One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.

Explanation

The suction pump creats the pressure difference, thus mercury rises in one limb of the U-tube. When it is removed, a net force acst on the liquid column due to the difference in levels of mercury in the two limbs and hence the liquid column executes S.H.M which can be expressed as consider the mercury contrained in a vertical U tube upto the level P and Q in its two limbs.,
Let P= density of the mercury
L=total length of the mercury column in both the limbs.
A=internal cross-sectional area of U-tube .m =mass of mercury in U-tube =LAP
Assume, the mercury be depressed in left limb to F by a small distance y, then it rises by the same amount in the right limb to position Q'.
`:.` Difference in levels in the two limbs =P'Q' =2y
`:.` Volume of mercury contained in the column of length 2y=A x 2y
`:. m-Axx 2y xxrho`
if W =weight of liquid contained in the column of length 2y
Then W=mg =A x 2y x `rho` x g
This weight products the restoring force (F) which tends to bring back the mercury to its equilibrium position .
`:. F=-2Ay rho g=-(2A rho g) y`
if a =acceleration produced in the liquid column, then
`a=F/m`
`=-((2A rho g)y)/(LArho) =-(2A rhog)/(LA)`
`=-(2rhog)/(2hrho) y.........(i) ( :. L=2h)`
Where =h height of mercury in each limb. now from eqn (i), it is clear that `a pro y` and -ve sign shows that it acts opposite to y,so the motion of mercury in u-tube is simple harmonic in nature having time period (T) given by
`T=2pi sqrt(y/a) =2pisqrt((2h rho)/(2rhog))=2pi sqrt((hrho)/(rhog))`
`T=2pisqrt(h/g)`