Correct Answer – A::B::D
Let `x_(0)` be the extension in the spring in equilibrium. Then equilibrium of `A` and `B` give,
`T = kx_(0) + mg sin theta` …(i)
and `2T = mg` …(ii)
Here, `T` is the tension in the string. Now, suppose `A` is further displaced by `(x)/(2)` and speed of `B` at this instant will be `(v)/(2)`. Total energy of the system in this position will be,
`E = (1)/(2) k(x + x_(0))^(2) + (1)/(2)m_(A)v^(2) + 1/2mB((v)/(2))^(2) + m_(A)gh_(A) – m(B)gh_(B)`
or `E = (1)/(2)k(x + x_(0))^(2) + 1/2mv^(2) + 1/8mv^(2) + mgxsin theta -mg x/2`
or `E = (1)/(2)k(x + x_(0))^(2) + 5/8mv^(2) + mgxsin theta – mg x/2`
Since, `E` is contant,
`(dE)/(dt) = 0`
or `0 = k(x + x_(0))(dx)/(dt) + 5/4mv ((dv)/(dt)) + mg (sin theta)((dx)/(dt)) – (mg)/(2)((dx)/(dt))`
Substituting, `(dx)/(dt) = v` rArr `(dv)/(dt) = a`
and `kx_(0) + mg sin theta = (mg)/(2)`
[From Eqs. (i)and(ii)]
We get, `(5)/(4)m a = – kx`
Since, `a prop – x`
Motion is simple harmonic, time period of which is,
`T = 2pi sqrt|(x)/(a)|`
`= 2pi sqrt((5m)/(4k))`
`:. omega = (2pi)/(T) = sqrt((4k)/(5m))`