$A$ body performs linear simple harmonic motion of amplitude $A$. At what displacement from the mean position is the potential energy of the body one-fourth of its total energy?

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

$A$ particle is executing linear simple harmonic motion of amplitude $A$. At what displacement is the energy of the particle half potential and half kinetic?

$A$ bob of a simple pendulum of mass $m$ performs $SHM$ with amplitude $A$ and period $T$. The kinetic energy of the pendulum at displacement $x = \frac{A}{2}$ will be:

The displacements of two particles of same mass executing $SHM$ are represented by the equations $x_1=4 \sin \left(10 t+\frac{\pi}{6}\right)$ and $x_2=5 \cos (\omega t)$. The value of $\omega$ for which the energies of both the particles remain same is (in $\text{ unit}$)

$A$ particle of mass $m$ performs $SHM$ along a straight line with frequency $f$ and amplitude $A.$