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Simple Harmonic Motion: Oscillator

  • Thread starter TJC747
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  • #1
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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.
 

Answers and Replies

  • #2
ideasrule
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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?
 
  • #3
AEM
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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?
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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