Correct Answer - B
As particle starts from rest, we have, `x=Acos omegat `
At `t=0, x=Acosomegaxx0=A`
When `t=tau, x=(A-a)=Acosomegatau` ..(i)
or `cos omegatau=(A-a)/(A)`
When `t=2tau, x=A-3a=Acos2omegatau`
`=2Acos^(2)omegatau-A`
or `A-3a=2A((A-a)/(A))^(2)-A`
or ` (A-3A)/(A)=2((A-a)/(A))^(2)-1`
or `A^(2)-3aA=2A^(2)+2a^(2)-4Aa-A^(2)`
`Aa=2a^(2) or A=2a`
From (i)`A-a=Acosomegatau`
`2a-a=2a cos omegatau or cos omega tau =1//2=cos pi//3`
or `(2pi)/(T)tau=(pi)/(3) or T=6tau`