Question

The displacement of a particle from its mean position (in metres) is given by $y=0.2\sin (10\pi t+1.5\pi )\cos (10\pi t+1.5\pi )$. The motion of the particle is SHM, with a time period of (in seconds)

Answer 1

Given, displacement
$y=0.2\sin (10\pi t+1.5\pi )\cos (10\pi t+1.5\pi )$
Considering $(10\pi t+1.5\pi )=\theta$
$\therefore y=0.2\sin \theta \cos \theta$
Multiplying and dividing by $2$,
$y=0.1\times \sin 2\theta (\because 2\sin \theta \cos \theta =\sin 2\theta )$
Putting value of $\theta$ back,
$y=0.1\times \sin 2(10\pi t+1.5\pi )$
$=0.1\sin (20\pi t+3\pi )$

Comparing with standard equation of SHM $y=A\sin (\omega t+\varphi )$,
we get, $\omega =20\pi$
Time period is given as $T=2\pi /\omega$
$\Rightarrow T=\frac{2\pi }{20\pi }$
$\Rightarrow T=0.1\text{s}$
Hence, motion of the particle is simple harmonic with time period $=0.1\text{s}$