Sine Wave Addition

  • Thread starter marla11
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  • #1
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Main Question or Discussion Point

If two sine waves have the same frequency and amplitude but have different phase shift do they still produce a standing wave?
Thanks for the help.
 

Answers and Replies

  • #3
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As far as i can see it doesn't say anything about phase shift so does that mean it doesn't affect anything?
 
  • #4
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wt is the phase shift

see http://en.wikipedia.org/wiki/Phase_(waves [Broken])
 
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  • #5
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I'm sorry but I'm still confused. If y1 = Asin(kx-wt+phi) and y2 = Asin(kx+wt)
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
4,662
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You have one forward-traveling wave (wt-kx) and one backward wave (wt+kx) of the same amplitude, which is a standing wave. My CRC Math Tables (10th Ed, 1954) on page 345 shows the sum

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

Bob S
 
  • #7
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Marla,

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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