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A particle undergoes simple harmonic motion with amplitude A. At what displacement from equilibrium is its kinetic energy equal to its potential energy?

1. A/3
2. A/2
3. A/sqrt(2)
4. A/(2*sqrt(2))

Correct answer: A/sqrt(2)

Solution

Total energy E = (1/2) k A². KE = PE means each equals E/2. PE = (1/2) k x² = E/2, so x² = A²/2, giving x = A/sqrt(2).

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