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Homework Help: Mathematical Description of a Wave

  1. May 6, 2012 #1
    Here is the question:

    Transverse waves on a string have a wave speed of 8.00 m/s, amplitude of 0.0700m, and a wavelength of 0.320m. The waves travel in the -x direction, and at t=0, the end of the string (x=0) has its maximum upward displacement.

    a)What is the frequency, period, and wave number of these waves?
    b) Write a wave function describing the wave.

    This is what I got for part a (if something is wrong please let me know):

    frequency = f = v/λ = 8.00m/s / 0.320m = 25 Hz
    period = T = 1/25 = 0.04 seconds
    wave number = k = 2π/λ = 2pi/0320m = 19.635 rad/m

    Here is where I had a few problems, on part B:

    I know the equation is Y(x,t) = A*cos(kx + wt) for this particular problem, but I am having troubles determining when I should use sin and when I should use cos as the trig function in the problem. I also am not too sure what the difference would be if the function would be if it were traveling in the positive-x direction instead of the negative-x direction.

    Could someone please explain these concepts to me so I can understand them? It would be greatly appreciated!

    Also, for the answer I got:

    y(x,t)= 0.700m*com(19.635rad/m*x + 157.08rad/s*t)

    Thanks for any help!!
  2. jcsd
  3. May 6, 2012 #2
    This choice is based on the initial conditions for the problem. The most general form of such a wave would be
    [tex]y(x,t) = A\ \text{cos}(\omega t + kx + \phi)[/tex]
    where [itex]\phi[/itex] would be a phase constant for your problem. Notice that if [itex]\phi = -\pi / 2[/itex], then you get a sine wave out of it (since cosine and sine are always 90 degrees out of phase). So in that sense, picking either sine or cosine for your function works, as long as you get the right phase constant for each function.

    And the phase constant is always based on the initial/boundary conditions. For example, consider what happens at the origin:
    [tex]y(0,0) = A\ \text{cos}(\phi).[/tex]
    If at the origin, at [itex]t = 0[/itex], you expect the function to have its maximum height, then the phase constant would be zero. But if you expect it to have some other value at that time, you would pick the phase constant to hold to that instead.

    You would have [itex]kx \rightarrow -kx[/itex] for motion in the positive [itex]x[/itex] direction.
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