A and B are two fixed points at ta distance `3l` apart. A particle of mass `m` placed at a point `P` experiences the force `2((mg)/(l))vecPA` and the force `((mg)/(l))vecPB` sumultaneously. Initially at `t=0`, the particle is projected from A towards B with speed `3sqrt(gl)`.
Q. The instant `t` at which the particle arrives at B in terms of the periodic time T will be

Options

A. `t=(T)/(2)`
B. `t=(T)/(3)`
C. `t=(T)/(4)`
D. `t=(2T)/(3)`

Answer

B

Explanation

`OB=2l`, `O` is the mean position, B is the amplitude point Also `|AP|=l=(a)/(2)`.
Hence the time t from A to O is given by `(a)/(2)=asinomegat`
Given `omegat=(pi)/(6)impliest=(pi)/(6omega)=(pi)/(6)xx(T)/(2pi)=(T)/(12)`
hence time from A to B `=(T)/(12)+(T)/(4)=(T)/(3)`