A particle is vibrating in S.H.M. with an amplitude of \(4 \mathrm{~cm}\). At what displacement from the equilibrium position is its energy half potential and half kinetic?

\(\begin{aligned} \text { K.E } & =\frac{1}{2} m \omega^2\left(\mathrm{~A}^2-\mathrm{x}^2\right) \text { and } \\ \text { P.E } & =\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2 \\ \therefore \quad \mathrm{T} . \mathrm{E} & =\mathrm{K} \cdot \mathrm{E}+\mathrm{P} \cdot \mathrm{E} \\ & =\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{~A}^2-\mathrm{x}^2\right)+\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2 \\ & =\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2 \\ & \text { Given: } \mathrm{K} \cdot \mathrm{E}=\mathrm{P} \cdot \mathrm{E}, \\ \therefore \quad \mathrm{A}^2 & -\mathrm{x}^2=\mathrm{x}^2 \\ \therefore \quad \mathrm{A}^2 & =2 \mathrm{x}^2 \\ \therefore \quad \mathrm{X} & =\sqrt{\frac{\mathrm{A}^2}{2}}=\frac{\mathrm{A}}{\sqrt{2}} \\ \therefore \quad \mathrm{x} & =\frac{4}{\sqrt{2}}=2 \sqrt{2} \mathrm{~cm}\end{aligned}\)