Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. Their phase difference is

Options

A. `(2pi)/(3)`
B. `(pi)/(3)`
C. `(4pi)/(3)`
D. `(3pi)/(4)`

Answer

A

Explanation

The given two particles pass each at time t, at `x_(1)=x_(2)=(1)/(2)x_(m)` Where
`x_(1)=x_(m)cos(omegat+phi_(1))`and `x_(2)=x_(m)cos(omegat+phi_(1)`
From this ,we conclude that `(omegat+phi_(1))=cos(omegat+phi_(2))=(1)/(2)`
And therfore the pases (the argment of the cosines ) are either both equal to `(pi)/(3)`or one is `(pi)/(3)`while the other is `-((pi)/(3))`.Also at this instant , we have `v_(1)=-v_(2)ne0`,where `f_91)=-x_(m)omegacos (omegat+phi_(1)=-sin (omegat+phi_(2))`
This leads us to m conclude that yhe phases have opposite sogh .Thus , one phase is `(pi)/(3)` and the other phase `-(pi)/(3)`,the `omega tau` trem can-cells if we take phase difference , which is seen to be `(pi)/(3)-(-(pi)/(3))=2(pi)/(3)`