SATHEE: Simple Harmonic Motion 5 Question 20

Simple Harmonic Motion 5 Question 20

25. Two non-viscous, incompressible and immiscible liquids of densities $ρ$ and $1.5 ρ$ are poured into the two limbs of a circular tube of radius $R$ and small cross-section kept fixed in a vertical plane as shown in figure. Each liquid occupies one-fourth the circumference of the tube.

$(1991, 4 + 4$ M)

(a) Find the angle $θ$ that the radius to the interface makes with the vertical in equilibrium position.

(b) If the whole liquid column is given a small displacement from its equilibrium position, show that the resulting oscillations are simple harmonic. Find the time period of these oscillations.

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Solution:

(a) In equilibrium, pressure of same liquid at same level will be same.

Therefore, $p_{1} = p_{2}$ or $p + (1.5 ρ g h_{1}) = p + (ρ g h_{2})$ ( $p =$ pressure of gas in empty part of the tube)

$\begin{aligned} ∴ 1.5 h_{1} = h_{2} \\ 1.5 [R \cos θ - R \sin θ] = ρ (R \cos θ + R \sin θ) \\ or 3 \cos θ - 3 \sin θ = 2 \cos θ + 2 \sin θ \\ or 5 \tan θ = 1 \\ θ = \tan^{- 1} \frac{1}{5} \end{aligned}$

(b) When liquids are slightly disturbed by an angle $β$ .

Net restoring pressure

$Δ p = 1.5 ρ g h + ρ g h = 2.5 ρ g h$

This pressure will be equal at all sections of the liquid. Therefore, net restoring torque on the whole liquid.

$\begin{aligned} τ & = - (Δ p) (A) (R) \\ or τ & = - 2.5 ρ g h A R \\ = - 2.5 ρ g A R [R \sin (θ + β) - R \sin θ] \\ = - 2.5 ρ g A R^{2} [\sin θ \cos β + \sin β \cos θ - \sin θ] \end{aligned}$