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Amplitude of a pendulum?

  1. Jan 12, 2009 #1
    How can one determine the amplitude, frequency and period of an amplitude? this is not homework, i was just curious, because i knew how to find the time using the 2 pi sqrt(l/g), but wanted to know about this, since i am learning about waves and harmonic motion! Help would be appreciated. Thanks!





    P.S. this is my first time using physics forums, so tell me if i should change the way i ask a question or if i made any kind of mistake :P
     
  2. jcsd
  3. Jan 13, 2009 #2

    atyy

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    Determine as in measure? Or as in calculate using Newton's laws?
     
  4. Jan 13, 2009 #3
    The [tex]2\pi \sqrt{\frac{l}{g}}[/tex] formula is only good for small amplitude displacements around 5 degrees, this is the harmonic approximation.

    Lets look how we get this. Newton says, for the tangential component:

    [tex]ml\ddot \varphi = - mg\sin\varphi[/tex]

    Where [tex]\varphi[/tex] is the angle between the vertical and the string of the pendulum.

    So the equation of motion for the pendulum:

    [tex]\ddot\varphi = -\frac{g}{l}\sin\varphi[/tex]

    Now as we see this is a non-linear differential equation. For small displacements (i.e. small angles) that is:
    [tex]\sin\varphi \approx \varphi [/tex]

    So:

    [tex]\ddot\varphi = -\frac{g}{l}\varphi[/tex]

    As we see this equation describes simple harmonic motion, and we can extract the frequency of oscillations:

    [tex]\omega^2=\frac{g}{l} \Longrightarrow T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{l}{g}}[/tex]

    So we obtained the formula for small displacements of the pendulum, and we can see that it (of course) doesnt depend on the amplitude...

    Now if the displacements arent small we cannot approximate the sine like we did.

    In this case after integrating the equation once and some manipulation, we obtain for the period:

    [tex]T(\varphi_0)=4\sqrt{\frac{l}{g}}\int_0^{\frac{\pi}{2}} \frac{d\psi}{\sqrt{1-k^2\sin^2\psi}}[/tex]

    Where [tex]k=\sin\left(\frac{\varphi_0}{2}\right)[/tex] Here \varphi_0 is the amplitude(maximum displacement) of the pendulum.

    As we see this is an elliptic integral of the first kind. So the period of the pendulum at arbitrary amplitudes cannot be given using elementary functions.
    We can however use a series approximation for the elliptic integral. Using this we get:

    [tex]T(\varphi_0)=2\pi\sqrt{\frac{l}{g}}\sum_{n=0}^{\infty}\left[\frac{(2n-1)!!}{(2n)!!}\sin^n\frac{\varphi_0}{2}\right][/tex]

    If the amplitude is still small but not that big, then we can further approximate the sines (now I do it upto fourth order of the amplitude):

    [tex]T(\varphi_0)=2\pi\sqrt{\frac{l}{g}}\left[1+\frac{1}{16}\varphi_0^2+\frac{11}{3072}\varphi_0^4 +\dots\right][/tex]

    So we can conclude that, we cant get an exact solution even for such trivial and simple configurations...
     
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