For a particle executing S.H.M., its potentia | vedclass

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(C) 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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$\frac{2\sqrt{2}}{3} A$

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(C) 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 $K = \frac{1}{2} k (A^2 - x^2)$, where $A$ is the amplitude.According to the problem, $U = 8K$.Substituting the expressions, we get $\frac{1}{2} k x^2 = 8 	imes \frac{1}{2} k (A^2 - x^2)$.Canceling $\frac{1}{2} k$ from both sides, we have $x^2 = 8(A^2 - x^2)$.$x^2 = 8A^2 - 8x^2$.$9x^2 = 8A^2$.$x^2 = \frac{8}{9} A^2$.Taking the square root of both sides, $x = \sqrt{\frac{8}{9}} A = \frac{2\sqrt{2}}{3} A$.

For a particle executing $S.H.M.$, its potential energy is $8$ times its kinetic energy at a certain displacement '$x$' from the mean position. If '$A$' is the amplitude of $S.H.M.$, the value of '$x$' is

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