Question

The end which contain mercury of a U-tube 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 $=\rho$
Acceleration due to gravity $=g$
Restoring force, $F=$ Weight of the mercury column of a certain height
$F=-(Volume\times Density\times g)$
$F=-(A\times 2h\times \rho \times g)=-2A\rho gh=-k\times$ 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 $k=-F/h=2A\rho g$

Time Period, $T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{m}{2A\rho g}}$

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 $\times$ Density of mercury $=Al\rho$

$\therefore T=2\pi \sqrt{\frac{Al\rho }{2A\rho g}}=2\pi \sqrt{\frac{l}{2g}}$