$A$ particle executing simple harmonic motion along a straight line with an amplitude $A$,attains maximum potential energy when its displacement from the mean position equals

In $SHM$,the restoring force is $F = -kx$,where $k$ is the force constant,$x$ is the displacement,and $A$ is the amplitude of motion. Then,the total energy depends upon:

In $S.H.M.$,the displacement of a particle at an instant is $Y = A \cos 30^{\circ}$,where $A = 40 \ cm$ and kinetic energy is $200 \ J$. If the force constant is $1 \times 10^{x} \ N/m$,then $x$ will be $(\cos 30^{\circ} = \sqrt{3}/2)$.

In the following table, the displacement of a $Simple \text{ } Harmonic \text{ } Oscillator \text{ } (SHO)$ is shown in column-$I$ and the kinetic energy is shown in column-$II$. Match them appropriately.

Column-$I$	Column-$II$
$(a)$ $y = \frac{A}{\sqrt{2}}$	$(i)$ $K = \frac{3E}{4}$
$(b)$ $y = \frac{\sqrt{3}A}{2}$	$(ii)$ $K = \frac{E}{4}$
	$(iii)$ $K = \frac{E}{2}$

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

The amplitude of a particle executing $S.H.M.$ is $3 \,cm$. The displacement at which its kinetic energy will be $25 \%$ more than the potential energy is (in $\,cm$)