OSCILLATIONS AND WAVES BYU 25

Solution – Problem 25

Problem 25 Solution

Gas V m S
1. Equilibrium Pressure

At equilibrium, the pressure inside the bottle \(P\) supports the atmospheric pressure \(P_0\) and the weight of the marble. \[ P S = P_0 S + mg \implies P = P_0 + \frac{mg}{S} \]

2. Adiabatic Process

Since the materials are perfect insulators, the process is Adiabatic (\(PV^\gamma = \text{constant}\)). Bulk Modulus \(B = \gamma P\). For a small downward displacement \(x\), the volume decreases by \(dV = -Sx\). The increase in pressure is \(dP = – \gamma \frac{P}{V} dV = \gamma \frac{P}{V} Sx\).

3. Restoring Force

The restoring force acting upwards on the marble is \(F = (dP)S\). \[ F = \left( \frac{\gamma P S x}{V} \right) S = \frac{\gamma P S^2}{V} x \] Substituting \(P = P_0 + \frac{mg}{S}\): \[ F = \frac{\gamma S^2}{V} \left( P_0 + \frac{mg}{S} \right) x = \frac{\gamma S}{V} (P_0 S + mg) x \]

4. Period of Oscillation

Spring constant \(k = \frac{\gamma S}{V} (P_0 S + mg)\). For monatomic gas, \(\gamma = 5/3\). \[ T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{m V}{\frac{5}{3} S (P_0 S + mg)}} \] \[ T = 2\pi \sqrt{\frac{3mV}{5S(P_0 S + mg)}} \]