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MHT CET 2024 (11 May, Shift 2) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from MHT CET 2024 (11 May, Shift 2)

Q21MHT CET 2024 · 11 MAY · SHIFT 2Easy

Velocity & Acceleration in SHM

A particle is performing S.H.M. about its mean position with an amplitude ' $a$ ' and periodic time ' $T$ '. The speed of the particle when its displacement from mean position is $\frac{a}{3}$ will be

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Answer: (b)

$v = \omega\sqrt{a^2 - x^2}$ 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 (b) $\frac{4\sqrt{2}\pi a}{3T}$ .

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