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Two general questions about wave functions

  1. Apr 17, 2013 #1
    In my Physics I class, we started learning about wave functions in the form:

    y(x,t) = sin(kx ± ωt ± ∅) or y(x,t) = cos(kx ± ωt ± ∅)

    1) I saw a question where the wave function was structured as:

    y(x,t) = sin(ωt - kx + ∅)

    and the answers for the direction of the wave was in the +x direction.

    I thought I could rewrite the equation as y(x,t) = sin(-kx + ωt + ∅), meaning the direction is in the -x direction, as the symbol preceding the "ω" is positive. Obviously that was wrong, so how does it actually work?


    2) Also, just a random question I was wondering:

    If the derivative with respect to t (holding x constant) of the equations above give you the speed of a particle in the wave, what does the derivative with respect to x give you?

  2. jcsd
  3. Apr 17, 2013 #2


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    Propagation velocity is related to [itex]k[/itex] and [itex]\omega[/itex] through:
    [itex]v = \omega/k[/itex]. So if you switch the sign of both [itex]\omega[/itex] and [itex]k[/itex], the sign of the velocity remains the same.
  4. Apr 18, 2013 #3

    Philip Wood

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    A subsidiary question: I've always preferred the 't first' version, y = A sin (wt - kx + phi), which is so clearly an oscillation (wrt time), with a phase that lags further and further behind with distance travelled by the wave. Yet most writers seem to prefer the 'x first' version. Why is this?
  5. Apr 18, 2013 #4
    oh, ok.
    so basically if either the k or ω is negative that would make it +x direction, so:

    y(x,t) = sin(ωt - kx + ∅) == y(x,t) = sin(kx - ωt + ∅)
  6. Apr 18, 2013 #5


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    Yeah, except that
    [itex] sin(\omega t - k x + \Phi) = sin(k x - \omega t + \Phi')[/itex]
    where [itex]\Phi' = \pi - \Phi[/itex].
  7. Apr 19, 2013 #6
    ok, i got it.
    thanks :)
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