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Group and Phase velocity of wave packet

  1. Sep 26, 2008 #1
    OK, I think I understand the problem, I'm just a little confused on some pieces in the middle.

    Consider a wave packet formed by the superposition of two waves
    psi1=cos(1.00x-2.00t) <--1.00=k, 2.00=w
    psi2=cos(1.01x-2.03t) <--1.01=k, 2.03=w
    where x and t are measured in meters and seconds respectively. What is the Phase velocity?
    What is the group velocity?

    We know:
    Vp= w/k

    -To find the Vp, can we simply say that Vp=2.00/1.00=2.00m/s?

    Now, to find the Group Velocity, we can use the suerposition principle and find the net wave by the sum of the two individual waves (psi1 + psi2) which gives us

    psi=2cos(1.00*1.01)x*cos(-2.00*-2.03)t = 2cos(1.01)x*cos(4.06)t

    thus Vg=dw/dk= -(2/1.01)sin(1.01)x*cos(4.06)t - (2/4.06)doc(1.01)x*sin(.06)t?

    Is this correct? Is the partial derivative correct?

  2. jcsd
  3. Sep 26, 2008 #2


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    Good enough. For the 2nd wave it's slightly different, but 2.00 m/s is fine.

    It doesn't make sense for Vg to be a function of x and t.

    There's a simpler approach to find Vg. You have w and k at two points. What's a good approximation for the slope of the w-k curve near those two points?
  4. Sep 26, 2008 #3
    Hmm...I'm not sure why I put that but I think I meant to say cos(1.01)x.

    I see what you mean by a slope being much easier, but I'm not sure how to apporximate the curve. Our text only shows that the Vg=delta(w)/delta(k) which can be approximated by differentials.

    -OK, w and k are at two points, one in the first wave and nother in the second. So can you say that the slope is just (-2.03--2.00)/(1.01-1.00)= -3?

    -Can group velocity be negative if the phase velocity is positive?
  5. Sep 26, 2008 #4


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    That's the right idea, but use the values of w you had in post #1: +2.03 and +2.00.
  6. Sep 26, 2008 #5
    Ah, I see. Thank you for your help!
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