A uniform cylinder of mass m, length l and cross sectional area A is suspended with its vertical from a fixed point, with the help of a massless spirng of force constant (K), The cylinder is then submerged in a liquid of density p. At equilibrium the cylinder remains submerged in the liquid with 1//4^(th) of its volume outside the liquid,as shown in the figure. The elongation of hte spring at this point is x_(0). WHen the cylinder is given a slight downward push into the liquid and then released, its starts oscillating with vertical S.H.M. of small amplitude. Time period of vertical S.H.M of the cylinder is

A uniform cylinder of mass m, length l and cross sectional area A is s

1.	A uniform cylinder of mass m, length l and cross sectional area A is suspended with its vertical from a fixed point, with the help of a massless spirng of force constant (K), The cylinder is then submerged in a liquid of density p. At equilibrium the cylinder remains submerged in the liquid with 1//4^(th) of its volume outside the liquid,as shown in the figure. The elongation of hte spring at this point is x_(0). WHen the cylinder is given a slight downward push into the liquid and then released, its starts oscillating with vertical S.H.M. of small amplitude. Time period of vertical S.H.M of the cylinder is
Answer» `2pi((m)/(K+APG))^(1//2)` `2pi((m)/(K-Apg))^(1//2)` `2pi((m)/(K-(1)/(4)Apg))^(1//2)` `2pi((m)/(K+(1)/(4)Apg))^(1//2)` SOLUTION :The ACCELERATION of the CYLINDER is `a. = (F)/(m) = -((k + Arhog)/(m))c` Comparing it with, `a. =-omega^(2)a` `omega^(2) = ((k + Arhog)/(m))` `omega = ((k + Arhog)/(m))^(1//2)` Time period, `T = (2pi)/(omega) = 2pisqrt((m)/(k + Arhog))^(1//2)`