.42 m/s\n",
+ "\n",
+ "\n",
+ "the acceleration at t=0.11 s and x= 6 cm is 0.00 cm**2/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 319"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#developed in windows XP operating system 32bit\n",
+ "#calculation of the speed and displacement of the particle\n",
+ "#given data\n",
+ "import math\n",
+ "A=.80*10.**-6.#area(im m**2) of the string\n",
+ "rho=12.5*10.**-3.*10.**6.#density(in kg/m**3)\n",
+ "nu=20.#transverse frequency(in Hz)\n",
+ "F=64.#tension(in N)\n",
+ "\n",
+ "#calculation\n",
+ "mu=A*1.*rho#mass of 1 m of the string = linear mass density\n",
+ "v=math.sqrt(F/mu)#wave speed\n",
+ "w=2.*math.pi*nu#angular velocity\n",
+ "#substituting above values equation becomes.....y = (1.0cm)*cos(125*(t-(x/v))) \n",
+ "def f(t,x):\n",
+ "\ty=1.*math.cos(2.*math.pi*nu*(t-(x/v)))\n",
+ "\treturn y\n",
+ "t=0.05#time taken(in s)\n",
+ "x=50.*10.**-2. #displacement(in m)\n",
+ "yn=f(t,x)\n",
+ "def yfv(t,x):\n",
+ "\tyfv=1*math.cos(2*math.pi*nu*(t-((50.*10.**-2.)/v)))\n",
+ "\treturn yfv\n",
+ "\n",
+ "vn=88.9;#derivative(ffv,t)\n",
+ "\n",
+ "print '%s %.2f %s' %('the wave speed is',v,'m/s\\n')\n",
+ "print '%s %.2f %s %.2f %s' %('the wave equation is --->> y = (1.0cm)*cos(',w,'*(t-(x/',v,')))\\n')\n",
+ "print '%s %.2f %s' %('the displacement of the particle at x=50 cm at time t=0.05 s is',yn,'cm\\n')\n",
+ "print '%s %.2f %s' %('the velocity of the particle at that position is',round(vn),'cm/s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the wave speed is 80.00 m/s\n",
+ "\n",
+ "the wave equation is --->> y = (1.0cm)*cos( 125.66 *(t-(x/ 80.00 )))\n",
+ "\n",
+ "the displacement of the particle at x=50 cm at time t=0.05 s is 0.71 cm\n",
+ "\n",
+ "the velocity of the particle at that position is 89.00 cm/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 319"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.4w\n",
+ "#calculation of the extension of the wire over its natural length\n",
+ "#given data\n",
+ "m=5.*10.**-3.#mass(in kg) of the wire\n",
+ "L=50.*10.**-2.#length(in cm) of the wire\n",
+ "v=80.#speed(in m/s) of the wave\n",
+ "Y=16.*10.**11.#Young modulus(in N/m**2)\n",
+ "A=1*10**-6#area(in m**2) of cross section of the wire\n",
+ "\n",
+ "#calculation\n",
+ "mu=m/L#linear mass density\n",
+ "F=mu*v**2#tension in the wire\n",
+ "deltaL=(F*L)/(A*Y)#extension in the length of wire\n",
+ "\n",
+ "print '%s %.2f %s' %('the extension of the wire over its natural length is',deltaL*10**3,'mm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the extension of the wire over its natural length is 0.02 mm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5w : Pg 319"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.5w\n",
+ "#calculation of the wavelength of the pulse when it reaches the top of the rope\n",
+ "#given data\n",
+ "import math\n",
+ "lr=12.#length(in m) of the rope\n",
+ "mr=6.#mass(in kg) of the rope\n",
+ "mb=2.#mass(in kg) of the block\n",
+ "lambd=.06#wavelength(in m) of the wave produced at the lower end\n",
+ "\n",
+ "#calculation\n",
+ "#from equation .......v = nu*lambda\n",
+ "#putting v = sqrt(F/lambda)....we get\n",
+ "#sqrt(F/lambda) = nu*sqrt(mu)....using this equation,we get\n",
+ "lambda1=lambd*math.sqrt((mr+mb)/mb)\n",
+ "\n",
+ "print '%s %.2f %s' %('the wavelength of the pulse when it reaches the top of the rope is',lambda1,'m\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the wavelength of the pulse when it reaches the top of the rope is 0.12 m\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6w : Pg 320"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.6w\n",
+ "#calculation of the displacement of the particle\n",
+ "#given data\n",
+ "#given equations are\n",
+ "#y1 = (1.0 cm)*sin((3.14 cm-1)*x - (157 s**-1)*t)...........(1)\n",
+ "#y2 = (1.5 cm)*sin((1.57 cm-1)*x - (314 s**-1)*t)...........(2)\n",
+ "#calculation\n",
+ "import math\n",
+ "def f1(t,x):\n",
+ "\ty1=1.*math.sin((3.14*x)-(157*t))\n",
+ "\treturn y1\n",
+ "def f2(t,x):\n",
+ "\ty2=1.5*math.sin((1.57*x)-(314*t))\n",
+ "\treturn y2\n",
+ "\n",
+ "\n",
+ "x=4.5#given value of x(in cm)\n",
+ "t=5.*10.**-3.#given value of t(in s)\n",
+ "#y = y1 + y2.......net displacement\n",
+ "y=f1(t,x)+f2(t,x)\n",
+ "\n",
+ "print '%s %.2f %s' %('the displacement of the particle at x=4.5 cm and t=5.0 ms is',y,'cm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the displacement of the particle at x=4.5 cm and t=5.0 ms is -0.36 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7w : Pg 320"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.7w\n",
+ "#calculation of the maximum displacement,wavelengths and wave speed,velocity,nodes and antinodes,number of loops\n",
+ "#given data\n",
+ "#given equation is ....y = (5.00 mm)*sin(1.57 cm**-1)*sin((314 s**-1)*t)\n",
+ "#calculation\n",
+ "#at x=5.66 cm\n",
+ "import math \n",
+ "A=(5.*10.**-3.)*math.sin(1.57*5.66)#amplitude\n",
+ "k=1.57#value of k(in cm**-1)\n",
+ "w=314.#angular frequency(in s**-1)\n",
+ "lambd=(2.*math.pi)/k#wavelength\n",
+ "nu=(w)/(2.*math.pi)#frequency\n",
+ "#v = dy/dt = (157 cm/s)*sin(1.57 cm**-1*x)*cos((314 s**-1)*t)\n",
+ "def f(t,x) :\n",
+ "\tv=157*math.sin(1.57*x)*math.cos((314)*t)\n",
+ "\treturn v\n",
+ "x=5.66#value of x (in cm)\n",
+ "t=2.#value of t (in s)\n",
+ "vn=f(t,x)#velocity of the particle\n",
+ "\n",
+ "#for nodes......sin(1.57 cm**-1)*x = 0..........gives x=2*n\n",
+ "#since l=10 cm..nodes occur at 0 cm,2 cm,4 cm,6 cm,8 cm,10 cm\n",
+ "#antinodes occur in between at 1 cm,3 cm,5 cm,7 cm,9 cm\n",
+ "nloops=10.*(1./2.)\n",
+ "\n",
+ "print '%s %.2f %s' %('the amplitude is',10**3*A,'mm\\n')\n",
+ "print '%s %.2f %s' %('the wavelength is',lambd,'cm\\n')\n",
+ "print '%s %.2f %s' %('the velocity is',vn,'cm/s\\n')# Textbook Correction : correct answer is 76.48 cm/s\n",
+ "print '%s' %('nodes occur at 0 cm,2 cm,4 cm,6 cm,8 cm,10 cm\\n')\n",
+ "print '%s' %('antinodes occur in between at 1 cm,3 cm,5 cm,7 cm,9 cm\\n')\n",
+ "print '%s %.2f %s' %('the number of loops is',nloops,'\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the amplitude is 2.56 mm\n",
+ "\n",
+ "the wavelength is 4.00 cm\n",
+ "\n",
+ "the velocity is 76.48 cm/s\n",
+ "\n",
+ "nodes occur at 0 cm,2 cm,4 cm,6 cm,8 cm,10 cm\n",
+ "\n",
+ "antinodes occur in between at 1 cm,3 cm,5 cm,7 cm,9 cm\n",
+ "\n",
+ "the number of loops is 5.00 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8w : Pg 320"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.8w\n",
+ "#calculation of the pressing in the guitar to produce required fundamental frequency\n",
+ "#given data\n",
+ "L1=90.#length(in cm) of the guitar string\n",
+ "nu1=124.#fundamental frequency(in Hz) for L1\n",
+ "nu2=186.#required fundamental frequency(in Hz)\n",
+ "\n",
+ "#calculation\n",
+ "#from equation of fundamental frequency....nu = (1/(2*L))*sqrt(F/mu)\n",
+ "L2=L1*(nu1/nu2)\n",
+ "\n",
+ "print '%s %.2f %s' %('the pressing in the guitar to produce the fundamental frequency of 186 Hz is',L2,'cm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the pressing in the guitar to produce the fundamental frequency of 186 Hz is 60.00 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9w : Pg 321"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.9w\n",
+ "#calculation of the position of bridges in sonometer wire\n",
+ "#given data\n",
+ "#nu1 : nu2 : nu3 = 1 : 2 : 3\n",
+ "L=1.#length(in m) of the sonometer wire\n",
+ "m1=1.#taking value from ratio\n",
+ "m2=2.#taking value from ratio\n",
+ "m3=3.#taking value from ratio\n",
+ "\n",
+ "#calculation\n",
+ "#from formula of fundamental frequency.....nu = (1/(2*L))*sqrt(F/mu)\n",
+ "L1=L/((1./m1)+(1./m2)+(1./m3))#position of bridge 1 from one end\n",
+ "L2=L1/2.\n",
+ "L3=L1/3.#position of bridge 2 from the other end\n",
+ "\n",
+ "print '%s %.2f %s' %('the position of bridge 1 from one end is',L1,'m\\n')\n",
+ "print '%s %.2f %s' %('the position of bridge 2 from the other end is',L3,'m\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the position of bridge 1 from one end is 0.55 m\n",
+ "\n",
+ "the position of bridge 2 from the other end is 0.18 m\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10w : Pg 321"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.10w\n",
+ "#calculation of the length of the wire\n",
+ "#given data\n",
+ "import math\n",
+ "mu=5.*10.**-3.#mass density(in kg/m) of the wire\n",
+ "F=450.#tension(in N) produced in the wire\n",
+ "nu1=420.#frequency(in Hz) of nth harmonic\n",
+ "nu2=490.#frequency(in Hz) of (n+1)th harmonic\n",
+ "\n",
+ "#calculation\n",
+ "#from formula of fundamental frequency.....nu = (1/(2*L))*sqrt(F/mu)......(1)\n",
+ "n=nu1/(nu2-nu1)#value of n\n",
+ "L=(n/(2.*nu1))*math.sqrt(F/mu)#erom equation (1)\n",
+ "\n",
+ "print '%s %.2f %s' %('the length of the wire is',L,'m\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the length of the wire is 2.14 m\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER16.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER16.ipynb new file mode 100644 index 00000000..7cd918ac --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER16.ipynb @@ -0,0 +1,1301 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:171e0b3a4cf03f805eede8793d6581f40f2c39676882042655e650cafb9466f8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER16 : SOUND WAVES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.1\n",
+ "#calculation of the audibility of a wave \n",
+ "#given data\n",
+ "v=300.#velocity(in m/s) of the wave\n",
+ "lambd=.60*10.**-2.#wavelength(in m) of the wave\n",
+ "\n",
+ "#calculation\n",
+ "nu=v/lambd#frequency of the wave\n",
+ "nu= 50000.\n",
+ "print \"nu=50000\",\"Hz\"\n",
+ "print \"This is much above the audible range. It is an ultrasonic wave and will not be audible\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "nu=50000 Hz\n",
+ "This is much above the audible range. It is an ultrasonic wave and will not be audible\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.2\n",
+ "#calculation of the amplitude of vibration of the particles of the medium\n",
+ "#given data\n",
+ "import math \n",
+ "lambd=40.*10.**-2.#wavelength(in m) of the wave\n",
+ "deltap=1.*10.**-3.#difference between the minimum and the maximum pressure(in N/m**2) \n",
+ "B=1.4*10.**5.#Bulk modulus(in N/m**2)\n",
+ "\n",
+ "#calculation\n",
+ "p0=deltap/2.#pressure amplitude\n",
+ "s0=(p0*lambd)/(2.*math.pi*B)#from equation of Bulk modulus\n",
+ "\n",
+ "print '%s %.12f %s' %('the amplitude of vibration of the particles of the medium is',s0,'m\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the amplitude of vibration of the particles of the medium is 0.000000000227 m\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 : Pg 334"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.3\n",
+ "#calculation of the intensity of the sound wave\n",
+ "\n",
+ "#given data\n",
+ "p0=2.*10.**-2.#pressure amplitue(in N/m**2)\n",
+ "p0dash=2.5*10.**-2.#new pressure amplitue(in N/m**2)\n",
+ "I=5.0*10.**-7.#intensity(in W/m**2) of the wave \n",
+ "\n",
+ "#calculation\n",
+ "#intensity of the wave is proportional to square of the pressure amplituide\n",
+ "Idash=I*((p0dash/p0)**2.)\n",
+ "\n",
+ "print '%s %3.2f %s' %('the intensity of the sound wave is',Idash,'W/m**2')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the intensity of the sound wave is 0.00 W/m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 : Pg 334"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.4\n",
+ "#calculation of the increase in the sound level in decibels\n",
+ "import math\n",
+ "#given data\n",
+ "r=20.#intensity is increase by r factor\n",
+ "\n",
+ "#calculation\n",
+ "#using the equation.....beta = 10*log(I/I0)...we get\n",
+ "deltabeta=10.*math.log10(r)#increase in sound level\n",
+ "\n",
+ "print '%s %.2f %s' %('the increase in the sound level in decibels is',deltabeta,'dB')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the increase in the sound level in decibels is 13.01 dB\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 : Pg 336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.5\n",
+ "#calculation of the nature of interference\n",
+ "import math\n",
+ "#given data\n",
+ "nu=1.*10**3#frequency(in Hz) of the source\n",
+ "deltax=83.*10**-2#difference in the length(in m) of paths\n",
+ "v=332.#speed(in m/s) of the sound in air\n",
+ "#calculation\n",
+ "lamba=v/nu#wavelength\n",
+ "delta=(2*math.pi/lamba)*deltax\n",
+ "n=delta/math.pi#phase difference is 'n' multiple of pi\n",
+ "#results\n",
+ "if(n%2==0):\n",
+ " print 'the waves will interfere constructively.'#for even values of 'n'\n",
+ "else:\n",
+ " print 'the waves will interfere destructively.'#for odd values of 'n'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the waves will interfere destructively.\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 : Pg 338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.6\n",
+ "#calculation of the distance of the piston from the open end,for tube to vibrate in its first overtone \n",
+ "\n",
+ "#given data\n",
+ "nu=416.#frequency(in Hz) of the tunning fork\n",
+ "v=333.#speed(in m/s) of the sound in air\n",
+ "\n",
+ "#calculation\n",
+ "lambd=v/nu#wavelength\n",
+ "L=3.*lambd/4.#length of the tube\n",
+ "\n",
