A particle performing linear S.H.M. of amplitude $0.1 \mathrm{~m}$ has displacement $0.02 \mathrm{~m}$ and…

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
  1. 0.05
  2. 0.50
  3. 0.01
  4. 0.25

Solution

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

Asked in: MHT CET 2021 (23 Sep Shift 1)

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