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(C) The velocity of a particle in $SHM$ is given by $v = \omega \sqrt{A^2 - x^2}$.The acceleration of a particle in $SHM$ is given by $a = \omega^2 x$

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(C) The velocity of a particle in $SHM$ is given by $v = \omega \sqrt{A^2 - x^2}$.The acceleration of a particle in $SHM$ is given by $a = \omega^2 x$.Given that the magnitude of acceleration equals the magnitude of velocity:$\omega^2 x = \omega \sqrt{A^2 - x^2}$$\omega x = \sqrt{A^2 - x^2}$ (Equation $i$)Given the time period $T = \frac{2 \pi}{\sqrt{3}} \,s$,we find the angular frequency:$\omega = \frac{2 \pi}{T} = \sqrt{3} \,rad/s$.Substituting $\omega$ into Equation $i$:$\sqrt{3} x = \sqrt{A^2 - x^2}$Squaring both sides:$3x^2 = A^2 - x^2$$4x^2 = A^2 \Rightarrow A = 2x$.The path length is $4 \,cm$,so the amplitude $A = \frac{	ext{path length}}{2} = 2 \,cm$.Substituting $A = 2 \,cm$ into $A = 2x$:$2 = 2x \Rightarrow x = 1 \,cm$.

$A$ particle performing $SHM$ has a time period of $\frac{2 \pi}{\sqrt{3}} \,s$ and a path length of $4 \,cm$. The displacement from the mean position at which the magnitude of acceleration is equal to the magnitude of velocity is (in $\,cm$)

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