For a simple pendulum, having time period $T$, the variation of kinetic energy $(K.E)$ with time $(t)$ is represented by:

The total energy of a particle,executing simple harmonic motion is

$A$ block is in simple harmonic motion $(S.H.M)$ at the end of a spring with position given by $x = 5 \cos \left(\omega t + \frac{\pi}{4}\right)$. If the total mechanical energy is $100 \ J$, then the potential energy at time $t = 0$ is: (in $J$)

Starting from the mean position,a body oscillates simple harmonically with a period $T$. After what time will its kinetic energy be $75 \%$ of the total energy? $(\sin 30^{\circ} = 0.5)$

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

$A$ particle of mass $10 \ g$ performs simple harmonic motion with an amplitude of $10 \ cm$ and a time period of $2 \ s$. What is its kinetic energy at a displacement of $5 \ cm$?