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JEE Main 2023 (12 Apr, Shift 1) · Energy in SHM

Energy in SHM question from JEE Main 2023 (12 Apr, Shift 1)

Q39JEE Main 2023 · 12 APR · SHIFT 1Easy

Energy in SHM

A particle is executing simple harmonic motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

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Answer: (d)

$PE = \frac{1}{2}kx^2$ , $KE = \frac{1}{2}k(A^2 - x^2)$ , so $\frac{PE}{KE} = \frac{x^2}{A^2 - x^2}$ .
At $x = A/2$ : $\frac{PE}{KE} = \frac{A^2/4}{3A^2/4} = \frac{1}{3}$
So the answer is (d) 1 : 3.

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