Question

Two particles are oscillating along the same line with the same frequency and same amplitude. They meet each other at a point mid-way between the mean position and extreme position while going in opposite direction. The phase difference between their motions of

• A. $\pi /3$
• B. $\pi /2$
• C. $2\pi /3$
• D. $5\pi /4$

Answer 1

The correct option is C $2\pi /3$

Two particles execute simple harmonic motion.
Let $\varphi$ be the phase difference between the two simple harmonic motions.
The two simple harmonic motions are represented by $y_1=a\sin \omega t$
$y_2=a\sin (\omega t+\varphi )$
Let the two particles meet one another, while moving in opposite direction at $t=t_1$
$\therefore$ At $t=t_1,y_1=y_2=\frac{a}{2}\therefore \frac{a}{2}=a\sin \omega t_1$
$\sin \omega t_1=\frac{1}{2}.....\left(i\right)$
$\frac{a}{2}=a\sin (\omega t_1+\varphi )\Rightarrow \sin (\omega t_1+\varphi )=\frac{1}{2}....\left(ii\right)$
$\frac{1}{2}=\sin \omega t_1\cos \varphi +\sin \varphi \cos \omega t_1$
$=\frac{1}{2}=\frac{1}{2}\cos \varphi +\sin \varphi \sqrt{1-{\sin }^2\omega t_1}$ [Using (i)]
$\frac{1}{2}=\frac{1}{2}\cos \varphi +\sin \varphi \sqrt{1-{\left(\frac{1}{2}\right)}^2}$
$\frac{1}{2}=\frac{1}{2}\cos \varphi +\frac{\sqrt{3}}{2}\sin \varphi \Rightarrow 1=\cos \varphi +\sqrt{3}\sin \varphi$
$1-\cos \varphi =\sqrt{3}\sin \varphi ....\left(iii\right)$
Square equation (iii), on both sides
$1+{\cos }^2\varphi -2\cos \varphi =3{\sin }^2\varphi$
$1+{\cos }^2\varphi -2\cos \varphi =3(1-{\cos }^2\varphi )$
$4{\cos }^2\varphi -2\cos \varphi -2=0$
$(2\cos \varphi +1)(\cos \varphi -1)-0....\left(iv\right)$
Equation (iv) is satisfied if either
$2\cos \varphi +1=0$ i.e. $\cos \varphi =-\frac{1}{2}$
or $\varphi ={120}^{\circ }=\frac{2\pi }{3}$ or $\cos \varphi -1=0$
$\cos \varphi =1$ or $\varphi =0$
$\varphi =0$ is not the solution to the problem.
Therefore $\varphi ={120}^{\circ }$.