$A$ particle of mass $m$ is executing simple harmonic motion about its mean position. If $A$ is the amplitude and $T$ is the period of $S$.$H$.$M$.,then the total energy of the particle is

The potential energy of a particle $(U_x)$ executing $S.H.M.$ is given by

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 amplitude of a particle executing $SHM$ is made three-fourth keeping its time period constant. Its total energy 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 executes simple harmonic motion along a straight line with an amplitude $A$. The potential energy is maximum when the displacement is