The U tube of uniform bore of cross-sectional area A has been set up vertically with open ends facing up. Now M gram of a liquid of density d is poured into it. The column of liquid in this tube will

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The U tube of uniform bore of cross-sectional area A has been set up vertically with open ends facing up. Now M gram of a liquid of density d is poured into it. The column of liquid in this tube will oscillate with a period T. The period is

TestHub · Accepted Answer

Let the liquid be displaced by a small distance $x$ in one arm of the U-tube.

The liquid level in the other arm rises by $x$ , creating a level difference of $2 x$ between the two arms.

The restoring force acting on the liquid is due to the weight of this excess liquid column.

$F = - (2 x) A d g$

The acceleration of the liquid column of mass $M$ is $a = \frac{F}{M}$ .

$a = - (\frac{2 A d g}{M}) x$

Comparing this with the standard equation of simple harmonic motion $a = - ω^{2} x$ , we get:

$ω^{2} = \frac{2 A d g}{M} \Rightarrow ω = \frac{2 A d g}{M}$

The time period of oscillation is $T = \frac{2 π}{ω}$ .

$T = 2 π \frac{M}{2 A g d}$

Answer: $2 π \frac{M}{2 A g d}$

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