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(C) The potential energy $(P.E.)$ of a particle performing linear $S.H.M.$ is given by $P.E. = \frac{1}{2} k x^2$, where $k = m \omega^2$ is the force

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$r\sqrt{\lambda}$

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(C) The potential energy $(P.E.)$ of a particle performing linear $S.H.M.$ is given by $P.E. = \frac{1}{2} k x^2$, where $k = m \omega^2$ is the force constant and $x$ is the displacement.The total energy $(T.E.)$ of the particle is given by $T.E. = \frac{1}{2} k r^2$, where $r$ is the amplitude.According to the problem, $P.E. = \lambda 	imes T.E.$Substituting the expressions, we get $\frac{1}{2} k x^2 = \lambda 	imes (\frac{1}{2} k r^2)$.Canceling $\frac{1}{2} k$ from both sides, we get $x^2 = \lambda r^2$.Taking the square root of both sides, we find $x = r\sqrt{\lambda}$.

For a particle performing linear $S.H.M.$ of amplitude '$r$',the potential energy is '$\lambda$' times its total energy. The displacement of the particle is

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