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 thhird their amplitude. What is the phase difference between them?

Explanation

Let `theta` be the phase difference between tow SHMs. Let these SHMs be represented by
`y=rsinomegat ` and `y=rsin(omegat+theta)`
When, `y=r//3, sinomega t=y//r=1//3`
and `cosomegat=sqrt(8)//3`
`:. (r)/(3)=rsin(omegat+theta)`
`=r(sinomegatcostheta+cosomega tsintheta)`
or `(1)/(3)=(1)/(3)costheta+(sqrt(8))/(3)sintheta`
or `costheta-1=-sqrt(8)sintheta`
Squaring both the sides we have
`cos^(2)theta-2costheta+1=8sin^(2)theta`
`=8(1-cos^(2)theta)`
or `9cos^(2)theta-2costheta-7=0.`
L On solving, `costheta=1 or -7//9`
When `costheta=1, theta=0^(@)` (It is not possible)
When `costheta=-7//9=-0.7778, `
then `theta=141^(@)4^(')`
Thus phase difference `=141^(@)4^(')`