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Phase speed of the wave

  1. Jun 20, 2016 #1
    1. The problem statement, all variables and given/known data
    transverse wave is traveling through a wire in a positive direction of the x-axes. Distance od the wire particles in the motion of the wave can be described as ##y(x,t)=53*10^{-6}sin(188t-3.14x)## Find the ratio of the phase wave speed and maximal speed of the wire particles in the motion of the wave.
    2. Relevant equations
    3. The attempt at a solution

    The phase speed is suppose to be ##w/k## right? So that makes it around 60. But what is the second term? How do i get that?
     
  2. jcsd
  3. Jun 20, 2016 #2

    jfizzix

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    The key here is to understand what the wire particles are doing.

    We know it's a transverse wave.
    Transverse waves are the ones where the individual vibrations move at 90 degree angles to the direction the overall wave moves in. The wire moves up and down, but the ripples in the wire travel along its length (90 degrees from the vertical direction).

    What does that mean for the individual wire particles?

    For any wire particle, we have the height [itex]y[/itex], the distance along the horizontal direction [itex]x[/itex], and the time [itex]t[/itex]
    How would we figure out the velocity of an individual wire particle from those variables?
     
  4. Jun 21, 2016 #3
    Well since we have the position, isnt the velocity just the derivative of that position? And is my phase velocity correct?
     
  5. Jun 21, 2016 #4

    jfizzix

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    Since this is a homework forum, I don't think I can just tell you what the answer is, since you should be able to prove it to yourself.

    As far as phase velocity goes, it's more than just a formula. It's the velocity of a point of constant phase. The phase of the wave is the number you take the sine of in a sine wave. Here, it's an expression in terms of [itex]x[/itex] and [itex]t[/itex].
    The point [itex]x[/itex] where [itex]sin(188t -3.14x)=0[/itex] is a point that changes with changing time since as [itex]t[/itex] increases, [itex]x[/itex] must increase as well to keep [itex]188t -3.14x[/itex] a constant value.
    The velocity of this point of constant phase is by definition the phase velocity.
     
  6. Jun 21, 2016 #5
    So the velocity of the constant phase point is then ##v(x)=188/3.14## ?
     
  7. Jun 21, 2016 #6

    jfizzix

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    If we look at the point [itex]\tilde{x}[/itex] of constant phase [itex]188t -3.14\tilde{x}=0[/itex]
    then we solve for [itex]\tilde{x}[/itex] , and find:
    [itex]\tilde{x}=\frac{188}{3.14}t[/itex]

    This point [itex]\tilde{x}[/itex] changes with time at a constant rate, which is the phase velocity.
     
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