For a particle executing S.H.M. having amplitude A, the speed of the article is…

For a particle executing S.H.M. having amplitude A, the speed of the article is $\left(\frac{1}{3}\right)^{\text {rd }}$ of its maximum speed when the displacement from the mean position is
  1. $\frac{3 \mathrm{~A}}{\sqrt{2}}$
  2. $\frac{2 \mathrm{~A}}{3}$
  3. $\frac{2 \sqrt{2}}{3} \mathrm{~A}$
  4. $\frac{\sqrt{2}}{3} \mathrm{~A}$

Solution

$\begin{aligned} & V=\omega \sqrt{A^2-x^2}\\ & V_{\max }=\omega A \\ & V=\frac{V_{\max }}{3}=\frac{\omega a}{3} \\ & \frac{\omega A}{3}=\omega \sqrt{A^2-x^2} \\ & \frac{A^2}{9}=\left(A^2-x^2\right) \\ & x^2=A^2-\frac{A^2}{9} \\ & x=\sqrt{\frac{8}{9} A^2} \\ & x=\frac{2 \sqrt{2}}{3} \cdot A \end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 2)

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