GSO001 Problem 15

1. Gas Pressure $P_1$ in State 1

When the gas expands from length $l_0$ to $l_1$ , the left liquid level drops by $(l_1 - l_0)$ and the right level rises by $(l_1 - l_0)$ . The height difference between liquid levels:

\Delta h = 2(l_1 - l_0)

Force balance at the left liquid surface:

P_1 = P_0 + \rho g \Delta h = P_0 + 2\rho g (l_1 - l_0)

2. Adiabatic Pressure Change in State 2

Let the displacement of the liquid column from equilibrium position be $y$ (positive = liquid rises on left side). The gas column length becomes $l_1 - y$ and volume $V = S(l_1 - y)$ .

By Poisson's law $P V^\gamma = \text{const}$ for adiabatic processes:

P = P_1 \left( \frac{l_1}{l_1 - y} \right)^\gamma = P_1 \left( 1 - \frac{y}{l_1} \right)^{-\gamma}

Applying the first-order Taylor expansion $(1 + x)^n \approx 1 + nx$ for small $y/l_1$ :

P \approx P_1 + \frac{\gamma P_1}{l_1} y

3. Equation of Motion of the Liquid Column

When displaced by $y$ , the left level rises by $y$ and the right level drops by $y$ , so the height difference becomes $2(l_1 - l_0) - 2y$ .

The net force on the liquid column (total mass $M = \rho L S$ ):

F = \left[ P_0 + \rho g (2(l_1 - l_0) - 2y) \right]S - P S

Substituting $P_1 = P_0 + 2\rho g (l_1 - l_0)$ and $P \approx P_1 + \frac{\gamma P_1}{l_1} y$ :

F = - \left( \frac{\gamma P_1}{l_1} + 2\rho g \right) S y

The equation of motion $M \ddot{y} = F$ with $M = \rho L S$ :

\ddot{y} = - \frac{\frac{\gamma P_1}{l_1} + 2\rho g}{\rho L} y

Comparing with $\ddot{y} = -\omega^2 y$ :

\omega^2 = \frac{\frac{\gamma P_1}{l_1} + 2\rho g}{\rho L}

4. Numerical Calculation

$l_1 - l_0 = 0.500 - 0.400 = 0.100 \text{ m}$

P_1 = 1.013 \times 10^5 + 2 \times (1.00 \times 10^3) \times 9.80 \times 0.100 = 101300 + 1960 = 103260 \text{ Pa}

Numerator (with $\gamma = 5/3$ ):

\text{Numerator} = \frac{5}{3} \times \frac{103260}{0.500} + 2 \times (1.00 \times 10^3) \times 9.80

= \frac{5}{3} \times 206520 + 19600 = 344200 + 19600 = 363800

Denominator: $\rho L = 1.00 \times 10^3 \times 2.00 = 2000$

\omega^2 = \frac{363800}{2000} = 181.9 \text{ rad}^2/\text{s}^2

The required answer is $181.9 \times 10 = 1819$ .

Final Answer: 1819

GSO001 Problem 15

Oscillation of a Liquid Column in a U-Tube with Adiabatic Change

Problem Statement

Constraints

Input Format

Solution

1. Gas Pressure $P_1$ in State 1

2. Adiabatic Pressure Change in State 2

3. Equation of Motion of the Liquid Column

4. Numerical Calculation

Oscillation of a Liquid Column in a U-Tube with Adiabatic Change

Problem Statement

Constraints

Input Format

Solution

1. Gas Pressure P1P_1P1​ in State 1

2. Adiabatic Pressure Change in State 2

3. Equation of Motion of the Liquid Column

4. Numerical Calculation

1. Gas Pressure $P_1$ in State 1