A particle performing linear S.H.M. of amplitude 0.1 m has displacement 0.02 m and acceleration 0.5 m/s². The maximum velocity of the particle in m/s is (a) 0.05 (b) 0.50 (c) 0.01 (d) 0.25

Answer: (b). a = ω² x ⇒ ω² = (0.5)/(0.02) = 25, ω = 5 rad/s. vmax = Aω = 0.1× 5 = 0.5 m/s. So the answer is (b) 0.50.

A particle performing linear S.H.M. of amplitude 0.1 m has… | MHT CET 2021 (24 Sep, Shift 1)

Q59MHT CET 2021 · 24 SEP · SHIFT 1Medium

Velocity & Acceleration in SHM

A particle performing linear S.H.M. of amplitude $0.1\text{ m}$ has displacement $0.02\text{ m}$ and acceleration $0.5\text{ m/s}^2$ . The maximum velocity of the particle in $\text{m/s}$ is

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Answer: (b)

$a = \omega^2 x \Rightarrow \omega^2 = \frac{0.5}{0.02} = 25$ , $\omega = 5$ rad/s.
$v_{max} = A\omega = 0.1\times 5 = 0.5$ m/s.
So the answer is (b) 0.50.

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