Starting from the mean position,a body oscillates simple harmonically with a period of $2\,s$. After what time will its kinetic energy be $75\%$ of the total energy?

$A$ particle undergoing simple harmonic motion has time-dependent displacement given by $x(t) = A \sin \left( \frac{\pi t}{90} \right)$. The ratio of kinetic energy to potential energy of the particle at $t = 210 \ s$ will be:

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. At time $t = \frac{T}{12}$,the ratio of its kinetic energy to potential energy is $\left[\sin \frac{\pi}{3} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \sin \frac{\pi}{6} = \cos \frac{\pi}{3} = \frac{1}{2}\right]$.

$A$ body performs $S.H.M.$ Its kinetic energy $K$ varies with time $t$ as indicated by which graph?

Write the coordinates of the points of intersection of the graph of kinetic energy $(K)$ and potential energy $(U)$ for a simple harmonic oscillator.

$A$ particle is executing linear simple harmonic motion of amplitude $A$. At what displacement is the energy of the particle half potential and half kinetic?