1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted