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(B) The equation of motion for a particle in $SHM$ passing through the origin at $t=2 \ s$ is $x(t) = A \sin(\omega(t - 2))$.Given $T = 16 \ s$, the a

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$\frac{32 \sqrt{2}}{\pi}$

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(B) The equation of motion for a particle in $SHM$ passing through the origin at $t=2 \ s$ is $x(t) = A \sin(\omega(t - 2))$.Given $T = 16 \ s$, the angular frequency is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{16} = \frac{\pi}{8} \ rad/s$.The velocity is $v(t) = \frac{dx}{dt} = A \omega \cos(\omega(t - 2))$.At $t = 4 \ s$, $v = 4 \ m/s$:$4 = A \left(\frac{\pi}{8}ight) \cos\left(\frac{\pi}{8}(4 - 2)ight)$$4 = A \left(\frac{\pi}{8}ight) \cos\left(\frac{\pi}{4}ight)$$4 = A \left(\frac{\pi}{8}ight) \left(\frac{1}{\sqrt{2}}ight)$$A = \frac{4 	imes 8 	imes \sqrt{2}}{\pi} = \frac{32 \sqrt{2}}{\pi} \ m$.

$A$ particle performs simple harmonic motion with a time period of $16 \ s$. At a time $t=2 \ s$, the particle passes through the origin and at $t=4 \ s$ its velocity is $4 \ m/s$. The amplitude of the motion is

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