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(D) For a particle executing $S.H.M.$, the potential energy $(U)$ at displacement $x$ is given by $U = \frac{1}{2} k x^2$.The kinetic energy $(K)$ at

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$8$

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(D) For a particle executing $S.H.M.$, the potential energy $(U)$ at displacement $x$ is given by $U = \frac{1}{2} k x^2$.The kinetic energy $(K)$ at displacement $x$ is given by $K = \frac{1}{2} k (A^2 - x^2)$.Given that $U = nK$, we have $\frac{1}{2} k x^2 = n \cdot \frac{1}{2} k (A^2 - x^2)$.This simplifies to $x^2 = n(A^2 - x^2)$.Given $x = (\frac{2\sqrt{2}}{3})A$, we have $x^2 = (\frac{8}{9})A^2$.Substituting $x^2$ into the equation: $(\frac{8}{9})A^2 = n(A^2 - \frac{8}{9}A^2)$.$(\frac{8}{9})A^2 = n(\frac{1}{9}A^2)$.Solving for $n$, we get $n = 8$.

For a particle executing $S.H.M.$, the potential energy is $n$ times the kinetic energy when its displacement from mean position is $(\frac{2\sqrt{2}}{3})A$, where $A$ is the amplitude of $S.H.M.$ The value of $n$ is

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