Let's denote the amplitude of the simple harmonic motion as A, and the displacement of the particle from the mean position as x. The given condition is that x = A/2.

For a particle in SHM, the potential energy (PE) is given by:

$$PE = \frac{1}{2} kx^2$$

And the total mechanical energy (E) of the particle remains constant and is given by:

$$E = \frac{1}{2} kA^2$$

Since the total mechanical energy is the sum of potential energy and kinetic energy (KE), we have:

$$E = PE + KE$$

Now, we need to find the ratio of potential energy to kinetic energy when x = A/2.

Calculate the potential energy at x = A/2:

$$PE = \frac{1}{2} k\left(\frac{A}{2}\right)^2 = \frac{1}{8} kA^2$$

Substitute the expression for total mechanical energy:

$$KE = E - PE = \frac{1}{2} kA^2 - \frac{1}{8} kA^2 = \frac{3}{8} kA^2$$

Now, find the ratio of potential energy to kinetic energy:

$$\frac{PE}{KE} = \frac{\frac{1}{8} kA^2}{\frac{3}{8} kA^2} = \frac{1}{3}$$

Therefore, the ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude is 1 : 3.