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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 - Physics

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.

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Solution 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)


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

k is a constant, given by  `k = - F/h = 2Arhog`

Time period, `T = 2pisqrt(m/k) = 2pisqrt(m/(2Arhog))`


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 = 2pisqrt((Alrho)/(2Arhog)) = 2pi sqrt(l/2g)` 

Hence, the mercury column executes simple harmonic motion with time period `2pi sqrt(1/(2g))`

Solution 2

The suction pump creates the pressure difference, thus mercury rises in one limb of the U-tube. When it is removed, a net force acts on the liquid column due to the difference in levels of mercury in the two limbs and hence the liquid column executes S.H.M. which can be expressed as:

Consider the mercury contained in a vertical U-tube upto the level P and Q in its two limbs.

Let P = density of the mercury.

L = Total length of the mercury column in both the limbs.

A = internal cross-sectional area of U-tube. m = mass of mercury in U-tube = LAP.

Assume, the mercury be depressed in left limb to F by a small distance y, then it rises by the same amount in the right limb to position Q’.

∴ Difference in levels in the two limbs = P’ Q’ = 2y.

∴ Volume of mercury contained in the column of length 2y = A X 2y

∴m – A x 2y x ρ.

If W = weight of liquid contained in the column of length 2y.

Then W = mg = A x 2y x ρ x g

This weight produces the restoring force (F) which tends to bring back the mercury to its equilibrium position.

`:. F = -2 Ayrhog = -  (2Arhog)y`

if a ~~ acceleration produced in the liquid column, then

`a =  F/m`

= `-(2Arhog)/LArho = - (Arhog)/LA`

`= - (2rhog)/(2hrho) y`   .....(i)  (∵ L = 2h)

where h = height of mercury in each li,mb. Now from equation (i), it is clear that `a prop y` and -ve sign shows that it acts opposite to y, so the motion of mercury in u-tube is simple harmonic in nature having time period (T) given by

`T = 2pisqrt(y/a) = 2pisqrt((2hrho)/(2rhog)) = 2pisqrt(hrho/rhog)`

`T = 2pi sqrt(h/g)`




Concept: Force Law for Simple Harmonic Motion
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NCERT Class 11 Physics Textbook
Chapter 14 Oscillations
Q 19 | Page 361
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