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MHT CET 2025 (5 May, Shift 2) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from MHT CET 2025 (5 May, Shift 2)

Q14MHT CET 2025 · 5 MAY · SHIFT 2Easy

Velocity & Acceleration in SHM

A simple pendulum starts oscillating simple harmonically from its mean position ( $x = 0$ ) with amplitude ' $a$ ' and periodic time ' $T$ '. The magnitude of velocity of pendulum at $x = \frac{a}{2}$ is

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Answer: (d)

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

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