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?

If a particle repeats its motion after a fixed time interval of $8 \,s$,then after how much time will its maximum value of $PE$ be attained after attaining its minimum value?

$A$ 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:

The position of a particle executing simple harmonic motion is given by $x(t) = 2 \cos \left(\frac{\pi}{15} t - \frac{\pi}{2}\right)$, where $x$ is in centimetre and $t$ is in seconds. The time period of the kinetic energy of the particle in seconds is

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):
$(A) E_1 \omega_1 = E_2 \omega_2$
$(B) \frac{\omega_2}{\omega_1} = n^2$
$(C) \omega_1 \omega_2 = n^2$
$(D) \frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$

If $< E >$ and $< U >$ denote the average kinetic and the average potential energies respectively of a mass describing a simple harmonic motion,over one period,then the correct relation is: