The ratio of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half its amplitude, the distance being measured from its equilibrium position is

Let the amplitude of SHM be A.

Now, potential energy of SHM, $\mathrm{U}=\frac{1}{2}{\mathrm{kx}}^2$

Here,  $\mathrm{x}=\frac{\mathrm{A}}{2}\Rightarrow \mathrm{U}=\frac{1}{2}\mathrm{k}\frac{{\mathrm{A}}^2}{4}$                                           …(i)

Kinetic energy $\mathrm{K}=\frac{1}{2}{\mathrm{kA}}^2-\frac{1}{2}\mathrm{k}\frac{{\mathrm{A}}^2}{4}\Rightarrow \mathrm{K}=\frac{3}{8}{\mathrm{kA}}^2$         …(ii)

On dividing Eq. (ii) by Eq. (i), we get

                     $\frac{\mathrm{K}}{\mathrm{U}}=\frac{3}{8}\times \frac{8}{1}\Rightarrow \frac{\mathrm{K}}{\mathrm{U}}=\frac{3}{1}$