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   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter1 - Free oscillations in one-dimension : Simple harmonic Oscillator"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1, page 9"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from math import sqrt, pi\n",
      "#  FREQUENCY AND TIME PERIOD\n",
      "#format('v',6)\n",
      "#ph=50*x**2+100 in joule/kg\n",
      "m=10 #mass in kg\n",
      "f=10**3/m #joule/kg\n",
      "w=sqrt(f) #oscillations\n",
      "fr=w/(2*pi) #oscillations/sec\n",
      "tp=1/fr #seconds\n",
      "print \"Frequency of oscillation = %0.1f oscillations/seconds \"%fr\n",
      "print \"Time period = %0.3f seconds \" %tp"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Frequency of oscillation = 1.6 oscillations/seconds \n",
        "Time period = 0.628 seconds \n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3, page 11"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# ENERGY\n",
      "ke=5 #joule\n",
      "pe=5 #joule\n",
      "rep=10 #joule\n",
      "eo=rep+ke+pe #joule\n",
      "print \"Energy of the oscillator = %0.f J\" %eo"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Energy of the oscillator = 20 J\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4, page 12"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#peroid ,maximum velocity and acceleration\n",
      "a=3 #cm\n",
      "b=4 #cm\n",
      "A=sqrt(a**2+b**2) #cm\n",
      "w=2 #sec**-1\n",
      "T=(2*pi)/w #seconds\n",
      "um=w*A #cm/s\n",
      "am=w**2*A #cm/s**2\n",
      "print \"Time period = %0.f seconds\" %T\n",
      "print \"Maximum velocity  = %0.f cm/s\" %um\n",
      "print \"Maximum acceleration = %0.f cm/s2 \" %am"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Time period = 3 seconds\n",
        "Maximum velocity  = 10 cm/s\n",
        "Maximum acceleration = 20 cm/s2 \n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5, page 12"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import pi\n",
      "# maximum velocity and acceleration\n",
      "A=5 #cm\n",
      "T=31.4#seconds\n",
      "w=(2*pi)/T #sec**-1\n",
      "um=w*A #cm/s\n",
      "am=w**2*A #cm/s**2\n",
      "print \"Maximum velocity  = %0.f cm/s\" %um\n",
      "print \"Maximum acceleration = %0.1f cm/s2 \" %am"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Maximum velocity  = 1 cm/s\n",
        "Maximum acceleration = 0.2 cm/s2 \n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6, page 20"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import pi, sqrt\n",
      "#  Period \n",
      "#given data :\n",
      "g=9.8 # constant\n",
      "l=1 # in m\n",
      "theta_m1=60 # in degree\n",
      "theta_m=pi/3 # in radians\n",
      "T0=round(2*pi*sqrt(l/g)) \n",
      "print \"(a) Time period for small displacement, T0 = %0.f seconds \" %T0\n",
      "T=T0*(1+(theta_m**2/16)) \n",
      "print \"(b) Time period, T = %0.1f seconds \" %T"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a) Time period for small displacement, T0 = 2 seconds \n",
        "(b) Time period, T = 2.1 seconds \n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 7, page 20"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# ENERGY\n",
      "es=1 #joule\n",
      "l=2 #metre\n",
      "am=3 #cm\n",
      "am1=5 #cm\n",
      "e1=(am1**2/am**2)*es #joules\n",
      "l2=1 #meter\n",
      "e2=(l/l2)*es #joules\n",
      "print \"Energy in first case = %0.3f J\" %e1\n",
      "print \"Energy in second case = %0.1f J\" %e2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Energy in first case = 2.778 J\n",
        "Energy in second case = 2.0 J\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8, page 28"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt, pi\n",
      "#  Period of motion\n",
      "#given data :\n",
      "x=0.16 # in m\n",
