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Home » 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.
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.
CBSE | Class 11 | NCERT | Oscillations | Physics
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.

Area of cross-section of the U-tube is given as A

Density of the mercury column is given as \rho

Acceleration due to gravity is given as g

Restoring force, F = Weight of the mercury column of a certain height

F=-(Volume\times Density\times g)
F=-(A\times 2h \times \rho \times g)
F=-2A\rho gh
F=-k \times Displacement in one of the arms

Where,

2h = height of the mercury column in the two arms

k is a constant, given by

k = – F / h

So,

k=2A\rho g

Time period will be T = 2π √(m / k)

On substituting k value, we get,

Time period, T = 2π √(m / 2Aρg)

Where,

m = mass of the mercury column

Let the length of the total mercury in the U-tube be l

Mass of mercury, m = volume of mercury x Density of mercury

=Al\rho

Hence,

T = 2π √(Alρ / 2Aρg)

T = 2π √(l / 2g)

As a result, the mercury column moves in a simple harmonic motion with a 2π √(l / 2g) time period.

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