A particle executes simple harmonic motion with an amplitude A and time period T. The displacement of the particle when its speed is half of its maxi…

The speed of a particle in SHM is given by $v = \omega \sqrt{A^2 - x^2}$. Maximum speed is $v_{max} = \omega A$. Given $v = v_{max}/2$, we get $\omega \sqrt{A^2 - x^2} = \frac{1}{2}\omega A \implies A^2 - x^2 = \frac{A^2}{4} \implies x^2 = \frac{3}{4}A^2 \implies x = \frac{\sqrt{3}}{2}A$.