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`

Options

A. `5 sec`
B. ` 2 sec`
C. `4 sec`
D. `6 sec`

Answer

B

Explanation

The cordirates of the particle are
`X_(1) = A_(1) cos omega t, x_(2) = A_(2) cos omega^(2)t`
seperation `= x_(1) - x_(2) = (A_(1)- A_(2)) cos omegat 12 cos omegat`
Now `= v_(1) - v_(2) = 6 = 12 cos omega t`
`rArr omegat = (pi)/(3) rArr (2pi)/(12).t = (pi)/(3)`
`rArr t = 2s`.