$A$ bob of a simple pendulum of mass $m$ performs $SHM$ with amplitude $A$ and period $T$. The kinetic energy of the pendulum at displacement $x = \frac{A}{2}$ will be:

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 total energy of a particle executing $S.H.M.$ is $80 \, J$. What is the potential energy when the particle is at a distance of $\frac{3}{4}$ of amplitude from the mean position?

The ratio between kinetic and potential energies of a body executing simple harmonic motion,when it is at a distance of $\frac{1}{N}$ of its amplitude from the mean position is

$A$ body is executing a linear $S.H.M.$ Its potential energies at the displacements $x$ and $y$ are $E_1$ and $E_2$ respectively. Its potential energy at displacement $(x+y)$ will be

The frequency at which kinetic energy changes into potential energy in a simple harmonic motion ($S$.$H$.$M$.) with frequency $f$ is: