A light inextensible string carrying a springs is passed over two smooth fixed pulles as shown. If the rod is slightly displaced about the hinge form is equilibrium position, then time period is.

Options

A. `pi sqrt((4ml)/(3(kl+2mg)))+pisqrt((2l)/(3g))`
B. `pi sqrt((2ml)/(kl+2mg)) +pi sqrt((2l)/(3))`
C. `pi sqrt((3ml)/(kl_2mg)) +pi sqrt((l)/(3))`
D. `pi sqrt((4ml)/(ml+2mg)) +pi sqrt((l)/(3))`

Answer

A

Explanation

Let the rod is displaced by `theta` towards left:
`:.` net torque on rod is
`-k((l)/(2)) sin theta.(l)/(2) cos theta + mg sin theta (l)/(2) +k(l)/(2) sin thetal cos theta`
as `theta` is small. `sin theta rarr theta, cos theta rarr 1`

`rArr tau = (kl^(2))/(2) theta +(mgl)/(2). theta - (kl^(2))/(4) theta`
`= [(kl^(2))/(4)+(mgl)/(2)] theta = ((ml^(2))/(3)) alpha`
`:. t_(1) =(T)/(2) = pi sqrt((4ml)/(3(kl+2mg))) ..........(1)`

When the rod is displaced right by an angle `theta` then string becomes slack i.e., rod becomes physical pendulum. `:. t_(22) = T =(1)/(2) = 2pi sqrt((I)/(mgd))`

where `I = (ml^(2))/(3), d = l//2`, time period `=t_(1) +t_(2) = ?`