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(C) The maximum speed of a particle in $S.H.M.$ is given by $v_{\max} = \omega A$.The speed $v$ at any displacement $y$ is given by $v = \omega \sqrt{

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$\frac{A\sqrt{3}}{2}$

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(C) The maximum speed of a particle in $S.H.M.$ is given by $v_{\max} = \omega A$.The speed $v$ at any displacement $y$ is given by $v = \omega \sqrt{A^2 - y^2}$.According to the problem,the speed is half of the maximum speed,so $v = \frac{v_{\max}}{2} = \frac{\omega A}{2}$.Equating the two expressions for speed:$\frac{\omega A}{2} = \omega \sqrt{A^2 - y^2}$Squaring both sides:$\frac{A^2}{4} = A^2 - y^2$Rearranging to solve for $y^2$:$y^2 = A^2 - \frac{A^2}{4} = \frac{3A^2}{4}$Taking the square root:$y = \frac{\sqrt{3}A}{2}$

$A$ particle starts $S.H.M.$ from the mean position as shown in the figure. Its amplitude is $A$ and its time period is $T$. At a certain instant,its speed is half of its maximum speed. What is its displacement at this instant?

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