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Homework Help: Max transverse speed on a sinusoidal wave

  1. May 6, 2014 #1
    Given the below equation of a sinusoidal wave, find the maximum transverse speed of a point on the string.

    y(x,t) = .00325m * sin(70x -3t)

    I am brand new to waves and trying to figure out what this question exactly means. The way I see it is that it might be at a max speed when 70x-3t = 1 because then the equation would just equal the amplitude.

    x is related to distance and t is related to time. the function is a position function I think so taking the derivative should give its velocity function.

    dy/dt = v = .2275m cos(70x-3t)

    Now my thinking has changed. Would velocity be at its max when 70x-3t = 0 since cos(0) = 1?

    70x = 3t
    t = 70x/3 or x = 3t/70

    Ha, I have no idea where I'm going with this. I'm just studying ahead.
  2. jcsd
  3. May 6, 2014 #2
    You did not calculate dy/dt correctly.

    The question did not ask you to calculate when or where the maximum speed occurs, only what the maximum speed is.
  4. May 7, 2014 #3
    Hmmm. How would I take the derivative in this case? Do I even have to take the derivative?

    Oh and wouldn't max speed be inbetween 2 amplitudes? Like if you look at a sin(x) graph it would be where y=0. Wouldn't this be where max speed would occur?
  5. May 7, 2014 #4


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    You did the derivative with respect to x instead of t.

  6. May 7, 2014 #5
    Ah I see. So then dy/dt = v = -.00975 cos(70x-3t)

    At its max, cos(70x-3t) = 1/2 since that will be inbetween two amplitudes where its Vmax is at (correct me if I'm wrong). This speed will be the same for any x/t combination that works with this idea.

    If cos(70x-3t) = 1/2 then v = -.00975m/s * .5 = -.004875 m/s

    Since it is negative, I have a hard time believing I'm on the right track. I think it may actually be when cos(...)=1 and it would be the absolute value of that answer. Or maybe when cos(..)= -1 then it would just negate the negative sign.

    So then the max speed would actually be .975 cm/s
    Last edited: May 7, 2014
  7. May 7, 2014 #6
    The latter answer is correct.
  8. May 7, 2014 #7


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    cos(.) varies between -1 and +1. If you are not sure about that, you need to review your trig courses.

    Pick the value in that range which gives you the maximum value for the velocity.

    I can't guess where your idea that "At its max,cos(70x-3t) = 1/2" came from.
  9. May 7, 2014 #8
    I think I was just stuck thinking about oscillations. From what I remember, when y=0 it's at a max velocity and at the top or bottom of its amplitudes, it's at its max acceleration. Something like that.

    Thanks though.
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