Question

Two particles move parallel to the $x$-axis about the origin with same amplitude ' $a$ ' and frequency $\omega$. At a certain instant they are found at a distance $a / 3$ from the origin on opposite sides but their velocities are in the same direction. What is the phase difference between the two?

• A. $\cos ^{-1} \frac{7}{9}$
• B. $\cos ^{-1} \frac{5}{9}$
• C. $\cos ^{-1} \frac{4}{9}$
• D. $\cos ^{-1} \frac{1}{9}$

Answer 1

Answer: (A)

Let $x_{1}=a \sin \omega t$ and $x_{2}=a \sin (\omega t+\delta)$ be two SHMs.
$\frac{a}{3}=a \sin \omega t$ and $-\frac{a}{3}=a \sin (\omega t+\delta)$
$\sin \omega t=1 / 3$ and $\sin (\omega t+\delta)=-1 / 3$
Eliminating $t, \frac{1}{3} \cos \delta+\sqrt{1-\frac{1}{9}} \sin \delta=-\frac{1}{3}$
$9 \cos ^{2} \delta+2 \cos \delta-7=0$
$\cos \delta=-1 $ or $ \frac{7}{9}$
i.e., $\delta=180^{\circ}$ or $\cos ^{-1}\left(\frac{7}{9}\right)$
If we put $180^{\circ}$, we find that $v_{1}$ and $v_{2}$ are of opposite signs.
Hence $\delta=180^{\circ}$ is not applicable.
$\therefore \delta=\cos ^{-1}\left(\frac{7}{9}\right)$