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. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `120^(@)`

Answer

D

Explanation

Equation of motion of simple harmonic oscillator in Y-axis
`y= a sin (omega t+phi)`
`y= a sin theta`
where, `omega t+phi = theta` (phase)
According to question,
`therefore (a)/(2) = a sin theta`
`therefore sin theta = (1)/(2)`
`theta = (pi)/(6)" or "(5pi)/(6)`
At `theta_(1)= (pi)/(6)`, particle is at P and move toward A.
At `theta_(2) = (5pi)/(6)`, particle is at P and move toward O.
Hence, phase difference between both particle at `P= theta_(2)- theta_(1)`
`=(5pi)/(2)-(pi)/(6)`
`=(4pi)/(6)`
`= (2pi)/(3)= (2xx 180^(@))/(3)`
`=120^(@)`.