Two particles execute simple harmonic motion of same amplitude and frequency along the same straight line. They pass on another, when going in opposite directions, each time their displacement is half of their amplitude. What is the phase difference between them?

Explanation

In SHM, `x=rsin(omegat +phi)`
velocity, `v=(dx)/(dt)=romega cos (omegat +phi)`
At `t=2, x=r//s,` then
`(r)/(2)=rsinphi or sin phi=(1)/(2)=sin((pi)/(6))or sin ((5pi)/(6))`
`:. Phi=(pi)/(6) or (5pi)/(6)`
If `phi=(pi)/(6),` displacement and velocity both are positive at `t=0` when `phi=(5pi)/(6)` , displacement is positive and velocity is negative, Figure,
Therefore, displacement-time equations of two particles will be `x_(1)=rsin(omegat-pi//6)`
`x^(2)=rsin(omegat+5pi//6)`
Phase difference
`dphi=(5pi)/(6)-(pi)/(6)=(4pi)/(6)=(2pi)/(3) rad`