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 direction 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:

 The phase difference is = 2∏/3 radians

Solution

Let us represent the particles by equation y1 and y2 respectively.

y1 = A sin ωt

Given A/2 = A sin ωt which implies ωt = 30º

Similarly, y2 = A sin ωt + ∏/2

Given A/2 = A sin ωt which implies ωt + ∏/2= 150º

Hence, the phase difference between y1 and y2

= 150º – 30º = 120º

= 2∏/3 radians

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