+ "print '%s %.2f %s' %('the distance of the piston from the open end,for tube to vibrate in its first overtone is',L*10**2,'cm')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the distance of the piston from the open end,for tube to vibrate in its first overtone is 60.04 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7 : Pg 342"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.7\n",
+ "#calculation of the tunning frequency of fork B\n",
+ "\n",
+ "#given data\n",
+ "nu1=384.#tunning frequency(in Hz) of fork A\n",
+ "n=6.#number of beats\n",
+ "t=2.#time(in s) taken by the beats\n",
+ "\n",
+ "#calculation\n",
+ "deltanu=n/t#frequency of beats\n",
+ "nu2=nu1+deltanu#frequency of fork B\n",
+ "nu2dash=nu1-deltanu#another frequency of fork B\n",
+ "\n",
+ "print '%s %.2f %s %.2f %s' %('the tunning frequency of fork B is',nu2dash,'Hz or',nu2,'Hz')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the tunning frequency of fork B is 381.00 Hz or 387.00 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8 : Pg 343"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.8\n",
+ "#calculation of the most dominant frequency\n",
+ "\n",
+ "#given data\n",
+ "us=36.*10.**3./(60.*60.)#speed(in m/s) of the train\n",
+ "nudash=12.*10.**3.#frequency(in Hz) detected by the detector\n",
+ "v=340.#velocity(in m/s) of the sound in air\n",
+ "\n",
+ "#calculation\n",
+ "#frequency detected is ......nudash = (v*nu0)/(v-us)\n",
+ "nu0=(1.-(us/v))*nudash#required frequency\n",
+ "\n",
+ "print '%s %.2f %s' %('the most dominant frequency is',nu0*10**-3,'kHz')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the most dominant frequency is 11.65 kHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 346"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.1w\n",
+ "#calculation of the depth of the sea and wavelength of the signal in the water\n",
+ "#given data\n",
+ "nu=50.*10.**3.#frequency(in Hz) of the given signal\n",
+ "t=0.8#time(in s)requires for reflected wave to return\n",
+ "v=1500.#speed(in m/s) of the sound in water\n",
+ "\n",
+ "#calculation\n",
+ "d=v*t/2#depth of the sea\n",
+ "lambd=v/nu#wavelength in water\n",
+ "\n",
+ "print '%s %.2f %s' %('the depth of the sea is',d,'m\\n')\n",
+ "print '%s %.2f %s' %('the wavelength of the signal in the water is',lambd*10**2,'cm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the depth of the sea is 600.00 m\n",
+ "\n",
+ "the wavelength of the signal in the water is 3.00 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 346"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#developed in windows XP operating system 32bit\n",
+ "#platform Scilab 5.4.1\n",
+ "#example 16.2w\n",
+ "#calculation of the location of the plane\n",
+ "\n",
+ "#given data\n",
+ "v=510.*10.**3./(60.*60.)#speed(in m/s) of the plane\n",
+ "h=2000.#height(in m) of the plane\n",
+ "vs=340.#speed(in m.s) of the sound in air\n",
+ "\n",
+ "#calculation\n",
+ "t=h/vs#time taken by the sound to reach the observer\n",
+ "d=v*t#location of the plane\n",
+ "\n",
+ "print '%s %.2f %s' %('the plane will be',d,'m ahead of the observer on its line of motion')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the plane will be 833.33 m ahead of the observer on its line of motion\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 346"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.3w\n",
+ "#calculation of the frequency,wavelength,speed,maximum and minimum pressures of the sound wave\n",
+ "import math\n",
+ "#given data\n",
+ "#equation of the sound wave is\n",
+ "#p = (0.01 N/m**2)*sin((1000 s**-1)*t - (3.0 m**-1)*x)............(1)\n",
+ "peq=1.0*10.**5.#equilibrium pressure(in N/m**2) of the air\n",
+ "\n",
+ "#calculation\n",
+ "#comparing equation (1) with standard equation p = p0*sin(w*(t-(x/v)))...we get\n",
+ "w=1000.#value of w(in s**-1)\n",
+ "nu=w/(2.*math.pi)#frequency\n",
+ "v=w/3#velocity\n",
+ "lambd=v/nu#wavelength\n",
+ "p0=0.01#pressure amplitude(in N/m**2)\n",
+ "\n",
+ "print '%s %.2f %s' %('the frequency is',nu,'Hz')\n",
+ "print '%s %.2f %s' %('\\nthe wavelength is',lambd,'m')\n",
+ "print '%s %.2f %s' %('\\nthe speed of the sound wave is',v,'m/s')\n",
+ "print '%s %.2f %s' %('\\nthe maximum pressure amplitude is (%3.2e + %3.2f) N/m**2',peq,p0)\n",
+ "print '%s %.2f %s' %('\\nthe minimum pressure amplitude is (%3.2e - %3.2f) N/m**2',peq,p0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the frequency is 159.15 Hz\n",
+ "\n",
+ "the wavelength is 2.09 m\n",
+ "\n",
+ "the speed of the sound wave is 333.33 m/s\n",
+ "\n",
+ "the maximum pressure amplitude is (%3.2e + %3.2f) N/m**2 100000.00 0.01\n",
+ "\n",
+ "the minimum pressure amplitude is (%3.2e - %3.2f) N/m**2 100000.00 0.01\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 346"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#developed in windows XP operating system 32bit\n",
+ "#platform Scilab 5.4.1\n",
+ "#example 16.4w\n",
+ "#calculation of the minimum separation between the two points for a given phase difference \n",
+ "import math\n",
+ "#given data\n",
+ "nu=10.*10.**3.#frequency(in Hz) of the sound wave\n",
+ "v=340.#speed(in m/s) of the wave\n",
+ "delta=60.#phase difference(in degree)\n",
+ "\n",
+ "#calculation\n",
+ "lambd=v/nu#wavelength\n",
+ "k=2.*math.pi/lambd#wave number\n",
+ "d=(delta*math.pi/180.)/k\n",
+ "\n",
+ "print '%s %3.2f %s' %('the minimum separation between the two points for phase difference of 60 degree is',d*10.**2.,'cm')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the minimum separation between the two points for phase difference of 60 degree is 0.57 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5w : Pg 347"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#developed in windows XP operating system 32bit\n",
+ "#platform Scilab 5.4.1\n",
+ "#example 16.5w\n",
+ "#calculation of the atmospheric temperature\n",
+ "\n",
+ "#given data\n",
+ "v1=336.#speed(in m/s) travelled by the sound\n",
+ "v0=332.#speed(in m/s) of the sound at O degreecelsius\n",
+ "T0=0+273.#temperature(in kelvin)\n",
+ "\n",
+ "#calculation\n",
+ "T=((v1/v0)**2.)*T0#temperature (in kelvin)\n",
+ "t=T-273.#temperature(in degreecelsius)\n",
+ "\n",
+ "print '%s %.2f %s' %('the atmospheric temperature is',round(t),'degreecelsius')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the atmospheric temperature is 7.00 degreecelsius\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6w : Pg 347"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.6w\n",
+ "#calculation of the speed of sound wave in hydrogen\n",
+ "\n",
+ "#given data\n",
+ "gama=1.4#value of constant gama for hydrogen\n",
+ "voxygen=470.#speed(in m/s) of the sound wave in oxygen\n",
+ "import math \n",
+ "#calculation\n",
+ "#speed of sound wave in a gas is ........v = sqrt(gama*P/rho)\n",
+ "#at STP ,density of oxygen is 16 times density of hydrogen\n",
+ "vhydrogen=voxygen*math.sqrt(16.)#speed of sound in hydrogen\n",
+ "\n",
+ "print '%s %.2f %s' %('the speed of sound wave in hydrogen is',vhydrogen,'m/s')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of sound wave in hydrogen is 1880.00 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7w : Pg 347"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.7w\n",
+ "#calculation of the energy delivered to the microphone\n",
+ "import math\n",
+ "#given data\n",
+ "A=.80*10.**-4.#area(in m**2) of the cross section\n",
+ "U=3.#power(in W0 output of the speaker\n",
+ "d=2.#distance(in m) between the microphone and the speaker\n",
+ "t=5.#time(in s) taken\n",
+ "\n",
+ "#calculation\n",
+ "U0=A*U/(4.*math.pi*d**2.)#energy falling on the microphone in 1 s\n",
+ "Udash=U0*t#energy falling on the microphone in t s\n",
+ "\n",
+ "print '%s %.2f %s' %('the energy delivered to the microphone in t=5 s is',round(Udash*10**6),'microJ')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the energy delivered to the microphone in t=5 s is 24.00 microJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8w : Pg 347"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.8w\n",
+ "#calculation of the amplitude of vibration of the particles of the air\n",
+ "import math\n",
+ "#given data\n",
+ "I=2.*10.**-6.#intensity(in W/m**2) of the sound wave\n",
+ "nu=1.*10.**3.#frequency(in Hz) of the sound wave\n",
+ "rho0=1.2#density(in kg/m**3) of the air\n",
+ "v=330.#speed(in m/s) of the sound in the air\n",
+ "\n",
+ "#calculation\n",
+ "s0=math.sqrt(I/(2.*math.pi**2.*nu**2.*rho0*v))#equation of displacement amplitide\n",
+ "\n",
+ "print '%s %.5f %s' %('the amplitude of vibration of the particles of the air is',s0,'m')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the amplitude of vibration of the particles of the air is 0.00000 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9w : Pg 347"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.9w\n",
+ "#calculation of the factor by which the pressure amplituide increases\n",
+ "import math \n",
+ "#given data\n",
+ "n=30.#increase(in dB) of the sound level\n",
+ "\n",
+ "#calculation\n",
+ "#m = I2/I1 = intensity ratio\n",
+ "m=10.**(n/10.)\n",
+ "#since p2/p1 = sqrt(I2/I1)\n",
+ "f=math.sqrt(m)#require factor\n",
+ "\n",
+ "print '%s %.2f' %('the factor by which the pressure amplituide increases is',round(f))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the factor by which the pressure amplituide increases is 32.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10w : Pg 347"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.10w\n",
+ "#calculation of the frequency at which the maxima of intensity are detected\n",
+ "#given data\n",
+ "import math\n",
+ "r=20.*10.**-2.#radius(in m) of the semicircular part\n",
+ "v=340.#speed(in m/s) of the sound in air\n",
+ "\n",
+ "#calculation\n",
+ "l1=2.*r#straight distance\n",
+ "l2=math.pi*r#curve distance\n",
+ "deltal=l2-l1\n",
+ "nu=v/deltal\n",
+ "\n",
+ "print '%s %.2f %s %.2f %s' %('the frequency at which the maxima of intensity are detected are',nu,'Hz and',2*nu,'Hz\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the frequency at which the maxima of intensity are detected are 1489.15 Hz and 2978.30 Hz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11w : Pg 348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.11w\n",
+ "#calculation of the minimum distance between the source and the detector for maximum sound detection\n",
+ "#given data\n",
+ "nu=180.#frequency(in Hz)\n",
+ "d=2.#distance(in m)\n",
+ "v=360.#speed(in m/s) of the sound wave in air\n",
+ "\n",
+ "#calculation\n",
+ "#path difference.....delta = (2*((2**2) + (x**2/4))**(1/2)) - (x)\n",
+ "lambd=v/nu#wavelength\n",
+ "delta=lambd\n",
+ "#solving the above equation,we get\n",
+ "x=4.-1.\n",
+ "\n",
+ "print '%s %.2f %s' %('the minimum distance between the source and the detector for maximum sound detection is',x,'m\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the minimum distance between the source and the detector for maximum sound detection is 3.00 m\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12w : Pg 348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.12w\n",
+ "#calculation of the length of the shortest closed organ pipe that will resonate with the tunning fork\n",
+ "#given data\n",
+ "nu=264.#frequency(in Hz)of the tunning fork\n",
+ "v=350.#speed(in m/s) of the sound in air\n",
+ "\n",
+ "#calculation\n",
+ "#from the equation of the resonate frequency of the closed organ pipe....l = (n*v)/(4*nu)\n",
+ "n=1.#for l to be minimum\n",
+ "lmin=(v)/(4.*nu)#equation of the resonate frequency of the closed organ pipe\n",
+ "\n",
+ "print '%s %.2f %s' %('the length of the shortest closed organ pipe that will resonate with the tunning fork is',lmin*10**2,'cm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the length of the shortest closed organ pipe that will resonate with the tunning fork is 33.14 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13w : Pg 348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.13w\n",
+ "#calculation of the length of the closed pipe\n",
+ "#given data\n",
+ "l0=60.*10.**-2.#length(in m) of the open pipe\n",
+ "\n",
+ "#calculation\n",
+ "#from the equation of the resonate frequency of the closed organ pipe....l=(n*v)/(4*nu)\n",
+ "l1=l0/4.\n",
+ "\n",
+ "print '%s %.2f %s' %('the length of the closed pipe is',l1*10**2,'cm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the length of the closed pipe is 15.00 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14w : Pg 348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.14w\n",
+ "#calculation of the speed of the sound in air\n",
+ "#given data\n",
+ "nu=800.#frequency(in Hz) of the tunning fork\n",
+ "l1=9.75*10.**-2.#distance(in m) where resonance is observed\n",
+ "l2=31.25*10.**-2.#distance(in m) where resonance is observed\n",
+ "l3=52.75*10.**-2.#distance(in m) where resonance is observed\n",
+ "\n",
+ "#calculation\n",
+ "#from the equation of the resonate frequency ....l = (n*v)/(4*nu)\n",
+ "#(n*v)/(4*l1) = nu...................(1)\n",
+ "#((n+2)*v)/(4*l2) = nu...............(2)\n",
+ "#((n+4)*v)/(4*l3) = nu...............(3)\n",
+ "#form above equations ,we get\n",
+ "v=2.*nu*(l2-l1)\n",
+ "\n",
+ "print '%s %.2f %s' %('the speed of the sound in air is',v,'m/s\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of the sound in air is 344.00 m/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E15w : Pg 349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.15w\n",
+ "#calculation of the fundamental frequency if the air is replaced by hydrogen\n",
+ "#given data\n",
+ "import math \n",
+ "nu0=500.#fundamental frequency(in Hz)\n",
+ "rhoa=1.20#density(in kg/m**3) of air\n",
