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Zero group velocity

  1. Jan 31, 2013 #1
    An infinitely long "mass-spring transmission line", consisting of masses (m) connected by springs (spring constant s) obeys the following dispersion relation:

    ω = \sqrt{4s/m} sin(kd/2).

    The group velocity is

    dω/dk = d/2 \sqrt{4s/m} cos(kd/2).

    What does zero group velocity "mean" for this system?

    This is a question from an exam i took recently. The answer was that you'd get a standing wave, which doensn't make sense to me. My answer was that you'd get stationary packets of traveling waves (like this: http://www.falstad.com/dispersion/groupzero.html ) (whereas you can get standing waves at any frequency. I still think I'm right, so if someone here can shed some light I'll know whether I should start arguing with my professor.
     
  2. jcsd
  3. Jun 15, 2013 #2
    To see why a pure standing wave has a zero phase velocity, let's look at the equation for the current, I, in a standing wave.

    I(x,t)=Imax[sin(kx)][cos(wt)] ;where w is omega=2*pi*freq and 0 < kx < 180 degrees

    At any time, t, the phase of the current is a constant, i.e. the phase of the current does not vary with position x as it does for a traveling wave. In a pure standing wave, the position, x, only affects the magnitude of the envelope of the current. Since the phase of the current is a constant over any 0-180 degrees, the difference in any two phase measurements at different locations along that 0-180 degree path is zero. Note that a phase delay of zero for a pure traveling wave would imply faster-than-light energy propagation.
     
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