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) The total energy of a $Simple \text{ } Harmonic \text{ } Oscillator$ is $E = \frac{1}{2}kA^2$.
The potential energy at displacement $y$ is $U = \frac{1}{2}ky^2$.
The kinetic energy is $K = E - U = \frac{1}{2}k(A^2 - y^2)$.
For $(a)$ $y = \frac{A}{\sqrt{2}}$:
$K = \frac{1}{2}k(A^2 - \frac{A^2}{2}) = \frac{1}{2}k(\frac{A^2}{2}) = \frac{1}{2}(\frac{1}{2}kA^2) = \frac{E}{2}$. Thus, $(a) - (iii)$.
For $(b)$ $y = \frac{\sqrt{3}A}{2}$:
$K = \frac{1}{2}k(A^2 - \frac{3A^2}{4}) = \frac{1}{2}k(\frac{A^2}{4}) = \frac{1}{4}(\frac{1}{2}kA^2) = \frac{E}{4}$. Thus, $(b) - (ii)$.
Therefore, the correct matching is $(a-iii, b-ii)$.