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MHT CET 2024 (16 May, Shift 1) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from MHT CET 2024 (16 May, Shift 1)

Q32MHT CET 2024 · 16 MAY · SHIFT 1Medium

Velocity & Acceleration in SHM

A particle executing S.H.M. has velocities ' $V_1$ ' and ' $V_2$ ' at distances ' $x_1$ ' and ' $x_2$ ' respectively, from the mean position. Its frequency is

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Answer: (c)

$V_1^2 = \omega^2(A^2 - x_1^2)$ , $V_2^2 = \omega^2(A^2 - x_2^2)$ .
Subtract: $\omega = \sqrt{\frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$ .
$f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$ .
So the answer is (c).

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