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Simple pendulum question

  1. Sep 28, 2015 #1
    1. The problem statement, all variables and given/known data

    A simple pendulum with l = 9.8m satisfies the equation:
    [tex]\ddot{\theta} + \sin{\theta} = 0[/tex]

    if [tex]\Theta_{0} = A[/tex]

    Show that the period T is given by:
    [tex]T = \int_0^\frac{\pi}{2}\left(\frac{1}{(1 - \alpha \sin^2{\phi})^\frac{1}{2} }\right)d\phi[/tex]

    where,
    [tex]\alpha=\sin^2{\frac{1}{2}\Theta_{0}}[/tex]

    2. Relevant equation

    It seems that I can use:
    [tex]t=\int\left(\frac{1}{\pm(\frac{2E}{m}-\frac{k}{m}x^2)^\frac{1}{2}}\right)dx[/tex]
    to find a T expression

    3. The attempt at a solution

    I start with:
    [tex]t=\int\left(\frac{1}{\pm(\frac{2E}{m}-\frac{k}{m}\theta^2)^\frac{1}{2}}\right)d\theta[/tex]

    I find the terms in the rational:
    [tex]\frac{2E}{m}=\frac{2}{2m}(m\dot{\theta}^2 + \omega_{0}^2 \theta^2)[/tex]
    [tex]k=\omega_{0}^2 m[/tex]
    Substitute back into t:
    [tex]t=\int\left(\frac{1}{\pm(\dot{\theta}^2-\omega_{0}\theta^2 + \omega_{0}\theta^2)^\frac{1}{2}}\right)d\theta[/tex]
    which is just [tex]\ln{\dot{\theta}}[/tex]
    at this point I am doubting whether I am on the right track, any insight would help tremendously!
     
    Last edited: Sep 28, 2015
  2. jcsd
  3. Sep 29, 2015 #2
    Drowning in alphabet soup. Can you define all your variables? A picture might help.
     
  4. Sep 29, 2015 #3
    Sure thing, here's the actual question.
    wkMKWaK.jpg
     
  5. Sep 29, 2015 #4

    Ray Vickson

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    Get rid of the ##\pm## sign in your expression for ##t##, The quantity ##( \cdots )^{1/2}## in the denominator is always ##> 0## (by definition of the 1/2-power function), and ##t > 0## also. Therefore, choose the + square root.
     
  6. Oct 2, 2015 #5

    vela

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    You might want to rethink this. The denominator comes from solving for the speed in ##\frac 12 mv^2 + \frac 12 kx^2 = E##, and the potential term corresponds to a force ##F = -kx##. That's not the same situation you have for the pendulum. The restoring torque isn't a linear function of ##\theta##.
     
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