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The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?
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Final Answer

5.91% of the mechanical energy is lost after each oscillation.

Solution Video

OpenStax College Physics Solution, Chapter 16, Problem 41 (Problems & Exercises) (1:38)

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Video Transcript

This is College Physics Answers with Shaun Dychko. The maximum potential energy of a simple harmonic oscillator is one half times the spring constant, times the amplitude squared. So in the first case we have one half <i>k</i> times <i>A one</i> squared. In the second case after there is some dampening after one oscillation, it'll be one half times <i>k</i> times the new amplitude <i>A two</i> squared. We're told that <i>A two</i> is going to be three percent less than <i>A one</i>. So amplitude two is <i>A one</i> minus 0.03 times <i>A one</i> which is 0.97 <i>A one</i>. We can substitute 0.97 <i>A one</i>, so amplitude two is going to be 97 percent of what amplitude one was. We'll substitute that in for <i>A two</i> here. This works out to 0.9409 which is what 0.97 squared is, multiplied by one half <i>k A one</i>. One half <i>k A one</i> squared is potential energy one. So I substitute that here. Potential energy two is 0.9409 times potential energy one. The percent change is going to be potential energy two minus potential energy one, divided by <i>P E one</i> times 100 percent. That's 0.9409 <i>P E one</i> that we figured out just now for <i>P E two</i>, minus <i>P E one</i> over <i>P E one</i>. The <i>P E one's</i> cancel and this turns into a number one and so 0.9409 minus one times 100 percent is negative 5.91 percent. So 5.91 percent of potential energy is lost during each oscillation.