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Homework Help: Amplitude of particle in SHM

  1. Nov 11, 2007 #1
    1. The problem statement, all variables and given/known data
    Adjacent antinodes of a standing wave on a string are a distance 15.0cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850cm and period 0.0750s. The string lies along the +x - axis and is fixed at x=0.
    At time t=0, all points on the string are at their minimum displacement.
    Find the amplitude at a point a distance 3.0cm to the right of an antinode.

    2. Relevant equations
    d^y(x,t) /dx^2 = 1/v^2 . d^2y(x,t)/(dt^2)

    3. The attempt at a solution
    wavlength is 2 x 15 cm = .30m
    f is 1/T = 13.333 Hz
    so v = 4 m/s, v^2 = 16 and 1/v^2 is 0.0625
    However, I' having trouble getting the second derivative of the wave equation using the point 3.0 cm to the right of the antinode.
  2. jcsd
  3. Nov 11, 2007 #2


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    What is the equation of a standing wave?

    why are you trying to get the second derivative?
  4. Nov 12, 2007 #3
    I thought the second derivative wrt x was equal to the second derivative wrt t by the inverse of v squared. Or maybe not.
  5. Nov 12, 2007 #4


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    Yes, that's the right equation.

    I still don't understand... why do you need the second derivative with respect to x, or with respect to time? you need the amplitude at a certain x value right?

    you've got the equation of the standing wave.


    The amplitude at any x value is simply Asinkx... because the maximum value that sinwt takes over time is 1... ie amplitude at x is Asinkx(1) = Asinkx.

    you have A. you can get k ... You need the amplitude at a point 0.03m to the right of an anti-node... so just choose any anti-node... you've got the wavelength = 0.30m. where is the first anti-node?

    what do you get for Asinkx, where x = 0.03 + position of anti-node ?
    Last edited: Nov 12, 2007
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