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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 3, Forced harmonic oscillator & resonance"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1, page 135"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from math import sqrt, degrees, atan, pi\n",
      "# Phase shift\n",
      "#given data :\n",
      "F0=25 # in N\n",
      "m=1 \n",
      "f0=F0/m \n",
      "K=1*10**3 # in N/m\n",
      "w0=sqrt(K/m) \n",
      "b=0.05 # in N-s/m\n",
      "r=b/(2*m) # in s**-1\n",
      "A=f0*10**3/sqrt(9*w0**4+(16*r**2*(w0)**2)) \n",
      "print \"The amplitude, A = %0.2f mm \" %A\n",
      "p=2*w0 \n",
      "fi=atan(2*r*p/(w0**2-p**2)) # radian \n",
      "fi = degrees(fi) # degree\n",
      "print \"Phase shift is\",round(fi,2),\"degree or\",round(fi*(pi/180),3),\"radian.\"\n",
      "#phase shift is converted wrong into radians"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The amplitude, A = 8.33 mm \n",
        "Phase shift is -0.06 degree or -0.001 radian.\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2, page 136"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from numpy import array\n",
      "# A/Amax\n",
      "x1=array([0.99,0.98,0.97]) #\n",
      "wt=50 #\n",
      "wo=1 #assume\n",
      "fo=1 #assume\n",
      "for x in x1:\n",
      "    a=((fo/((wo**2)*((1-x**2)**2+((1/wt**2)*x**2))**(1/2)))) #\n",
      "    am=fo/((wo**2)*(1/wt**2)**(1/2)) #\n",
      "    z=a/am #\n",
      "    print \"For p/wo\",x,\", value of A/Amax is\",round(z,2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "For p/wo 0.99 , value of A/Amax is 0.71\n",
        "For p/wo 0.98 , value of A/Amax is 0.45\n",
        "For p/wo 0.97 , value of A/Amax is 0.32\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3, page 154"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import pi, sqrt\n",
      "# Reactance and impedence\n",
      "#given data :\n",
      "n=50 # in cycles\n",
      "w=2*pi*n # in rad/sec\n",
      "L=1/pi # in H\n",
      "XL=w*L \n",
      "print \"The reactance, XL = %0.0f ohm \" %XL\n",
      "R=100 # in ohm\n",
      "Z=sqrt(R**2+XL**2) \n",
      "print \"The impedence, Z = %0.1f ohm \" %Z"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The reactance, XL = 100 ohm \n",
        "The impedence, Z = 141.4 ohm \n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4, page 155"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt, pi\n",
      "# Current and Capacity\n",
      "#given data :\n",
      "E=110 # in V\n",
      "R=10 # in ohm\n",
      "L=1*10**-3 # in H\n",
      "C=1*10**-6 # in F\n",
      "n=10000 # in Hz\n",
      "w=2*pi*n \n",
      "I=E/sqrt(R**2+((w*L)-(1/(w*C)))**2) \n",
      "print \"The current, I = %0.2f A \" %I\n",
      "L1=1/(w**2*C) \n",
      "print \"The value of capacity, L1 = %0.2e F \" %L1\n",
      "#Capacitance  is calculated wrong in the textbook"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The current, I = 2.29 A \n",
        "The value of capacity, L1 = 2.53e-04 F \n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5, page 155"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt\n",
      "# Resonent frequency and Separation\n",
      "#given data :\n",
      "L=1*10**-3 # in H\n",
      "C=0.1*10**-6 # in F\n",
      "w0=1/sqrt(L*C) \n",
      "print \"Resonant frequency, w0 = %0.e rad/s \" %w0\n",
      "R=10 # in ohm\n",
      "w2_w1=R/L \n",
      "print \"The separation = %0.e rad/s \" %w2_w1\n",
      "S=w0/w2_w1 \n",
      "print \"The sharpness = %0.f \" %S"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Resonant frequency, w0 = 1e+05 rad/s \n",
        "The separation = 1e+04 rad/s \n",
        "The sharpness = 10 \n"
       ]
      }
     ],
     "prompt_number": 16
    }
   ],
   "metadata": {}
  }
 ]
}