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JEE Main 2021 (26 Feb, Shift 2) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from JEE Main 2021 (26 Feb, Shift 2)

Q52JEE Main 2021 · 26 FEB · SHIFT 2Easy

Velocity & Acceleration in SHM

A particle executes S.H.M. with amplitude 'a', and time period 'T'. The displacement of the particle when its speed is half of maximum speed is $\frac{\sqrt{x}a}{2}$ . The value of x is __________.

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

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Answer: 3

$v = \omega\sqrt{a^2 - y^2}$ , $v_{max} = \omega a$
$\sqrt{a^2 - y^2} = \frac{a}{2} \Rightarrow y = \frac{\sqrt{3}a}{2}$
Comparing with $\frac{\sqrt{x}a}{2}$ : $x = 3$

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