An object of mass $0.5\, \text{kg}$ is executing simple harmonic motion. Its amplitude is $5\, \text{cm}$ and time period $T$ is $0.2\, \text{s}$. What will be the potential energy of the object at an instant $t = \frac{T}{4}\, \text{s}$ starting from the mean position? Assume that the initial phase of the oscillation is zero. (In $\text{J}$)

The $K.E.$ and $P.E.$ of a particle executing $SHM$ with amplitude $A$ will be equal when its displacement is

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)

$A$ particle starting from mean position executes simple harmonic motion with a period $8 \ s$. The minimum time in which its potential energy becomes half of the total energy is . . . . . . . (in $s$)

The total energy of a simple harmonic oscillator is $E$. What is its kinetic energy at half the amplitude?

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):
$(A) E_1 \omega_1 = E_2 \omega_2$
$(B) \frac{\omega_2}{\omega_1} = n^2$
$(C) \omega_1 \omega_2 = n^2$
$(D) \frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$