A particle executes S.H.M. starting from the mean position. Its amplitude is 'a' and its periodic time is 'T'. At a certain instant, its speed ' $u$ ' is half that of maximum speed $V_{\text {max }}$. The displacement of the particle at that instant is

In simple harmonic motion, the velocity at displacement $x$ is $u = \omega \sqrt{a^2 - x^2}$, with maximum velocity $V_{\text{max}} = \omega a$ at $x = 0$.

Given $u = \frac{V_{\text{max}}}{2}$, substitution yields $\omega \sqrt{a^2 - x^2} = \frac{\omega a}{2}$.

Canceling $\omega$ and squaring both sides gives $a^2 - x^2 = \frac{a^2}{4}$.

Solving for $x^2$: $x^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}$.

Thus $x = \frac{\sqrt{3} a}{2}$, corresponding to option D.