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Elastic Waves. Acoustics.

Problem 4.185

A plane longitudinal harmonic wave propagates in a medium with density ρ\rho. The velocity of the wave propagation is vv. Assuming that the density variations of the medium, induced by the propagating wave, Δρ≪ρ,\Delta \rho \ll \rho, demonstrate that (a) the pressure increment in the medium Δp=−ρv2(∂ξ/∂x)\Delta p=-\rho v^{2}(\partial \xi / \partial x), where ∂ξ/∂x\partial \xi / \partial x is the relative deformation (b) the wave intensity is defined by Eq. (4.3i).

Reveal Answer
4.185. (a) Let us consider the motion of a plane element of the medium of thickness dxd x and unit area of cross-section. In accordance with Newton's second law ρdxξ¨=−dp,\rho d x \ddot{\xi}=-d p, where dpd p is the pressure increment over the length dxd x. Recalling the wave equation ξ¨=\ddot{\xi}= =v2(∂2ξ/∂x2),=v^{2}\left(\partial^{2} \xi / \partial x^{2}\right), we can write the foregoing equation as ρv2∂2ξ∂x2dx=−dp \rho v^{2} \frac{\partial^{2} \xi}{\partial x^{2}} d x=-d p Integrating this equation, we get Δp=−ρv2∂ξ∂x+ const.  \Delta p=-\rho v^{2} \frac{\partial \xi}{\partial x}+\text { const. } In the absence of a deformation (a wave) the surplus pressure is Δp=0.\Delta p=0 . Hence, const =0=0
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