Question

Two particles execute S.H.M of same amplitude and frequency along the same straight line from same mean position. They cross one another without collision, when going in opposite direction, each time their displacement is half of their amplitude. The phase-difference between them is

• A. $0^{\circ }$
• B. ${120}^{\circ }$
• C. ${180}^{\circ }$
• D. ${135}^{\circ }$

Answer 1

The correct option is B ${120}^{\circ }$

Since both particles have same amplitude and frequency, its equation is

$x_1=x_m\cos \omega t{x}_2=x_m\cos (\omega t+\varphi )$

Crossing happens when $x_1=\frac{x_m}{2}$

So $x_1=x_m\cos \omega t\frac{x_m}{2}=x_m\cos \omega t\omega t=\frac{\pi }{3}$

Similarly for $x_2$, but it is moving in different direction so the closest value of $\omega t+\varphi =\frac{2\pi }{3}$

$\omega t+\varphi =\frac{2\pi }{3}\frac{\pi }{3}+\varphi =\frac{2\pi }{3}\varphi =\frac{\pi }{3}$

Ans: 120 degreee