      "m1=4 # in kg\n",
      "g=9.8 \n",
      "K=m1*g/x \n",
      "m=0.50 # in kg\n",
      "T=2*pi*sqrt(m/K) # \n",
      "print \"The period of motion, T = %0.2f seconds \" %T\n",
      "# answer is wrong in textbook"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The period of motion, T = 0.28 seconds \n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9, page 28"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt\n",
      "#foce constant,displacement , acceleration and energy\n",
      "#given data :\n",
      "x1=.10 # in m\n",
      "F1=4 # in N\n",
      "K=F1/x1 \n",
      "x2=0.12 # in m\n",
      "print \"(a) The force constant, K = %0.2f N/m\" %K\n",
      "F=-K*x2 \n",
      "print \"(b) The force, F = %0.2f N\" %F\n",
      "m=1.6 # in kg\n",
      "T=2*pi*sqrt(m/K) \n",
      "print \"(c) Period of oscillation, T = %0.3f s \" %T\n",
      "A=x2 \n",
      "print \"(d) Amplitude of motion, A = %0.2f m \" %A\n",
      "alfa=A*K/m \n",
      "print \"(e) Maximum acceleration, alfa = %0.2f m/s2 \" %alfa\n",
      "x=A/2 # in m\n",
      "w=sqrt(K/m) \n",
      "v=w*sqrt(A**2-x**2) \n",
      "a=w**2*x # in m/s**2\n",
      "KE=(1/2)*m*v**2 # in J\n",
      "PE=(1/2)*K*x**2 # in J\n",
      "TE=KE+PE \n",
      "print \"(f) Velocity = %0.2f m/s \" %v\n",
      "print \"(f) Acceleration = %0.2f m/s2 \" %a\n",
      "print \"(f) Kinetic energy = %0.2f J \" %KE\n",
      "print \"(f) Potential energy = %0.2f J\" %PE\n",
      "print \"(g) Total energy of the oscillating system, TE = %0.2f J\" %TE\n",
      "# In textbook part f is inculded in the part e so their is the numbeing error in parts"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a) The force constant, K = 40.00 N/m\n",
        "(b) The force, F = -4.80 N\n",
        "(c) Period of oscillation, T = 1.257 s \n",
        "(d) Amplitude of motion, A = 0.12 m \n",
        "(e) Maximum acceleration, alfa = 3.00 m/s2 \n",
        "(f) Velocity = 0.52 m/s \n",
        "(f) Acceleration = 1.50 m/s2 \n",
        "(f) Kinetic energy = 0.22 J \n",
        "(f) Potential energy = 0.07 J\n",
        "(g) Total energy of the oscillating system, TE = 0.29 J\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 10, page 30"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sin\n",
      "from sympy import symbols, pi\n",
      "# ENERGY\n",
      "t=8/3 #seconds\n",
      "v=-10*pi*sin((35*pi)/6)#cm/s\n",
      "print \"Velocity =\",v,\"cm\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Velocity = 5.0*pi cm\n"
       ]
      }
     ],
     "prompt_number": 31
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11, page 30"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt, pi\n",
      "#given data :\n",
      "K1=3 # in N/m\n",
      "K2=2 # in N/m\n",
      "m=0.050 # in kg\n",
      "w=sqrt((K1+K2)/m) \n",
      "n=w/(2*pi) \n",
      "print \"(i) The frequency, n = %0.2f oscillations/sec \" %n\n",
      "A=0.004 # in m\n",
      "E=(1/2)*A**2*(K1+K2) \n",
      "print \"(ii) The energy, E = %0.e J \" %E\n",
      "v=sqrt(2*E/m) \n",
      "print \"(iii) The velocity, v = %0.2f m/s\" %v"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i) The frequency, n = 1.59 oscillations/sec \n",
        "(ii) The energy, E = 4e-05 J \n",
        "(iii) The velocity, v = 0.04 m/s\n"
       ]
      }
     ],
     "prompt_number": 35
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 12, page 33"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Rotational inertia\n",
      "#given data :\n",
      "M=0.1 # in m\n",
      "l=0.1 # in m\n",
      "I1=M*l**2/12 # in kg-m**2\n",
      "T1=2 # in s\n",
      "T2=6 # in s\n",
      "I2=(I1*T2**2)/T1**2 \n",
      "print \"Rotational inertia, I2 = %0.1e kg-m2 \" %I2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Rotational inertia, I2 = 7.5e-04 kg-m2 \n"
       ]
      }
     ],
     "prompt_number": 38
