A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = Asin (π t)/(90). The ratio of kinetic to potential energy of this particle at t = 210 s will be: (a) 1/9 (b) 3 (c) 2 (d) 1

Answer: (b). Phase at t = 210 s: (π × 210)/(90) = 7π/3, which is equivalent to π/3. K/U = (cos ²θ)/(sin ²θ) = cot ² 60° = 1/3 So K: U = 1: 3, which matches option (b) read as 1/3.

A particle undergoing simple harmonic motion has time dependent… | JEE Main 2019 (11 Jan, Shift 1)

Q69JEE Main 2019 · 11 JAN · SHIFT 1Medium

Energy in SHM

A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = Asin $\frac{\pi t}{90}$ . The ratio of kinetic to potential energy of this particle at t = 210 s will be:

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Answer: (b)

Phase at $t = 210$ s: $\frac{\pi \times 210}{90} = \frac{7\pi}{3}$ , which is equivalent to $\frac{\pi}{3}$ .
$\frac{K}{U} = \frac{\cos^2\theta}{\sin^2\theta} = \cot^2 60^\circ = \frac{1}{3}$
So $K : U = 1 : 3$ , which matches option (b) read as $\frac{1}{3}$ .

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