A particle starts oscillating simple harmonically from its equilibrium position with time period 'T'. What is the ratio of potential energy to kinetic energy of the particle at time \(t=\frac{T}{12}\) ?
\(\left(\sin \left(\frac{\pi}{6}\right)=\frac{1}{2}\right)\)

For SHM,
\(\begin{aligned}
& \text { P.E. }=\frac{1}{2} m \omega^2 x^2 ...(i)\\
& \text { and K.E. }=\frac{1}{2} m \omega^2\left(A^2-x^2\right)...(ii)
\end{aligned}\)
Given \(\mathrm{t}=\frac{\mathrm{T}}{12}\)
\(\begin{aligned}
\therefore \quad x & =A \sin \omega t=A \sin \frac{2 \pi}{T} \cdot t \\
\therefore \quad x & =A \sin \frac{2 \pi}{T} \times \frac{T}{12} \\
& =A \sin \left(\frac{\pi}{6}\right) \\
& =\frac{A}{2}...(iii)
\end{aligned}\)
Substituting equation (iii) in equation (i),
\(\begin{aligned}
\text { P.E. } & =\frac{1}{2} \mathrm{~m} \omega^2\left(\frac{\mathrm{~A}^2}{4}\right) \\
& =\frac{\mathrm{m} \omega^2 \mathrm{~A}^2}{8}
\end{aligned}\)
Substituting equation (iii) in equation (ii),
\(\begin{aligned}
\text { K.E. } & =\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{~A}^2-\frac{\mathrm{A}^2}{4}\right) \\
& =\frac{1}{2} \mathrm{~m} \omega^2\left(\frac{3 \mathrm{~A}^2}{4}\right) \\
\therefore \quad \frac{\text { P.E. }}{\text { K.E. }} & =\frac{\frac{1}{2} m \omega^2 A^2 \times \frac{1}{4}}{\frac{1}{2} m \omega^2 A^2 \times \frac{3}{4}}=\frac{1}{3}
\end{aligned}\)