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.

Asked by Abhisek | 1 year ago |  83

##### Solution :-

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 x Density x g)

F = – (A x 2h x ρ x g)

= – 2Aρgh

= – k x 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 =$$\dfrac{-F}{h}$$

We get,

k = 2Aρg

Time period, T = 2π ($$\sqrt{\dfrac{m}{k}}$$)

On substituting k value, we get,

Time period, T = 2π ($$\sqrt{\dfrac{m}{2Aρg}}$$)

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 x Density of mercury

= Alρ

Hence,

T = 2π ($$\sqrt{\dfrac{Alρ}{2Aρg}}$$)

T = 2π ($$\sqrt{\dfrac{I}{2g}}$$)

Therefore, the mercury column executes simple

harmonic motion with time period 2π $$\sqrt{\dfrac{I}{2g}}$$

Answered by Pragya Singh | 1 year ago

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