Answer
$\pm\frac{a}{\sqrt 2}$
Work Step by Step
Let at a displacement x(t) from the equilibrium position, the energy of the system is half kinetic energy and half potential energy.
At at a displacement x(t) from the equilibrium position,
potential energy$=\frac{1}{2} m\omega^2x^2$
kinetic energy$=\frac{1}{2} mv^2=\frac{1}{2} m\omega^2(a^2-x^2)$
According to the given condition,
$\frac{1}{2} m\omega^2x^2=\frac{1}{2} m\omega^2(a^2-x^2)$
or, $m\omega^2x^2=\frac{1}{2} m\omega^2a^2$
or, $\boxed{x=\pm\frac{a}{\sqrt 2}}$
