The differential equation of a particle performing S.H.M. is (d²x)/(dt²) + 64x = 0, where x is in meter. The period of oscillation of the particle is (in second) (a) π/3 (b) π/4 (c) π/5 (d) π/8

Answer: (b). Compare with (d²x)/(dt²) + ω² x = 0: ω² = 64, ω = 8 rad/s. T = (2π)/(ω) = 2π/8 = π/4 s. So the answer is (b) π/4.

The differential equation of a particle performing S.H.M. is… | MHT CET 2026 (19 May, Shift 1)

Q1MHT CET 2026 · 19 MAY · SHIFT 1Easy

Periodic Motion & SHM Basics

The differential equation of a particle performing S.H.M. is $\frac{d^2x}{dt^2} + 64x = 0$ , where $x$ is in meter. The period of oscillation of the particle is (in second)

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Answer: (b)

Compare with $\frac{d^2x}{dt^2} + \omega^2 x = 0$ : $\omega^2 = 64$ , $\omega = 8\ \text{rad/s}$ .
$T = \frac{2\pi}{\omega} = \frac{2\pi}{8} = \frac{\pi}{4}\ \text{s}$ .
So the answer is (b) $\frac{\pi}{4}$ .

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