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(D) The velocity of a particle in $S.H.M.$ at displacement $y$ is given by $v = \omega \sqrt{a^2 - y^2}$,where $a$ is the amplitude and $\omega$ is th

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$3$

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(D) The velocity of a particle in $S.H.M.$ at displacement $y$ is given by $v = \omega \sqrt{a^2 - y^2}$,where $a$ is the amplitude and $\omega$ is the angular frequency.Maximum speed is $v_{\max} = a\omega$.Given that the speed $v = \frac{v_{\max}}{2}$,we have $\frac{a\omega}{2} = \omega \sqrt{a^2 - y^2}$.Squaring both sides: $\frac{a^2}{4} = a^2 - y^2$.Rearranging for displacement $y$: $y^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}$.Thus,$y = \frac{\sqrt{3} a}{2}$.Comparing this with the given expression $\frac{\sqrt{x} a}{2}$,we find $x = 3$.

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

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