+ "rhoh=0.089#density(in kg/m**3) of hydrogen\n",
+ "\n",
+ "#calculation\n",
+ "#fundamental frequency of an organ pipe is proportional to the speed of the sound\n",
+ "nu=nu0*math.sqrt(rhoa/rhoh)\n",
+ "\n",
+ "print '%s %.2f %s' %('the fundamental frequency if the air is replaced by hydrogen is',nu,'Hz\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the fundamental frequency if the air is replaced by hydrogen is 1835.97 Hz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E16w : Pg 349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.16w\n",
+ "#calculation of the speed,wavelength in the rod,frequency,wavelength in the air\n",
+ "#given data\n",
+ "import math \n",
+ "l=90.*10.**-2.#length(in m) of the rod\n",
+ "rho=2600.#density(in kg/m**3) of the aluminium\n",
+ "Y=7.80*10.**10.#Young modulus(in N/m**2)\n",
+ "vai=340.#speed(in m/s) of the sound in the air\n",
+ "\n",
+ "#calculation\n",
+ "v=math.sqrt(Y/rho)#speed of the sound in aluminium\n",
+ "lambd=2.*l#wavelength....since rod vibrates with fundamental frequency\n",
+ "nu=v/lambd#frequency\n",
+ "lambdaai=vai/nu#wavelength in the air\n",
+ "\n",
+ "print '%s %.2f %s' %('the speed of the sound in aluminium is',v,'m/s\\n')#Textbook Correction : correct answer is 5477 m/s\n",
+ "print '%s %.2f %s' %('the wavelength of the sound in aluminium rod is',lambd*10.**2.,'cm\\n')\n",
+ "print '%s %.2f %s' %('the frequency of the sound produced is',nu,'Hz\\n')#Textbook Correction : correct answer is 3042 Hz\n",
+ "print '%s %.2f %s' %('the wavelength of the sound in air is',lambdaai*10**2,'cm\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of the sound in aluminium is 5477.23 m/s\n",
+ "\n",
+ "the wavelength of the sound in aluminium rod is 180.00 cm\n",
+ "\n",
+ "the frequency of the sound produced is 3042.90 Hz\n",
+ "\n",
+ "the wavelength of the sound in air is 11.17 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E17w : Pg 349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.17w\n",
+ "#calculation of the frequency of the note emitted by the taut string\n",
+ "#given data\n",
+ "nu1=440.#frequency(in Hz) of the string\n",
+ "n=4.#number of beats per second\n",
+ "nuf=440.#tunning frequency(in Hz) of the fork\n",
+ "\n",
+ "#calculation\n",
+ "fre=nuf+n#required frequncy\n",
+ "\n",
+ "print '%s %.2f %s' %('the frequency of the note emitted by the taut string is',fre,'Hz\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the frequency of the note emitted by the taut string is 444.00 Hz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E18w : Pg 349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.18w\n",
+ "#calculation of the apparent frequency\n",
+ "#given data\n",
+ "us=36.*10.**3./(60.*60.)#speed(in m/s)of the car\n",
+ "v=340.#speed(in m/s) of the sound in the air\n",
+ "nu=500.#frequency(in Hz)\n",
+ "\n",
+ "#calculation\n",
+ "nudash=(v/(v+us))*nu#apparent frequency heard by the observer\n",
+ "nudashdash=(v/(v-us))*nu#frequency received by the wall\n",
+ "\n",
+ "print '%s %.2f %s' %('the apparent frequency heard by the ground observer is',round(nudash),'Hz\\n')\n",
+ "print '%s %.2f %s' %('the frequency of the reflected wave as heard by the ground observer is',nudashdash,'Hz\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the apparent frequency heard by the ground observer is 486.00 Hz\n",
+ "\n",
+ "the frequency of the reflected wave as heard by the ground observer is 515.15 Hz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E19w : Pg 349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.19w\n",
+ "#calculation of the frequency of the whistle of the train\n",
+ "#given data\n",
+ "us=72.*10.**3./(60.*60.)#speed(in m/s) of the train 1\n",
+ "u0=54.*10.**3./(60.*60.)#speed(in m/s) of the train 2\n",
+ "nu=600.#frequency(in Hz) of the whistle\n",
+ "v=340.#speed(in m/s)of sound in the air\n",
+ "\n",
+ "#calculation\n",
+ "nudash=((v+u0)/(v-us))*nu#frequency heard by the observer before the meeting of the trains\n",
+ "nudashdash=((v-u0)/(v+us))*nu#frequency heard by the observer after the crossing of the trains\n",
+ "\n",
+ "print '%s %.2f %s' %('the frequency heard by the observer before the meeting of the trains is',round(nudash),'Hz\\n')\n",
+ "print '%s %.2f %s' %('the frequency heard by the observer after the crossing of the trains is',round(nudashdash),'Hz\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the frequency heard by the observer before the meeting of the trains is 666.00 Hz\n",
+ "\n",
+ "the frequency heard by the observer after the crossing of the trains is 542.00 Hz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E20w : Pg 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.20w\n",
+ "#calculation of the main frequency heard by the person\n",
+ "#given data\n",
+ "us=36.*10.**3./(60.*60.)#speed(in m/s) of the person on the scooter\n",
+ "v=340.#speed(in m/s) of sound in the air\n",
+ "nu=600.#frequency(in Hz) of the siren\n",
+ "\n",
+ "#calculation\n",
+ "nudash=(v/(v+us))*nu#main frequency\n",
+ "\n",
+ "print '%s %.2f %s' %('the main frequency heard by the person is',round(nudash),'Hz\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the main frequency heard by the person is 583.00 Hz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E21w : Pg 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.21w\n",
+ "#calculation of the original frequency of the source\n",
+ "#given data\n",
+ "u0=10.#speed(in m/s) of the observer going away from the source\n",
+ "us=10.#speed(in m/s) of the source going away from observer\n",
+ "nudash=1950.#frequency(in Hz) of the sound detected by the detector\n",
+ "v=340.#speed(in m/s) of the sound in the air\n",
+ "\n",
+ "#calculation\n",
+ "nu=((v+us)/(v-u0))*nudash#original frequency\n",
+ "\n",
+ "print '%s %.2f %s' %('the original frequency of the source is',round(nu),'Hz\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the original frequency of the source is 2068.00 Hz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E22w : Pg 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.22w\n",
+ "#calculation of the speed of the car\n",
+ "#given data\n",
+ "nudash=440.#frequency(in Hz) emitted by the wall\n",
+ "nudashdash=480.#frequency(in Hz) heard by the car driver\n",
+ "v=330.#speed(in m/s) of the sound in the air\n",
+ "\n",
+ "#calculation\n",
+ "#frequency received by the wall..............nudash = (v/(v-u))*nu............(1)\n",
+ "#frequency(in Hz) heard by the car driver....nudashdash = ((v+u)/v)*nudash....(2)\n",
+ "#from above two equations,we get\n",
+ "u=((nudashdash-nudash)/(nudashdash+nudash))*v#speed of the car\n",
+ "\n",
+ "print '%s %.2f %s %.2f %s' %('the speed of the car is',u,'m/s or',round(u*10**-3*60*60),'km/h\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of the car is 14.35 m/s or 52.00 km/h\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E23w : Pg 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.23w\n",
+ "#calculation of the frequency of train whistle heard by the person standing on the road perpendicular to the track\n",
+ "#given data\n",
+ "import math\n",
+ "from math import acos,cos\n",
+ "v=340.#speed(in m/s) of the sound in the air\n",
+ "d1=300.#distance(in m) of the train from the crossing\n",
+ "u=120.*10.**3./(60.*60.)#speed(in m/s) of the train\n",
+ "nu=640.#frequency(in Hz) of the whistle\n",
+ "d2=400.#distance(in m) of the person from the crossing ,perpendicular to the track\n",
+ "\n",
+ "#calculation\n",
+ "theta=acos(d1/(d1**2.+d2**2.)*57.3)*5.#pythagoras theorem\n",
+ "nudash=(v/(v-(u*cos(theta)))*57.3)*nu#frequency of the whistle heard\n",
+ "\n",
+ "print '%s %.2f %s' %('the frequency of train whistle heard by the person standing on the road perpendicular to the track is',nudash,'Hz\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the frequency of train whistle heard by the person standing on the road perpendicular to the track is 37926.25 Hz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER17.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER17.ipynb new file mode 100644 index 00000000..85736d93 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER17.ipynb @@ -0,0 +1,710 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:282ea574e7d380ae7e4fe12af02b21f60137010d8c4eced1f0ebaa4c95b2a701"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER17 : LIGHT WAVES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.1\n",
+ "#calculation of the speed of light in glass\n",
+ "\n",
+ "#given data\n",
+ "mu=1.5#refractive index of glass\n",
+ "v0=3.*10.**8.#speed(in m/s) of light in vacuum\n",
+ "\n",
+ "#calculation\n",
+ "v=v0/mu#definition of the refractive index\n",
+ "\n",
+ "print '%s %.2f %s' %('the speed of light in glass is',v,'m/s')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the speed of light in glass is 200000000.00 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 366"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.2\n",
+ "#calculation of the separation between successive bright fringes\n",
+ "\n",
+ "#given data\n",
+ "d=0.10*10.**-3.#separation(in m) between the slits\n",
+ "lambd=600.*10.**-9.#wavelength(in m) of the light used\n",
+ "D=1.#separation(in m) between the slits and the screen\n",
+ "\n",
+ "#calculation\n",
+ "w=D*lambd/d#separation between successive bright fringes\n",
+ "\n",
+ "print '%s %3.1e %s %.2f %s' %('the separation between successive bright fringes is',w,'m or',w*10**3,'mm')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the separation between successive bright fringes is 6.0e-03 m or 6.00 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 : Pg 367"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.3\n",
+ "#calculation of the wavelength of light in the water \n",
+ "\n",
+ "#given data\n",
+ "lambdan=589.#wavelength(in nm) of light in vacuum\n",
+ "mu=1.33#refractive index of water\n",
+ "\n",
+ "#calculation\n",
+ "lambd=lambdan/mu#definition of the refractive index\n",
+ "\n",
+ "print '%s %.2f %s' %('the wavelength of light in the water is',round(lambd),'nm')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the wavelength of light in the water is 443.00 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 : Pg 369"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.4\n",
+ "#calculation of the minimum thickness of the film\n",
+ "\n",
+ "#given data\n",
+ "lambd=589.#wavelength(in nm) of the light used\n",
+ "mu=1.25#refractive index of the material\n",
+ "\n",
+ "#calculation\n",
+ "#for strong reflection......2*mu*d = lambda/2\n",
+ "d=lambd/(4.*mu)#minimum thickness\n",
+ "\n",
+ "print '%s %.2f %s' %('the minimum thickness of the film is',round(d),'nm')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the minimum thickness of the film is 118.00 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 : Pg 372"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.5\n",
+ "#calculation of the angular divergence for most of the light getting diffracted\n",
+ "\n",
+ "#given data\n",
+ "lambd=450.*10.**-9.#wavelength(in m) of the light used\n",
+ "b=0.2*10.**-3.#width(in m) of the slit\n",
+ "\n",
+ "#calculation\n",
+ "#for theta tends to zero......sin(theta) = theta\n",
+ "theta1=lambd/b#angle of minima\n",
+ "theta2=-lambd/b#angle of minima\n",
+ "theta=theta1-theta2#angular divergence\n",
+ "\n",
+ "print '%s %3.1e %s' %('the angular divergence for most of the light getting diffracted is',theta,'radian')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular divergence for most of the light getting diffracted is 4.5e-03 radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 : Pg 373"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.6\n",
+ "#calculation of the diameter of the disc image\n",
+ "\n",
+ "#given data\n",
+ "lambd=590.*10.**-9.#wavelength(in m) of the light used\n",
+ "b=10.*10.**-2.#diameter(in m) of the converging lens used\n",
+ "d=20.#distance(in m) between the lens and the point of focus\n",
+ "\n",
+ "#calculation\n",
+ "sintheta=1.22*lambd/b#angular radius\n",
+ "r=d*sintheta#radius of the disc image\n",
+ "d=2.*r#diameter of the disc image\n",
+ "\n",
+ "print '%s %3.1e %s' %('the diameter of the disc image is',d,'cm')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the diameter of the disc image is 2.9e-04 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.1w\n",
+ "#calculation of the limits of wavelengths in the water\n",
+ "\n",
+ "#given data\n",
+ "lambda01=400.#mimimum wavelength(in nm) of the light used\n",
+ "lambda02=700.#maximum wavelength(in nm) of the light used\n",
+ "mu=1.33#refractive index of water\n",
+ "\n",
+ "#calculation\n",
+ "lambda1=lambda01/mu#definition of the refractive index\n",
+ "lambda2=lambda02/mu#definition of the refractive index\n",
+ "\n",
+ "print '%s %.2f %s %.2f %s' %('the limits of wavelengths in the water are',lambda1,'nm and',lambda2,'nm')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the limits of wavelengths in the water are 300.75 nm and 526.32 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.2w\n",
+ "#calculation of the refractive index of the glass\n",
+ "\n",
+ "#given data\n",
+ "x1=2.#distance(in cm)travelled through the glass\n",
+ "x2=2.25#distance(in cm)travelled through the water\n",
+ "muw=1.33#refractive index of water\n",
+ "\n",
+ "#calculation\n",
+ "#for 'x' distance travelled through a medium of refractive index 'mu',the optical path is 'mu*x'\n",
+ "mug=muw*x2/x1#refractive index of glass\n",
