Calculate the energy density, the energy flux density and the momentum density: a) for a progressive plane sound wave, whose velocity potential is $Ψ = f(t - e.r/v_s)$ ; and b) for a progressive spherical diverging sound wave whose velocity potential is $Ψ =(R/r)Ψ_m \cos (wt-kr).$ Show that, at a large distance in this last case, $E_v$ and the intensity decrease like 1/r² . Interpret this result.

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a) Writing [latex] \Psi =f(\xi ) ext { where } \xi \equiv t-e.r/v_s, ext {the velocity is} \ \dot{u} = - abla .\Psi =(e/v_s) f^\prime (\xi ). [/latex]. Integrating with respect to time, we obtain the displacement [latex] u=(1/v_s) f(\xi )e. [/latex] The acoustic pressure is [latex] p_a=-B( abla .u) =m_v f^\prime(\xi ) . [/latex] Using expressions [5.53], we obtain

[latex] E_v=E_{(C)v}+E_{(P)v}=½ m_v \dot{u}^2 +½B \sum_{ij} (\partial_iu_j)(\partial_iu_j), \ S^i=-B\sum\limits_{j} \dot{u}_j (\partial_iu_j), \ \ \ \ \ \ P_v=-m_v\sum\limits_{j} ( abla u_j)\dot{u}_j [/latex] [sound waves]. [5.53]

[latex] E_{(C)v}=½(m_v/v_s^2)f^ \prime(\xi )^2 \ E_{(P)v}=½B\sum\limits_{ij} \left[(e_ie_j/v_s^2) f^ \prime (\xi ) ight] \left[(e_ie_j/v_s^2)f^ \prime (\xi ) ight] =½(m_v/v_s^2) f^ \prime (\xi )^2 \ E_v=E_{(C)v}+E_{(P)v}=(m_v/v_s^2) f^ \prime (\xi )^2. \ S_i(r,t)=-B\sum\limits_{j} \left[-(e_j/v_s)f^ \prime (\xi ) ight]\left[(e_ie_j/v_s^2)f^ \prime(\xi ) ight] =(m_v/v_s)f^ \prime(\xi )^2e_i=E_vv_s \ e_i. \ P_v^i=-m_v\sum\limits_{j} \left[(-e_i/v_s^2)f^ \prime (\xi ) e_j ight] (e_j/v_s)f^ \prime (\xi )=(m_v/v_s^3)f^ \prime (\xi )^2e_i=(E_v/v_s)e_i . [/latex] [5.74]

b) If the velocity potential is [latex] \Psi =(R/r)\Psi _m \cos(ωt-kr) \ ext {where} \ k=ω/v_s [/latex], the velocity, displacement and acoustic pressure at large distance are the 1/r terms in [5.42], i.e.

[latex] \partial_tu(r,t)=R\Psi _m\left[(1/r^2) \cos(ωt\pm kr)\pm (k/r) \sin(ωt\pm kr) ight] e_r , \ u(r,t)=(R\Psi _m/ω)\left[(1/r^2) \sin(ωt\pm kr)\mp (k/r) \cos(ωt\pm kr) ight] e_r , \ P_a(r,t)=(Bk^2R\Psi _m/ωr) \cos(ωt\pm kr+\pi /2), [/latex] [5.42]

[latex] \partial_tu(r,t)=-(ωR/v_sr)\Psi _m \sin(ωt-kr)]e_r, \ u=(R/v_sr)\Psi _m \cos(ωt-kr)]e_r \ P_a(r,t)=(BωR/v_s^2r)\Psi _m \cosωt\pm kr+\pi /2). [/latex] [5.75]

Using the expressions of [5.53], we obtain the dominant terms

[latex] E_{(c)v}=½ m_V(ωR\Psi _m/v_sr)^2 \sin^2(ωt-kr) \ E_{(P)v}=½m_vv_s^2(r\Psi _m/v_s)^2 \sin^2(ωt-kr)\sum\limits_{ij} (kr^ir^j/r^3)(kr^ir^j/r^3)=E_{(C)V} \ E_v=E_{(C)v}+E_{(P)V}=m_v(ωr\Psi _m/v_sr)^2 \sin^2(ωt-kr) \ S_i=-B\sum\limits_{j}(-ωkR^2\Psi _m^2/v_s^2r^3) \sin^2(ωt-kr)\sum\limits_{j}(r^i/r)r^j(r^j/r)=v_sE_ve_r^i \ P_v^i=m_v\sum\limits_{j} (kωR^2/v_s^2r^2)\Psi _m^2 \sin^2(ωt-kr)](r^j/r)(r^i/r)(r^j/r)=(E_v/v_s)e_r^i [/latex] .[5.76]

The spherical wave, as specified by the velocity potential or the acoustic pressure [latex] p_a [/latex] , propagates without a change of profile but its amplitude decreases like 1/r. This decrease of the amplitude can be explained by the distribution of the emitted sound energy on a spherical surface of radius r. If P is the mean emitted power, the sound intensity, which is received at a point of this surface, is P/4πr². Thus, it decreases like 1/r² and the wave amplitude decreases like 1/r. Alternatively, if the wave is specified by u or [latex] \dot{u} [/latex] in the equations of [5.42], it is a superposition of a term, which decreases like 1/r and a term, which decreases like 1/r². The wave decreases as it propagates and its profile varies.

Question 5.2: Calculate the energy density, the energy flux density and th...

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