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MHT CET 2020 (16 Oct, Shift 2) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from MHT CET 2020 (16 Oct, Shift 2)

Q56MHT CET 2020 · 16 OCT · SHIFT 2Medium

Periodic Motion & SHM Basics

A particle performs simple harmonic motion with period of 3 s. The time taken by it to cover a distance equal to half the amplitude from mean position is [ $\sin 30^\circ = 0.5$ ]

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Answer: (a)

$\sin\omega t = \frac{1}{2} \Rightarrow \omega t = \frac{\pi}{6} \Rightarrow t = \frac{T}{12}$
$t = \frac{3}{12} = \frac{1}{4}\text{ s}$
So the answer is (a) $\frac{1}{4}\text{ s}$ .

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