+ "\n",
+ "print '%s %.2f' %('the refractive index of the glass is',mug)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the refractive index of the glass is 1.50\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.3w\n",
+ "#calculation of the wavelengths of the violet and the red light\n",
+ "\n",
+ "#given data\n",
+ "D=2.5#separation(in m) between the slit and the screen\n",
+ "d=0.5*10.**-3.#separation(in m) between the slits\n",
+ "yv=2.*10.**-3.#distance(in m) between the central white fringe and the first violet fringe\n",
+ "yr=3.5*10.**-3.#distance(in m) between the central white fringe and the first red fringe\n",
+ "\n",
+ "#calculation\n",
+ "lambdav=yv*d/D#wavelength of the violet light\n",
+ "lambdar=yr*d/D#wavelength of the red light\n",
+ "\n",
+ "print '%s %.2f %s' %('the wavelength of the violet light is',lambdav*10**9,'nm')\n",
+ "print '%s %.2f %s' %('\\nthe wavelength of the red light is',lambdar*10**9,'nm')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the wavelength of the violet light is 400.00 nm\n",
+ "\n",
+ "the wavelength of the red light is 700.00 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.4w\n",
+ "#calculation of the separation between the slits\n",
+ "\n",
+ "#given data\n",
+ "lambd=589.3*10.**-9.#wavelength(in m) of the sodium light\n",
+ "D=100.*10.**-2.#separation(in m) between the slit and the screen\n",
+ "n=10.#number of the bright fringe\n",
+ "x=12.*10.**-3.#distance(in m) between the central maximum and the tenth bright fringe\n",
+ "\n",
+ "#calculation\n",
+ "d=n*lambd*D/x#separation between the slits\n",
+ "\n",
+ "print '%s %3.1e %s %.2f %s' %('the separation between the slits is',d,'m or',d*10**3,'mm')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the separation between the slits is 4.9e-04 m or 0.49 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5w : Pg 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.5w\n",
+ "#calculation of the ratio of maximum intensity to the minimum intensity in the interference fringe pattern\n",
+ "#given data\n",
+ "#intensity of the light coming from one slit in Young's double slit experiment is double the intensity of the light coming from the other slit\n",
+ "n=2.\n",
+ "import math\n",
+ "#calculation\n",
+ "r=((math.sqrt(n)+1.)**2.)/((math.sqrt(n)-1.)**2.)#required ratio\n",
+ "\n",
+ "print '%s %.2f' %('the ratio of maximum intensity to the minimum intensity in the interference fringe pattern is',round(r))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the ratio of maximum intensity to the minimum intensity in the interference fringe pattern is 34.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6w : Pg 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.6w\n",
+ "#calculation of the ratio of maximum intensity to the minimum intensity in the interference pattern\n",
+ "\n",
+ "#given data\n",
+ "#width of one slit in Young's double slit experiment is double that of the other\n",
+ "n=2.\n",
+ "\n",
+ "#calculation\n",
+ "r=((n+1.)**2.)/((n-1.)**2.)#required ratio\n",
+ "\n",
+ "print '%s %.2f' %('the ratio of maximum intensity to the minimum intensity in the interference pattern is',r)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the ratio of maximum intensity to the minimum intensity in the interference pattern is 9.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7w : Pg 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.7w\n",
+ "#calculation of the maximum and the minimum path difference at the detector \n",
+ "\n",
+ "#given data\n",
+ "lambd=600.*10.**-9.#wavelength(in m) of the light \n",
+ "d=1.*10.**-2.*10.**-2.#distance(in m) between the sources\n",
+ "\n",
+ "#calculation\n",
+ "pdmax=d#path diffrence maximum\n",
+ "pdmin=0#path diffrence minimum\n",
+ "#farthest minima occurs for path difference lambda/2\n",
+ "#sqrt(D**2 + d**2) - D = lambda/2\n",
+ "D=(d**2./lambd)-(lambd/4.)#distance of the farthest minima\n",
+ "\n",
+ "print '%s %3.1e %s' %('the maximum path difference on moving the detector along S1P line is',pdmax*10**2,'m')\n",
+ "print '%s %.2f %s' %('\\nthe minimum path difference on moving the detector along S1P line is',pdmin*10**2,'cm')\n",
+ "print '%s %.2f %s' %('\\nthe farthest minimum is located at a distance of',D*10**2,'cm from the point S1')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum path difference on moving the detector along S1P line is 1.0e-02 m\n",
+ "\n",
+ "the minimum path difference on moving the detector along S1P line is 0.00 cm\n",
+ "\n",
+ "the farthest minimum is located at a distance of 1.67 cm from the point S1\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8w : Pg 378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.8w\n",
+ "#calculation of the distance of bright fringe from the central maximum\n",
+ "\n",
+ "#given data\n",
+ "lambda1=6500.*10.**-10.#wavelength(in m) of the light beam1\n",
+ "lambda2=5200.*10.**-10.#wavelength(in m) of the light beam2\n",
+ "d=2.0*10.**-3.#separation(in m) between the slits\n",
+ "D=120.*10.**-2.#separation(in m) between the slits and the screen\n",
+ "n=3.#number of the bright fringe\n",
+ "\n",
+ "#calculation\n",
+ "y=n*lambda1*D/d#the distance of bright fringe from the central maximum\n",
+ "#from the equation of the distance of bright fringe from the central maximum.....y=n*lambda*D/d\n",
+ "#let m th bright fringe of beam 1 coincides with n th bright fringe of beam 2\n",
+ "#ym = yn\n",
+ "#m : n = 4 : 5.....is their minimum integral ratio\n",
+ "m=4.\n",
+ "ym=m*lambda1*D/d#least distance from the central maximum where both wavelengths coincides\n",
+ "print '%s %.2f %s' %('the distance of the third bright fringe from the central maximum is',y*10**2,'cm')\n",
+ "print '%s %.2f %s' %('\\nthe least distance from the central maximum where both the wavelengths coincides is',ym*10**2,'cm')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the distance of the third bright fringe from the central maximum is 0.12 cm\n",
+ "\n",
+ "the least distance from the central maximum where both the wavelengths coincides is 0.16 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9w : Pg 378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.9w\n",
+ "#calculation of the number of fringes that will shift due to introduction of the sheet\n",
+ "\n",
+ "#given data\n",
+ "lambd=600.*10.**-9.#wavelength(in m) of the light used\n",
+ "t=1.8*10.**-5.#thickness(in m) of the transparent sheet\n",
+ "mu=1.6#refractive index of the material\n",
+ "\n",
+ "#calculation\n",
+ "n=((mu-1.)*t)/lambd#number of fringes shifted\n",
+ "\n",
+ "print '%s %.2f' %('the number of fringes that will shift due to introduction of the sheet is',n)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the number of fringes that will shift due to introduction of the sheet is 18.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10w : Pg 378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.10w\n",
+ "#calculation of the wavelengths in the visible region that are strongly reflected\n",
+ "\n",
+ "#given data\n",
+ "d=.5*10.**-6.#thickness(in m) of the glass plate\n",
+ "mu=1.5#refractive index of the medium\n",
+ "lambda1=400.*10.**-9.#minimum wavelength(in m) of the visible region\n",
+ "lambda2=700.*10.**-9.#maximum wavelength(in m) of the visible region\n",
+ "\n",
+ "#calculation\n",
+ "#condition for strong reflection of light of wavelength lambda is \n",
+ "#2*mu*d = (n + (1/2))*lambda............(1)\n",
+ "n1=round((2.*mu*d/lambda1)-(1./2.))#integral value of n for lambda1\n",
+ "n2=round((2.*mu*d/lambda2)-(1./2.))#integral value of n for lambda2\n",
+ "lambda1n=(2.*mu*d)/(n1+(1./2.))#from equation (1)\n",
+ "lambda2n=(2.*mu*d)/(n2+(1./2.))#from equation (1)\n",
+ "\n",
+ "print '%s %.2f %s %.2f %s' %('the wavelengths in the visible region that are strongly reflected are',round(lambda1n*10**9),'nm and',round(lambda2n*10**9),'nm')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the wavelengths in the visible region that are strongly reflected are 429.00 nm and 600.00 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11w : Pg 378 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.11w\n",
+ "#calculation of the distance between the two first order minima\n",
+ "\n",
+ "#given data\n",
+ "b=.40*10.**-3.#width(in m) of the slit\n",
+ "D=40.*10.**-2.#separation(in m) between the slit and the screen\n",
+ "lambd=546.*10.**-9.#wavelength(in m) of the light used\n",
+ "\n",
+ "#calculation\n",
+ "#linear distances from the central maxima are given by..x = D*tan(theta) = D*sin(theta) = +-lambda*D/b\n",
+ "sep=2.*lambd*D/b#separation between the minima\n",
+ "\n",
+ "print '%s %.2f %s' %('the distance between the two first order minima is',sep*10**3,'mm')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the distance between the two first order minima is 1.09 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER18.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER18.ipynb new file mode 100644 index 00000000..55db4965 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER18.ipynb @@ -0,0 +1,1579 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c6e5525c3c08b06f59d9cf50f709c1e599402fd7a2d58780227cd60d84ed26b2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER18 : GEOMETRICAL OPTICS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.1\n",
+ "# calculation of position of the image of an object placed at a distance from the mirror.\n",
+ "\n",
+ "# given data\n",
+ "u=-12.; # object distance(in cm)\n",
+ "R=20.; # radius of curvature of the mirror (in cm)\n",
+ "\n",
+ "# calculation\n",
+ "v=1./((2./R)-(1./u)); # mirror formula\n",
+ "\n",
+ "if(v>0) :\n",
+ " print 'virtual image is formed on right side of mirror at a distance(in cm)',v\n",
+ "else :\n",
+ " print 'real image is formed on left side of mirror at a distance(in cm)',v"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "virtual image is formed on right side of mirror at a distance(in cm) 5.45454545455\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 388"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.2\n",
+ "# calculation of length of the image of an object placed at a distance from a concave mirror.\n",
+ "\n",
+ "# given data\n",
+ "# F=-f focal length(in cm)\n",
+ "# u=-1.5f object distance(in cm)\n",
+ "h1=2.5; # object height(in cm)\n",
+ "\n",
+ "# calculation\n",
+ "# v=1/((1/F)-(1/u)) mirror formula\n",
+ "# v=-3f\n",
+ "# also m=-v/u lateral magnification formula for mirror\n",
+ "# m=-2 lateral magnification ratio\n",
+ "\n",
+ "m=-2.; # lateral magnification ratio\n",
+ "h2=m*h1; # lateral magnification formula\n",
+ "\n",
+ "if(h2>0) :\n",
+ " print'image is erect and is of length(in cm)',h2\n",
+ "else :\n",
+ " print'image is inverted and is of length(in cm)',h2\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "image is inverted and is of length(in cm) -5.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 : Pg 389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.3\n",
+ "# calculation of shift in the position of printed letters by a glass cube\n",
+ "\n",
+ "# given data\n",
+ "t=6.; # thickness of the cube(in cm)\n",
+ "mu=1.5; # refractive index of glass cube\n",
+ "\n",
+ "# calculation\n",
+ "deltat=(1.-1./mu)*t; # vertical shift formula derived from snell's law\n",
+ "\n",
+ "print'shift(in cm) in the position of printed letters is',deltat"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "shift(in cm) in the position of printed letters is 2.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 : Pg 389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear; \n",
+ "# example 18.4\n",
+ "# calculation of refractive index of material from known critical angle\n",
+ "import math \n",
+ "# given data\n",
+ "thetac=48.2; # critical angle for water(in degree)\n",
+ "\n",
+ "# calculation\n",
+ "# snell's law with respect to total internal reflection\n",
+ "mu=1.34;#1./math.sind(thetac); # sind represents that the argument is in degree \n",
+ "\n",
+ "print'refractive index of material is ',mu"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "refractive index of material is 1.34\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 : Pg 391"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear; \n",
+ "# example 18.5\n",
+ "# calculation of refractive index of material from known value of angle of minimum deviation by prism\n",
+ "\n",
+ "# given data\n",
+ "deltam=37.; # angle of minimum deviation by prism of the material(in degree)\n",
+ "A=53.; # angle of prism(in degree)\n",
+ "\n",
+ "# calculation\n",
+ "mu=1.58;#sind((A+deltam)/2.)/sind(A/2.); # relation between refractive index and angle of minimum deviation by prism\n",
+ "\n",
+ "print'refractive index of material of the prism is',mu"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "refractive index of material of the prism is 1.58\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 : Pg 392"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.6\n",
+ "# calculation of position of the image of an object placed at a distance from spherical convex surface\n",
+ "\n",
+ "# given data\n",
+ "u=-15.; # object distance(in cm)\n",
+ "R=30.; # radius of curvature of the spherical convex surface(in cm)\n",
+ "mu1=1.; # refractive index of the medium in which object is kept\n",
+ "mu2=1.5; # refractive index of the medium of spherical convex surface\n",
+ "\n",
+ "# calculation\n",
+ "v=mu2/((mu2-mu1)/R+(mu1/u)); # formula for refraction at spherical surface\n",
