A particle of mass `'m'` moves on a horizont smooth line `AB` of length `'a'` such that when particle is at any general point `P` on the line two force act on it. A force `(mg(AP))/(a)` towards `A` and another force `(2mg(BP))/(a)` towards `B`. particle released from rest from mid point of AB.
Find the minimum distance of the particle from `B` during the motion.

Options

A. `(sqrt(2ga))/(3)`
B. `(sqrt(2ga))/(6)`
C. `(sqrt(2ga))/(5)`
D. `(sqrt(ga))/(3)`

Answer

B

Explanation

`(i)` force on particle at point `P`
`f = (2mg(a-x))/(a)-(mg(x))/(a)`
`F = (mg)/(a)(2a - 3x)`
m `F = (-3mg)/(a)(x - 2a//3)`

`(ii)` So this is equation of `S.H.M. (F = momega^(2)x)` so particle
Perform `S.H.M.` with mean position `x - 2a//3 = 0`
So `omega^(2) = (3g)/(a) rArr omega = sqrt(3g//a)`
So Time period `T = 2pi sqrt(a//3g)`
and amplitude `= 2a//3 - a//2 = a//6`
`(iii)` minimum distance from `B`
`= a - (2a//3 + a//6) = a//6`
`(iv)` at point `P` velocity of partivle `= 0`
and force `= (2mgx)/(a)`
(at point `q`)
acc. `= (2gx)/(a) rArr overset(v)underset(o)(int)vdv = overset(a//6)underset(o)(int)(2gxdx)/(a) rArr V = (sqrt(2ga))/(6)`