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

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$\pm \frac{A}{\sqrt{2}}$

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(D) The potential energy $(U)$ of a particle executing $S.H.M.$ at a displacement $x$ is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant.The maximum potential energy $(U_{max})$ occurs at the extreme positions where $x = \pm A$,so $U_{max} = \frac{1}{2} k A^2$.According to the problem,the potential energy is half of its maximum value: $U = \frac{1}{2} U_{max}$.Substituting the expressions: $\frac{1}{2} k x^2 = \frac{1}{2} (\frac{1}{2} k A^2)$.Simplifying the equation: $x^2 = \frac{1}{2} A^2$.Taking the square root on both sides: $x = \pm \frac{A}{\sqrt{2}}$.Thus,the displacement is $\pm \frac{A}{\sqrt{2}}$.

$A$ particle is executing $S.H.M.$ of amplitude $A$. When the potential energy of the particle is half of its maximum value during the oscillation,its displacement from the equilibrium position is

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