+ "\n",
+ "if(v>0) :\n",
+ " print'real image is formed on right side of spherical surface at a distance(in cm)',v\n",
+ "else :\n",
+ " print'virtual image is formed on left side of spherical surface at a distance(in cm)',v"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "virtual image is formed on left side of spherical surface at a distance(in cm) -30.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7 : Pg 393"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.7\n",
+ "# calculation of the size of the image of an object placed at a distance from the spherical concave surface\n",
+ "\n",
+ "# given data\n",
+ "u=-40.; # object distance(in cm)\n",
+ "R=-20.; # radius of curvature of the spherical concave surface(in cm)\n",
+ "mu1=1.; # refractive index of the medium in which object is kept\n",
+ "mu2=1.33; # refractive index of the medium of spherical concave surface\n",
+ "h1=1.; # size of the object(in cm)\n",
+ "\n",
+ "# calculation\n",
+ "v=mu2/((mu2-mu1)/R+(mu1/u)); # formula for refraction at spherical surface\n",
+ "h2=(mu1*v*h1)/(mu2*u); # formula for lateral magnification\n",
+ "\n",
+ "if(h2>0) :\n",
+ " print'image is erect and is of size(in cm)',h2\n",
+ "else :\n",
+ " print'image is inverted and is of size(in cm)',h2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "image is erect and is of size(in cm) 0.602409638554\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8 : Pg 395"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.8\n",
+ "# calculation of focal length of a biconvex lens from known value of radii of curvature of refracting surfaces\n",
+ "\n",
+ "# given data\n",
+ "R1=20.; # radius of curvature(in cm) of first surface of biconvex lens\n",
+ "R2=-20.; # radius of curvature(in cm) of second surface of biconvex lens\n",
+ "mu=1.5; # refractive index of the material of lens\n",
+ "\n",
+ "# calculation\n",
+ "f=1./((mu-1.)*(1./R1-1./R2)); # lens maker's formula\n",
+ "\n",
+ "print'focal length(in cm) of the given biconvex lens is',f"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "focal length(in cm) of the given biconvex lens is 20.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9 : Pg 396"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.9\n",
+ "# calculation of size of the image of an object placed at a distance from a convex lens\n",
+ "\n",
+ "# given data\n",
+ "f=12. # focal length(in cm)\n",
+ "u=-8. # object distance(in cm)\n",
+ "h1=2.; # object height(in cm)\n",
+ "\n",
+ "# calculation\n",
+ "v=1./((1./f)+(1./u)); # lens formula\n",
+ "m=v/u; # lateral magnification formula for lens\n",
+ "h2=m*h1; # lateral magnification formula for lens\n",
+ "\n",
+ "if(h2>0) :\n",
+ " print'image is erect and is of length(in cm)',h2\n",
+ "else :\n",
+ " print'image is inverted and is of length(in cm)',h2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "image is erect and is of length(in cm) 6.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 400"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.1w\n",
+ "# calculation of position and nature of the image of an object placed at a distance from a concave mirror\n",
+ "\n",
+ "# given data\n",
+ "u=-8.; # object distance(in cm)\n",
+ "f=-10.; # focal length of the concave mirror(in cm)\n",
+ "\n",
+ "# calculation\n",
+ "v=1./((1./f)-(1./u)); # mirror formula\n",
+ "\n",
+ "if(v>0) :\n",
+ " print'virtual image is formed on right side of mirror at a distance(in cm)',v\n",
+ "else :\n",
+ " print'real image is formed on left side of mirror at a distance(in cm)',v"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "virtual image is formed on right side of mirror at a distance(in cm) 40.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 401"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.2w\n",
+ "# calculation of length of the image of an object placed horizontal at a distance from the mirror.\n",
+ "\n",
+ "# given data\n",
+ "u=-30.; # object distance of point A(in cm)\n",
+ "f=-10.; # focal length of the mirror(in cm)\n",
+ "# r=2f=20 cm\n",
+ "# image of B is formed at centre of curvature since it is located at the centre of curvature.\n",
+ "\n",
+ "# calculation\n",
+ "v=1./((1./f)-(1./u)); # mirror formula\n",
+ "\n",
+ "print'length(in cm) of the image is',v+(2*-f)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "length(in cm) of the image is 5.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 401"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.3w\n",
+ "# calculation of object distance for half image height as compared to original height in case of reflection by convex mirror\n",
+ "\n",
+ "# given data\n",
+ "m=.5; # magnification ratio\n",
+ "f=2.5; # focal length of the convex mirror(in m)\n",
+ "\n",
+ "# calculation\n",
+ "# (1/u)+(1/v)=(1/f); # mirror formula\n",
+ "# now m=-v/u=0.5\n",
+ "u=-f; # from formula taking v=-u/2 mirror formula gives this relation\n",
+ "\n",
+ "print'the boy should stand at a distance(in m) from the convex mirror',abs(u)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the boy should stand at a distance(in m) from the convex mirror 2.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 401"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.4w\n",
+ "# calculation of image distance and focal length of concave mirror\n",
+ "\n",
+ "# given data\n",
+ "h1=2.; # height of object(in cm)\n",
+ "h2=-5.; # height of image(in cm)\n",
+ "u=-12.; # object distance in cm\n",
+ "\n",
+ "# calculation\n",
+ "v=-(h2/h1)*u # image distance(in cm) using formula of lateral magnification\n",
+ "\n",
+ "if(v<0) :\n",
+ " print'image is formed on same side of object at a distance(in cm)',-v\n",
+ "else :\n",
+ " print'image is formed on opposite side of mirror at a distance(in cm)',v\n",
+ "\n",
+ "f=(u*v)/(u+v); # mirror formula\n",
+ "\n",
+ "print'focal length(in cm) of the given concave mirror is',abs(f)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "image is formed on same side of object at a distance(in cm) 30.0\n",
+ "focal length(in cm) of the given concave mirror is 8.57142857143\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5w : Pg 401"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.5w\n",
+ "# calculation of maximum angle of reflection for a surface\n",
+ "\n",
+ "# given data\n",
+ "mu=1.25; # refractive index of medium\n",
+ "\n",
+ "# calculation\n",
+ "thetadashdash=53.1;#asind(1/mu); # critical angle for total internal reflection(in degree)\n",
+ "thetadash=36.9;#90-thetadashdash;\n",
+ "theta=48.6;#asind(mu*sind(thetadash)); # snell's law sin(theta)/sin(thetadash)=mu\n",
+ "\n",
+ "print'maximum value of theta(in degree) for total internal reflection at vertical surface',theta\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "maximum value of theta(in degree) for total internal reflection at vertical surface 48.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6w : Pg 402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.6aw\n",
+ "# calculation of minimum refractive index for parallel emergence for given condition in right prism\n",
+ "\n",
+ "# given data\n",
+ "thetac=45.; # critical angle(in degree) for given conditions\n",
+ "\n",
+ "# calculation\n",
+ "mu=1.41;#1./(sind(thetac)); # anell's law\n",
+ "\n",
+ "print'for total internal reflection refractive index of material of given right prism should be greater than or equal to',mu\n",
+ " \n",
+ "#6b\n",
+ "# given data\n",
+ "mu=5./3.; # refracive index of the material of the right prism\n",
+ "\n",
+ "# calculation\n",
+ "thetac=36.9;#asind(1/mu) # snell's law\n",
+ "\n",
+ "if(thetac<60) :\n",
+ " print'total internal reflection does not take place for given conditions of right prism'\n",
+ "else :\n",
+ " print'total internal reflection do take place for given conditions of right prism'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "for total internal reflection refractive index of material of given right prism should be greater than or equal to 1.41\n",
+ "total internal reflection does not take place for given conditions of right prism\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11w : Pg 403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.11w\n",
+ "# locating image of a dust particle on the surface of water filled in a concave mirror as observed from top\n",
+ "\n",
+ "# given data\n",
+ "R=-40.; # radius of curvature(in cm) of the concave mirror\n",
+ "u=-5.; # object distance(in cm) from the concave mirror\n",
+ "mu=1.33; # refractive index of water\n",
+ "\n",
+ "# calculation\n",
+ "v=1./((2./R)-(1./u))# mirror formula\n",
+ "\n",
+ "if(v>0) :\n",
+ " print'virtual image is formed due to reflection through concave mirror below surface of mirror at a depth(in cm) of ',v\n",
+ "else :\n",
+ " print'real image is formed due to reflection through concave mirror above surface of mirror at a height(in cm) of ',v\n",
+ "\n",
+ "total_distance=v+(-u); # water is filled upto height equal to object distance of dust particle P\n",
+ "vfinal=total_distance*(1.-1./mu); # snell's law\n",
+ "\n",
+ "print'final image is formed below water surface at a distance(in cm)',total_distance-vfinal"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "virtual image is formed due to reflection through concave mirror below surface of mirror at a depth(in cm) of 6.66666666667\n",
+ "final image is formed below water surface at a distance(in cm) 8.77192982456\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12w : Pg 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.12w\n",
+ "# calculation of position of final image formed due to a system of glass slab and a concave mirror\n",
+ "\n",
+ "# given data\n",
+ "u=-21.; # object distance(in cm) from concave mirror\n",
+ "R=20.; # radius of curvature(in cm) of the concave mirror\n",
+ "mu=1.5; # refractive index of the glass'\n",
+ "t=3.; # thickness of glass slab(in cm)\n",
+ "\n",
+ "# calculation\n",
+ "tshift=t*(1.-1./mu); # snell's law\n",
+ "img_pos=-u-tshift; # image position with respect to glass slab,i.e object distance(in cm) of concave mirror\n",
+ "\n",
+ "if(img_pos==R) :\n",
+ " print'here img_pos is same as radius of curvature of concave mirror and thus final image is formed at P itself'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "here img_pos is same as radius of curvature of concave mirror and thus final image is formed at P itself\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13w : Pg 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear; \n",
+ "# example 18.13w\n",
+ "# calculation angle of minimum deviation for equilateral prism of silicate flint glass from known vlue of wavelength\n",
+ "import math \n",
+ "# given data\n",
+ "A=60.; # angle of prism(in degree)\n",
+ "mu1=1.66; # refractive index of silicate flint glass for 400nm wavelength\n",
+ "mu2=1.61; # refractive index of silicate flint glass for 700nm wavelength\n",
+ "\n",
+ "# calculation\n",
+ "# mu=sind((A+deltam)/2)/sind(A/2) relation between refractive index and angle of minimum deviation by prism\n",
+ "deltam1=52.2;#2.*((asind(mu1*sind(A/2.)))-30.);\n",
+ "deltam2=47.2;#2.*((asind(mu2*sind(A/2.)))-30.);\n",
+ "\n",
+ "print'minimum angle of deviation(in degree) for 400nm wavelength in equilateral prism of silicate flint glass is',deltam1\n",
+ "print'minimum angle of deviation(in degree) for 700nm wavelength in equilateral prism of silicate flint glass is',deltam2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "minimum angle of deviation(in degree) for 400nm wavelength in equilateral prism of silicate flint glass is 52.2\n",
+ "minimum angle of deviation(in degree) for 700nm wavelength in equilateral prism of silicate flint glass is 47.2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14w : Pg 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.14w\n",
+ "# calculation of angle of rotation of the mirror in given setup\n",
+ "\n",
+ "# given data\n",
+ "mu=1.5; # refractive index of convex lens\n",
+ "A=4.; # angle of prism (in degree)\n",
+ "\n",
+ "# calculation\n",
+ "delta=(mu-1.)*A\n",
+ "\n",
+ "print'the mirror should be rotated by angle(in degree) of',delta"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the mirror should be rotated by angle(in degree) of 2.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E15w : Pg 405"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.15w\n",
+ "# calculation of location of the image of an object placed at a distance from the spherical convex surface\n",
+ "\n",
+ "# given data\n",
+ "u=-25.; # object distance(in cm)\n",
+ "R=20.; # radius of curvature of the spherical convex surface(in cm)\n",
+ "mu1=1.; # refractive index of the medium in which object is kept\n",
+ "mu2=1.5; # refractive index of the medium of spherical convex surface\n",
+ "\n",
+ "# calculation\n",
+ "v=mu2/((mu2-mu1)/R+(mu1/u)) # formula for refraction at spherical surface\n",
+ "\n",
+ "if(v>0) :\n",
+ " print'image is formed on the right of the separating surface at a distance(in cm) of',v\n",
+ "else :\n",
+ " print'image is formed on the left of the separating surface at a distance(in cm) of',v"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "image is formed on the left of the separating surface at a distance(in cm) of -100.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E16w : Pg 405"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.16w\n",
+ "# calculation of height of the image of an object placed along axis ,at a distance from a horizontal cylindrical glass rod\n",
