An elastic string of constant `k` is attached to a block of mass `m` as shown. The block is given an extension of `(2mg)/(k)` from the equilibrium position and released. The time period of oscillations of the block is

Options

A. `2pi sqrt((m)/(k))`
B. `4sqrt((m)/(k)) ((pi)/(3)+sqrt(3))`
C. `2pi sqrt((m)/(k)) +2sqrt(3)sqrt((m)/(k))`
D. `pi sqrt((m)/(k)) +sqrt(3) sqrt((m)/(k))`

Answer

B

Explanation

Here string becomes slack when acceleration of body becomes more than g above equilibrium position. Hence during that period, `SHM` disappears:

upto `P_(3), SHM` appears and between `P_(3) & P_(4), SHM` disapears. `= (2T_(1))/(3) +(t_(P_(3)P_(4))+t_(P_(4)P_(3))):`
`:. T = t_(OP_(2))+t_(P_(2)O) +t_(OP_(3))+t_(P_(3)P_(4)) +t_(P_(4)P_(3)) +t_(P_(3)O)`
where: `T_(1) = 2pi sqrt((,)/(k))` velocity at `P_(3)` is:
`v = omega sqrt(A^(2)-x^(2)) = omega sqrt(4b^(2)-b^(2)) = b omega sqrt(3)`
`t_(P_(3)P_(4)) +t_(P_(4)P_(3)) = (2v)/(g)`