The displacement of a particle from its mean position (in metre) is given by
$y = 0.2\sin \left( {10\pi t + 1.5\pi } \right)\cos \left( {10\pi t + 1.5\pi } \right)$
The motion of particle is
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$\mathrm{y}=0.2 \sin (10 \pi \mathrm{t}+1.5 \pi) \cos (10 \pi \mathrm{t}+1.5 \pi)$
$=0.1 \sin 2(10 \pi \mathrm{t}+1.5 \pi)$
$[\because \sin 2 A=2 \sin A \cos A]$
$=0.1 \sin (20 \pi t+3.0 \pi)$
Time period, $T=\frac{2 \pi}{\omega}=\frac{2 \pi}{20 \pi}=\frac{1}{10}=0.1 \mathrm{sec}$
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