A particle starts oscillating simple harmonically from its mean position with time period ' T '. At time t = T/6, 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) 1: 3 (c) 2: 1 (d) 3: 1

Answer: (d). x = Asin ω t; at t = T/6, ω t = 2π/T/6 = π/3 = 60°. U ∝ sin ²ω t and K ∝ cos ²ω t, so U/K = tan ² 60° = 3. So the answer is (d) 3: 1.

A particle starts oscillating simple harmonically from its mean… | MHT CET 2025 (25 Apr, Shift 2)

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Energy in SHM

A particle starts oscillating simple harmonically from its mean position with time period ' $T$ '. At time $t = \frac{T}{6}$ , 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)

$x = A\sin\omega t$ ; at $t = \frac{T}{6}$ , $\omega t = \frac{2\pi}{T}\cdot\frac{T}{6} = \frac{\pi}{3} = 60^\circ$ .
$U \propto \sin^2\omega t$ and $K \propto \cos^2\omega t$ , so $\frac{U}{K} = \tan^2 60^\circ = 3$ .
So the answer is (d) $3 : 1$ .

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