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Am I understanding what phase velocity is?

  1. May 31, 2007 #1
    ok so the picture I get in my head is: if I took a wave, and stretched it so that it was a straight line, the phase velocity would be the speed at which a point travels along that line. is that an accurate analogy or am I way off?
     
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  3. Jun 1, 2007 #2

    FredGarvin

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    In all of my acoustics classes, they used wave speed, c, and phase velocity interchangeably. It is the speed of the wave propagation that is perpendicular to the particle displacement and velocity. I am not quite sure if this is applicable to all other aspects of physics though.
     
  4. Jun 1, 2007 #3
    well it's mentioned briefly in the book I'm reading, it says that sometimes for X rays in a vacuum, phase velocity can exceed C. It's only mentioned for about three lines, but from the equation it gives, that's the picture I get:

    If I were to take a wave, and mark, say, each crest with a dot. Then I would "stretch" the wave so that it's a straight line (as if it was a string—the distance between each "crest dot" would increase), and have that line move so that the difference in t between each crest-dot passing through point x remains the same as it was before the wave was "stretched", the new velocity of each dot is the phase velocity.

    Am I understanding correctly? ... I don't know if my explanation makes sense.
     
    Last edited: Jun 2, 2007
  5. Jun 1, 2007 #4
    im pretty sure phase velocity is the velocity of the propagation of the wave
     
  6. Jun 2, 2007 #5

    cepheid

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    The phase velocity is the velocity of a point of constant phase on the wave. It corresponds to the propagation velocity for a monochromatic wave.
     
  7. Jun 2, 2007 #6

    rcgldr

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  8. Jun 2, 2007 #7
    that seems like it's describing what I understood. sorry if my explanation made no sense; I just always need a mental picture of something or I don't get it :smile:, I'm not very confident in my understandings from just looking at a formula yet.
     
    Last edited: Jun 2, 2007
  9. Jun 2, 2007 #8

    FredGarvin

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    That makes perfect sense. That would account for what I am used to seeing for single frequencies. Cool.
     
  10. Jun 2, 2007 #9

    cepheid

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    Glad I could clarify it. In fact, a really simple way to show it mathematically just occured to me (i.e. my memory was jogged). Say we have a wave of the form:

    [tex] \cos(kx - \omega t) [/tex]

    Then a point of constant phase (or a whole "plane" of them corresponding to a particular wavefront) has a trajectory x(t) satisfying the following relation:

    [tex] kx - \omega t = \ \textrm{const.} [/tex]

    [tex] kx = \omega t + \ \textrm{const.} [/tex]

    [tex] x(t) = \frac{\omega}{k} t + \ \frac{\textrm{const.}}{k} [/tex]

    [tex] \frac{dx(t)}{dt} = v_{\textrm{phase}} = \frac{\omega}{k} [/tex]

    [tex] = c \ \textrm{(for light)} [/tex]
     
  11. Jun 2, 2007 #10

    cepheid

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    or you can say that the phase [itex] \phi(x,t) = kx-\omega t [/itex] will be constant for some x(t) such that:

    [tex] \frac{d}{dt} \phi(x(t),t)) = 0 [/tex]

    and you'll get the same answer.
     
  12. Jun 2, 2007 #11

    cepheid

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    Also,

    [tex] \frac{\omega}{k} = 2 \pi \nu \frac{\lambda}{2 \pi} = \nu \lambda [/tex]

    [tex] = \frac{\lambda}{T} [/tex]

    which is obviously the propagation velocity.
     
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