Two particle are oscillating along two close parallel straight lines side by side, with the same frequency and amplitude. They pass each other, moving on 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.........

Options

A. `0`
B. `(2pi)/(3)`
C. `pi`
D. `(pi)/(6)`

Answer

B

Explanation

For SHO `x= A sin (omega t+ phi), omega t+ phi` is phase
For first particle `x= (A)/(2)`
`A sin (omega t+phi_(1))= (A)/(2)`
`therefore sin (omega t+ phi_(1))= (1)/(2) = sin (pi)/(6)`
`therefore phi_(1) = (pi)/(6)" ""……"(1)`
For second particle moving in opposite motion
`therefore phi_(2) = pi -(pi)/(6)`
`= (5pi)/(6)" ""........."(2)`
Phase difference `phi_(2)- phi_(1) = (5pi)/(6)- (pi)/(6)`
`=(4pi)/(6)= (2pi)/(3)`.