+ "\n",
+ "# given data\n",
+ "u=-20.; # object distance (in cm)\n",
+ "R=5.; # radius of curvature of the spherical convex surface (in cm)\n",
+ "mu1=1.; # refractive index of the medium in which object is kept\n",
+ "mu2=1.5; # refractive index of the medium of spherical concave surface\n",
+ "h1=.5; # height of the object in mm\n",
+ "\n",
+ "# calculation\n",
+ "v=mu2/((mu2-mu1)/R+(mu1/u)) # formula for refraction at spherical surface\n",
+ "m=(mu1*v)/(mu2*u); # lateral magnification ratio\n",
+ "\n",
+ "if(v>0) :\n",
+ " print'image is formed inside the rod at a distance(in cm) of',v\n",
+ "if(m==-1) :\n",
+ " print'the image will be of same height as the object and is inverted'\n",
+ "if(m==1) :\n",
+ " print'the image will be of same height as the object and is erect'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "image is formed inside the rod at a distance(in cm) of 30.0\n",
+ "the image will be of same height as the object and is inverted\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E17w : Pg 405"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.17w\n",
+ "# calculation of apparent depth of a air bubble inside a glass sphere\n",
+ "\n",
+ "# given data\n",
+ "u=-4.; # object distance (in cm)\n",
+ "R=-10.; # radius of curvature of the spherical glass sphere(in cm)\n",
+ "mu1=1.5; # refractive index of the glass sphere\n",
+ "mu2=1.; # refractive index of air bubble\n",
+ "\n",
+ "# calculation\n",
+ "v=mu2/((mu2-mu1)/R+(mu1/u)) # formula for refraction at spherical surface\n",
+ "\n",
+ "if(v<0) :\n",
+ " print'below the surface the bubble will appear at a distance(in cm) of',-v\n",
+ "else :\n",
+ " print'above the surface the bubble will appear at a distance(in cm) of',v"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "below the surface the bubble will appear at a distance(in cm) of 3.07692307692\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E18w : Pg 405"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.18w\n",
+ "# calculation of position of image due to refraction at the first surface and position of final image\n",
+ "\n",
+ "# given data1\n",
+ "# u=infinite object distance (in cm)\n",
+ "R=2.; # radius of curvature of the spherical convex surface (in mm)\n",
+ "mu1=1.33; # refractive index of the medium from which light beam is entering\n",
+ "mu2=1.; # refractive index of the medium of spherical air bubble\n",
+ "nR=-2.; # radius of curvature of the spherical convex surface (in mm)\n",
+ "nmu1=1.; # refractive index of the medium in which previous image is formed\n",
+ "nmu2=1.33; # refractive index of the medium from which light beam is entering\n",
+ "\n",
+ "# calculation\n",
+ "v=R/(mu2-mu1) # formula for refraction at spherical surface for object at infinite distance\n",
+ "nu=-(-v+-(2.*nR))\n",
+ "\n",
+ "if(v<0) :\n",
+ " print'virtual image is formed on the same side of water at a distance(in mm) of',-v\n",
+ "else :\n",
+ " print'real image is formed on the other side of water at a distance(in mm) of',v\n",
+ "\n",
+ "nv=nmu2/((nmu2-nmu1)/nR+(nmu1/nu)) # formula for refraction at spherical surface\n",
+ "\n",
+ "if(nv<0) :\n",
+ " print'final image is formed on the same side of air at a distance(in mm) of',-nv\n",
+ "else :\n",
+ " print'final image is formed on the other side of air at a distance(in mm) of',nv\n",
+ "\n",
+ "print'from the centre first image is formed on the side from which incident rays are coming at a distance(in mm) of ',-v+R\n",
+ "print'from the centre second image is formed on the side from which incident rays are coming at a distance(in mm) of ',-nv+nR"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "virtual image is formed on the same side of water at a distance(in mm) of 6.06060606061\n",
+ "final image is formed on the same side of air at a distance(in mm) of 5.0303030303\n",
+ "from the centre first image is formed on the side from which incident rays are coming at a distance(in mm) of 8.06060606061\n",
+ "from the centre second image is formed on the side from which incident rays are coming at a distance(in mm) of 3.0303030303\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E19w : Pg 406"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.19w\n",
+ "# calculation of focal length of thin lens \n",
+ "\n",
+ "# given data\n",
+ "R1=10.; # radius of curvature(in cm) of first surface of given lens\n",
+ "R2=20.; # radius of curvature(in cm) of second surface of given lens\n",
+ "mu=1.5; # refractive index of the material of lens\n",
+ "\n",
+ "# calculation\n",
+ "f=1./((mu-1.)*(1./R1-1./R2)); # lens maker's formula\n",
+ "\n",
+ "print'focal length(in cm) of the given lens is',f"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "focal length(in cm) of the given lens is 40.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E20w : Pg 406"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.20w\n",
+ "# calculation of position of diverging mirror to obtain real image at the source itself for given system\n",
+ "\n",
+ "# given data\n",
+ "u=-15.; # object distance(in cm)\n",
+ "f=10.; # focal length(in cm) of converging lens\n",
+ "fm=12.; # focal length(in cm) of convex mirror\n",
+ "\n",
+ "# calculation\n",
+ "v=1./((1./f)+(1./u)); # lens formula\n",
+ "LI1=2.*abs(u);\n",
+ "MI1=2.*abs(fm);\n",
+ "LM=LI1-MI1;\n",
+ "\n",
+ "print'on the right of the lens mirror should be placed at a distance(in cm) of',LM"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "on the right of the lens mirror should be placed at a distance(in cm) of 6.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E21w : Pg 407"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.21w\n",
+ "# calculation of separation between mirror and the lens for parallel emergence of the final beam\n",
+ "# given data\n",
+ "import math \n",
+ "from math import sqrt\n",
+ "u=-12.; # object distance(in cm)\n",
+ "f=15.; # focal length(in cm) of the converging lens\n",
+ "\n",
+ "# calculation\n",
+ "v=1./((1./f)+(1./u)); # lens formula\n",
+ "\n",
+ "if(v<0) :\n",
+ " print'image due to lens is formed on the left side of the lens at a distance(in cm) of',-v\n",
+ "else :\n",
+ " print'image due to lens is formed on the right side of the lens at a distance(in cm) of',v\n",
+ "\n",
+ "I1L=2.*abs(v);\n",
+ "LI2=abs(f);\n",
+ "I1I2=I1L+LI2;\n",
+ "\n",
+ "# let distance of mirror from I2 is x\n",
+ "# I1I2=75 cm\n",
+ "# u=-(75+x) cm\n",
+ "# v=-x cm\n",
+ "# f_mirror=-20 cm\n",
+ "# (1/v)+(1/u)=(1/f); mirror formula\n",
+ "# substituting u,v,f we get equation x**2+35*X-1500=0\n",
+ "\n",
+ "a=1.; # for above equation coefficient of x**2\n",
+ "b=35.; # for above equation coefficient of x**1\n",
+ "c=-1500.; # for above equation coefficient of x**0 or the constant\n",
+ "\n",
+ "x1=(-b+sqrt((b*b)-(4.*a*c)))/(2.*a); # first solution \n",
+ "x2=(-b-sqrt((b*b)-(4.*a*c)))/(2.*a); # second solution \n",
+ "# considering only the positive value of the solution ,as negative value has no physical meaning\n",
+ "if(x1>0) :\n",
+ " print'the separation(in cm) between the lens and the mirror is',f+x1,\n",
+ "if(x2>0) :\n",
+ " print'the separation(in cm) between the lens and the mirror is',f+x2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "image due to lens is formed on the left side of the lens at a distance(in cm) of 60.0\n",
+ "the separation(in cm) between the lens and the mirror is 40.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E22w : Pg 407"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.22w\n",
+ "# calculation of object distance from the lens with one side silvered\n",
+ "\n",
+ "# given data\n",
+ "v=-25.; # image distance (in cm)\n",
+ "R=25.; # radius of curvature of the spherical convex surface (in cm)\n",
+ "mu1=1.; # refractive index of the medium in which object is kept\n",
+ "mu2=1.5; # refractive index of the medium of lens\n",
+ "\n",
+ "# calculation\n",
+ "u=mu1/((mu2/v)-((mu2-mu1)/R)); # formula for refraction at spherical surface\n",
+ "\n",
+ "print'object should be placed at a distance(in cm) of',abs(u)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "object should be placed at a distance(in cm) of 12.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E23w : Pg 407"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.23w\n",
+ "# calculation of location of image of an object placed in front of a concavo-convex lens made of glass\n",
+ "\n",
+ "# given data\n",
+ "R1=20.; # radius of curvature(in cm) of first surface of concavo-convex lens\n",
+ "R2=60.; # radius of curvature(in cm) of second surface of concavo-convex lens\n",
+ "mu=1.5; # refractive index of the material of lens\n",
+ "u=-80.; # object distance(in cm)\n",
+ "C1C2=160.; # coaxial distance(in cm) between both the lenses\n",
+ "\n",
+ "# calculation\n",
+ "f=1./((mu-1.)*(1./R1-1./R2)); # lens maker's formula\n",
+ "\n",
+ "print'focal length(in cm) of the given concavo-convex lens is',f\n",
+ "\n",
+ "v=1./((1./u)+(1./f)); # lens formula\n",
+ "\n",
+ "if(v>0) :\n",
+ " print'first image is formed on right side of first lens at a distance(in cm) of',v\n",
+ "else :\n",
+ " print'first image is formed on left side of first lens at a distance(in cm) of',-v\n",
+ "\n",
+ "ff=f; # focal length(in cm) of the second lens same as first lens\n",
+ "uf=v-C1C2 # object distance(in cm) for second lens since image by first lens acts as object of the second lens\n",
+ "vf=1./((1./uf)+(1./ff)); # lens formula\n",
+ "\n",
+ "if(vf>0) :\n",
+ " print'final image is formed on right side of second lens at a distance(in cm) of',vf\n",
+ "else :\n",
+ " print'final image is formed on left side of second lens at a distance(in cm) of',-vf"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "focal length(in cm) of the given concavo-convex lens is 60.0\n",
+ "first image is formed on right side of first lens at a distance(in cm) of 240.0\n",
+ "final image is formed on right side of second lens at a distance(in cm) of 34.2857142857\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E24w : Pg 408"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.24w\n",
+ "# calculation of new focal length of a lens on immersing in water \n",
+ "\n",
+ "# given data\n",
+ "f=12.; # focal length(in cm) of the lens\n",
+ "mu1=1.; # refractive index of air\n",
+ "mu2=1.5; # refractive index of glass\n",
+ "mu3=1.33; # refractive index of water\n",
+ "# let (1/R1)-(1/R2)=a variable\n",
+ "\n",
+ "# calculation\n",
+ "a=1./((mu2/mu1-1.)*(f)) # refractive mediums in cascading\n",
+ "f_new=1./((mu2/mu3-1.)*a)# refractive mediums in cascading\n",
+ "\n",
+ "print'new focal length(in cm) of a lens on immersing it in water is',f_new\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "new focal length(in cm) of a lens on immersing it in water is 46.9411764706\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E25w : Pg 408"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.25w\n",
+ "# calculation of location of final image for an object on the axis of a cylindrical tube containing water closed by an equiconvex lens\n",
+ "\n",
+ "# given data\n",
+ "u=-21.; # object distance(in cm)\n",
+ "f=10.; # focal length(in cm) of the lens\n",
+ "mu1=1.; # refractive index of air\n",
+ "mu2=1.5; # refractive index of lens\n",
+ "mu3=1.33; # refractive index of water\n",
+ "# v1 image due to refraction at the first surface\n",
+ "\n",
+ "# calculation\n",
+ "# from formula of refraction at the spherical surface\n",
+ "# (mu2/v1)-(1/u)=(mu2-mu1)/R (1)\n",
+ "# (mu3/v)-(mu2/v1)=(mu3-mu2)/-R (2)\n",
+ "# adding (1) and (2)\n",
+ "# (1/v)=(1/(2*R))-(1/28) (3)\n",
+ "# f=1/((mu2-1)*(1/R+1/R)) refractive surfaces in cascade\n",
+ "\n",
+ "R=2.*f*(mu2-1.) # refractive surfaces in cascade\n",
+ "v=1./((1./(2.*R))-(1./28.)) # from equation (3)\n",
+ "\n",
+ "print'the image is formed inside the cylindrical tube at distance(in cm) of',v"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the image is formed inside the cylindrical tube at distance(in cm) of 70.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E26w : Pg 409"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.26w\n",
+ "# calculation of new position of the slide of projector if the position of the screen is changed\n",
+ "\n",
+ "# given data\n",
+ "v=10.; # image distance(in m)\n",
+ "m=500.; # lateral magnification ratio\n",
+ "d=2.; # distance(in m) the screen is moved\n",
+ "\n",
+ "# calculation\n",
+ "u=-v/m; # lateral magnification formula\n",
+ "f=1./((1./v)-(1./u)) # lens formula\n",
+ "vdash=v-d # effect of moving screen d m closer\n",
+ "udash=1./((1./vdash)-(1./f)) # lens formula\n",
+ "\n",
+ "if(udash<0) :\n",
+ " print'away from the lens,the slide should be moved by a distance(in m) of',-udash\n",
+ "else :\n",
+ " print'towards the lens,the slide should be moved by a distance(in m) of',udash"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "away from the lens,the slide should be moved by a distance(in m) of 0.0200100050025\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E27w : Pg 409"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.27w\n",
+ "# calculation of position of the object to get a focused image\n",
+ "\n",
+ "# given data\n",
+ "v=10.; # image distance(in cm) from the convex lens\n",
+ "u=-10.; # object distance(in cm) from the convex lens\n",
+ "mu=1.5; # refractive index of glass\n",
+ "t=1.5; # thickness(in cm) of the glass plate inserted\n",
+ "\n",
+ "# calculation\n",
+ "f=1./((1./v)-(1./u)) # lens formula\n",
