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MHT CET 2021 (23 Sep, Shift 1) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from MHT CET 2021 (23 Sep, Shift 1)

Q51MHT CET 2021 · 23 SEP · SHIFT 1Easy

Velocity & Acceleration in SHM

A particle executing S.H.M starts from the mean position. Its amplitude is 'A' and time period ' $T$ '. At what displacement its speed is one-fourth of the maximum speed?

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Answer: (d)

$v = \omega\sqrt{A^2 - x^2}$ , $v_{max} = \omega A$
$\sqrt{A^2 - x^2} = \frac{A}{4} \Rightarrow x^2 = \frac{15A^2}{16}$
$x = \frac{A\sqrt{15}}{4}$
So the answer is (d) $\frac{A\sqrt{15}}{4}$ .

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