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  4. The time period of a particle undergoing SHM is 16 s. It starts motion from the mean position. After 2 s, its velocity is 0.4 m s -1. The amplitude is

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Q. The time period of a particle undergoing SHM is $16\, s$. It starts motion from the mean position. After $2\, s$, its velocity is $0.4\, m s ^{-1}$. The amplitude is

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Solution:

Velocity, $v = r \omega \cos \omega t$
$04=r \times \frac{2 \pi}{16} \cos \frac{2 \pi}{16} \times 2=r \times \frac{2 \pi}{16} \times \frac{1}{16} \times \frac{1}{\sqrt{2}}$
or $r=\frac{0.4 \times 16 \times \sqrt{2}}{2 \pi}=\frac{3.2 \sqrt{2}}{\pi}=1.44\, m$