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MHT CET 2026 (18 Apr, Shift 1) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from MHT CET 2026 (18 Apr, Shift 1)

Q3MHT CET 2026 · 18 APR · SHIFT 1Easy

Velocity & Acceleration in SHM

A particle is performing simple harmonic motion about $x = 0$ with an amplitude ' $a$ ' and periodic time T. The speed of the particle at $x = \frac{a}{3}$ will be

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Answer: (c)

$v = \omega\sqrt{a^2 - x^2}$ with $\omega = \frac{2\pi}{T}$
$\sqrt{a^2 - \frac{a^2}{9}} = \frac{2\sqrt{2}a}{3}$
$v = \frac{2\pi}{T}\cdot\frac{2\sqrt{2}a}{3} = \frac{4\sqrt{2}\pi a}{3T}$
So the answer is (c) $\frac{4\sqrt{2}\pi a}{3T}$ .

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