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Small amplitude oscillations from a potential function. watch

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    Hi guys,

    This should be fairly straightforward, im just a bit rusty on how to go about it.

    So we've got a potential function U(x)=k(x^2 - 4xl) with the constnat k > 0.

    First part of the problem is to show we've got a stable equilibirium when x=2l. Which is easy enough, find -\frac{dU}{dx}, shove in x=2l and it comes out as 0. (Feel free to point this out if its wrong, but im pretty sure its not.)

    Then it wants the frequency of small amplitude oscillations about this equilibirum position. Not entirely sure how to proceed, I went with

    m\ddot{x} = -\frac{dU}{dx} = -k(2x - 4l) (the double dot represents acceleration if anyone hasnt seen the notation before) so we've basically F = ma = -dU/dx

    Then I decided that x=e^{i\omega t}, did the differentiation, and solve for \omega^2, then put  x = e^{i\omega t} = 2l, but this results in  \omega = 0 , which I'm not sure is right, my guess is it probably isn't, but I'm not sure why...

    Anyway, rep is available to anyone who can help.
    Enjoy :p:

    (This shouldn't be too hard, as the frequency bit is only a short 4-marker on an exam question. This is from an exam on Lagrangian dynamics, but I can't really see where Lagrangian would be needed, if at all as the equation of motion is easy enough to get from F=-dU/dx)
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    It makes it neater to say X = x - 2l, so your equation becomes mX'' = -2kX, so X'' = -2kX/m, which you can just read off omega squared as 2k/m.
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    As above, you're basically just expanding about the stationary point so x becomes  x = 2l + \epsilon and as above you'll end up with a SHM differential equation for epsilon. You use the Lagrangian to find the equation of motion:

     \mathcal{L} = \frac{1}{2} m \dot{x}^2 - k(x^2 - 4xl) \to m\ddot{x} = -k(2x-4l) \to m \ddot{\epsilon} = -2k\epsilon
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    Cheers both
 
 
 
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