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Sinusoidal Wave Problem

  1. Apr 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider the sinusoidal wave in the figure, with the wave function below.

    At a certain instant, let point A be at the origin and point B be the first point along the x axis where the wave is 60.0° out of phase with point A. What is the coordinate of point B?

    y = (15.0 cm) cos(0.157x - 50.3t)

    and I am not allowed to post the image but heres some helpful information:
    its a cosine graph, and [tex]\lambda[/tex] = 40 cm, T = 0.125 s, f = 8.00 hz, k = 0.157 rads/cm

    2. Relevant equations

    Some equations that I have are:
    k=2[tex]\pi[/tex]/[tex]\lambda[/tex]
    v=[tex]\lambda[/tex]*f
    y=Asin(kx-[tex]\omega[/tex]t)

    3. The attempt at a solution
    What I have tried so far is:
    y=15*sin((.157)x-(50.3)t + [tex]\pi[/tex]/3)
    The example says " The vertical position of an element of the medium at t = 0 and x = 0 is also 15.0 cm "
    The only problem is that I am really stuck and I cannot seem what to set the 60.0[tex]\circ[/tex] equal to. Help would be much appreciated! Thank you!
     
  2. jcsd
  3. Apr 18, 2008 #2

    cepheid

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    Everything in the argument of the cosine is the phase of the wave. So the phase is a function of both x and t. However, we're looking at this wave at a particular "instant", which means that t is constant. So, the only thing that would cause a phase change is x. So how far from the origin would the x coordinate of point B have to be in order for 0.157x - 50.3t to differ from 0.157*(0) - 50.3t by 60 degrees, given that t is constant?
     
  4. Apr 18, 2008 #3
    Would x = 6.37 * [tex]\pi[/tex] because that would make it ([tex]\pi[/tex]/3 -50.3t)
     
  5. Apr 18, 2008 #4

    cepheid

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    I think you might be off by a factor of 3. Check your work again.
     
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