Question

The displacement of a particle from its mean position (in $\text{m}$) is given by $y=0.2\sin (10\pi t+1.5\pi )\cos (10\pi t+0.5\pi )$. The motion of the particle is

• A. Periodic but not S.H.M
• B. Non-periodic
• C. S.H.M with period $0.1\text{s}$
• D. S.H.M with period $0.2\text{s}$

Answer 1

The correct option is C S.H.M with period $0.1\text{s}$

Given ,
Displacement of the particle $y=0.2\sin (10\pi t+1.5\pi )\cos (10\pi t+0.5\pi )$
From the above equation, we get
$y=0.1\left[2\sin \right(10\pi t+1.5\pi \left)\cos \right(10\pi t+0.5\pi \left)\right]$
Using, $\text{2 sin A cos B = sin (A + B) + sin (A - B)}$
we get ,
$y=0.1\left[\sin \right(20\pi t+2.0\pi )+\sin (\pi \left)\right]$
$\Rightarrow y=0.1\sin (20\pi t+2.0\pi )$
$\Rightarrow y=0.1\sin \left(20\pi t\right)$
$\therefore$ The particle starts from mean position and executes SHM.
Comparing this with $y=A\sin \left(\omega t\right)$, we get $\omega =20\pi ;A=0.1\text{m}$
Time period of oscillation, $T=\frac{2\pi }{\omega }=\frac{2\pi }{20\pi }\Rightarrow T=\frac{1}{10}=0.1\text{s}$
Thus, option (c) is the correct answer.