Two particles are executing SHM in a straight line. Amplitude A and the time period T of both the particles are equal. At time t=0, one particle is at displacement `x_(1)=+A` and the other `x_(2)=(-A/2)` and they are approaching towards each other. After what time they across each other? `T/4`

Options

A. `T//3`
B. `T//4`
C. `5T//6`
D. `T//6`

Answer

D

Explanation

`x = A cos omegat`
`x_(2) = Asin (omegat - (pi)/(6))`
`omega = (2pi)/(T)`
`omega = (2pi)/(T)`
`x_(1) = x_(2)` we get `t = (T)/(6)`