Question

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

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

Answer 1

Answer: (B)

$\pi /2$

Let equation of motion to particles be given by,

$x_1=asin\left(\omega t\right)$ and $x_2=asin(\omega t+\varphi )$

So initial phase difference between two particles is $\varphi$.

Now,

$x_2-x_1=a\left[sin\right(\omega t+\varphi )-sin(\omega t\left)\right]=2asin(\frac{\omega t+\varphi -\omega t}{2})cos(\frac{\omega t+\varphi +\omega t}{2})$

$=2asin\left(\varphi /2\right)cos(\omega t+\varphi /2)$

Maximum value of above expression is $2asin\left(\varphi /2\right)=\sqrt{2}a$

$⟹sin\left(\varphi /2\right)=1/\sqrt{2}⟹\varphi /2=\pi /4$

$⟹\varphi =\pi /2$