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

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$8 \sqrt{2} \ m$

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(D) The displacement of a particle performing $S.H.M.$ starting from the equilibrium position is given by $x = A \sin(\omega t)$.The velocity of the particle is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.Given: $v = \pi \ m \ s^{-1}$,$T = 16 \ s$,and $t = 2 \ s$.The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{16} = \frac{\pi}{8} \ rad \ s^{-1}$.Substituting the values into the velocity equation:$\pi = A 	imes \frac{\pi}{8} 	imes \cos\left(\frac{\pi}{8} 	imes 2ight)$$\pi = A 	imes \frac{\pi}{8} 	imes \cos\left(\frac{\pi}{4}ight)$$1 = \frac{A}{8} 	imes \frac{1}{\sqrt{2}}$$A = 8\sqrt{2} \ m$.

$A$ particle performing $S.H.M.$ starts from the equilibrium position and its time period is $16 \ s$. After $2 \ s$,its velocity is $\pi \ m \ s^{-1}$. The amplitude of oscillation is (Given: $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$).

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