Two particles execute SHM of same amplitude and frequency on parallel lines. They pass one another when moving in opposite directions each time their displacement is one third their amplitude. What is the phase difference between them?

Options

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

Answer

C

Explanation

Let `x = A "sin" (omega t + alpha)` be the equation of S.H.M. and it is given that `x=(A)/(2)`
`therefore (A)/(2)=A "sin" (omega t+ alpha) therefore "sin" (omega t+alpha)=(1)/(2)`
`therefore (omega t+ alpha)=(pi)/(6)`
The particles is going from O to Q and is at R.
The other particle crosses the first particle at R, but it is the opposite direction and it is going towards O.
`because "sin" (180-theta)="sin" theta or "sin" (pi-theta)= "sin" theta`
`therefore` The other angle will be `(pi-(pi)/(6))=(5pi)/(6)`
`therefore` The phase difference
` =(5pi)/(6)-(pi)/(6)=(4pi)/(6)=(2pi)/(3)=120^(@)`