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(C) The total energy $(E)$ of a particle performing $S.H.M.$ is given by $E = \frac{1}{2} k A^2$.The kinetic energy $(K)$ at a displacement $x$ is giv

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$4$

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(C) The total energy $(E)$ of a particle performing $S.H.M.$ is given by $E = \frac{1}{2} k A^2$.The kinetic energy $(K)$ at a displacement $x$ is given by $K = \frac{1}{2} k (A^2 - x^2)$.Given that the displacement $x = \frac{\sqrt{3}}{2} A$.Substituting the value of $x$ into the kinetic energy formula:$K = \frac{1}{2} k (A^2 - (\frac{\sqrt{3}}{2} A)^2) = \frac{1}{2} k (A^2 - \frac{3}{4} A^2) = \frac{1}{2} k (\frac{1}{4} A^2) = \frac{1}{8} k A^2$.According to the problem,$E = n 	imes K$.Substituting the expressions for $E$ and $K$:$\frac{1}{2} k A^2 = n 	imes (\frac{1}{8} k A^2)$.Dividing both sides by $\frac{1}{2} k A^2$:$1 = n 	imes \frac{1}{4}$.Therefore,$n = 4$.

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

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