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(D) The speed of a particle in $S.H.M.$ at a displacement '$x$' is given by $u = \omega \sqrt{a^2 - x^2}$.The maximum speed of the particle is $V_{	e

Vedclass · Accepted Answer

$\frac{\sqrt{3} a}{2}$

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(D) The speed of a particle in $S.H.M.$ at a displacement '$x$' is given by $u = \omega \sqrt{a^2 - x^2}$.The maximum speed of the particle is $V_{	ext{max}} = a\omega$.Given that the speed '$u$' is half of the maximum speed,we have $u = \frac{V_{	ext{max}}}{2} = \frac{a\omega}{2}$.Equating the two expressions for speed: $\frac{a\omega}{2} = \omega \sqrt{a^2 - x^2}$.Canceling $\omega$ from both sides: $\frac{a}{2} = \sqrt{a^2 - x^2}$.Squaring both sides: $\frac{a^2}{4} = a^2 - x^2$.Rearranging for $x^2$: $x^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}$.Taking the square root: $x = \frac{\sqrt{3}a}{2}$.

$A$ particle executes $S.H.M.$ starting from the mean position. Its amplitude is '$a$' and its periodic time is '$T$'. At a certain instant,its speed '$u$' is half that of maximum speed $V_{\text{max}}$. The displacement of the particle at that instant is

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