A point particle of mass 200 gram is executing S.H.M. of amplitude 0.2 m. When the particle passes through the mean position, its kinetic energy is 16 × 10⁻³ J. The equation of motion of this particle is (Initial phase of oscillation = 0°) (a) Y = 0.2sin (4t) (b) Y = 0.2sin (t/4) (c) Y = 0.2sin (t/2) (d) Y = 0.2sin (2t)

Answer: (d). KE at mean position = total energy: 1/2mω²A² = 16 × 10⁻³ J. 1/2(0.2)ω²(0.2)² = 0.004 ω² = 0.016 ⇒ ω = 2 rad/s Y = Asin (ω t) = 0.2sin (2t) So the answer is (d) Y = 0.2sin (2t).

A point particle of mass 200 gram is executing S.H.M. of amplitude… | MHT CET 2025 (5 May, Shift 2)

Q21MHT CET 2025 · 5 MAY · SHIFT 2Medium

Energy in SHM

A point particle of mass 200 gram is executing S.H.M. of amplitude 0.2 m. When the particle passes through the mean position, its kinetic energy is $16 \times 10^{-3}\text{ J}$ . The equation of motion of this particle is (Initial phase of oscillation $= 0^\circ$ )

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Answer: (d)

KE at mean position = total energy: $\frac{1}{2}m\omega^2A^2 = 16 \times 10^{-3}$ J.
$\frac{1}{2}(0.2)\omega^2(0.2)^2 = 0.004\,\omega^2 = 0.016 \Rightarrow \omega = 2$ rad/s
$Y = A\sin(\omega t) = 0.2\sin(2t)$
So the answer is (d) $Y = 0.2\sin(2t)$ .

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Energy in SHM question from MHT CET 2025 (5 May, Shift 2)

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