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Period of oscillation (potential energy equation given)

  1. Nov 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Let the potential energy of particle depend upon coordinate x as:
    U(x) = U0(1-cos(ax)). Where "U0" and "a" are constants. Find the period of small oscillations that particle performs about its equilibrium position.

    2. Given Answer

    T = 2∏√(m/a2U0)

    3. The attempt at a solution

    It can be seen from the equation that equilibrium will be at x = 0, where forces acting are zero. Also after integrating the equation with dx, it is seen that the motion is relevant to forced oscillations.
     
  2. jcsd
  3. Nov 9, 2011 #2

    ehild

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    U(x) is the potential function. Why do you integrate it?
    The motion is not forced oscillation. For that, an external time dependent force is needed.

    ehild
     
  4. Nov 9, 2011 #3
    There a two sufficient conditions to prove something is simple harmonic oscillator
    F = -kx
    or
    Etotal = [itex]\frac{1}{2}[/itex]Av2+[itex]\frac{1}{2}[/itex]Bq2. Where A and B are some constants and, q is some coordinate (in your case ax). Then the period is T = 2π√(A/B) Also since it says small oscillations, I would use the small angle approximation for cosine.
     
  5. Nov 9, 2011 #4
    Thanks for that. But this questions has two parts which I didn't mention.
    The part-2 of the problem has the equation:
    U(x)= a/x2 - b/x
    How to deal with this one ?
    I mean in above equation x has powers -2 and -1. So how to form its differential equation?
     
  6. Nov 9, 2011 #5
    Well assuming the kinetic energy is of the form: [itex]\frac{1}{2}[/itex]mv2, the the period for small oscillations is 2π√(k/m) where, k is the second derivative of the potential energy function evaluated at the equilibrium position (you might have to take the limit as it goes to the equilibrium position in this case...)
     
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