Two general questions about wave functions

  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. stevendaryl

    stevendaryl 3,525
    Science Advisor

    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. Philip Wood

    Philip Wood 1,180
    Gold Member

    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. 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. stevendaryl

    stevendaryl 3,525
    Science Advisor

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