Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions, when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendic…

Question

Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions, when their displacement 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

Accepted Answer

(3) (2π)/3. Given: both particles at x=A/2 at the same instant, moving in opposite directions. Idea: write each as x=A sin θ and let the direction of travel pick which solution for θ belongs to which particle. sin θ=1/2 gives θ=π/6 or θ=(5π)/6. At θ=π/6 the particle is moving outwards (cos θ>0); at θ=(5π)/6 it is moving back inwards (cos θ<0). Δθ=(5π)/6-π/6=(4π)/6 Δθ=(2π)/3

Step-by-step solution

Why the other options are wrong