Question

Two particles are executing simple harmonic motion of the same amplitude $A$ and angular frequency $\omega$ along the $x-$axis. Their mean position is separated by distance $X_o(X_o>A)$. If the maximum separation between them is $(X_o+A)$, the phase difference between their motion is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{\pi }{3}$

Equation of motion of particle $1$,
$x_1=A\sin (\omega t+{\varphi }_1)...........\left(1\right)$
Equation of motion of particle $2$,
$x_2=A\sin (\omega t+{\varphi }_2)...........\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$x_2-x_1=A\sin (\omega t+{\varphi }_2)-A\sin (\omega t+{\varphi }_1)$
$=A\left[\sin \right(\omega t+{\varphi }_2)-\sin (\omega t+{\varphi }_1\left)\right]$
Using, $\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right).\sin \left(\frac{C-D}{2}\right)$
$x_2-x_1=2A\cos (\omega t+\frac{{\varphi }_1+{\varphi }_2}{2}).\sin \left(\frac{{\varphi }_2-{\varphi }_1}{2}\right)$
Given that, $(x_o+x_2-x_1{)}_{max}=x_o+A$
$\Rightarrow (x_2-x_1{)}_{max}=A$
To get $(x_2-x_1{)}_{max}$, we assume $\cos (\omega t+\frac{{\varphi }_1+{\varphi }_2}{2})=1$
$\Rightarrow 2A\sin \left(\frac{{\varphi }_2-{\varphi }_1}{2}\right)=A$
$\Rightarrow \sin \left(\frac{{\varphi }_2-{\varphi }_1}{2}\right)=\frac{1}{2}$
$\Rightarrow \frac{{\varphi }_2-{\varphi }_1}{2}=\frac{\pi }{6}\text{or}\frac{5\pi }{6}$
$\Rightarrow {\varphi }_2-{\varphi }_1=\frac{\pi }{3}\text{or}\frac{5\pi }{3}$
The phase difference between their motion is $\frac{\pi }{3}$