A particle executes linear S.H.M. with amplit | vedclass

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(B) Given: Amplitude $A = 4 \ cm$,Displacement $x = 3 \ cm$.Velocity in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$.Acceleration in $S.H.M.$ is

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$\frac{6 \pi}{\sqrt{7}} \ s$

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(B) Given: Amplitude $A = 4 \ cm$,Displacement $x = 3 \ cm$.Velocity in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$.Acceleration in $S.H.M.$ is given by $a = \omega^2 x$.According to the problem,the magnitude of velocity equals the magnitude of acceleration: $|v| = |a|$.$\omega \sqrt{A^2 - x^2} = \omega^2 x$.$\sqrt{A^2 - x^2} = \omega x$.Substitute the values: $\sqrt{4^2 - 3^2} = \omega (3)$.$\sqrt{16 - 9} = 3 \omega$.$\sqrt{7} = 3 \omega$.$\omega = \frac{\sqrt{7}}{3} \ rad/s$.The time period $T$ is given by $T = \frac{2 \pi}{\omega}$.$T = \frac{2 \pi}{\frac{\sqrt{7}}{3}} = \frac{6 \pi}{\sqrt{7}} \ s$.

$A$ particle executes linear $S.H.M.$ with amplitude $4 \ cm$. The magnitude of velocity and acceleration is equal when it is at $3 \ cm$ from the mean position. The time period of oscillation is:

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