The graph shows the variation of displacement of a particle executing $S$.$H$.$M$. with time. We infer from this graph that

$A$ 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 displacement-time graph of a particle executing $SHM$ is as shown in the figure. The corresponding graph between potential energy $(PE)$ and time is:

An object of mass $0.5\, \text{kg}$ is executing simple harmonic motion. Its amplitude is $5\, \text{cm}$ and time period $T$ is $0.2\, \text{s}$. What will be the potential energy of the object at an instant $t = \frac{T}{4}\, \text{s}$ starting from the mean position? Assume that the initial phase of the oscillation is zero. (In $\text{J}$)

Starting from the origin,a particle oscillates simple harmonically with a time period of $2 \ s$. After what time will its kinetic energy be $75 \%$ of the total energy?

$A$ body performs simple harmonic oscillations along the straight line $ABCDE$ with $C$ as the midpoint of $AE.$ Its kinetic energies at $B$ and $D$ are each one-fourth of its maximum value. If $AE = 2R,$ the distance between $B$ and $D$ is