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Thanks for the help.

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- Thread starter marla11
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- #1

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Thanks for the help.

- #2

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- #3

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- #4

- 5,601

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wt is the phase shift

see http://en.wikipedia.org/wiki/Phase_(waves [Broken])

see http://en.wikipedia.org/wiki/Phase_(waves [Broken])

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- #5

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the addition is y= 2Acos(wt-phi/2)sin(kx+phi/2) where phi is the phase shift between 0 and 2pi. Does this still fit the standing wave equation y=(2Asin(kx))cos(wt) meaning its a standing wave or does the difference in phase shift mean they do not create a standing wave?

- #6

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sin(x) + sin(y) = 2·sin[(x+y)/2]·cos[(x-y)/2]

Bob S

- #7

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Your equations will be easier to read if you typeset them in LaTeX.

Yes, the equation you give is a standing wave. If you start with

[itex] \Psi(x,t) = A\cos(\omega \left[t-t_0\right]) \sin (k\left[x-x_0\right]) [/itex]

you can just define a new time coordinate and new space coordinate by

[itex] t' = t - t_0 [/itex]

[itex] x' = x - x_0 [/itex].

Then your original equation is just

[itex] \Psi(x',t') = A \cos(\omega t')\sin(k x')[/itex],

showing that the waveform is exactly the same as the standing wave you're used to.

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