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JEE Main 2021 (18 Mar, Shift 1) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from JEE Main 2021 (18 Mar, Shift 1)

Q47JEE Main 2021 · 18 MAR · SHIFT 1Easy

Periodic Motion & SHM Basics

A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a}$ s. The value of 'a' to the nearest integer is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 6

From the mean position: $x = A\sin\frac{2\pi t}{T}$ with $T = 2\ \text{s}$ .
$\frac{A}{2} = A\sin(\pi t) \Rightarrow \pi t = \frac{\pi}{6}$ .
$t = \frac{1}{6}\ \text{s} = \frac{1}{a}\ \text{s}$ .
So $a = 6$ .

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