Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 70 2 a Solution Created 2026-10-03 Updated 2026-10-07
Use the prescribed harmonic convention . For a propagating acoustic plane wave, let and . The incident and reflected pressure amplitudes in the upper half-space have vertical factors and respectively:The linear homentropic acoustic equations imply . At the surface, the normal velocity is therefore , while the pressure amplitude is . The surface acoustic impedance condition givesHere is the normal acoustic impedance; the angle in this question is measured from the horizontal, not the normal. For a passive acoustic impedance, the mean power absorbed per unit area is . The four limiting cases have distinct meanings:
- If , . This is a pressure-release boundary: the pressure perturbation vanishes, while the normal velocity is generally nonzero. The reflected pressure has equal amplitude and a phase reversal.
- If , . The boundary is acoustically rigid, with zero normal velocity and doubled total surface pressure. There is no pressure phase reversal.
- If , . This is a matched boundary, taking up the incoming wave without reflection. Its pressure and normal velocity are those of the incident wave.
- Formally, means . A nonzero outgoing field can then exist with vanishing incoming amplitude. For a passive boundary at a real propagating incidence angle, a negative-real-part surface acoustic impedance cannot describe ordinary absorption: such a scattering pole must be interpreted through an active source or an continued by analytic continuation free-mode resonance. The sheet calculation below identifies the relevant free modes.