For a particle in SHM with amplitude $A$, the total energy is $E = \frac{1}{2}kA^2$ where $k$ is the force constant.

At displacement $x$, the potential energy is $U = \frac{1}{2}kx^2$

The kinetic energy at displacement $x$ is $K = E - U = \frac{1}{2}kA^2 - \frac{1}{2}kx^2 = \frac{1}{2}k(A^2 - x^2)$

Given that the displacement is half of amplitude: $x = \frac{A}{2}$

Substitute into potential energy: $U = \frac{1}{2}k\left(\frac{A}{2}\right)^2 = \frac{1}{2}k\cdot \frac{A^2}{4} = \frac{1}{8}kA^2$

Substitute into kinetic energy: $K = \frac{1}{2}k\left(A^2 - \frac{A^2}{4}\right) = \frac{1}{2}k\cdot \frac{3A^2}{4} = \frac{3}{8}kA^2$

Therefore, the ratio of potential energy to kinetic energy is $U:K = \frac{1}{8}kA^2 : \frac{3}{8}kA^2 = 1:3$

So the correct option is D.