Question

The period of SHM of a particle is $12 \,s$. The phase difference between the positions at $ t=3\,s $ and $ t=4\,s $ will be:

• A. $ \pi /4 $
• B. $ 3\pi /5 $
• C. $ \pi /6 $
• D. $ \pi /2 $

Answer 1

Answer: (C)

We know simple harmonic motion has equation
$\Rightarrow$ $y=a \sin \omega t$
$(a\to$ amplitude; ($\omega t) \rightarrow$ phase)
Time period, $T=12\, s$
So at $t=3 \,s $
$\Rightarrow$ phase $=\omega \times 3=\frac{2 \pi}{T} \times 3$
$=\frac{6 \pi}{12}=\frac{\pi}{2}$
$ \Rightarrow \phi_{1}$
[We know $ \omega=2 \pi / T]$
So at $t=4 \,s $
$\Rightarrow$ phase $=\omega \times 4$
$=\frac{2 \pi}{T} \times 4=\frac{8 \pi}{12}$
$\Rightarrow$ phase $=\frac{2 \pi}{3} \Rightarrow \phi_{2}$
So phase difference $=\phi_{2}-\phi_{1}=\frac{2 \pi}{3}-\frac{\pi}{2}$
$\Rightarrow$ phase difference $=\frac{4 \pi-3 \pi}{6}=\frac{\pi}{6}$