The variations of kinetic energy $K(x)$,potential energy $U(x)$,and total energy $E$ as a function of displacement $x$ of a particle in $SHM$ are as shown in the figure. The value of $|x_0|$ 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):
$(A) E_1 \omega_1 = E_2 \omega_2$
$(B) \frac{\omega_2}{\omega_1} = n^2$
$(C) \omega_1 \omega_2 = n^2$
$(D) \frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$

The displacement of a particle executing $SHM$ is given by $y = 5 \sin \left(4t + \frac{\pi}{3}\right)$. If $T$ is the time period and the mass of the particle is $2 \text{ g}$, the kinetic energy of the particle when $t = \frac{T}{4}$ is given by (in $\text{ J}$)

$A$ particle of mass $m$ is executing $S.H.M.$ about the origin on the $x$-axis with frequency $f = \frac{\sqrt{Ka}}{\pi m}$, where $K$ is a constant and $a$ is the amplitude of $S.H.M.$ If $x$ is the displacement of the particle at time $t$, the potential energy of the particle will be:

$A$ particle is vibrating in $S.H.M.$ with an amplitude of $4 \,cm$. At what displacement from the equilibrium position is its energy half potential and half kinetic?

The potential energy of a particle executing $S.H.M.$ is $2.5 \ J$,when its displacement is half of its amplitude. The total energy of the particle is .... $J$.