$A$ body performs linear simple harmonic motion of amplitude $A$. At what displacement from the mean position is the potential energy of the body one-fourth of its total energy?

Assertion $(A)$: In $S.H.M$,kinetic and potential energy become equal when the distance is $1/\sqrt{2}$ times its amplitude. Reason $(R)$: The potential energy of a particle executing $S.H.M$ is periodic with time period being maximum at the extreme displacement.

For a body performing simple harmonic motion,its potential energy is $E_x$ at displacement $x$ and $E_y$ at displacement $y$ from the mean position. The potential energy $E_0$ at displacement $(x+y)$ is

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

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. What is the ratio of potential energy to kinetic energy of the particle at time $t = \frac{T}{12}$? (Given: $\sin(\frac{\pi}{6}) = \frac{1}{2}$)

$A$ linear harmonic oscillator of force constant $6 \times 10^5 \, N/m$ and amplitude $4 \, cm$ has a total energy of $600 \, J$. Select the correct statement.