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True of false?

  1. Apr 25, 2005 #1

    quasar987

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    Every "spacially periodic" function [i.e. s.t there exist P s.t. f(x+P,t) = f(x,t)] of the form f(x,t) = X(x)cos(wt) is a solution of the wave equation.
     
    Last edited: Apr 25, 2005
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  3. Apr 25, 2005 #2

    James R

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    True..........
     
  4. Apr 25, 2005 #3
    quasar,

    False.

    For one thing, if f(x,t) is periodic (and it doesn't have to be periodic), then there's a strict relation between the periodicity in x and in t. In other words, in an example using your format, if f(x,t) = sin(kx)cos(wt), then w/k = v.

    It's not too hard to see what the wave equation is saying about f(x,t) if you think about it. Within a multiplicative constant, the two partial derivatives are the same. That means f has to depend on x and t in very similar ways. I think the most general form for f(x,t) that satisfies the W.E. is f(kx-wt). Certainly any function of that form will work. Although that's not really what you were asking.
     
  5. Apr 25, 2005 #4
    Uh oh!

    James R., I posted before I saw yours. Why do you say it's true?
     
  6. Apr 26, 2005 #5

    James R

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    Hmm... Here are my thoughts. I think I might have changed my mind!

    At first, I thought that any function of the form f(x,t) = X(x)cos wt will describe a standing wave, and so it is necessarily a solution of the wave equation.

    But then...

    The wave equation, in 1 dimension, is:

    [tex]\frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{v^2 \partial t^2} = 0[/tex]

    where v is a constant (the wave speed).

    Using the function given, we have:

    [tex]\frac{\partial^2 f}{\partial x^2} = \frac{d^2 X}{dx^2}\cos \omega t[/tex]
    [tex]\frac{\partial^2 f}{\partial t^2} = -\omega^2 X(x)\cos \omega t[/tex]

    Therefore, we require:

    [tex]\frac{d^2 X}{dx^2} + \frac{X}{v^2} = 0[/tex]

    This restricts X(x) to harmonic functions of the form:

    [tex]X(x) = A \sin kx + B \cos kx[/tex]

    where A and B are arbitrary constants, but k is restricted:

    [tex]k=\omega / v[/tex]

    So, it seems that the most general functions of the given form which satisfy the wave equation are:

    [tex]f(x,t) = [A \sin kx + B \cos kx]\cos \omega t[/tex]

    with the above restriction on k.

    Does that seem right?
     
  7. Apr 26, 2005 #6

    Galileo

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    You don't you just work out the simplest case: [itex]f(x,t)=\cos(wt)[/itex].
    We clearly have [itex]f(x+P,t)=f(x,t)[/itex] for any t (and any P).
    The wave equation clearly doesn't hold in this case (unless [itex]\omega=0[/itex], but I understand [itex]\omega[/itex] can be arbitrary).
     
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