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Sine Wave Addition

  1. Mar 7, 2010 #1
    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.
  2. jcsd
  3. Mar 7, 2010 #2
  4. Mar 7, 2010 #3
    As far as i can see it doesn't say anything about phase shift so does that mean it doesn't affect anything?
  5. Mar 7, 2010 #4
    wt is the phase shift

    see http://en.wikipedia.org/wiki/Phase_(waves [Broken])
    Last edited by a moderator: May 4, 2017
  6. Mar 7, 2010 #5
    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?
  7. Mar 7, 2010 #6
    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
  8. Mar 9, 2010 #7

    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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