(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Calculate the ratio of the kinetic energy to the potential energy of a simple harmonic oscillator when its displacement is half its amplitude.

2. Relevant equations

KE=1/2mv^{2}= 1/2kA^{2}sin^{2}(wt)

U=1/2kx^{2}= 1/2kA^{2}cos^{2}(wt)

KE_{max}=1/2kA^{2}

U_{max}=1/2KA^{2}

3. The attempt at a solution

What I'm thinking is that when the displacement is half the amplitude, the simple harmonic oscillator is halfway to the equilibrium point. When the displacement is half of the amplitude, is the potential energy half of its maximum value? At this point, wouldn't the kinetic energy be half of its maximum value as well? I assumed this is true and attempted to answer the question, which yielded an answer of 1, which is incorrect. Is there something wrong with my logic?

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# Homework Help: Ratio of Kinetic to Potential Energy of Simple Harmonic Oscillator

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