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Writing a function for a wave.

  1. Jan 18, 2013 #1
    1. The problem statement, all variables and given/known data

    A wave travels along a string in the positive x-direction at 27 m/s. The frequency of the wave is 46 Hz. At x = 0 and t = 0, the wave velocity is 2.6 m/s and the vertical displacement is y = 3 mm. Write the function y(x, t) for the wave. (Use the following as necessary: x and t. Assume x and y are in m, and that t is in s.)

    2. Relevant equations

    y(x,t) = Asin(κx - ωt + ∂)

    ω = 2pif = 2pi(46) = 289.03

    κ = ω/v = 289.03/27 = 10.70

    3. The attempt at a solution

    I determined that:

    y(x,t) = Asin(10.70x - 289.03t + ∂)
    y(0,0) = Asin(∂) = 0.003 m

    Here is where I have trouble. I tried to take a partial derivative of the first function in order to solve for the velocity, since velocity is given. I've never encountered partial derivatives before, so I am not sure if what I did is correct. It doesn't give me the right answer.

    yt= -289.03Acos(10.70x - 289.03t + ∂)

    If (x,t) = (0,0)
    yt = -289Acos(∂) = 2.6 m/s

    I tried to combine this equation with

    Asin(∂) = 0.003 m

    to solve for ∂ and A, but I wasn't close to the final answer.

    Is this a matter of taking the partial derivative wrong, or am I making some other mistake?
  2. jcsd
  3. Jan 19, 2013 #2

    Simon Bridge

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    the wave velocity is 2.6 m/s
    ... the wave velocity is the speed of the wave in the x direction.

    the partial derivative: ∂y(x,t)/∂t is the speed of point x in the y direction.

    If the wave at t=0 has form ##y(x,0)=f(x)## and it is travelling in the +x direction with speed c,
    then at time t it will have form: ##y(x,t)=f(x-ct)##.

    In your case, f(x) is the wave equation at t=0.
  4. Jan 19, 2013 #3
    Thanks for the response.

    At t = 0, x = 0.

    y(0,0) = Asin(∂)

    According to the problem, at this point, y(0,0) = 0.003 m.


    Asin(∂) = 0.003 m.

    I now have two variables, A and ∂. How do I solve for them?
  5. Jan 19, 2013 #4


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    You did it right. Show your further work.

  6. Jan 19, 2013 #5

    Simon Bridge

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    Looking at it again and squinting this time:
    ...this could be interpreted either way ... this wave velocity could be the rate of change of displacement y of whatever is at point x or it could be the speed of the whole wave-form (and if the latter, why mention the x position?) ... but it has already been said that:
    ... OK, so c=27m/s and vy(0,0) = 2.6m/s

    So what was the final answer supposed to be?
    What did you get?
  7. Jan 19, 2013 #6
    Equation #1: -289Acos(∂) = 2.6 m/s
    Equation #2: Asin(∂) = 0.003 m

    Equation #1 (solved for A): A = 2.6/[-289.03cos(∂)]

    Equation #2 (with substitution): (-2.6/289.03)tan(∂) = 0.003
    (cont.) tan(∂) = -(0.003*289.03)/2.6
    ∂ = tan-1[-(0.003*289.03)/2.6]
    ∂ = -0.32

    Back to Equation #2: Asin(-0.32) = 0.003
    A = -0.0095 (Amplitude is always positive, so can I just disregard the negative sign?)

    The correct answer is: 0.0095 sin (10.70x - 289.03t + 2.82)

    My answer as it is currently: 0.0095 sin (10.70x - 289.03t - 0.32)

    My phase does not equal 2.82. Yet, using -0.32 as the phase resulted in the correct amplitude (assuming that it is alright to disregard the negative sign.)
  8. Jan 19, 2013 #7


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    You got A correctly. The easier way to calculate it is:


    Square both equations and add them together: A2cos2δ+A2sin2δ=A2=90*10-6, A=0.0095.

    From the first equations you see that sin(δ) >0 and cos(δ)<0. So δ is in which quadrant?

  9. Jan 19, 2013 #8
    I see now. It is in the second quadrant. I need to add pi to -0.32 which results in 2.82.

    Thanks! I think I can go to bed now...
  10. Jan 19, 2013 #9


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    Good night! :)

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