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

Let a U-tube contain some liquid up to levels `A_(1), A_(2)`. Let L be the total length of liquid in the pipe and `rho` be the density of liquid.
Mass of liquid in the tube m = AL p, where A is the area cross A is the area cross section of U-tube.
If liquid is depressed through distance y in left side of the tube, the liquid will rise by y in the other tube and then release.
Before, release, the difference in level in the two tubes = 2y
Weight of liquid in column of length 2y in U tube will provide the restoring force to liquid.
Restoring force, F = -(2yA) p g = - (2A p g)y
We find that `F prop y`
Therefore, the liquid will execute S.H.M.
Spring factor, k = 2Apg
`T = 2pi sqrt(("Inertial factor")/("Spring factor"))`
`T = 2pi sqrt((LAp)/(2Apg))`
`T = 2pi sqrt((L)/(2g))`
Frequency, `n = (1)/(T) = (1)/(2pi) sqrt((2g)/(L))`
If h is the height of the liquid column in each limb, then L = 2h
`T = 2pi sqrt((2h)/(2g)) = 2pi sqrt((h)/(g))`
Hence the motion of liquid column in the U-tube is S.H.M.