For a single relaxation process the absorption per wavelength is

PHYSICS · Multi-part · Fermi University Physics · Oscillations, Waves & Acoustics · Acoustics of fluids: generation, propagation and attenuation · Acoustics of fluids: generation, propagation and attenuation

Signal Speeds in Fluids, Ch Sound in a moving, absorbing fluid: sources, refraction and loss

Problem

For a single relaxation process the absorption per wavelength is αλ=πε ωτ/(1+ω2τ2)\alpha\lambda=\pi\varepsilon\,\omega\tau/(1+\omega^2\tau^2), with relaxation frequency fr=1/2πτf_r=1/2\pi\tau. For the oxygen relaxation in air at 20∘C20^\circ\mathrm{C} take ε=6.7×10−4\varepsilon=6.7\times10^{-4} and c0=343.2 m s−1c_0=343.2\,\mathrm{m\,s^{-1}}. The ISO 9613-1 formula puts the oxygen frf_r at 35 kHz35\,\mathrm{kHz} at 5050 per cent relative humidity and at 3.8 kHz3.8\,\mathrm{kHz} at 1010 per cent. ((a)) Show that the absorption per metre is α=(ε/2c0τ) ω2τ2/(1+ω2τ2)\alpha=(\varepsilon/2c_0\tau)\,\omega^2\tau^2/(1+\omega^2\tau^2), so that, unlike αλ\alpha\lambda, it has no peak in frequency. What happens to it at ωτ=1\omega\tau=1? ((b)) Find the oxygen contribution to the absorption of a 4 kHz4\,\mathrm{kHz} tone, in dB km−1\mathrm{dB\,km^{-1}}, at each humidity. ((c)) Water vapour is what speeds the oxygen relaxation, yet the drier air absorbs the 4 kHz4\,\mathrm{kHz} tone more. Explain why in one or two sentences.

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