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

In SHM, at displacement $$x$$ from the mean position:

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

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

When $$KE = PE$$:

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

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

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

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

The distance from mean position is $$\frac{A}{\sqrt{2}}$$.

This matches option 1.

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