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JEE Main 2019 (10 Jan, Shift 2) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from JEE Main 2019 (10 Jan, Shift 2)

Q64JEE Main 2019 · 10 JAN · SHIFT 2Medium

Velocity & Acceleration in SHM

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is -

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Answer: (d)

$\omega\sqrt{A^2 - x^2} = \omega^2 x$ with $A = 5$ , $x = 4$ (same units cancel).
$\sqrt{25 - 16} = 4\omega \Rightarrow \omega = \frac{3}{4}$ rad/s.
$T = \frac{2\pi}{\omega} = \frac{8\pi}{3}$ s.
So the answer is (d) $\frac{8\pi}{3}$ .

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