$A$ body of mass $1\,kg$ is executing simple harmonic motion. Its displacement $y$ (in $cm$) at $t$ seconds is given by $y = 6\sin(100t + \pi/4)$. Its maximum kinetic energy is ..... $J$.

As a body performs $S.H.M.$,its potential energy $U$ varies with time $t$ as indicated in:

The quantity which does not vary periodically for a particle performing $S.H.M.$ is

The variations of potential energy $(U)$ with position $x$ for three simple harmonic oscillators $A, B$ and $C$ are shown in the figure. The oscillators have the same mass. The time period of oscillation is greatest for

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 displacement of a particle of mass $2 \text{ g}$ executing simple harmonic motion is $x = 8 \cos \left(50 t + \frac{\pi}{12}\right) \text{ m}$,where $t$ is time in seconds. The maximum kinetic energy of the particle is (in $\text{ J}$)