Determine the period of small oscillations of a mathematical pendulum, that is a ball suspended by a thread `l=20 cm` in length, if it is located in a liquuid whose density is `eta=3.0` times less than that of the ball. The resistance of the liquid is to be neglected.

Explanation

Let us depict the forces acting on the osciallating ball at an arbitraty angular position ` theta` (figure), relative to equilibrium position where `F_(B)` is the force of buoyancy. For the ball from the equation `:`
`N_(Z)=Ibeta_(Z)`, (where we have taken the positive sense of `Z` axis in the direction of angular velocity `i.e., theta` of the ball and passes through the point of suspension of the pendulum `O)`, we get :
`- mgl sin theta + F_(B) sin theta=m l^(2)ddot(theta) ...(1)`
Using `m=(4)/(3) pi r^(3) sigma, F_(B)=(4)/(3) pi r^(3) rho` and` sin theta~= theta` for small `theta`, for Eqn `(1)` , we get `:`
`ddot(theta)=-(g)/(l)(1-(rho)/(sigma))theta`
Thus the sought time period
`T=2pi(1)/(sqrt((g)/(l)(1-(rho)/(sigma))))=2 alpha sqrt((l//g)/(1-(1)/(eta)))`
Hence `t=2sqrt((etal)/(g(eta-1)))=1.1s`