MCQ
A particle executing simple harmonic motion with amplitude of $0.1 m$. At a certain instant when its displacement is $0.02 m$, its acceleration is 0.5 $m / s$. The maximum velocity of the particle is (in $m / s$ )
  • A
    0.01
  • B
    0.05
  • 0.5
  • D
    0.25

Answer

Correct option: C.
0.5
(c)$\begin{aligned}& \text { Acceleration } A=\omega^2 y \Rightarrow \omega=\sqrt{\frac{A}{y}}=\sqrt{\frac{0.5}{0.02}}=5 \\& \text { Maximum velocity } v_{\max }=a \omega=0.1 \times 5=0.5\end{aligned}$

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