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MHT CET 2026 (18 Apr, Shift 2) · Energy in SHM

Energy in SHM question from MHT CET 2026 (18 Apr, Shift 2)

Q13MHT CET 2026 · 18 APR · SHIFT 2Medium

Energy in SHM

A particle of mass ' $m$ ' is executing S.H.M. about the origin on x-axis with frequency $\sqrt{\frac{Ka}{\pi m}}$ , where K is a constant and a is the amplitude of S.H.M. If ' $x$ ' is the displacement of a particle at time ' $t$ ', the potential energy of a particle will be

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Answer: (c)

$\omega = 2\pi f \Rightarrow \omega^2 = 4\pi^2 \cdot \frac{Ka}{\pi m} = \frac{4\pi Ka}{m}$
$U = \frac{1}{2}m\omega^2x^2 = \frac{1}{2}m \cdot \frac{4\pi Ka}{m}x^2$
$U = 2\pi Kax^2$
So the answer is (c) $2\pi Kax^2$ .

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