$A$ particle of mass $4 \,kg$ is executing $SHM$. Its displacement is given by the equation $y=8 \cos [100 t+\pi / 4] \,cm$. Its maximum kinetic energy is (in $\,J$)

$A$ particle of mass $m$ is hanging vertically by an ideal spring of force constant $K$. If the mass is made to oscillate vertically,its total energy is

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

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:

An object of mass $0.2 \,kg$ executes simple harmonic motion along the $x$-axis with a frequency of $(\frac{25}{\pi}) \,Hz$. At the position $x=0.04 \,m$, the object has a kinetic energy of $0.5 \,J$ and a potential energy of $0.4 \,J$. The amplitude of oscillation is ............ $cm$.

$A$ particle starts oscillating simple harmonically from its mean position with time period $T$. At time $t = T/6$,the ratio of the potential energy to kinetic energy of the particle is $\left[\sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2\right]$