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Question about the wave function of a travelling wave

  1. Jan 3, 2016 #1
    Hi guys,
    I have a confusion about the wave function of a travelling wave. This is the wave function of a travelling wave travelling towards the positive direction of x axis

    u(x,t)=Acos[ω(t-x/v)+φ0], where v is the velocity of the wave, ω is the angular velocity, φ0 is the initial phase.
    Consider u as the displacement of a particle in y direction perpendicular to the x direction, that is, a longitudinal wave.
    in the textbook, the above wave function is derived by first considering a particle oscillating at x=0 with an oscillation function u(0,t)=Acos(ωt+φ0). then when the oscilaltion spreads towards the positive x direction, it takes the oscillation x/v to arrive at x. then the oscillation at x is x/v left behind that of x=0, so we have ω(t-x/v)+φ0 the phase of the oscillation at x with respect to x=0.

    my question is, for the oscillation of x at t=x/v (just at the time the wave arrived at x), according to the wave function, the displacement should be u(x,x/v)=Acos[φ0]. but since the wave has just been arrived, the starting point for the particle shold be its equilibrium point, with u(x,x/v)=0 in this case. So is there a contradiction? I have some thoughts about this, and i will post it in the next floor. I dont know whether it is right. would anyone please give me some instruction? Thanks a lot for your kind help!
    Last edited: Jan 3, 2016
  2. jcsd
  3. Jan 3, 2016 #2
    i think that maybe because the wave function u(x,t)=Acos[ω(t-x/v)+φ0] is the function of a wave that is steady in the space. so the derivation in the textbook gets the right wave function, but it is wrong to think like that.
  4. Jan 4, 2016 #3
    The wave function you gave, u(x,t)=Acos[ω(t-x/v)+φ0], assumes that at t = 0, x = 0, the oscillation is Acos[φ0]. This is an initial condition, and it is in your hand. If you want the wave to start from 0, you just put φo = π/2.

    Incidentally, A wave that oscillates along the y direction while traveling in the x direction is a transverse wave, not a longitudinal one.

    Reference https://www.physicsforums.com/threads/question-about-the-wave-function-of-a-travelling-wave.850689/
  5. Jan 5, 2016 #4

    Thanks Prayaga, it really helps!
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