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Difference between two wave functions?

  1. Jan 16, 2012 #1
    Hi, I'm looking through my teacher's notes and he says that a wave function for a sinusoidal wave can be written: y = A sin (ω t – k x) or y = A sin (k x - ω t)

    The textbook gives it in the second form. I think that using one over the other gives the same answer but in the opposite sign. I would like to know which one to use over the other. Any clarification is appreciated.
     
  2. jcsd
  3. Jan 16, 2012 #2

    jtbell

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    Staff: Mentor

    Pick numbers for A, k and ω, and draw some graphs of y versus x, for t = 0 and for t = (some fairly small number). Then you can see explicitly the differences (if any) in the behavior of the two functions.
     
  4. Jan 16, 2012 #3
    The difference between those wave functions is the phase but they are the same. You can check it with some graphics.
     
  5. Jan 16, 2012 #4
    I just started studying this stuff but on a couple occasions my answer was a different sign than the teacher's answer key, so I used the wrong function. This is what confuses me, how to know which to use in order to get the right sign. In the question, an end of a string is given sinusoidal motion and starts at t = 0 with zero displacement. Is that what you are supposed to look at to decide? I understand in SHM that the initial phase angle is zero if an object is released a certain displacement away from its equilibrium with zero speed, but I don't know how to interpret this wave function in the same way.

    Tbh it's also frustrating that I find 3-4 clear mistakes on every page of the teacher's notes. I much prefer when the teacher lets the students know the exact sections from the official textbook that are covered because at least I trust the textbook.
     
  6. Jan 16, 2012 #5
    The second one can be witten as
    [tex]Asin(\omega t - kx +\pi)[/tex]
    so it's a phase difference of 180 degrees.
    In general you can have any initial phase:
    [tex]Asin(\omega t - kx +\phi)[/tex]
    The value of the phase can be found from the initial configuration of the wave. Or we can take as "initial" the configuration that gives phase of zero (or 180 degree).
    Could you show a problem for which your answer is not as in the book?
     
  7. Jan 16, 2012 #6
    4.
    a) A certain string has a linear mass density of 0.25 kg/m and is stretched with a tension of 25 N. One end is given a sinusoidal motion with frequency 5 Hz and amplitude 0.01m. At time t=0, the end has zero displacement and is moving in the positive y-direction. Find the wave speed, amplitude, angular frequency, period, wavelength, and wave number.

    I found all of those, and then I am asked for:

    b) Write the wave function describing the wave.

    And my answer, based on the textbook, is: Asin(kx - ωt) (I'm using the symbols rather than the actual values to simplify things.

    But the answer key is: Asin(ωt - kx)

    This leads to opposite answers in the subsequent parts c and d of the question. For example:

    d) Find the transverse velocity of the point, x=0.25 m at time t=0.1

    so I write my answer as v = -Aωcos(kx - ωt) because I take the derivative of the textbook function with respect to t, and I get the answer 0.22 m/s, but the answer key gives it as -0.22 m/s.
     
  8. Jan 16, 2012 #7

    jtbell

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    Staff: Mentor

    At x = 0 your answer reduces to A sin (-ωt). At t = 0 this obviously gives a displacement of zero. A short time later, does it give a positive displacement as the problem statement implies?
     
  9. Jan 16, 2012 #8
    The initial phase is determined by the distribution of displacement and velocities at t=0 (and not just displacements). It does not matter what format you use as long as you find the right initial phase.
    The velocity of a point is given by the derivative of the displacement in respect to time.
    If we start with the format
    [tex]y(x,t)=Asin(\omega t -kx +\phi)[/tex]
    then
    [tex]v(x,t)=\omega A cos(\omega t -kx +\phi)[/tex]
    At t=0, x=0,
    [tex]y=A sin(\phi)=0[/tex]
    This can happen for a phase of 0 or 180.
    For
    [tex]v=\omega A cos(\phi)[/tex]
    to be positive you need phase zero so the equation will be
    [tex]y(x,t)=Asin(\omega t -kx)[/tex]

    If we start with the format
    [tex]y(x,t)=Asin( kx -\omega t+\phi)[/tex]
    The velocity will be
    [tex]v(x,t)=-\omega A cos(kx -\omega t +\phi)[/tex]
    The condition fo v(0,0) to be positive will require a phase of 180 so the final formula will be
    [tex]y(x,t)=Asin( kx -\omega t+180)[/tex] which is the same as
    [tex]y(x,t)=Asin(\omega t -kx)[/tex]
     
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