A particle starts oscillating simple harmonically form its mean position with time period \(T\). At time \(t=\frac{T}{12}\), the ratio of the potential energy to kinetic energy of the particle is \(\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right)\)

Let, the equation of the particle performing SHM is given by \(x=A \sin \omega t\), where \(\omega=\frac{2 \pi}{T}\)
At \(t=\frac{T}{12}, x=A \sin \left(\frac{2 \pi}{T} \cdot \frac{T}{12}\right)=A \sin \left(\frac{\pi}{6}\right)=\frac{A}{2}\)
The potential energy of the particle at \(T=\frac{T}{12}\)

The kinetic energy of the particle at \(t=\frac{T}{12}\)
\(\begin{aligned} & K=\frac{1}{2} m \omega^2 x^2\left(A^2-x^2\right)=\frac{1}{2} m \omega^2\left[A^2-\left(\frac{A}{2}\right)^2\right] \\ & K=\frac{1}{2} m \omega^2\left(\frac{3 A^2}{4}\right)---(2)\end{aligned}\)
From equation (1) and (2)
\(\frac{K}{U}=\frac{3}{1}\)