If $< E >$ and $< U >$ denote the average kinetic and the average potential energies respectively of a mass describing a simple harmonic motion,over one period,then the correct relation is:

$A$ particle is performing $S.H.M.$ with energy of vibration $90 \,J$ and amplitude $6 \,cm$. When the particle reaches a distance of $4 \,cm$ from the mean position,it is stopped for a moment and then released. The new energy of vibration will be ........... $J$.

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 $x=$ . . . . . . .

The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal,when displacement (amplitude = $a$) is

The displacement of a simple harmonic motion of amplitude $6 \text{ cm}$ when its kinetic energy is equal to its potential energy is:

An object of mass $0.2 \ kg$ executes simple harmonic oscillations along the $X$-axis with a frequency of $(\frac{25}{\pi}) \ Hz$. At the position $x=0.04 \ m$,the object has kinetic energy $1 \ J$ and potential energy $0.6 \ J$. The amplitude of oscillation is (in $m$)