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13, page 34"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt\n",
      "# Time period\n",
      "#given data :\n",
      "M=4 # in kg\n",
      "R=0.10 # in m\n",
      "I=(2/5)*M*R**2 # in kg.m**2\n",
      "C=4*10**-3 # in Nm/radian\n",
      "T=2*pi*sqrt(I/C) \n",
      "print \"Time period, T = %0.2f s \" %T\n",
      "# answer is wrong in textbook"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Time period, T = 12.57 s \n"
       ]
      }
     ],
     "prompt_number": 39
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 15, page 41"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt, pi\n",
      "# Energy\n",
      "#given data :\n",
      "L=10*10**-3 # in H\n",
      "C=20*10**-6 # in F\n",
      "n=1/(2*pi*sqrt(L*C)) \n",
      "V=10 #in V\n",
      "U=(1/2)*C*V**2 \n",
      "print \"Frequency, n = %0.2f cycles/s \" %n\n",
      "print \"Energy of oscillations,U = %0.1e J \" %U\n",
      "#answer of frequency is calculated wrong in textbook"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Frequency, n = 355.88 cycles/s \n",
        "Energy of oscillations,U = 1.0e-03 J \n"
       ]
      }
     ],
     "prompt_number": 41
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 16, page 47"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# distance,binding energy and force constant\n",
      "print \"Equilibrium inter-nuclear distance correspondes to lowest potential enegy is ro= 2*\u00c5\"\n",
      "pet=0 #eV\n",
      "peb=-4 #eV\n",
      "be=pet-peb #eV\n",
      "x1=-2 #eV\n",
      "x2=-4 #eV\n",
      "V=x1-x2 #eV\n",
      "e=1.6*10**-19 #electronic charge\n",
      "x=0.5 #armstrong\n",
      "K=((2*V)/x**2) #eV/\u00c5**2\n",
      "k1=(K*e)/(10**-10)**2 #joule/m**2\n",
      "print \"Binding energy = %0.2f eV \" %be\n",
      "print \"Force constant = %0.2f N/m \" %k1"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Equilibrium inter-nuclear distance correspondes to lowest potential enegy is ro= 2*\u00c5\n",
        "Binding energy = 4.00 eV \n",
        "Force constant = 256.00 N/m \n"
       ]
      }
     ],
     "prompt_number": 42
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 17, page 48"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# possible values and energy\n",
      "r1=2 #from graph\n",
      "r2=4.5 #units from graph\n",
      "print \"Possible values of r are\",r1,\"units and\",r2,\"units.\"\n",
      "osc=1-(-2.5) #units\n",
      "print \"Maximum energy of oscillations for r=2 units is\",osc,\"units.\"\n",
      "osc1=0.5-(-1) #units\n",
      "print \"Maximum energy of oscillations for r=4.5 units is\",osc1,\"units.\"\n",
      "t=1 #from graph\n",
      "v=0 #from graph\n",
      "e=t+v #\n",
      "print \"Total energy = %0.2f unit \" %e\n",
      "print \"At infinity V =\",v,\"therefore T =\",t,\"unit.\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Possible values of r are 2 units and 4.5 units.\n",
        "Maximum energy of oscillations for r=2 units is 3.5 units.\n",
        "Maximum energy of oscillations for r=4.5 units is 1.5 units.\n",
        "Total energy = 1.00 unit \n",
        "At infinity V = 0 therefore T = 1 unit.\n"
       ]
      }
     ],
     "prompt_number": 43
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 19, page 67"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt, pi\n",
      "# Frequency\n",
      "#given data :\n",
      "m1=10 # in g\n",
      "m2=90 # in g\n",
      "K=10**3 # in N/m\n",
      "mu=m1*m2*10**-3/(m1+m2) \n",
      "n=round(sqrt(K/mu)/(2*pi)) \n",
      "print \"The frequency, n = %0.2f oscillations/sec \" %n\n",
      "x1=0 #\n",
      "x2=10 #cm\n",
      "xb=((m1*x1+m2*x2)/(m1+m2)) #cm\n",
      "mo=(m1*10**-3)*(xb*10**-2)**2+(m2*10**-3)*(1*10**-2)**2 #\n",
      "print \"Moment of inertia = %0.1e kg-m2 \" %mo"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The frequency, n = 53.00 oscillations/sec \n",
