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For a particle executing simple harmonic motion, what is the ratio of the kinetic energy at the equilibrium position to the potential energy when its displacement is equal to half the amplitude?
- 4/1
- 2/3
- 4/3
- 1/2
Correct answer: 4/1
Solution
KE at equilibrium = total energy = (1/2)kA^2. PE at x=A/2 = (1/2)k(A/2)^2 = (1/8)kA^2. Ratio = (1/2)/(1/8) = 4/1.
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