Question

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?

Final Answer

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

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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

*k*times*A one*squared. In the second case after there is some dampening after one oscillation, it'll be one half times*k*times the new amplitude*A two*squared. We're told that*A two*is going to be three percent less than*A one*. So amplitude two is*A one*minus 0.03 times*A one*which is 0.97*A one*. We can substitute 0.97*A one*, so amplitude two is going to be 97 percent of what amplitude one was. We'll substitute that in for*A two*here. This works out to 0.9409 which is what 0.97 squared is, multiplied by one half*k A one*. One half*k A one*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*P E one*times 100 percent. That's 0.9409*P E one*that we figured out just now for*P E two*, minus*P E one*over*P E one*. The*P E one's*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.