A particle executes SHM of amplitude A. The distance from the mean position when its's kinetic energy becomes equal to its potential energy is :

The total energy of a particle executing simple harmonic motion (SHM) is given by: <br/><br/> $E = \frac{1}{2}m\omega^2A^2$ <br/><br/> where $m$ is the mass of the particle, $\omega$ is the angular frequency of the SHM, and $A$ is the amplitude of the motion. <br/><br/> At any point during SHM, the kinetic energy of the particle is given by: <br/><br/> $K = \frac{1}{2}m\omega^2(x^2 + A^2\cos^2\omega t)$ <br/><br/> where $x$ is the displacement of the particle from the mean position. <br/><br/> The potential energy of the particle at the same point is given by: <br/><br/> $U = \frac{1}{2}m\omega^2(x^2 + A^2\sin^2\omega t)$ <br/><br/> When the kinetic energy becomes equal to the potential energy, we have: <br/><br/> $K = U$ <br/><br/> $$\frac{1}{2}m\omega^2(x^2 + A^2\cos^2\omega t) = \frac{1}{2}m\omega^2(x^2 + A^2\sin^2\omega t)$$ <br/><br/> Simplifying this equation, we get: <br/><br/> $x^2 = \frac{1}{2}A^2$ <br/><br/> $x = \pm\frac{1}{\sqrt{2}}A$ <br/><br/> Therefore, the distance from the mean position when the kinetic energy becomes equal to the potential energy is $\boxed{\frac{1}{\sqrt{2}}A}$