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Simple Harmonic Motion - Getting Acceleration from Velocity

  1. Aug 29, 2016 #1
    1. The problem statement, all variables and given/known data
    I am doing an experiment where I am measuring the force a speaker is exerting when it is driven by a certain voltage and frequency, so my voltage and frequency values are known. I am assuming the speaker is undergoing SHM and I am measuring its peak to peak velocity.

    2. Relevant equations
    I understand that for SHM:

    x = Asin(wt)
    x' = wAcos(wt)
    x'' = -w^2Asin(wt)

    3. The attempt at a solution
    Now, because I am measuring the peak to peak velocity, is it correct to say cos(wt) and sin(wt) are equal and therefore x'' = -w*x', where w = 2*pi*f and x' = my peak to peak velocity measurements divided by 2.

    Furthermore, if I were not to take this approach my acceleration equation would be x'' = -w^2Asin(wt).
    Where,

    w = 2*pi*f
    t = period of 1 cycle
    A = (?)

    I am not quite sure what the amplitude would be in my situation. What kind of amplitude is required? The amplitude of the voltage driving the speaker? Or the amplitude of the displacement of the speaker? (Which in this case is unknown as only the velocity measurements are taken).

    Any help is appreciated. Thanks.
     
  2. jcsd
  3. Aug 29, 2016 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Look at graphs of ##\sin(\theta)## and ##\cos(\theta)##; you will see that where one of them is at a peak the other is half-way between a peak and a through; that is, when ##\sin(\theta) = \pm 1## we have ##\cos(\theta) = 0##, and vice-versa. You can also see this without drawing a graph if you remember that ##\sin^2(\theta) + \cos^2(\theta) = 1## for all angles ##\theta##.
     
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