Question

Determine the period of oscillation of mercury of mass m=200g poured into a bent tube whose right arm forms an angle $\theta ={30}^o$ with the vertical. the cross -sectional area of the tube $S=0.50c{m}^2$ the viscosity of mercury is to be neglected.

Answer 1

If the mercury rises in the left arm by $x$ it must fall by a slanting length equal to $x$ in the other arm. Total pressure difference in the two arms will then be
$\rho gx+\rho gx\cos \theta =\rho gx(1+\cos \theta )$
This will give rise to a restoring force
$\rho gSx(1+\cos \theta )$
This must equal mass time acceleration which can be obtained from work energy principle.
The K.E. of the mercury in the tube is clearly : $\frac{1}{2}m{\dot{x}}^2$
So mass times acceleration must be : $m\ddot{x}$
Hence $m\ddot{x}+\rho gS(1+\cos \theta )x=0$
This is S>H>M. with a time period
$T=2\pi \sqrt{\frac{m}{\rho gS(1+\cos \theta )}}$