Two particle executing SHM along a straight linehave same amplitude A and time period T. At t=0 one particle is at a displacement +A and another is at a displacement `-A//2` and they are approaching towards each other. They cross each other after a tme.

Explanation

Equation can be written as : `x_(1)=Acos omega t` and
`x_(2)=Asin(omegat-pi//6)" " "Here" omega=(2pi)/T`
Equating `x_(1)=x_(2)" "Sinx(omegat-(pi)/6)=cos omegat`
`sinomegat"cos"(pi)/6-"sin"(pi)/6cosomegast=cos omegat`
`tan omega t=sqrt(3), omegat=(2pi)/T t=(pi)/3`, we get `t=T//6`