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(C) Given: Amplitude $A = 2 \,cm = 2 	imes 10^{-2} \,m$,displacement $x = 1 \,cm = 1 	imes 10^{-2} \,m$.The magnitude of acceleration in $S.H.M.$ is

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

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(C) Given: Amplitude $A = 2 \,cm = 2 	imes 10^{-2} \,m$,displacement $x = 1 \,cm = 1 	imes 10^{-2} \,m$.The magnitude of acceleration in $S.H.M.$ is given by $|a| = \omega^2 x$.The magnitude of velocity in $S.H.M.$ is given by $|v| = \omega \sqrt{A^2 - x^2}$.According to the problem,$|a| = |v|$,so:$\omega^2 x = \omega \sqrt{A^2 - x^2}$Dividing both sides by $\omega$ (assuming $\omega 
eq 0$):$\omega x = \sqrt{A^2 - x^2}$Substituting the values:$\omega (1 	imes 10^{-2}) = \sqrt{(2 	imes 10^{-2})^2 - (1 	imes 10^{-2})^2}$$\omega (1 	imes 10^{-2}) = \sqrt{(4 - 1) 	imes 10^{-4}}$$\omega (1 	imes 10^{-2}) = \sqrt{3} 	imes 10^{-2}$$\omega = \sqrt{3} \,rad/s$.The time period $T$ is given by:$T = \frac{2 \pi}{\omega} = \frac{2 \pi}{\sqrt{3}} \,s$.

$A$ particle executes $S.H.M.$ along a straight line. The amplitude of oscillation is $2 \,cm$. When the displacement of the particle from the mean position is $1 \,cm$,the magnitude of its acceleration is equal to the magnitude of its velocity. The time period of oscillation is ........

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