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(B) The kinetic energy $K$ of a particle performing $SHM$ is given by $K = \frac{1}{2} m \omega^{2} (A^{2} - y^{2})$.At the mean position,the displace

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$4$

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(B) The kinetic energy $K$ of a particle performing $SHM$ is given by $K = \frac{1}{2} m \omega^{2} (A^{2} - y^{2})$.At the mean position,the displacement $y = 0$.Therefore,the kinetic energy at the mean position is $K_{mean} = \frac{1}{2} m \omega^{2} A^{2}$.The potential energy $U$ is given by $U = \frac{1}{2} m \omega^{2} y^{2}$.At $y = A / 2$,the potential energy is $U = \frac{1}{2} m \omega^{2} (A / 2)^{2} = \frac{1}{2} m \omega^{2} (A^{2} / 4) = \frac{1}{8} m \omega^{2} A^{2}$.Taking the ratio of kinetic energy at the mean position to the potential energy at $y = A / 2$:$\frac{K_{mean}}{U} = \frac{\frac{1}{2} m \omega^{2} A^{2}}{\frac{1}{8} m \omega^{2} A^{2}} = \frac{1/2}{1/8} = \frac{8}{2} = 4$.Thus,the ratio is $4: 1$.

Column-$I$	Column-$II$
$(a)$ $K = 10\,J$	$(i)$ $U = 40\,J$
$(b)$ $K = 60\,J$	$(ii)$ $U = 90\,J$
	$(iii)$ $U = 50\,J$

What is the ratio of the kinetic energy at the mean position to the potential energy at $y = A / 2$ for a particle performing $SHM$ (in $: 1$)?

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