A particle is performing S.H.M. with an amplitude 4 cm . At the mean position the velocity of the particle…

A particle is performing S.H.M. with an amplitude 4 cm . At the mean position the velocity of the particle is $12 \mathrm{~cm} / \mathrm{s}$. When the speed of the particle becomes $6 \mathrm{~cm} / \mathrm{s}$, the distance of the particle from mean position is
  1. $\sqrt{3} \mathrm{~cm}$
  2. $\sqrt{6} \mathrm{~cm}$
  3. $2 \sqrt{3} \mathrm{~cm}$
  4. $2 \sqrt{6} \mathrm{~cm}$

Solution

At mean position, $\mathrm{v}_{\max }=\mathrm{A} \omega$ $\begin{array}{ll} \therefore & \omega=\frac{v_{\max }}{A}=\frac{12}{4}=3 \mathrm{rad} / \mathrm{s} \\ & \text { Now, } v=\omega \sqrt{A^2-\mathrm{x}^2} \\ \therefore & \mathrm{v}^2=\omega^2\left(\mathrm{~A}^2-\mathrm{x}^2\right) \\ \therefore & x^2=A^2-\frac{v^2}{\omega^2} \\ \therefore & x=\sqrt{16-\frac{36}{9}}=\sqrt{12}=2 \sqrt{3} \mathrm{~cm} \end{array}$

Asked in: MHT CET 2024 (09 May Shift 1)

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