A particle executes $S.H.M.$ with amplitude $'a'$ and time period $V$. The displacement of the particle when its speed is half of maximum speed is $\frac{\sqrt{ x } a }{2} .$ The value of $x$ is $\ldots \ldots \ldots$
JEE MAIN 2021, Medium
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$V =\omega \sqrt{ A ^{2}- x ^{2}} \quad V _{\max }= A\omega$
$\frac{ A\omega }{2}=\omega \sqrt{ A ^{2}- x ^{2}}$
$\frac{ A ^{2}}{4}= A ^{2}- x ^{2}$
$x ^{2}=\frac{3 A ^{2}}{4}$
$x =\frac{\sqrt{3}}{2} \,A$
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