Correct Answer - A
Let `theta` be the phase difference betweent two simple harmonic motions, r and `omega` be the amplitude and angular frequency of motion of the particles. The two SHMs are represented by the equations.
`y_(1)=rsin omegat and y_(2)=rsin (omegat+theta)`.
Let the two particles cross one another, moving in opposite directions at time `t=t_(1)` (say) when displacement of each particle is `r//2`, then
`y_(1)=y_( 2)=r//2` at time ` t=t_(1)`.
`:. (r)/(2)=rsin omegat_(1) or sinomegat_(1)=(1)/(2):. cos omega t_(1)=(sqrt(3))/(2)`
Also `(r)/(2)=rsin(omegat_(1)+theta) or (1)/(2)=sin (omegat_(1)+theta)`
`(1)/(2)=sin omegat_(1)costheta+cosomegat_(1)sintheta`
or `(1)/(2)=(1)/(2)costheta+(sqrt(3))/(2)sintheta`
or `1-costheta+sqrt(3) sin theta or 1-cos theta =sqrt(3) sin theta`
squaring both sides, we get
`1+cos^(2)theta-2cos theta=3sin^(2)theta=3(1-cos^(2)theta)`
`4cos^(2)theta-2cos theta-2=0`
or `(4cos theta +2)(cos theta-1)=0`
when `4cos theta -1=0, cos theta =1, theta=0^(@)`, this is true when both particles start together.
`:. `Phase difference, `theta=120^(@)`
