Question

Two particle are executing simple harmonic motion of the same amplitude $A$ and 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

Answer: (B)

$\frac{\pi }{3}$

The SHM equation of $1^{st}$ particle $\to x_1=A\sin (\omega t+{\varphi }_1)$
The SHM equation of $2^{nd}$ particle $\to x_2=x_o+A\sin (\omega t+{\varphi }_2)$
Separation $\to |x_1-x_2|=x_o+A\sin (\omega t+{\varphi }_1)-A\sin (\omega t+{\varphi }_2)$
$x_o+A=x_o+2A\sin \left(\frac{{\varphi }_1-{\varphi }_2}{2}\right)$
$\sin \left(\frac{{\varphi }_1-{\varphi }_2}{2}\right)=\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$
$\therefore {\varphi }_1-{\varphi }_2=\frac{\pi }{3}$