In simple harmonic motion (SHM), the total mechanical energy of the system is conserved and is a combination of kinetic energy (KE) and potential energy (PE). The total energy (E) of the particle can be expressed as:

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

where $k$ is the spring constant, and $A$ is the amplitude of motion.

At a displacement $x$, the potential energy (PE) and kinetic energy (KE) of the particle are given by:

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

$$\text{KE} = E - \text{PE} = \frac{1}{2}kA^2 - \frac{1}{2}kx^2 = \frac{1}{2}k(A^2 - x^2)$$

We need to find the displacement where the potential and kinetic energies are equal. This implies:

$$\text{PE} = \text{KE}$$

Thus, we get:

$$\frac{1}{2}kx^2 = \frac{1}{2}k(A^2 - x^2)$$

By simplifying, we get:

$x^2 = A^2 - x^2$

Adding $x^2$ to both sides:

$2x^2 = A^2$

Now, solving for $x$:

$$x = \frac{A}{\sqrt{2}}$$

Therefore, the displacement at which the potential and kinetic energies are equal is:

Option C:

$\frac{A}{\sqrt{2}}$