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Calculating functions for wave problems

  1. Jan 12, 2012 #1
    1. The problem statement, all variables and given/known data
    A wave travels along a string in the positive x-direction at 30 m/s. The frequency of the wave is 50 Hz. At x = 0 and t = 0, the wave velocity is 2.5 m/s and the vertical displacement is y = 4 mm. Write the function y(x,t) for the wave.

    2. Relevant equations
    velocity = (wavelength)(frequency) = (angular velocity)/(wave number)
    angular velocity = 2*pi*frequency
    What the function should look like:
    y(x,t) = Asin(kx - wt + θ)
    A = amplitude
    k = wave number
    w = angular velocity
    t = time (s)
    θ = initial phase shift

    3. The attempt at a solution
    Figuring out the wave number (k) as well as the wavelength and angular velocity was the easy part:
    w = 2*pi*(50 Hz) = 100*pi
    wavelength = velocity/frequency = (30 m/s)/(50 Hz) = .6 m
    wave number (k) = w/v = (100*pi)/(30 m/s) = 10.47 m^-1

    So far the function looks like:
    y(x,t) = Asin(10.47(x) - 314.16(t) + θ)

    The only variables left to find are the amplitude (A) and initial phase shift (θ). I tried plugging in the initial conditions the problem gave me, but I end up with:
    y(x,t) = Asin(θ) = .004
    differentiating the function y(x,t) = Asin(10.47(x) - 314.16(t) + θ) with respect to time gives the velocity equation, which is:
    v(x,t) = -A(314.16)cos(10.47(x) - 314.16(t) + θ)

    and plugging in the initial conditions for velocity gives:
    -A(314.16)cos(θ) = 2.5

    I don't know where to go from here to find amplitude and the initial phase shift. I have tried setting the velocity equation equal to zero which would give me a critical point (amplitude), but that didn't help much. I've also tried using a substitution method between these two equations:
    y(x,t) = Asin(θ) = .004
    -A(314.16)cos(θ) = 2.5

    but that didn't get me anywhere either. Any help would be greatly appreciated!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 12, 2012 #2
    Divide one of the equations by the other. So we have,

    A sinθ = .004
    -314.16Acosθ = 2.5


    -(1/314.16)tanθ = .004/2.5

    A is gone, so this can be solved for θ immediately, and then of course you can plug this back in to get A.
  4. Jan 12, 2012 #3
    So the initial phase shift is -.466 rad and the amplitude is .0089 m. In order to get the correct initial phase shift from the book I have to add -.466 and 3.14 (pi) which yields 2.67 rads (this is the answer given in the book). Why do I have to add my answer in order to get the desired answer from the book?

    The amplitude is correct as is, though. At any rate thanks for the help!
    Last edited: Jan 12, 2012
  5. Jan 12, 2012 #4
    The phase shift problem is due to the multivalued-ness of the arctan "function" -- there are many angles that give the same tangent, even within the same 2π "period." In particular, for every solution in the first quadrant there is one in the third, and for every solution in the second quadrant there is one in the fourth (which you can verify by thinking about the definition of the tangent function, or looking at a graph of tan(x)). This means that even if you know tanθ, you need additional information to find θ itself.

    In this problem, we need to decide between θ = -.466 and 2.67, which, as you can check with a calculator, have the same tangent. So is the choice arbitrary? It would be if we didn't have more information, but in this problem we do. θ = 2.67 is correct because we can plug it back into the original two equations involving sine and cosine and get a true statement. But plugging in θ = -.466 doesn't work, as you can check.

    Sorry I didn't notice this before. It's actually a common problem when solving for angles; arcsine and arccosine do this too. In general, evaluating an inverse trig function will give you a couple of possibilities, and you need to pick the right one using additional information from the problem. Unfortunately, calculators will blithely give you a single number like nothing is wrong, so you need to be alert.
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