Ncert Exercises & Solutions

14.19 One end of a U-tube containing mercury is connected to a suction pump and the other end to the 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.

Let the area of a cross-section of the U-tube = A
And the 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 and k is a constant

$k = - \frac{F}{h} = 2 Aρg$
Time period,

$T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{m}{2 Aρg}}$

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

$∴ T = 2 π \sqrt{\frac{Alρ}{2 Aρg}} = 2 π \sqrt{\frac{l}{2 g}}$

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

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