Question

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.

Answer 1

It is given that one end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere.

When the pump is removed then, a net force acts on the liquid column because of the difference in the levels of mercury in the two limbs, hence the liquid column executes simple harmonic motion.

Let the level of mercury in a vertical U-tube up to P and Q in its two limbs.

Let the density of mercury is ρ, the total length of the mercury column in both limbs is L, the internal cross-sectional area of U-tube is A.

The mass of mercury is given as,

m=LAρ

Let assumed that the mercury be depressed in left limb to F ′ by a small distance y after this it rises by the same amount in the right limb to position Q'.

The difference in levels in the two limbs is given as,

P'Q'=2y

The volume of mercury contained in the column of length 2y is given as,

V=A×2y

The weight of liquid contained in the column of length 2y is given as,

W=mg =A×2y×ρ×g

The restoring force is given as,

F=−2Ayρg

The acceleration is given as,

a= F m

By substituting the given values in the above expression, we get

a=− 2Aρgy LAρ =− 2ρg 2hρ y

Where, the height of mercury in each limb is h.

The time period is given as,

T=2π y a

By substituting the given values in the above expression, we get

T=2π 2hρ 2ρg =2π h g

Thus, the time period is 2π h g so, the column of mercury in the U-tube executes simple harmonic motion.