Force on the liquid = pressure ? area of the cross-section of the tube

**Oscillation of liquid column in a U - tube **

Consider a non
viscous liquid column of length l of uniform cross-sectional area A (Fig.).
Initially the level of liquid in the limbs is the same. If the liquid on one
side of the tube is depressed by blowing gently the levels of the liquid
oscillates for a short time about their initial positions O and C, before
coming to rest. If the liquid in one of the limbs is depressed by y , there
will be a difference of 2 y in the liquid levels in the two limbs. At some
instant, suppose the level of the liquid on the left side of the tube is at D,
at a height y above its original position O, the level B of the liquid on the
other side is then at a depth y below its original position C. So the excess
pressure P on the liquid due to the restoring force is excess height ? density
? g

(i.e) pressure =
2 y ρ g

∴
Force on the liquid = pressure ? area of the cross-section of the tube

= ? 2 y ρ g ? A ???.(1)

The negative sign
indicates that the force towards O is opposite to the displacement measured
from O at that instant.

The mass of the
liquid column of length l is volume ? density

(i.e) m = l A ρ

∴
F = l A ρ a .... (2)

From equations
(1) and (2)

l A ρ a = - 2 y
A ρ g

∴
a = ? 2g /l . y

We know that a =
?ω^{2} y

(i.e) a = ? 2 g/
l . y= ?ω^{2} y

where ω =
rt(2g/l)

Here, the
acceleration is proportional to the displacement, so the motion is simple
harmonic and the period T is

T=[ 2π / ω ]
- [2π rt(l/2g)]

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11th 12th std standard Class Physics sciense Higher secondary school College Notes : Oscillation of liquid column in a U - tube |

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