Question

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

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

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{7}{9}\right)$

Let the positions of two particles at any instant is given by,
$x_1=A\sin \omega t...\left(1\right)$
and $x_2=A\sin (\omega t+\delta )...\left(2\right)$
At a certain instant $x_1=\frac{A}{3}$ and $x_2=-\frac{A}{3}$
Substituting in the Eq.$\left(1\right),\left(2\right)$ respectively we get,
$\frac{A}{3}=A\sin \omega t$
and $-\frac{A}{3}=A\sin (\omega t+\delta )$
This gives,
$\sin \omega t=\frac{1}{3}...\left(3\right)$
$\sin (\omega t+\delta )=-\frac{1}{3}...\left(4\right)$
On expanding Eq.$\left(4\right)$,
$\Rightarrow \sin \omega t\cos \delta +\cos \omega t\sin \delta =-\frac{1}{3}$
substituting for $\sin \omega t$ and $\cos \omega t=\sqrt{1-{\sin }^2\omega t}$,
$\Rightarrow \frac{1}{3}\cos \delta +\left(\sqrt{1-\frac{1}{9}}\right)\sin \delta =-\frac{1}{3}$
Again substituting $\sin \delta =\sqrt{1-{\cos }^2\delta }$ and simplifying we get,
$\Rightarrow 9{\cos }^2\delta +2\cos \delta -7=0$
On solving the quadratic equation, we get the two roots:
$\Rightarrow \cos \delta =-1\&\cos \delta =\frac{7}{9}$
$\therefore \delta ={180}^{\circ }\text{or}\delta ={\cos }^{-1}\frac{7}{9}$
If we consider $\delta ={180}^{\circ }$, then from Eq.$\left(1\right)\&\left(2\right)$ we can conclude that velocity of particle $v_1$ and $v_2$ are in opposite direction.
But it is given that velocities are in same direction, hence $\delta ={\cos }^{-1}\left(\frac{7}{9}\right)$ is the correct answer.