The expression for displacement of an object in SHM is $\mathrm{x}=\mathrm{Acos}\mathrm{\omega t}$ . The potential energy at t = T / 4 is

Potential energy of an object executing SHM is given by
$\mathrm{U}\left(\mathrm{x}\right)=\frac{1}{2}{\mathrm{kx}}^2$
 Given, $\mathrm{x}=\mathrm{Acos}\mathrm{\omega t}$
$\begin{array}{l}\Rightarrow \mathrm{U}\left(\mathrm{x}\right)=\frac{1}{2}{\mathrm{kA}}^2{\cos }^2\mathrm{\omega t}\\ \text{ At }\mathrm{t}=\left(\frac{\mathrm{T}}{4}\right),\\ \mathrm{U}\left(\mathrm{x}\right)=\frac{1}{2}{\mathrm{k}}^2{\mathrm{A}}^2{\cos }^2\left(\frac{2\mathrm{\pi }}{\mathrm{T}}\times \frac{\mathrm{T}}{4}\right)\left(\therefore \mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{T}}\right)\\ =\frac{1}{2}{\mathrm{k}}^2{\mathrm{A}}^2{\cos }^2\left(\frac{\mathrm{\pi }}{2}\right)=0\end{array}$
So, PE at t = T / 4 is zero.