The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where $E$ is the total energy)

For a particle executing $S.H.M.$, its potential energy is $8$ times its kinetic energy at a certain displacement '$x$' from the mean position. If '$A$' is the amplitude of $S.H.M.$, the value of '$x$' 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):
$(A) E_1 \omega_1 = E_2 \omega_2$
$(B) \frac{\omega_2}{\omega_1} = n^2$
$(C) \omega_1 \omega_2 = n^2$
$(D) \frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$

$A$ particle of mass $m$ is hanging vertically by an ideal spring of force constant $K$. If the mass is made to oscillate vertically,its total energy is

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?

$A$ spring-mass system performs $S.H.M.$ If the mass is doubled keeping the amplitude same,then the total energy of $S.H.M.$ will become: