$A$ linear harmonic oscillator has a total mechanical energy of $300 \ J$. If its potential energy at the mean position is $100 \ J$,find its kinetic energy at $x = +\frac{A}{\sqrt{2}}$. (in $J$)

$A$ body is performing $SHO$ with a total energy of $100\,J$. In the table below, column-$I$ shows the kinetic energy $(K)$ at a specific time, and column-$II$ shows the potential energy $(U)$ at that same time. Match them appropriately.

Column-$I$	Column-$II$
$(a)$ $K = 10\,J$	$(i)$ $U = 40\,J$
$(b)$ $K = 60\,J$	$(ii)$ $U = 90\,J$
	$(iii)$ $U = 50\,J$

The total energy of a body executing simple harmonic motion is $E$. When the displacement is half of the amplitude,the kinetic energy is:

$A$ mass $0.4 \,kg$ performs $S.H.M.$ with a frequency $\frac{16}{\pi} \,Hz$. At a certain displacement, it has kinetic energy $2 \,J$ and potential energy $1.2 \,J$. The amplitude of oscillation is (in $m$)

If a particle repeats its motion after a fixed time interval of $8 \,s$,then after how much time will its maximum value of $PE$ be attained after attaining its minimum value?

$A$ 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