Question

Two particles are oscillating along two close parallel straight lines side by side. With the same frequency and amplitude . They pass each other moving in opposite directions when their displacements is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is?

Answer 1

X=a sinθ
a/2=a sinθ
θ=30°,150°
∴ difference = 150°–30°=120°
                     =$2 \pi /3$

Answer 2

The phase difference of the particles is $\frac{2 \pi}{3}$.

Solution:

The given situation states that the two particles oscillates along two close parallel straight lines having same amplitude as well as frequency and they happen to pass each other when their displacement is half of amplitude, in opposite directions.  

Also, the mean positions of the two particles lies on a straight line which is perpendicular to the paths of those two particles. Thereby, we have the displacement formula as $x=a \sin \theta$.

And we know that $x=\frac{a}{2}$

Therefore, $x=a \sin \theta$

$\begin{array}{c}{\frac{a}{2}=a \sin \theta} \\ {\sin \theta=\frac{1}{2}} \\ {\theta=\sin -1\left(\frac{1}{2}\right)} \\ {\theta=30^{\circ} \text { or } 150^{\circ}} \\ {\text { Phase difference, } \phi=150-30=120^{\circ}} \\ {\text { Phase difference is } \frac{2 \pi}{3} \text { . }}\end{array}$