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Group velocity of two superimposed sine waves

  1. Jun 18, 2014 #1
    Hi all,
    I understand the concept of group velocity when applied to superimposed sine waves of the same amplitude, and even when applied to wave packets (in which case you get the well-known expression ∂ω/∂k).
    My question is what happens when you add two sine waves of different amplitudes? So something like:
    y(x,t) = A1Sin[k1x-ω1t]+A2Sin[k2x-ω2t]

    I tried to work out whether the concept of group velocity even has a meaning here, and if so, what it would be.
    If the amplitudes were the same, one can use trigonometric identities to express the wave as the product of an envelope and something else. Then, the envelope can be seen to travel at the velocity Δω/Δk. But, again, I don't know how to proceed if the amplitudes are not the same.

    Any comments are very appreciated.
     
  2. jcsd
  3. Jun 18, 2014 #2

    UltrafastPED

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    Variation in the amplitudes is common; do a fourier analysis of an arbitrary blob ....
     
  4. Jun 18, 2014 #3

    olivermsun

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    Do the same manipulations with different amplitudes. What do you get that's different from the "usual" case?
     
  5. Jun 19, 2014 #4

    You get two terms of the same form as in the usual case. So, you get something like: envelope1*(sine wave) + envelope2*(cosine wave)

    Is the group velocity still delta omega/ delta k ?
    That's what my instinct told me but I somehow feel it's not right...
     
  6. Jun 19, 2014 #5

    olivermsun

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    Okay, maybe one way you can think of it is this:
    y(x,t)
    = A1Sin[k1x-ω1t]+A2Sin[k2x-ω2t]
    = A1(Sin[k1x-ω1t]+Sin[k2x-ω2t]) +(A2-A1)Sin[k2x-ω2t],
    i.e., the usual packet of two waves (with a group velocity, etc.), superposed with a plane wave (no group velocity).
     
  7. Jun 19, 2014 #6
    But then, is there a group velocity to the whole thing? The first two terms you wrote have a group velocity Δω/Δk, and the last one is just a plane wave. But what is the group velocity of all three terms put together?
     
  8. Jun 19, 2014 #7

    olivermsun

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    The envelope should still travel at the same Δω/Δk... It's just that the envelope no longer goes to zero, right?
     
  9. Jun 19, 2014 #8
    Sorry I don't think I'm getting it. What would the expression for the envelope be?
     
  10. Jun 19, 2014 #9

    olivermsun

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    If you look at the expression in post #5, you see the wave field is expressed as the sum of (a) two superposed sine waves of the same amplitude, for which you have a clear interpretation of the group velocity, and (b) a monochromatic wave, which does not have an unambiguous group velocity. You can calculate the envelope for (a), and (b) has a constant envelope. So my interpretation would be that the total envelope moves at the same speed as that for (a). (I believe you can decompose this wave such that it's the product of the carrier with a complex envelope, but I'm not sure this helps much with the interpretation).

    If the problem arose in a physical context, you would likely know ∂ω/∂k and there would be not be this ambiguity.
     
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