A particle undergoing simple harmonic motion has time dependent displacement given by  $\mathrm{x}\left(\mathrm{t}\right)=\mathrm{Asin}\frac{\mathrm{\pi t}}{90}.$ The ratio of kinetic to potential energy of this particle at time  $t=30 \mathrm{s}$ will be:

$$\begin{gathered} \mathrm{Let} \mathrm{displacement} \mathrm{equation} \mathrm{of} \mathrm{an} \mathrm{oscillator} \mathrm{be} \mathrm{y}=\mathrm{Asin\omega t}, \mathrm{here} \mathrm{\omega } \mathrm{is} \mathrm{angular} \mathrm{velocity}, \mathrm{A} \mathrm{is} \mathrm{Amplitude}. \\ \mathrm{differentiate} \mathrm{y} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{time} \mathrm{t}, \mathrm{to} \mathrm{get} \mathrm{velocity} \mathrm{v}=\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{A}\left(\mathrm{cos\omega t}\right)\mathrm{\omega } \\ \mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{PE}=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{y}}^2 \\ \mathrm{Substitute} \mathrm{displacement} \mathrm{and} \mathrm{velocity} \mathrm{in} \mathrm{PE}, \mathrm{KE} \mathrm{and} \mathrm{take} \mathrm{ratio}, \\ \frac{\mathrm{KE}}{\mathrm{PE}}=\frac{\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{A}}^2{\cos }^2\mathrm{\omega t}}{\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{A}}^2{\sin }^2\mathrm{\omega t}} \\ \frac{\mathrm{KE}}{\mathrm{PE}}={\cot }^2\mathrm{\omega t}, \\ \mathrm{by} \mathrm{the} \mathrm{question} \mathrm{\omega }=\frac{\mathrm{\pi }}{90} \\ \left(\text{at time\hspace{0.17em}\hspace{0.17em}t\hspace{0.17em}=\hspace{0.17em}30\hspace{0.17em}}\right)\Rightarrow \frac{\mathrm{KE}}{\mathrm{PE}}={\cot }^2\left(\frac{\mathrm{\pi }}{90}\times 30\right)={\cot }^2\left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{3} \end{gathered}$$