$A$ body executes simple harmonic motion with an amplitude $A$. At what displacement,from the mean position,is the potential energy of the body one fourth of its total energy?

When a longitudinal wave propagates through a medium,the particles of the medium execute simple harmonic oscillations about their mean positions. These oscillations of a particle are characterised by an invariant

$A$ body is executing simple harmonic motion. At a displacement $x$,its potential energy is $E_1$ and at a displacement $y$,its potential energy is $E_2$. The potential energy $E$ at a displacement $(x + y)$ is

$A$ particle starts oscillating simple harmonically from its mean position with time period $T$. At time $t = T/6$,the ratio of the potential energy to kinetic energy of the particle is $\left[\sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2\right]$

When the displacement of a simple harmonic oscillator is one third of its amplitude,the ratio of total energy to the kinetic energy is $\frac{x}{8}$,where $x=$ . . . . . . .

An object of mass $3 \,kg$ is executing simple harmonic motion with an amplitude $\frac{2}{\pi} \,m$. If the kinetic energy of the object when it crosses the mean position is $6 \,J$, the time period of oscillation of the object is (in $\,s$)