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. Tha particle moves simple harmonically with period `T` and amplitude `A`.

Options

A. `A=2l`,`T=2pisqrt((l)/(g))`
B. `A=3l`,`T=2pisqrt((l)/(2g))`
C. `A=2l`,`T=2pisqrt((l)/(3g))`
D. `A=l`,`T=2pisqrt((l)/(g))`.

Answer

C

Explanation

Take x axis along AB and origin at A.
let `AP=x`, then `PB=(3l-x)`
The force on the particle of mass m at P is `vecF` given by
`vecF=2((mg)/(l))(-vecx)+(mg)/(l)((3l-x)/(x))vecx`
`=(3mg-(3mg)/(l)x)` directed towards A
`=(3mg)/(l)(l-x)` directed towards A
Hence equation of motion of mass m is
`m(d^2x)/(dt^2)=-(3mg)/(l)(l-x)=momega^2(l-x)`
where `omega^2=(3g)/(l)`
This is a simple harmonic motion of period `T=2pisqrt((l)/(2g))`.
The mean position is at the point `x=l` (i.e.) at `O` where `AO=l`.
Now for point A displacement `=-l` and hence
`v_A^2=omega^2(a^2-(-l)^2)=omega^2(a^2-l^2)`
where `v_A^2=omega^2(a^2-(-l)^2)=omega^2(a^2-l^2)` and a is the amplitude.
now `n_A=sqrt(gl)` (given) and hence `v_A^2=9gm=(3g)/(l)(a^2-l^2)`
`a^2=4l^2` or `a=2l`