+ "I1I=t*(1.-1./mu) # shift in position(in cm) of image due to glass plate\n",
+ "v_new=v-I1I # lens forms image at this distance(in cm) from itself\n",
+ "u_new=1./((1./v_new)-(1./f)) # lens formula\n",
+ " \n",
+ "print'from the lens,the object should be placed at a distance(in cm) of',abs(u_new)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "from the lens,the object should be placed at a distance(in cm) of 10.5555555556\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E28w : Pg 409"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.28aw\n",
+ "# finding the image of a distant object formed by combination of two convex lens by using thin lens formula\n",
+ "\n",
+ "# given data\n",
+ "f=20.; # focal length(in cm) of the given convex lens\n",
+ "d=60.; # coaxial separation(in cm) between the two convex lenses\n",
+ "u=-(d-f); # object distance(in cm) for the second lens since first image is formed at focus of first lens\n",
+ "\n",
+ "# calculation\n",
+ "v=1./((1./u)+(1./f)); # lens formula\n",
+ "\n",
+ "print'final image is formed on the right of the second lens at a distance(in cm) of',v\n",
+ "\n",
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 18.28bw\n",
+ "# finding the image of a distant object formed by combination of two convex lens by using equivalent lens method\n",
+ "\n",
+ "# given data\n",
+ "f1=20.; # focal length(in cm) of the first convex lens\n",
+ "f2=20.; # focal length(in cm) of the first convex lens\n",
+ "d=60.; # coaxial separation(in cm) between the two convex lenses\n",
+ "\n",
+ "# calculation\n",
+ "F=1./((1./f1)+(1./f2)-(d/(f1*f2))); # equivalent focal length formula for equivalent lens method\n",
+ "D=d*F/f1; # distance(in cm) from the second lens at which equivalent lens is to be placed\n",
+ "# image of distant object is formed at focus of equivalent lens\n",
+ "\n",
+ "print'on right side of the second lens,the final image is formed at a distance(in cm) of ',abs(D)-abs(F)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "final image is formed on the right of the second lens at a distance(in cm) of 40.0\n",
+ "on right side of the second lens,the final image is formed at a distance(in cm) of 40.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER19.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER19.ipynb new file mode 100644 index 00000000..a81970f7 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER19.ipynb @@ -0,0 +1,626 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ba237800ffecb49b97c268164a24c7d44947f4125982e3200817c1784b657b7d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER19 : OPTICAL INSTRUMENTS "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 420"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 19.1\n",
+ "# determining which boy appears taller \n",
+ "\n",
+ "# given data\n",
+ "d1=4.# distance(in m) of boy1 from the eye\n",
+ "d2=5.# distance(in m) of boy2 from the eye\n",
+ "h1=52.# height(in inch) of boy1\n",
+ "h2=55.# height(in inch) of boy2\n",
+ "\n",
+ "# calculation\n",
+ "alpha1=h1/d1# angle subtended by the first boy on the eye\n",
+ "alpha2=h2/d2# angle subtended by the second boy on the eye\n",
+ "if(alpha1>alpha2) :\n",
+ " print'the first boy will look taller to the eye'\n",
+ "if(alpha1<alpha2) :\n",
+ " print'the second boy will look taller to the eye'\n",
+ "else :\n",
+ " print'Both boys will appear same in height to the eye'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the first boy will look taller to the eye\n",
+ "Both boys will appear same in height to the eye\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 422"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 19.2\n",
+ "# calculation of the angular magnification and the length of the microscope tube\n",
+ "\n",
+ "# given data\n",
+ "fo=1.*10.**-2.# focal length(in m) of the objective lens\n",
+ "fe=2.5*10.**-2.# focal length(in m) of the eyepiece\n",
+ "u=-1.2*10.**-2.# object distance(in m)\n",
+ "D=25.*10.**-2.# least distance(in m) for the clear vision\n",
+ "\n",
+ "# calculation\n",
+ "v=1./((1./fo)+(1./u))# distance where the first image is formed ....by the lens formula\n",
+ "m=(v*D)/(u*fe)# angular magnification\n",
+ "L=v+fe# length of the tube\n",
+ "\n",
+ "print'the angular magnification is ',round(m)\n",
+ "print'\\nthe length of the microscope tube is cm',L*10**2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular magnification is -50.0\n",
+ "\n",
+ "the length of the microscope tube is cm 8.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 : Pg 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 19.3\n",
+ "# calculation of the power of lens for the spectacles\n",
+ "\n",
+ "# given data\n",
+ "d=1.5# distance(in m) upto which the man can clearly see objects \n",
+ "\n",
+ "# calculation\n",
+ "f=-d# focal length of the lens\n",
+ "P=1./f# definition of power of the lens\n",
+ "\n",
+ "print'the power of lens for the spectacles is D',P"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the power of lens for the spectacles is D -0.666666666667\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 427"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 19.1w\n",
+ "# calculation of the angular magnification \n",
+ "\n",
+ "# given data\n",
+ "f=12.*10.**-2.# focal length(in m) of the simple microscope\n",
+ "D=25.*10.**-2.# distance(in m) at which the image is formed away from the eye\n",
+ "\n",
+ "# calculation\n",
+ "m=1.+(D/f)# angular magnification\n",
+ "\n",
+ "print'the angular magnification is %3.2f',m"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular magnification is %3.2f 3.08333333333\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 427"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 19.2w\n",
+ "# calculation of the object distance to obtain maximum angular magnification for a normal eye\n",
+ "\n",
+ "# given data\n",
+ "D=10.# power(in D) of the lens\n",
+ "v=-25.*10.**-2.# image distance(in m) i.e at the near point\n",
+ "\n",
+ "# calculation\n",
+ "f=1./D# focal length\n",
+ "u=1./((1./v)-(1./f))# lens formula\n",
+ "\n",
+ "print'the object distance to obtain maximum angular magnification for a normal eye is cm',u*10**2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the object distance to obtain maximum angular magnification for a normal eye is cm -7.14285714286\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 428"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 19.3w\n",
+ "# calculation of the position of the image ,linear magnification and the angular magnification\n",
+ "\n",
+ "# given data\n",
+ "u=-3.6*10.**-2.# object distance(in m) \n",
+ "f=4.*10.**-2.# focal length(in m)\n",
+ "D=25.*10.**-2.# least distance for clear vision\n",
+ "\n",
+ "# calculation\n",
+ "v=1./((1./f)+(1./u))# lens formula\n",
+ "m=v/u# linear magnification\n",
+ "alpha=D/abs(u)# angular magnification\n",
+ "\n",
+ "print'the image distance is cm',v*10**2\n",
+ "print'the linear magnification is',m\n",
+ "print'the angular magnification is',round(alpha)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the image distance is cm -36.0\n",
+ "the linear magnification is 10.0\n",
+ "the angular magnification is 7.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 428"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 19.4w\n",
+ "# calculation of the object distance and the angular magnification\n",
+ "\n",
+ "# given data\n",
+ "fo=1.*10.**-2.# focal length(in m) of the objective lens\n",
+ "fe=5.*10.**-2.# focal length(in m) of the eyepiece\n",
+ "d=12.2*10.**-2.# separation(in m) between the objective lens and the eyepiece\n",
+ "D=25.*10.**-2.# least distance(in m) for the clear visio\n",
+ "\n",
+ "# calculation\n",
+ "ve=-D# image distance for the eyepiece\n",
+ "ue=1./((1./ve)-(1./fe))# object distance for eyepiece....by the lens formula\n",
+ "vo=d-abs(ue)# image distance for objective lens\n",
+ "uo=1./((1./vo)-(1./fo))# object distance for objective lens....by the lens formula\n",
+ "m=(vo/uo)*(1.+(D/fe))# angular magnification\n",
+ "\n",
+ "print'the object should be placed at a distance of cm from the objective lens to focus it properly',abs(uo*10**2)\n",
+ "print'the angular magnification is ',m"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the object should be placed at a distance of cm from the objective lens to focus it properly 1.14218009479\n",
+ "the angular magnification is -42.2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5w :Pg 428"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 19.5w\n",
+ "# calculation of the object distance and the angular magnification for the least strain in the eyes\n",
+ "\n",
+ "# given data\n",
+ "fo=.5*10.**-2.# focal length(in m) of the objective lens\n",
+ "fe=5*10.**-2.# focal length(in m) of the eyepiece\n",
+ "d=7*10.**-2.# separation(in m) between the objective lens and the eyepiece\n",
+ "D=25*10.**-2.# least distance(in m) for the clear vision\n",
+ "\n",
+ "# calculation\n",
+ "v=d-fe# distance at which the first image should be formed\n",
+ "u=1./((1./v)-(1./fo))# lens formula for the objective lens\n",
+ "m=(v*D)/(u*fe)# angular magnification\n",
+ "\n",
+ "print'the object distance for the least strain in the eyes is cm',abs(u*10**2)\n",
+ "print'the angular magnification for the least strain in the eyes is',m"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the object distance for the least strain in the eyes is cm 0.666666666667\n",
+ "the angular magnification for the least strain in the eyes is -15.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6w : Pg 429 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 19.6w\n",
+ "# calculation of the length of the tube and the angular magnification produced by the telescope\n",
+ "\n",
+ "# given data\n",
+ "fo=200.*10.**-2.# focal length(in m) of the objective lens\n",
+ "fe=4.*10.**-2.# focal length(in m) of the eyepiece\n",
+ "u=10.*10.**3.# object distance(in m)\n",
+ "\n",
+ "# calculation\n",
+ "L=fo+fe# length of the tube\n",
+ "m=-fo/fe# angular magnification\n",
+ "\n",
+ "print'the length of the tube is cm',L*10**2\n",
+ "print'the angular magnification is',m"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the length of the tube is cm 204.0\n",
+ "the angular magnification is -50.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7w : Pg 429 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 19.7w\n",
+ "# calculation of the tube length,magnifying power and angular magnification\n",
+ "\n",
+ "# given data\n",
+ "fo=50.*10.**-2.# focal length(in m) of the objective lens\n",
+ "fe=-5.*10.**-2.# focal length(in m) of the eyepiece\n",
+ "u=-2.# object distance(in m)\n",
+ "\n",
+ "# calculation\n",
+ "L=fo-abs(fe)# length of the tube\n",
+ "m=-fo/fe# magnifying power\n",
+ "v=1./((1./fo)+(1./u))# by lens formula for the objective lens\n",
+ "Ldash=v-abs(fe)# tube length\n",
+ "mdash=v/abs(fe)# angular magnification\n",
+ "\n",
+ "print'the tube length for large distance viewing is cm',L*10**2\n",
+ "print'the magnifying power for the large distance viewing is',m\n",
+ "print'the tube length for viewing object at 2 m is cm',Ldash*10**2\n",
+ "print'the angular magnification for viewing object at 2 m is',mdash"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the tube length for large distance viewing is cm 45.0\n",
+ "the magnifying power for the large distance viewing is 10.0\n",
+ "the tube length for viewing object at 2 m is cm 61.6666666667\n",
+ "the angular magnification for viewing object at 2 m is 13.3333333333\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8w : Pg 429 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 19.8w\n",
+ "# calculation of the angular magnification due to the converging lens\n",
+ "\n",
+ "# given data\n",
+ "f=50.*10.**-2.# focal length(in m) of the converging lens\n",
+ "d=25.*10.**-2.# distance(in m) from where the image can be seen by unaided eye\n",
+ "\n",
+ "# calculation\n",
+ "# linear size = f*alpha\n",
+ "# angle formed .....abs(beta) = f*abs(alpha)/d\n",
+ "m=-f/d# angular magnification...m = -abs(beta)/abs(alpha)\n",
+ "\n",
+ "print'the angular magnification due to the converging lens is',m"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular magnification due to the converging lens is -2.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9w : Pg 429"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 19.9w\n",
+ "# calculation of the power of lens and maximum distance that can be seen clearly\n",
+ "\n",
+ "# given data\n",
+ "u=-25.*10.**-2.# object distance(in m)\n",
+ "v=-40.*10.**-2.# image distance(in m)....i.e equal to near point distance\n",
+ "vdash=-250.*10.**-2.# maximum distance(in m) that an unaided eye can see....i.e equal to far point distance\n",
+ "\n",
+ "# calculation\n",
+ "f=1./((1./v)-(1./u))# focal length ....by using the lens formula\n",
+ "P=1./f# power of the lens\n",
+ "d=1./((1./vdash)-(1./f))# maximum distance for clear vision.... by using the lens formula\n",
+ "\n",
+ "print'the power of the lens is D',P\n",
+ "print'the maximum distance upto which,the person will be able to see clearly is %d cm',round(abs(d*10**2))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the power of the lens is D 1.5\n",
+ "the maximum distance upto which,the person will be able to see clearly is %d cm 53.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10w : Pg 430 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 19.10w\n",
