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. `(4T)/(6)`
D. `(T)/(6)`

Answer

D

Explanation

From phasor diagram,
`omegat_0-(pi)/(6)+omegat_0=(pi)/(2)`
or `2omegat_0=(pi)/(6)+(pi)/(2)=(pi+3pi)/(6)=(4pi)/(6)=(2pi)/(3)`
`t_0=(2pi)/(6omega)=(2pi)/(6xx(2pi)/(T))=(T)/(6)`