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(D) For a particle executing $SHM$,the velocity $v$ at a displacement $y$ from the mean position is given by $v = \omega \sqrt{A^2 - y^2}$,where $A$ i

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$2$

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(D) For a particle executing $SHM$,the velocity $v$ at a displacement $y$ from the mean position is given by $v = \omega \sqrt{A^2 - y^2}$,where $A$ is the amplitude.At the mean position $(y = 0)$,the velocity is maximum,given by $v_{max} = A\omega$.Given $A = 4 \,cm$ and $v_{max} = 16 \,cm/s$,we have $16 = 4\omega$,which implies $\omega = 4 \,rad/s$.Now,we need to find the distance $y$ where $v = 8\sqrt{3} \,cm/s$.Using the formula $v = \omega \sqrt{A^2 - y^2}$:$8\sqrt{3} = 4 \sqrt{4^2 - y^2}$Divide both sides by $4$:$2\sqrt{3} = \sqrt{16 - y^2}$Square both sides:$(2\sqrt{3})^2 = 16 - y^2$$4 	imes 3 = 16 - y^2$$12 = 16 - y^2$$y^2 = 16 - 12 = 4$$y = 2 \,cm$.

The amplitude of a particle executing $SHM$ is $4 \,cm$. At the mean position,the speed of the particle is $16 \,cm/s$. The distance of the particle from the mean position at which the speed of the particle becomes $8\sqrt{3} \,cm/s$ will be .... $cm$.

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