Question

Two particles P and Q describe SHM of same amplitude a and frequency v along the same straight line. The maximum distance between two particles is $\sqrt{2}$ a. The initial phase difference between particles is:

• A. 0
• B. $\frac{\pi }{2}$
  
• C. $\frac{\pi }{6}$
  
• D. $\frac{\pi }{3}$

Answer 1

The correct option is B

$\frac{\pi }{2}$

Let the equations of particles executing SHM be $x_1=Asin\omega t$ and $x_2=Asin(\omega t+\varphi )$.
So, distance between them, $\Delta x=x_2-x_1=Asin(\omega t+\varphi )-Asin\omega t=2Acos(\omega t+\frac{\varphi }{2})sin\left(\frac{\varphi }{2}\right)$
For $\Delta x$ to be maximum, $cos(\omega t+\frac{\varphi }{2})=1$
$\therefore \Delta x_{max}=2Asin\left(\frac{\varphi }{2}\right)\Rightarrow \sqrt{2}A=2Asin\left(\frac{\varphi }{2}\right)\therefore \varphi =\frac{\pi }{2}$