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