1. Write expressions for energy in SHM

For a particle executing SHM with amplitude $A$:

• Total energy, $E = \frac{1}{2}kA^2$
• Potential energy at displacement $x$ from mean position, $U = \frac{1}{2}kx^2$
• Kinetic energy at displacement $x$, $K = E-U = \frac{1}{2}k(A^2-x^2)$

2. Use the condition given

We are told that kinetic energy becomes equal to potential energy: $K=U$

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

Cancel $\frac{1}{2}k$ from both sides: $A^2-x^2=x^2$

3. Solve for $x$

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

Since the question asks for the distance from mean position, we take the positive value.

4. Match with options

The required distance is: $\boxed{\frac{A}{\sqrt{2}}}$

This matches Option A.