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Q. If the displacement equation of a particle from its mean position is given as $y=0.2\, \sin (10 \pi t+1.5 \pi) \cos (10 \pi t+1.5 \pi)$ then, the motion of particle is

Bihar CECEBihar CECE 2012Oscillations

Solution:

Given,
$y=0.2 \sin (10 \pi t+1.5 \pi) \cos (10 \pi t+1.5 \pi)$
$=0.1 \sin 2(10 \pi t+1.5 \pi)$
$[\sin 2 A=2 \sin A \cos A]$
$=0.1\, \sin (20 \pi t+3.0 \pi)$
So, the motion of particle is SHM Time period,
$T=\frac{2 \pi}{\omega}=\frac{2 \pi}{20 \pi}=\frac{1}{10}=0.1\, s$