OSCILLATIONS AND WAVES BYU 24

Solution – Problem 24

Problem 24 Solution

P, V1, T1 P, V2, T2 m x
1. Process Analysis

The problem states the chambers are “maintained” at temperatures \(T_1\) and \(T_2\). This implies an Isothermal Process (\(PV = \text{constant}\)). For small displacement \(x\) to the right:

  • Chamber 1 (Left): Expands. Volume becomes \(V_1 + Ax\). Pressure decreases.
  • Chamber 2 (Right): Compresses. Volume becomes \(V_2 – Ax\). Pressure increases.

2. Restoring Force

Bulk modulus for isothermal process is \(B = P\). Change in pressure \(dP = – \frac{P}{V} dV\).
For Chamber 1: \(dP_1 = – \frac{p}{V_1} (Ax)\).
For Chamber 2: \(dP_2 = – \frac{p}{V_2} (-Ax) = \frac{p}{V_2} Ax\).
Net restoring force \(F = (dP_2 – dP_1) A\) (opposing displacement): \[ F = – \left( \frac{p A x}{V_2} + \frac{p A x}{V_1} \right) A = – p A^2 \left( \frac{1}{V_1} + \frac{1}{V_2} \right) x \] \[ F = – p A^2 \left( \frac{V_1 + V_2}{V_1 V_2} \right) x \]

3. Time Period

The effective spring constant \(k = p A^2 \frac{V_1 + V_2}{V_1 V_2}\). Period \(T = 2\pi \sqrt{\frac{m}{k}}\). \[ T = 2\pi \sqrt{\frac{m V_1 V_2}{p A^2 (V_1 + V_2)}} \]