JEE Main · PhysicsHard

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

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Correct answer: B — $A / 2$

Solution

The total energy is E = \frac{1}{2} k A^{2} . Setting potential energy U = \frac{1}{2} k x^{2} equal to kinetic energy K = E - U, we get U = E /2 . Thus, \frac{1}{2} k x^{2} = \frac{1}{4} k A^{2}, which simplifies to x = A / 2 .