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Simple Harmonic Motion: Oscillator

  1. Dec 16, 2009 #1
    An oscillator with a mass of 600 g and a period of 0.50 s has an amplitude that decreased by 2.0% during each complete oscillation. If the initial amplitude is 6 cm, what will be the amplitude after 25 oscillations?

    I should most likely be using T = 2pi*sqrt L/g, as well as the many derivations of Hooke's laws, yet I cannot piece them all together. Help would be appreciated. Thanks.
  2. jcsd
  3. Dec 16, 2009 #2


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    Homework Helper

    You don't need to use any formulas. Just think. After one oscillation, amplitude = 0.98 * initial amplitude, correct? After two oscillations, A=0.98*0.98*Ai. After three, 0.98*0.98*0.98*Ai. After 25?
  4. Dec 16, 2009 #3


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    Are you sure it's a constant decrease of .02? Wouldn't it be a decrease by 2.0% of the previous complete oscillation, so the amplitude as

    1.00 - 1.00*.02
    .98 - .98*.02
    .9604 - .9604*.02
    .9401 - .9401*.02

    and so on.

    This looks rather like compound interest but in "reverse".
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