When a body is in $SHM$, then match the following.

$\mathrm{List} - \mathrm{I}$		$\mathrm{List} - \mathrm{II}$
$\left(\mathrm{A}\right)$	Velocity is maximum	$\left(\mathrm{I}\right)$	Acceleration is maximum
$\left(\mathrm{B}\right)$	$\mathrm{KE}$ is ${\left(\frac{3}{4}\right)}^{\text{th }}$ of total energy	$\left(\mathrm{II}\right)$	At mean position
$\left(\mathrm{C}\right)$	$\mathrm{PE}$ is ${\left(\frac{3}{4}\right)}^{\mathrm{th}}$ of total energy	$\left(\mathrm{III}\right)$	At half of the amplitude
$\left(\mathrm{D}\right)$	Acceleration is maximum	$\left(\mathrm{IV}\right)$	At $\frac{\sqrt{3}}{2}$ times the amplitude

 

In $\mathrm{S}.\mathrm{H}.\mathrm{M}.$
If equation of $\mathrm{S}.\mathrm{H}.\mathrm{M}.$ is given by$y=A\sin \omega t$
Velocity $v=\omega \sqrt{A^2-x^2}$
And Acceleration $a=-{\omega }^{2}x$
Here $\omega$ is angular frequency and $x$ is displacement from mean position.

From velocity equation we can see that velocity will be max when $x=0$ i.e. particle is at mean position.

From acceleration equation we can see that acceleration is maximum at when particle is at extreme position i.e. $x=A$.
Kinetic energy is $\frac{1}{2}m{v}^2=\frac{m{\omega }^2}{2}\left(A^2-x^2\right)$
Hence, Kinetic energy will be $\frac{3}{4}th$ of maximum kinetic energy$\frac{1}{2}m{\omega }^2{A}^2$ when displacement is equal to half of amplitude$\left(A\right).$

Potential energy is given by$\frac{1}{2}m{\omega }^2{x}^2$. 
Hence potential energy will be $\frac{3}{4}th$ of the maximum potential energy$\left(\frac{1}{2}m{\omega }^2{A}^2\right)$ , when displacement from mean position is equal to $\frac{\sqrt{3}}{2}A.$