Two particles `P_(1)` and `P_(2)` are performing `SHM` along the same line about the same meabn position , initial they are at their position exterm position. If the time period of each particle is `12` sec and the difference of their amplitude is `12 cm` then find the minimum time after which the seopration between the particle becomes `6 cm`

Explanation

The x coordinates of the particles are
`x_(1)=A_(1) cos omegat, x_(2)=A_(2) cos omegat`
separation `=x_(1)-x_(2)=(A_(1)-A_(2))cos omegat=12 cos omegat`
`rArr omegat=(pi)/3rArr (2pi)/12.t=(pi)/3rArr t =2s` Ans.