Physics - SHM Question with Solution | TestHub

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 $t=\frac{T}{12}$ $\begin{array}{rl}& K=\frac{1}{2}m{\omega }^2{x}^2\left(A^2-x^2\right)=\frac{1}{2}m{\omega }^2\left[A^2-{\left(\frac{A}{2}\right)}^2\right]\\ & K=\frac{1}{2}m{\omega }^2\left(\frac{3A^2}{4}\right)---(2)\end{array}$ From equation (1) and (2) $\frac{K}{U}=\frac{3}{1}$