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MHT CET 2025 (23 Apr, Shift 1) · Energy in SHM

Energy in SHM question from MHT CET 2025 (23 Apr, Shift 1)

Q20MHT CET 2025 · 23 APR · SHIFT 1Easy

Energy in SHM

A particle is executing linear S.H.M. starting from mean position. The ratio of the kinetic energy to the potential energy of the particle at a point of half the amplitude is

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Answer: (b)

$\dfrac{KE}{PE} = \dfrac{A^2 - x^2}{x^2}$
At $x = A/2$ : $\dfrac{A^2 - A^2/4}{A^2/4} = \dfrac{3/4}{1/4} = 3$
So the answer is (b) $3 : 1$ .

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