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Anharmonic Pendulum

  1. Aug 1, 2004 #1

    cj

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    Suppose I have a simple pendulum whose period was designed to be exactly 2 seconds. I am able to time the period reliably to within 0.03s. What is the smallest amplitude, in degrees, I would need to use before I could see the anharmonic effects in this pendulum?
     
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  3. Aug 1, 2004 #2

    Gokul43201

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    If you are 'seeing' the anharmonic effects only through a measurement of T, I can't see how the amplitude plays any role at all.

    Perhaps, I don't understand exactly what you're trying to do here...
     
  4. Aug 1, 2004 #3

    cj

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    There's a point, beyond which, it won't behave
    harmonically -- rather anharmonically.

    The value of T is related to this.

     
  5. Aug 2, 2004 #4

    Galileo

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    The question is basically when the 'small angle' approximation is no longer valid.
    We can't solve the the differential equation without this approximation:
    [tex]\ddot\theta+\frac{mgl}{I}\sin\theta =0[/tex]
    Where [itex]\theta[/itex] is the angle with the vertical, m is the mass of your pendulum,l is the distance from the center of mass to the axis of rotation and I is the moment of intertia about the axis of rotation.

    When the angle is small, then [itex]\sin\theta\approx \theta[/itex] and we can solve de diferrential equation.

    I think it must be solved numerically. A way to measure when visible anharmonic effects arise is to see when the period of the oscillation is greater than 0.03 s off with the expected value. Don't know how to calculate it though...
     
  6. Aug 2, 2004 #5

    arildno

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    We may go a few steps further, before resorting to numerical means:
    We start with Galileo's equation, and add a couple of initial conditions:
    [tex]\ddot{\theta}+\omega^{2}\sin\theta=0,\theta(0)=\theta_{0},\dot{\theta}(0)=0[/tex]
    We multiply the equation with [tex]\dot{\theta}[/tex] integrate, rearrange, and utilize intitial conditions, and gain:
    [tex]\dot{\theta}=\pm\omega\sqrt{2(\cos\theta-\cos\theta_{0})}[/tex]

    The negative root is used on time intervals where [tex]\theta\to{-\theta_{0}}[/tex]
    (assuming [tex]\theta_{0}>0[/tex])

    We thereby gain:
    [tex]T=\frac{2}{\omega}\int_{-\theta_{0}}^{\theta_{0}}\frac{d\theta}{\sqrt{2(\cos\theta-\cos\theta_{0})}}[/tex]

    In the harmonic case, we have [tex]T_{h}=\frac{2\pi}{\omega}[/tex]

    Hence, given a relative error bound [tex]\epsilon[/tex] we gain the bound of the initial angle as:
    [tex]|1-\frac{1}{\pi}\int_{-\theta_{0}}^{\theta_{0}}\frac{d\theta}{\sqrt{2(\cos\theta-\cos\theta_{0})}}|<\epsilon[/tex]
     
    Last edited: Aug 2, 2004
  7. Aug 3, 2004 #6

    cj

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    Thanks very much arildno.

    Question: does the [tex]\epsilon[/tex] in the equation

    [tex]|1-\frac{1}{\pi}\int_{-\theta_{0}}^{\theta_{0}}\frac{d\theta}{\sqrt{2(\cos\theta-\cos\theta_{0})}}|<\epsilon[/tex]

    specifically refer to the uncertainty (in my case, 0.03)?

     
  8. Aug 3, 2004 #7

    arildno

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    No, it does not!
    [tex]\epsilon[/tex] is the relative error bound, that is:

    [tex]\epsilon=|\frac{(2\pm0.03)-2}{2}|=0.015[/tex]
     
  9. Aug 3, 2004 #8

    cj

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    Got it -- thanks again!

     
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