        "Moment of inertia = 9.0e-05 kg-m2 \n"
       ]
      }
     ],
     "prompt_number": 45
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 20, page 68"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt, pi\n",
      "# frequency and amplitude\n",
      "c=10**-4 #N-m\n",
      "m1=9 #gm\n",
      "m2=1 #gm\n",
      "mu=((m1*m2)/(m1+m2))*10**-3 #kg\n",
      "r=20 #cm\n",
      "I=mu*(r*10**-2)**2 #kg-m**2\n",
      "fr=((1/(2*pi))*sqrt(c/I)) #vibrations/sec\n",
      "print \"Frequency of vibration = %0.2f vibrations/s \" %fr\n",
      "e=10**-2 #joule\n",
      "thmax=sqrt((2*e)/c) #radians\n",
      "print \"Amplitude = %0.2f radians \" %thmax"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Frequency of vibration = 0.27 vibrations/s \n",
        "Amplitude = 14.14 radians \n"
       ]
      }
     ],
     "prompt_number": 46
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 21, page 69"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt\n",
      "# frequency ,energy and maximum velocity\n",
      "c=1 #N-m \n",
      "m1=6 #gm\n",
      "m2=2 #gm\n",
      "mu=((m1*m2)/(m1+m2))*10**-3 #kg\n",
      "fr=((1/(2*pi))*sqrt(c/mu)) #vibrations/sec\n",
      "print \"Frequency of oscillations = %0.1f vibrations/s \" %fr\n",
      "td= 1+(1/3) #cm\n",
      "e=((1/2)*c*(td*10**-2)**2) #joule\n",
      "print \"Energy = %0.1e J \" %e\n",
      "y=((1/2)*m2*10**-3)+((1/2)*(1/3)**2*m1*10**-3) #\n",
      "v1=sqrt((e/y)) #m/sec\n",
      "print \"Maximum velocity of smaller mass = %0.2f m/s\" %v1\n",
      "#velocity is calculated wrong in the book"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Frequency of oscillations = 4.1 vibrations/s \n",
        "Energy = 8.9e-05 J \n",
        "Maximum velocity of smaller mass = 0.26 m/s\n"
       ]
      }
     ],
     "prompt_number": 48
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 22, page 70"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt, pi\n",
      "# frequency\n",
      "k=100 #N/m\n",
      "m=100 #gm\n",
      "n1=((1/(2*pi))*sqrt(k/(m*10**-3))) #sec**-1\n",
      "m1=100 #gm\n",
      "m2=200 #gm\n",
      "mu=((m1*m2)/(m1+m2))*10**-3 #kg\n",
      "fr=((1/(2*pi))*sqrt(k/mu)) #sec**-1\n",
      "print \"In first case frequency = %0.f sec^-1 \"%n1\n",
      "print \"In second case frequency = %0.1f sec^-1 \"%fr"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "In first case frequency = 5 sec^-1 \n",
        "In second case frequency = 6.2 sec^-1 \n"
       ]
      }
     ],
     "prompt_number": 50
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 23, page 73"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# force constant and work done\n",
      "m1=1 #assume\n",
      "m2=19 #assume\n",
      "mh=1.66*10**-27 #kg\n",
      "mu=((m1*m2)/(m1+m2))*mh #kg\n",
      "w=7.55*10**14 #radians/sec\n",
      "k=mu*(w)**2 #N/m\n",
      "print \"Force constant = %0.1f N/m \" %k\n",
      "x=0.5 #arngstrom\n",
      "wh=((1/2)*k*(x*10**-10)**2) #joule\n",
      "print \"Work done = %0.3e J\" %wh"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Force constant = 898.9 N/m \n",
        "Work done = 1.124e-18 J\n"
       ]
      }
     ],
     "prompt_number": 52
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 24, page 74"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt\n",
      "# frequency\n",
      "m1=1 #a.m.u\n",
      "m2=35 #a.m.u\n",
      "mu1=((m1*m2)/(m1+m2)) #a.m.u\n",
      "m3=2 #\n",
      "mu2=((m3*m2)/(m3+m2)) #a.m.u\n",
      "n1=8.99*10**13 #cycle/sec\n",
      "n2=(sqrt(mu1/mu2))*n1 #c/s\n",
      "print \"Frequecy of vibrations = %0.2e c/s \" %n2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Frequecy of vibrations = 6.44e+13 c/s \n"
       ]
      }
     ],
     "prompt_number": 53
    }
   ],
   "metadata": {}
  }
 ]
}