A particle starts oscillating simple harmonically from its mean position with time period ' T '. At time t = T/12, the ratio of the potential energy to kinetic energy of the particle is (sin 30° = cos 60° = 0.5, cos 30° = sin 60° = √(3)/2) (a) 1: 2 (b) 3: 1 (c) 2: 1 (d) 1: 3

Answer: (d). At t = T/12, ω t = π/6 = 30°. U/K = tan ² 30° = ((0.5)²)/((√(3)/2)²) = 1/3 So the answer is (d) 1: 3.

A particle starts oscillating simple harmonically from its mean… | MHT CET 2022 (11 Aug, Shift 2)

Q48MHT CET 2022 · 11 AUG · SHIFT 2Medium

Energy in SHM

A particle starts oscillating simple harmonically from its mean position with time period ' $T$ '. At time $t = \frac{T}{12}$ , the ratio of the potential energy to kinetic energy of the particle is

$\left(\sin 30^\circ = \cos 60^\circ = 0.5, \cos 30^\circ = \sin 60^\circ = \sqrt{3}/2\right)$

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Answer: (d)

At $t = \frac{T}{12}$ , $\omega t = \frac{\pi}{6} = 30^\circ$ .
$\frac{U}{K} = \tan^2 30^\circ = \frac{(0.5)^2}{(\sqrt{3}/2)^2} = \frac{1}{3}$
So the answer is (d) $1 : 3$ .

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Energy in SHM question from MHT CET 2022 (11 Aug, Shift 2)

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