The total energy of a particle executing simple harmonic motion (SHM) is given by:

$$E = \frac{1}{2}m\omega^2A^2$$

where $$m$$ is the mass of the particle, $$\omega$$ is the angular frequency of the SHM, and $$A$$ is the amplitude of the motion.

At any point during SHM, the kinetic energy of the particle is given by:

$$K = \frac{1}{2}m\omega^2(x^2 + A^2\cos^2\omega t)$$

where $$x$$ is the displacement of the particle from the mean position.

The potential energy of the particle at the same point is given by:

$$U = \frac{1}{2}m\omega^2(x^2 + A^2\sin^2\omega t)$$

When the kinetic energy becomes equal to the potential energy, we have:

$$K = U$$

$$\frac{1}{2}m\omega^2(x^2 + A^2\cos^2\omega t) = \frac{1}{2}m\omega^2(x^2 + A^2\sin^2\omega t)$$

Simplifying this equation, we get:

$$x^2 = \frac{1}{2}A^2$$

$$x = \pm\frac{1}{\sqrt{2}}A$$

Therefore, the distance from the mean position when the kinetic energy becomes equal to the potential energy is $$\boxed{\frac{1}{\sqrt{2}}A}$$