- #1
fazio93
- 3
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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
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