1. In simple harmonic motion, the total energy is constant:

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

where $A$ is the amplitude.

2. At a displacement $x$ from the mean position, the potential energy is

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

3. The kinetic energy at that position is

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

4. For kinetic energy and potential energy to be equal:

$K = U$

So,

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

5. Cancel $\frac{1}{2}k$ from both sides:

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

$A^2 = 2x^2$

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

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

6. Hence, the correct option is:

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

This corresponds to Option C.