A particle executes S.H.M. with amplitude $$A$$ and time period $$T$$. The displacement of the particle when its speed is half of maximum speed is $$\frac{\sqrt{x}A}{2}$$. The value of $$x$$ is

For a particle executing S.H.M. with amplitude $$A$$ and time period $$T$$, the velocity at displacement $$y$$ is given by $$v = \omega\sqrt{A^2 - y^2}$$, where $$\omega = \frac{2\pi}{T}$$.

The maximum speed is $$v_{max} = \omega A$$. We are given that the speed is half the maximum speed, so $$v = \frac{\omega A}{2}$$.

Substituting into the velocity equation: $$\frac{\omega A}{2} = \omega\sqrt{A^2 - y^2}$$.

Dividing both sides by $$\omega$$ and squaring: $$\frac{A^2}{4} = A^2 - y^2$$, which gives $$y^2 = A^2 - \frac{A^2}{4} = \frac{3A^2}{4}$$.

Therefore, $$y = \frac{\sqrt{3} \, A}{2} = \frac{\sqrt{3}}{2} A$$.

Comparing with $$\frac{\sqrt{x} \, A}{2}$$, we get $$x = 3$$.