A bob of mass `2m` hanges by a string attached to the block of mass `m` of a spring blocks syetem. The whole arrangement is in a state of equilibrium. The bob of mass `2m` is pulled down slowely by a distance `x_(0)` and released.

Options

A. For `x_(0) = (3mg)/(k)`, maximum tension in string is `4mg`
B. For `x_(0) gt (3mg)/(k)`, minimum tension in string is mg
C. Frequency of oscillation of system is `(1)/(2pi) sqrt((k)/(3m))`, for all non-zero values of `x_(0)`
D. The motion will remain simple harmonic for `x_(0) ge (3mg)/(k)`

Answer

A, D

Explanation

(A) `2(6 mg-T-mg-ma)-(T-2mg-2ma) = 0`
`12 mg - 3T = 0`
`T = 4mg`
(B) `Kx +T + mg = ma`
`2mg - T = 2ma` for `T = 0, a = g`
`rArr x =0`
As particle is released `x_(0)` below equilibriums so it will go `x_(0)` above equilibrium i.e. at `x =0 rArr T_(min) = 0`
(C ) & (D) for `x_(0) gt (3mg)/(k)` it will be longer `SHM` as string will slack