1. For a particle in S.H.M. with amplitude $a$ and angular frequency $\omega$, the maximum speed is $v_{\max} = a\omega.$

2. The speed at displacement $x$ from mean position is $v = \omega\sqrt{a^2 - x^2}.$

3. Given that the speed is half of the maximum speed: $v = \frac{v_{\max}}{2} = \frac{a\omega}{2}.$

4. Substitute into the speed-displacement relation: $\omega\sqrt{a^2 - x^2} = \frac{a\omega}{2}.$

5. Cancel $\omega$: $\sqrt{a^2 - x^2} = \frac{a}{2}.$

6. Squaring both sides: $a^2 - x^2 = \frac{a^2}{4}.$

7. Therefore, $x^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}.$

8. Hence the magnitude of displacement is $x = \frac{\sqrt{3}a}{2}.$

9. Comparing with the given form $\frac{\sqrt{x}\,a}{2},$ we get $\sqrt{x} = \sqrt{3} \implies x = 3.$

Therefore, the required integer value is $3$.