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Wave Equation

  1. May 25, 2010 #1

    bon

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    1. The problem statement, all variables and given/known data

    I'm given that the motion of an infinite string is described by the wave equation:

    (let D be partial d)

    D^2 y /Dx^2 - p/T D^2/Dt^2 = 0

    I'm asked for what value of c is Ae^[-(x-ct)^2] a solution (where A is constant)

    Then im asked to show that the potential and KE of the wave packet are equal..

    2. Relevant equations



    3. The attempt at a solution

    So im guessing the value of c is root(T/p)?since the solution is a function of (x-ct) so this corresponds to D'Alembert..But then PE and KE dont seem equal...

    KE = integral from -infinity to + infinity of 1/2 p A^2 e^[4c^2(x-ct)] while PE = integral from - inf to + inf of 1/2 p A^2 c^2 e^[-(4x-ct)]..and these dont seem equal..

    any help?

    thanks!
     
  2. jcsd
  3. May 29, 2010 #2

    bon

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    Any ideas on this?
    To me it doesn't even seem to obey the wave equation - though it is of the form (x-ct) which corresponds to the d'alembert solution.

    yxx = Ae^-2 whereas ytt = Ae^(-2c^2)

    Any ideas? Thanks
     
  4. May 29, 2010 #3
    Hi
    As you say one solution is [tex]c=\sqrt{T/p}[/tex] but the kinetic energy is given by
    [tex]E_k=p\frac{1}{2}\int_{-\infty}^{\infty}(\frac{dy}{dt})^2 dx[/tex] and the potential energy by [tex]E_p=T\frac{1}{2}\int_{-\infty}^{\infty}(\frac{dy}{dx})^2 dx[/tex].

    I hope this helps.
     
  5. May 29, 2010 #4

    bon

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    But if you look at my post above..it doesn't seem that c = root T/p will satisfy the equation..

    Also i cant get hte KE and PE to be equal...
     
  6. May 30, 2010 #5
    Okay, you have done something wrong in the calculations of [tex]\frac{d^2 y}{dx^2}[/tex] and [tex]\frac{d^2 y}{dt^2} [/tex]. I have
    [tex]\frac{dy}{dx}=-2A(x-ct)e^{-(x-ct)^2}[/tex],
    [tex]\frac{d^2}{dx^2}=-2A(1-2(x-ct)^2)e^{-(x-ct)^2}[/tex],
    [tex]\frac{dy}{dt}=2Ac(x-ct)e^{-(x-ct)^2}[/tex]
    and
    [tex]\frac{d^2}{dt^2}=-2Ac^2(1-2(x-ct)^2)e^{-(x-ct)^2}[/tex].
    Thus for [tex]c=\pm \sqrt{T/p}[/tex] is
    [tex]\frac{d^2}{dx^2}-\frac{p}{T}\frac{d^2 y}{dt^2}=0[/tex]
    satisfied. With the expressions above it's also easily seen that [tex]E_k=E_p[/tex].
     
  7. May 30, 2010 #6

    bon

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    Oh i see where I went wrong.. Thank you so much for your help! :)
     
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