The displacement of a particle in simple harmonic motion $(SHM)$ is given by $y = \sqrt{3 \pi} \sin \left(\frac{100}{\pi} t + \frac{\pi}{4}\right)$. What will be the displacement of the particle from the mean position when its kinetic energy is eight times that of its potential energy?

Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies $\omega_1$ and $\omega_2$ and have total energies $E_1$ and $E_2$,respectively. The variations of their momenta $p$ with positions $x$ are shown in the figures. If $\frac{a}{b}= n^2$ and $\frac{a}{R}= n$,then the correct equation$(s)$ is(are):
$(A) E_1 \omega_1 = E_2 \omega_2$
$(B) \frac{\omega_2}{\omega_1} = n^2$
$(C) \omega_1 \omega_2 = n^2$
$(D) \frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$

The kinetic energy and the potential energy of a particle executing $S.H.M.$ are equal. The ratio of its displacement and amplitude will be

Show that for a particle in linear $SHM$,the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

$A$ particle starts its oscillation from the equilibrium position with time period $T$. Find the ratio of kinetic energy to potential energy of the particle at time $t = \frac{T}{6}$.

$A$ particle of mass $m$ oscillates with simple harmonic motion between points $X_1$ and $X_2$,the equilibrium position being $O$. Its potential energy will be as shown in the following graph: