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(B) The displacement of a particle performing $S.H.M.$ starting from the equilibrium position is given by $x = A \sin(\omega t)$.Velocity of the parti

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$12 \ m$

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(B) The displacement of a particle performing $S.H.M.$ starting from the equilibrium position is given by $x = A \sin(\omega t)$.Velocity of the particle is $v = \frac{dx}{dt} = A \omega \cos(\omega t) \dots (i)$.Given: $v = \pi \ m/s$,$T = 12 \ s$,and $t = 2 \ s$.Angular frequency $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{12} = \frac{\pi}{6} \ rad/s$.Substituting these values into equation $(i)$:$\pi = A 	imes \frac{\pi}{6} 	imes \cos\left(\frac{\pi}{6} 	imes 2ight)$$\pi = A 	imes \frac{\pi}{6} 	imes \cos\left(\frac{\pi}{3}ight)$Since $\cos(60^{\circ}) = 0.5 = \frac{1}{2}$,we have:$1 = \frac{A}{6} 	imes \frac{1}{2}$$1 = \frac{A}{12}$$A = 12 \ m$.

$A$ particle performing $S.H.M.$ starts from the equilibrium position and its time period is $12 \ s$. After $2 \ s$,its velocity is $\pi \ m/s$. The amplitude of the oscillation is: $[\sin 30^{\circ}=\cos 60^{\circ}=0.5, \sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3}/2]$

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