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)

The total mechanical energy of a particle executing simple harmonic motion is $E$. When the displacement is half the amplitude,its kinetic energy will be

$A$ body performing simple harmonic motion has potential energy $P_{1}$ at displacement $x_{1}$. Its potential energy is $P_{2}$ at displacement $x_{2}$. The potential energy $P$ at displacement $(x_{1}+x_{2})$ 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

When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation,the displacement of the particle from the equilibrium position in terms of its amplitude $a$ 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):
$(A) E_1 \omega_1 = E_2 \omega_2$
$(B) \frac{\omega_2}{\omega_1} = n^2$
$(C) \omega_1 \omega_2 = n^2$
$(D) \frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$