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

Correct answer: (B) 0.50. Acceleration a=ω^2 x ∴ ω^2=a/x=0.5/0.02=25 ∴ ω=5 rad / s V_max =A ω=0.1 × 5=0.5 ~m / s

Q116 / 150 · MHT CET 2021 (23 Sep Shift 1) · Physics

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

A
0.05
Correct answer
B
0.50
C
0.01
D
0.25

Solution

Acceleration $\mathrm{a}=\omega^2 \mathrm{x}$ $$ \begin{aligned} & \therefore \omega^2=\frac{\mathrm{a}}{\mathrm{x}}=\frac{0.5}{0.02}=25 \\ & \therefore \omega=5 \mathrm{rad} / \mathrm{s} \\ & \mathrm{V}_{\max }=\mathrm{A} \omega=0.1 \times 5=0.5 \mathrm{~m} / \mathrm{s} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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