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)/(6)`
B. `(5T)/(6)`
C. `(T)/(3)`
D. `(T)/(4)`

Answer

A

Explanation

By considering projection of uniform circular motion on diameter of circle :
Let the particle 1 starts from point P at t = 0 and reaches at point Q in time t and particle 2 starts from R at t = 0 and reaches at point S in time t and they meet at M.
Angular displacement
`theta - 30^(@) = 90^(@) - theta`
(By geometry alternate interior angle)
` 2theta = 120^(@)`
`theta = 60^(2) = (pi)/(3) rArr omegat = (pi)/(3)`
`t = (T)/(6)`