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:

$A$ body is performing $S.H.M.$ of amplitude $A$. The displacement of the body from a point where kinetic energy is maximum to a point where potential energy is maximum,is

The potential energy of a simple harmonic oscillator at the mean position is $2\,J$. If its mean kinetic energy $(K.E.)$ is $4\,J$,its total energy will be .... $J$

The potential energy of a particle executing $S.H.M.$ is $2.5 \ J$,when its displacement is half of its amplitude. The total energy of the particle is .... $J$.

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. At time $t = \frac{T}{12}$,the ratio of its kinetic energy to potential energy is $\left[\sin \frac{\pi}{3} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \sin \frac{\pi}{6} = \cos \frac{\pi}{3} = \frac{1}{2}\right]$.

$A$ particle starts from the mean position and performs $S.H.M.$ with a period of $6 \ s$. At what time is its kinetic energy $50 \%$ of its total energy (in $s$)? $\left(\cos 45^{\circ} = \frac{1}{\sqrt{2}}\right)$