Question

Two particles $P$ and $Q$ describe $\text{SHM}$ of same amplitude $A$, same frequency $f$ along the same straight line. The maximum distance between the two particles is $A\sqrt{2}$. The phase difference between the particle is

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

Answer 1

The correct option is B $\frac{\pi }{2}$

Equation of motion of particle $1$,
$x_1=A\sin (\omega t+{\varphi }_1)$
Equation of motion of particle $2$,
$x_2=A\sin (\omega t+{\varphi }_2)$
At a particular instant, separation between two particles is given by,
$|x_1-x_2|=A\left[\sin \right(\omega t+{\varphi }_1)-\sin (\omega t+{\varphi }_2\left)\right]$
$\because [\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)]$
$|x_1-x_2|=2A\cos (\omega t+\frac{{\varphi }_1+{\varphi }_2}{2})\sin \left(\frac{{\varphi }_1-{\varphi }_2}{2}\right)$
Given maximum distance between two particle is $A\sqrt{2}$
$\therefore$ To maximize $|x_1-x_2|$, we choose
$\cos (\omega t+\frac{{\varphi }_1+{\varphi }_2}{2})=1$
$\Rightarrow A\sqrt{2}=2A\sin \left(\frac{{\varphi }_1-{\varphi }_2}{2}\right)$
$\Rightarrow \sin \left(\frac{{\varphi }_1-{\varphi }_2}{2}\right)=\frac{1}{\sqrt{2}}$
$\Rightarrow \sin \left(\frac{{\varphi }_1-{\varphi }_2}{2}\right)=\sin \left(\frac{\pi }{4}\right)$
$\Rightarrow \frac{{\varphi }_1-{\varphi }_2}{2}=\frac{\pi }{4}$
$\Rightarrow {\varphi }_1-{\varphi }_2=\frac{\pi }{2}$