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Transverse Waves

  1. Apr 18, 2007 #1
    Please help!!!

    Having a little difficulty with a homework question and concept

    1. The problem statement, all variables and given/known data
    y = 2.28sin(0.0276pi x + 2.42pi t)

    Find the amplitude, wavelength, frequency, speed and maximum transverse speed.

    3. The attempt at a solution

    Definitely correct answers

    A = 2.28cm
    Wavelength = 72.46 cm
    Frequency = 1.21 Hz

    Questionable answers

    Speed = 87.68 cm/s
    Max transverse speed = ??????


    The speed should be w/k which would give 2.42pi/0.0276pi = 87.68 cm/s

    The max transverse velocity though....would that be when the trig function of cos is equal to 1 in this equation u = -wAcos(kx-wt) ????
    Last edited: Apr 18, 2007
  2. jcsd
  3. Apr 18, 2007 #2


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    The speed is correct. Transverse speed is when dy/dt is a max. I think you are pretty much nailing that as well.
  4. Apr 19, 2007 #3
    Apologies...still not quite sure.....would I use the w and A i know multiplied with cos(kx-wt) to get my answer.....so if cos(kx-wt)=1, then max transverse speed would be -wA???
  5. Apr 19, 2007 #4


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    If your displacement function is a sin wave, it's not possible that your velocity "function" be a constant of 87.68cm/s as you calculated. Your speed would be a function as well.

    As you know, velocity is the change of displacement over time (i.e. dy/dt), so by differentiating the function for y, you will get a the velocity function (which in this case is a cosine function). That cosine function will allow you to calculate the speed at any location x and time t.

    To calculate the maximum speed, we take a look at the velocity function. If you do your calculus right, the equation should look something like:

    v = B cos (Cx + Dt)

    There is an infinite possibility for the values of x and t, so we can assume that hthe cosine can return any value. The maximum value however of a cosine function is 1. So what does that tell you about your maximum velocity?
  6. Apr 19, 2007 #5


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    Yes. Except I'd call the maximum speed +wA. Why would you say minus?
  7. Apr 19, 2007 #6
    Thanks much....makes a lot more sense.....appreciate the help
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