For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

For a simple harmonic oscillator with amplitude $$A$$, the kinetic energy at displacement $$x$$ is $$K = \frac{1}{2}m\omega^2(A^2 - x^2)$$ and the potential energy is $$U = \frac{1}{2}m\omega^2 x^2$$.

Setting $$K = U$$ gives $$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2 x^2$$, which simplifies to $$A^2 - x^2 = x^2$$, so $$2x^2 = A^2$$, yielding $$x = \pm \frac{A}{\sqrt{2}}$$.

The correct answer is option 3: $$x = \pm \frac{A}{\sqrt{2}}$$.