Question

The period of SHM of a particle is $12s$. The phase difference between the positions at $t=3s$ and $t=4s$ 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; ($\omega t)\to$ phase)
Time period, $T=12s$
So at $t=3s$
$\Rightarrow$ phase $=\omega \times 3=\frac{2\pi }{T}\times 3$
$=\frac{6\pi }{12}=\frac{\pi }{2}$
$\Rightarrow {\varphi }_1$
[We know $\omega =2\pi /T]$
So at $t=4s$
$\Rightarrow$ phase $=\omega \times 4$
$=\frac{2\pi }{T}\times 4=\frac{8\pi }{12}$
$\Rightarrow$ phase $=\frac{2\pi }{3}\Rightarrow {\varphi }_2$
So phase difference $={\varphi }_2-{\varphi }_1=\frac{2\pi }{3}-\frac{\pi }{2}$
$\Rightarrow$ phase difference $=\frac{4\pi -3\pi }{6}=\frac{\pi }{6}$