A particle executes S.H.M. of period frac{2 p | vedclass

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(B) Given: Period $T = \frac{2 \pi}{\sqrt{3}} 	ext{ s}$, total path length $= 4 	ext{ cm}$.Amplitude $A = \frac{	ext{path length}}{2} = \frac{4}{2}

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(B) Given: Period $T = \frac{2 \pi}{\sqrt{3}} 	ext{ s}$, total path length $= 4 	ext{ cm}$.Amplitude $A = \frac{	ext{path length}}{2} = \frac{4}{2} = 2 	ext{ cm}$.Angular frequency $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{2 \pi / \sqrt{3}} = \sqrt{3} 	ext{ rad/s}$.Magnitude of acceleration $a = \omega^2 |x|$ and magnitude of velocity $v = \omega \sqrt{A^2 - x^2}$.Given $v = a$, so $\omega \sqrt{A^2 - x^2} = \omega^2 |x|$.Squaring both sides: $\omega^2 (A^2 - x^2) = \omega^4 x^2$.Dividing by $\omega^2$: $A^2 - x^2 = \omega^2 x^2$.Substituting values: $2^2 - x^2 = (\sqrt{3})^2 x^2$.$4 - x^2 = 3x^2 \implies 4x^2 = 4 \implies x^2 = 1$.Thus, $x = 1 	ext{ cm}$.

$A$ particle executes $S.H.M.$ of period $\frac{2 \pi}{\sqrt{3}} \text{ s}$ along a straight line $4 \text{ cm}$ long. The displacement of the particle at which the velocity is numerically equal to the acceleration is (in $\text{ cm}$)

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