Two general questions about wave functions

by fazio93
Tags: functions, wave
 P: 3 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? Thanks
P: 1,410
 Quote by fazio93 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?
Propagation velocity is related to $k$ and $\omega$ through:
$v = \omega/k$. So if you switch the sign of both $\omega$ and $k$, the sign of the velocity remains the same.
 P: 749 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?
P: 3

Two general questions about wave functions

 Quote by stevendaryl Propagation velocity is related to $k$ and $\omega$ through: $v = \omega/k$. So if you switch the sign of both $\omega$ and $k$, the sign of the velocity remains the same.
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 + ∅)
P: 1,410
 Quote by fazio93 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 + ∅)
Yeah, except that
$sin(\omega t - k x + \Phi) = sin(k x - \omega t + \Phi')$
where $\Phi' = \pi - \Phi$.
 P: 3 ok, i got it. thanks :)

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