A particle starts from mean position and performs S.H.M. with period 4 second. At what time its kinetic energy is \(50 \%\) of total energy?

The total energy is given by T.E \(=\frac{1}{2} \mathrm{kA}^2\) Kinetic energy is given by K.E \(=\frac{1}{2} k\left(A^2-x^2\right)\)
\(
\begin{array}{ll}
& \text { K.E }=\frac{1}{2} P . E \\
\therefore \quad & \frac{1}{2} k\left(A^2-x^2\right)=\frac{1}{2}\left(\frac{1}{2} \mathrm{kA}^2\right) \\
\therefore \quad & \left(\mathrm{A}^2-\mathrm{x}^2\right)=\frac{1}{2} \mathrm{~A}^2 \\
\therefore \quad & \left(\mathrm{A}^2-\mathrm{x}^2\right)=\frac{1}{2} \mathrm{~A}^2 \\
\therefore \quad & \mathrm{x}=\frac{\mathrm{A}}{\sqrt{2}}
\end{array}
\)
The equation of displacement in SHM is
\(
\begin{array}{ll}
& \mathrm{x}=\mathrm{A} \sin \frac{2 \pi \mathrm{t}}{\mathrm{T}} \\
\therefore \quad & \frac{\mathrm{A}}{\sqrt{2}}=\mathrm{A} \sin \frac{2 \pi \mathrm{t}}{4} \\
\therefore \quad \frac{1}{\sqrt{2}}=\sin \frac{\pi \mathrm{t}}{2} & \quad \ldots .(\because \mathrm{T}=4 \mathrm{sec}) \\
\therefore \quad \frac{\pi \mathrm{t}}{2} & =\sin ^{-1} \frac{1}{\sqrt{2}} \\
& \frac{\pi \mathrm{t}}{2}=\frac{\pi}{4} \\
\therefore \quad & \mathrm{t}=0.5 \mathrm{~s}
\end{array}
\)