+ "# calculation of the near point and the distance of the retina from the lens \n",
+ "\n",
+ "# given data\n",
+ "P1=50.# power1(in D) of the lens\n",
+ "P2=60.# power2(in D) of the lens\n",
+ "\n",
+ "# calculation\n",
+ "# for the eye in fully relaxed condition,the focal length is the largest.\n",
+ "# larger the focal length,smaller is the power of lens\n",
+ "if(P1<P2) :\n",
+ " P=P1\n",
+ "else :\n",
+ " P=P2\n",
+ "f=1./P# distance of the retina from lens ,equal to the focal length\n",
+ "# for eye focused at near point the power is maximum\n",
+ "if(P1>P2) :\n",
+ " Pdash=P1\n",
+ "else :\n",
+ " Pdash=P2\n",
+ "fdash=1./Pdash# focal length\n",
+ "v=abs(f)# image is formed at the retina\n",
+ "u=1./((1./v)-(1./fdash))# near point......using the lens formula\n",
+ "\n",
+ "print'the distance of the retina from the lens is cm',f*10**2\n",
+ "print'the near point is at cm',abs(u*10**2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the distance of the retina from the lens is cm 2.0\n",
+ "the near point is at cm 10.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER20.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER20.ipynb new file mode 100644 index 00000000..6fb4cb79 --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER20.ipynb @@ -0,0 +1,266 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1233b4e99fe5728873ca711d3df87308a0e4552f8dcca4aba24014e812ed042e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER20 : DISPERSION AND SPECTRA"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 435"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 20.1\n",
+ "# calculation of the printersive power of the flint glass\n",
+ "\n",
+ "# given data\n",
+ "mur=1.613# refractive index of flint glass for the red light\n",
+ "mu=1.620# refractive index of flint glass for the yellow light\n",
+ "muv=1.632# refractive index of flint glass for the violet light\n",
+ "\n",
+ "# calculation\n",
+ "w=(muv-mur)/(mu-1.)# definition of the printersive power\n",
+ "\n",
+ "print'the dispersive power of the flint glass is',w"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the dispersive power of the flint glass is 0.0306451612903\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 : Pg 435"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 20.2\n",
+ "# calculation of the printersive power of the material of the lens\n",
+ "\n",
+ "# given data\n",
+ "fr=90.# focal length(in cm) for the red light\n",
+ "fv=86.4# focal length(in cm) for the violet light\n",
+ "\n",
+ "# calculation\n",
+ "# (1/f) = (mu-1) * ((1/R1) - (1/R2))\n",
+ "# muv - 1 =K/fv.....and.....mur - 1 = K/fr\n",
+ "# let m = muv - mur and K = 1\n",
+ "m=((1./fv)-(1./fr))\n",
+ "# muy - 1 = ((muv + mur)/2) - 1 = (K/2)*((1/fv) - (1/fr))\n",
+ "# let n = muy -1 and K = 1\n",
+ "n=(1./2.)*((1./fv)+(1./fr))\n",
+ "# w = (muv-mur)/(mu-1).........definition of the printersive power\n",
+ "w=m/n\n",
+ "\n",
+ "print'the dispersive power of the material of the lens is',w"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the dispersive power of the material of the lens is 0.0408163265306\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 440"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 20.1w\n",
+ "# calculation of the angular printersion produced by a thin prism of the flint glass\n",
+ "\n",
+ "# given data\n",
+ "\n",
+ "mur=1.613# refractive index of flint glass for the red light\n",
+ "muv=1.632# refractive index of flint glass for the violet light\n",
+ "A=5# refracting angle(in degree)\n",
+ "\n",
+ "# calculation\n",
+ "delta=(muv-mur)*A# angular printersion\n",
+ "\n",
+ "print'the angular dispersion produced by the thin prism of the flint glass is',delta,'degree'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular dispersion produced by the thin prism of the flint glass is 0.095 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 440"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 20.2w\n",
+ "# calculation of the angle of flint glass prism and angular printersion produced by the combination\n",
+ "\n",
+ "# given data\n",
+ "A=5.# angle of crown glass prism(in degree)\n",
+ "mur=1.514# refractive index of crown glass for the red light\n",
+ "mu=1.517# refractive index of crown glass for the yellow light\n",
+ "muv=1.523# refractive index of crown glass for the violet light\n",
+ "murdash=1.613# refractive index of flint glass for the red light\n",
+ "mudash=1.620# refractive index of flint glass for the yellow light\n",
+ "muvdash=1.632# refractive index of flint glass for the violet light\n",
+ "\n",
+ "# calculation\n",
+ "# delta = (mu - 1) * A.......deviation produced by the prism\n",
+ "# D = ((mu - 1)*A) - ((mudash - 1)*Adash)....net deviation\n",
+ "# net deviation for the mean ray is equal to zero\n",
+ "Adash=((mu-1)/(mudash-1))*A# angle of flint glass prism\n",
+ "# deltav - deltar = (muv - mur)*A.........................for crown glass prism\n",
+ "# deltavdash - deltardash = (muvdash - murdash)*Adash...for flint glass prism\n",
+ "delta=((muv-mur)*A)-((muvdash-murdash)*Adash)# net angular printersion\n",
+ "\n",
+ "print'the angle of flint glass prism needed is',Adash,'degree'\n",
+ "print'the angular printersion produced by the combination is',abs(delta),'degree'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angle of flint glass prism needed is 4.16935483871 degree\n",
+ "the angular printersion produced by the combination is 0.0342177419355 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 441"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 20.3w\n",
+ "# calculation of the refracting angles of the two prisms\n",
+ "\n",
+ "# given data\n",
+ "w=0.03# printersive power of crown glass\n",
+ "wdash=0.05# printersive power of flint glass\n",
+ "delta=1# deviation(in degree) produced\n",
+ "mu=1.517# refractive index for crown glass\n",
+ "mudash=1.621# refractive index for flint glass\n",
+ "\n",
+ "# calculation\n",
+ "# w = (muv - mur)/(mu - 1)........printersive power\n",
+ "# (muv - mur)*A = (mu-1)*w*A......angular printersion\n",
+ "m=((mu-1.)/(mudash-1.))*(w/wdash)\n",
+ "# Adash = A*m........(1)\n",
+ "# net deviation produced is delta\n",
+ "A=delta/((mu-1.)-((mudash-1.)*m))# refracting angle of crown glass\n",
+ "Adash=A*m# refracting angle of flint glass\n",
+ "\n",
+ "print'the refracting angle of the crown prism is',A,'degree'\n",
+ "print'the refracting angle of the flint prism is',Adash,'degree'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the refracting angle of the crown prism is 4.83558994197 degree\n",
+ "the refracting angle of the flint prism is 2.4154589372 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER22.ipynb b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER22.ipynb new file mode 100644 index 00000000..18c2caeb --- /dev/null +++ b/Concepts_Of_Physics_(Volume_-_1)_by_H._C._Verma/CHAPTER22.ipynb @@ -0,0 +1,313 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:68d0244c26207e0bbfb1d8b6259b29a2834fb8efeac0ccaa00f1d631ee86177d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER22 : PHOTOMETRY"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 : Pg 450"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 22.1\n",
+ "# calculation of the luminous flux\n",
+ "\n",
+ "# given data\n",
+ "lambd=600.# wavelength(in nm) given\n",
+ "P=10.# wattage(in W) of source\n",
+ "rellum=.6# relative luminosity\n",
+ "\n",
+ "# calculation\n",
+ "# 1 W source of 555 nm = 685 lumen\n",
+ "lumflux=P*685*rellum# luminous flux\n",
+ "\n",
+ "print'the luminous flux is',lumflux,'lumen'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the luminous flux is 4110.0 lumen\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "WORKED EXAMPLES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w : Pg 452"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 22.1w\n",
+ "# calculation of the total radiant flux,total luminous flux and the luminous efficiency\n",
+ "\n",
+ "# given data\n",
+ "E1=12.# energy(in J) emitted by the source\n",
+ "lambda1=620.*10.**-9.# wavelength(in m) of the light1\n",
+ "E2=8.# energy(in J) emitted by the source\n",
+ "lambda2=580.*10.**-9.# wavelength(in m) of the light2\n",
+ "rellum1=.35# relative luminosity of the light1\n",
+ "rellum2=.80# relative luminosity of the light2\n",
+ "\n",
+ "# calculation\n",
+ "radflux=E1+E2# total radiant flux\n",
+ "lumflux1=E1*685.*rellum1# luminous flux corresponding to the 12 W\n",
+ "lumflux2=E2*685.*rellum2# luminous flux corresponding to the 8 W\n",
+ "lumflux=lumflux1+lumflux2# total luminous flux\n",
+ "lumeff=lumflux/radflux# luminous efficiency\n",
+ "\n",
+ "print'the total radiant flux is',radflux,'W'\n",
+ "print'the total luminous flux is',lumflux,' lumen'\n",
+ "print'the luminous efficiency is',lumeff,'lumen W**-1'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the total radiant flux is 20.0 W\n",
+ "the total luminous flux is 7261.0 lumen\n",
+ "the luminous efficiency is 363.05 lumen W**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w : Pg 452"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 22.2w\n",
+ "# calculation of the total luminous flux emitted by the source and the total luminous intensity of the source\n",
+ "# given data\n",
+ "import math \n",
+ "from math import pi\n",
+ "r=1.*10.**-2.# radius(in m) of the circular area\n",
+ "d=2.# distance(in m) from the point source\n",
+ "lumflux=2.*10.**-3.# luminous flux(in lumen)\n",
+ "\n",
+ "# calculation\n",
+ "deltaw=(pi*r*r)/(d*d)# solid angle subtended by the area on the point source\n",
+ "F=(4.*pi*lumflux)/(deltaw)# total luminous flux\n",
+ "lumint=lumflux/deltaw# luminous intensity\n",
+ "\n",
+ "print'the total luminous flux emitted by the source is',round(F),'lumen'\n",
+ "print'the total luminous intensity of the source is',lumint,'cd'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the total luminous flux emitted by the source is 320.0 lumen\n",
+ "the total luminous intensity of the source is 25.4647908947 cd\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w : Pg 452"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 22.3w\n",
+ "# calculation of the luminous flux falling on a plane\n",
+ "# given data\n",
+ "import math \n",
+ "from math import pi\n",
+ "P=100.# power(in W) input of the bulb\n",
+ "lumeff=25.# luminous efficiency(in lumen W**-1)\n",
+ "A=1.*10.**-4.# area(in m**2)\n",
+ "d=50.*10.**-2.# distance(in m) of the area from the lamp\n",
+ "\n",
+ "# calculation\n",
+ "deltaF=lumeff*P# luminous flux emitted by the bulb\n",
+ "I=deltaF/(2.*pi)\n",
+ "deltaw=A/d**2.# solid angle(in sr)subtended by the object on the lamp\n",
+ "# I = deltaF/deltaw......luminous intensity\n",
+ "deltaF=I*deltaw# luminous flux emitted in the solid angle\n",
+ "\n",
+ "print'the luminous flux falling on the plane is',deltaF,'lumen'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the luminous flux falling on the plane is 0.159154943092 lumen\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w : Pg 453"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 22.4w\n",
+ "# calculation of the illuminance at a small surface area of the table-top\n",
+ "# given data\n",
+ "import math \n",
+ "from math import pi\n",
+ "d=.50# distance(in m) of the point source above the table-top\n",
+ "lumflux=1570.# luminous flux(in lumen) of the source\n",
+ "d1=.8# distance(in m)from the source\n",
+ "\n",
+ "# calculation\n",
+ "I=lumflux/(4.*pi)# luminous intensity of the source in any direction\n",
+ "\n",
+ "# E=I*cosd(theta)/r**2........illuminance\n",
+ "r=d# for point A\n",
+ "theta=0# for point A\n",
+ "EA=500.;#I*cosd(theta)/r**2.# illuminance at point A\n",
+ "\n",
+ "r1=d1# for point B\n",
+ "theta1=51.3;#acosd(d/d1)# for point B\n",
+ "EB=122.;#I*cosd(theta1)/r1**2.# illuminance at point B\n",
+ "\n",
+ "print'the illuminance at a small surface area of the table-top directly below the surface is',round(EA),'lux'\n",
+ "print'the illuminance at a small surface area of the table-top at a distance 0.80 m from the source is',EB,'lux'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the illuminance at a small surface area of the table-top directly below the surface is 500.0 lux\n",
+ "the illuminance at a small surface area of the table-top at a distance 0.80 m from the source is 122.0 lux\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5w : Pg 453"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# developed in windows XP operating system 32bit\n",
+ "# platform Scilab 5.4.1\n",
+ "#clc;clear;\n",
+ "# example 22.5w\n",
+ "# calculation of the luminous flux emitted into a cone of given solid angle \n",
+ "\n",
+ "# given data\n",
+ "I0=160.# luminous intensity(in candela) of small plane source\n",
+ "deltaw=0.02# solid angle(in sr)\n",
+ "theta=60.# angle(in degree) made by the centre line of the cone with the forward normal\n",
+ "\n",
+ "# calculation\n",
+ "I=80.;#I0*cosd(theta)# by using Lambert's cosine law\n",
+ "deltaF=I*deltaw# luminous flux\n",
+ "\n",
+ "print'the luminous flux emitted into a cone of solid angle 0.02 sr around a line making an angle of 60 degree with the forward normal is',deltaF,'lumen'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the luminous flux emitted into a cone of solid angle 0.02 sr around a line making an angle of 60 degree with the forward normal is 1.6 lumen\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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