
#1
Apr1713, 04:11 PM

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 



#2
Apr1713, 05:59 PM

P: 1,657

[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. 



#3
Apr1813, 08:41 AM

P: 860

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
Apr1813, 08:54 AM

P: 3

Two general questions about wave functionsso 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
Apr1813, 09:20 AM

P: 1,657

[itex] sin(\omega t  k x + \Phi) = sin(k x  \omega t + \Phi')[/itex] where [itex]\Phi' = \pi  \Phi[/itex]. 



#6
Apr1913, 05:35 PM

P: 3

ok, i got it.
thanks :) 


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