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

Question

Dr. Abubacker · Accepted Answer

(c) $y = 0.2\sin{(10\pi\ t + 1.5\pi)}\cos{(10\pi\ t + 1.5\pi)}$
$= 0.1\sin{2(10\pi t + 1.5\pi)}\ \ \ \lbrack\because\sin{2A} = 2\sin{A\cos A}brack$
$= 0.1\sin(20\pi\ t + 3.0\pi)$
$	herefore Time\ period,\ T = \frac{2\pi}{\omega} = \frac{2\pi}{20\pi} = \frac{1}{10} = 0.1\ sec$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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