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

Area of cross-section of the U-tube=A
Density of the mercury column = ρ
Acceleration due to gravity =g
Restoring force, F = Weight of the mercury column of a certain height
F = –(Volume × Density × g)
F = –(A × 2h × ρ ×g) = –2Aρgh = –k × Displacement in one of the arms (h)
Where,
2h is the height of the mercury column in the two arms
k is a constant, given by $k=\frac{-F}{h}$=2Aρg

Time period,$T=2π\sqrt\frac{m}{k}=2π\sqrt\frac{m}{2Aρg}$
Where,
m is the mass of the mercury column
Let l be the length of the total mercury in the U-tube
Mass of mercury, m = Volume of mercury × Density of mercury
=Alρ
$∴ T=2π\sqrt\frac{Alρ}{2Aρg}=2π\sqrt\frac{l}{2g}$

Hence, the mercury column executes simple harmonic motion with time period $2π\sqrt\frac{l}{2g}.$