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.
$\therefore \delta ={\cos }^{-1}\left(\frac{7}{9}\right)$