# Two general questions about wave functions

1. ### fazio93

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?

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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

2. ### stevendaryl

3,343
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.

3. ### Philip Wood

1,175
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?

4. ### fazio93

3
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 + ∅)

5. ### stevendaryl

3,343
Yeah, except that
$sin(\omega t - k x + \Phi) = sin(k x - \omega t + \Phi')$
where $\Phi' = \pi - \Phi$.

6. ### fazio93

3
ok, i got it.
thanks :)