Four identical bars of mass `m`, length `l` are connected by pins at `A,B,C` and `D`. The bars are attached to four springs of same stiffness as shown. The entire sysetm can move in horizontal plane. In the equilibrium position shown, `theta = 45^(@)`. if the corners `A` and `C` are given small displacements towards each other and released, then time period of vibration is

Options

A. `2pi sqrt((2M)/(3K))`
B. `2pi sqrt((4m)/(K))`
C. `2pi sqrt((M)/(4K))`
D. `2pi sqrt((M)/(K))`

Answer

A

Explanation

Let the angle turned by the barse be '`theta`' after being displaced form equilibrium position. Compression in springs `A,C` is `Deltax_(1) = l [cos (45^(@) - theta)-cos 45^(@)]`extension i springs `B,D` is
`Deltax_(2) = l [cos 45^(@)-cos (45^(@)+theta)]`
`:.` Potential energy
`=2 [(1)/(2)K Deltax_(1)^(2)] +2 [(1)/(2)K Deltax_(2)^(2)]` ..............(1)

`~~ Kl^(2) theta^(2) ( :'` after simplication)
Kinetic energy =
`(1)/(2)I ((d theta)/(dt))`, where `I = (ml^(2))/(12) +m ((l)/(2)tan 45^(@))^(2) = (ml^(2))/(3)`
`:. dE//dt = 0 rArr` solve for `T`.