A Novel Theoretical Modeling for Predicting the Sound Absorption of Woven Fabrics Using Modification of Sound Wave Equation and Genetic Algorithm

Suppose two acoustic media 2 and 1 interfacing through a plane surface and characterized by the surface impendence Z₂ and Z₁ (Figure 1). The acoustic wave moves from medium 1 to medium 2. The sound wave moves from medium 1 is partly reflected and partly refracted (transmitted) as well as partly absorbed.

Figure 1

If in the case the medium 1 is the air, its surface impedance can be written as Eq. (14) (14) Z=ρoco Z = {\rho _o}{c_o}

Where the terms ρ_o and c_o are the density of the ambient air in the unit (kg/m³) and the speed of sound in the ambient air in the unit (m/s²). The continuity equation of sound wave propagation can be formulated as follows Eq. (15) (15) ∇.ρv=−∂ρ∂t \nabla .\rho v = - {{\partial \rho } \over {\partial t}}

Where the terms ρ and v are the density of sound in specific medium in the unit (kg/m³) and the speed of sound in specific medium in the unit (m/s²). With the Cauchy equation, it can be formulated as follows Eq. (16) and Eq. (17) (16) σij,j+fi=ρai {\sigma ^{ij}}_{,j} + {f^i} = \rho {a^i} (17) ∇⋅σ↔+f¯=ρa¯ \nabla \cdot \mathop {\boldsymbol \sigma} \limits^ \leftrightarrow + \overline {\boldsymbol {f}} = \rho \overline a

Where ρ is the volume mass density, σ↔ \mathop {\sigma} \limits^ \leftrightarrow , is a tensor stress which has a general form (18) σij=δijλe+2μeij=λtr(e↔)I+2μe↔ {\sigma ^{ij}} = {\delta ^{ij}}\lambda e + 2\mu {e^{ij}} = \lambda tr\left( {\mathop e\limits^ \leftrightarrow } \right){\boldsymbol {I}} + 2\mu \mathop e\limits^ \leftrightarrow

Where the terms σ^ij and e^ij are the stress tensor and the strain tensor. λ and μ are the elasticity constant. If the magnitude of the gravitational force can be ignored, then by defining that ∇⋅σ↔=∇¯(−P) \nabla \cdot \mathop {\sigma} \limits^ \leftrightarrow = \overline \nabla ( - P) and the sound wave moves in the z direction, then the wave motion equation can be written as in Eq. (19) and Eq. (20) (19) ∇⋅σ↔+f¯=ρa¯ \nabla \cdot \mathop \sigma \limits^ \leftrightarrow + \overline {\boldsymbol {f}} = \rho \overline a (20) ∇⋅σ↔+ρg+f¯ext=ρa \nabla \cdot \mathop \sigma \limits^ \leftrightarrow + \rho {\boldsymbol {g}} + {\overline {\boldsymbol {f}} _{{\boldsymbol {ext}}}} = \rho a

Where the terms ρ and g are the density of sound in specific medium in the unit (kg/m³) and the constant of gravity in the unit (m/s²). If there is no external force f̅_ext (can be ignored), then it can be written as in Eq. (21) and Eq. (22) (21) ∇¯(−P)=ρdvdt \overline \nabla \left( { - P} \right) = \rho {{dv} \over {dt}} (22) ∇¯(P)=−ρdvdt \overline \nabla \left( P \right) = - \rho {{dv} \over {dt}}

Divergent of the two segments, we get Eq. (23) to Eq. (25) (23) ∇¯⋅[∇¯(P)]=∇¯⋅[−ρdvdt]=[−ρd(∇¯⋅v)dt] \overline \nabla \cdot \left[ {\overline \nabla \left( P \right)} \right] = \overline \nabla \cdot \left[ { - \rho {{dv} \over {dt}}} \right] = \left[ { - \rho {{d\left( {\overline \nabla \cdot v} \right)} \over {dt}}} \right] (24) ∇2P=−ddt(∇.ρv) {\nabla ^2}P = - {d \over {dt}}\left( {\nabla .\rho v} \right) (25) ∇2P=ddt(∂ρ∂t) {\nabla ^2}P = {d \over {dt}}\left( {{{\partial \rho } \over {\partial t}}} \right)

Adiabatic expansion process is a process in which there is no change in Q heat in the system on the environment. In the case of adiabatic can be formulated as follows Eq. (26) and Eq. (27) (26) CdT=−PdV CdT = - PdV (27) NcvdT=−PdV N{c_v}dT = - PdV

Where C is the total heat capacity, C = Nc_v and pressure, P=kBTV P = {{{k_B}T} \over V} [30], Where the terms k_B, T and V are Thermodynamic const, temperature in the unit (K) and volume of matter in the unit (m³), hence we get Eq. (28) to Eq. (30) (28) cvdT=−kBTdVV {c_v}dT = - {{{k_B}TdV} \over V} (29) ∫dTT =−kBcv∫dVV \int {{dT} \over T}\; = - {{{k_B}} \over {{c_v}}}\int {{dV} \over V} (30a) cvkB∫dTT =−∫dVV {{{c_v}} \over {{k_B}}}\int {{dT} \over T}\; = - \int {{dV} \over V} (30b) cvkBln(TTo)=ln(VoV) {{{c_v}} \over {{k_B}}}\ln \left( {{T \over {{T_o}}}} \right) = \ln \left( {{{{V_o}} \over V}} \right)

To simplify the calculations, suppose that the ratio cvkB=32 {{{c_v}} \over {{k_B}}} = {3 \over 2} [30], then it can be described as follows Eq. (31) and Eq. (32) (31) 32ln(TTo)=ln(VoV) {3 \over 2}\ln \left( {{T \over {{T_o}}}} \right) = \ln \left( {{{{V_o}} \over V}} \right) (32) (TTo)3/2=VoV {\left( {{T \over {{T_o}}}} \right)^{3/2}} = {{{V_o}} \over V}

To determine the form of P-V, it can be described as follows Eq. (33) and Eq. (34) (33) (PVPoVo)3/2=VoV {\left( {{{PV} \over {{P_o}{V_o}}}} \right)^{3/2}} = {{{V_o}} \over V} (34) (PPo)3/2=(VoV)1+32 {\left( {{P \over {{P_o}}}} \right)^{3/2}} = {\left( {{{{V_o}} \over V}} \right)^{1 + {3 \over 2}}}

Where P=NkBTV=ρkBT P = {{N{k_B}T} \over V} = \rho {k_B}T , then the relationship between ρ and V is inversely proportional, so we get the following equation (Eq. (35) to Eq. (38)) (35) (PPo)3/2=(ρρo)1+32 {\left( {{P \over {{P_o}}}} \right)^{3/2}} = {\left( {{\rho \over {{\rho _o}}}} \right)^{1 + {3 \over 2}}} (36) ρρo=(PPo)3/5 {\rho \over {{\rho _o}}} = {\left( {{P \over {{P_o}}}} \right)^{3/5}} (37) ρρo=(PPo)γ {\rho \over {{\rho _o}}} = {\left( {{P \over {{P_o}}}} \right)^\gamma } (38) ρ=ρo(PPo)γ \rho = {\rho _o}{\left( {{P \over {{P_o}}}} \right)^\gamma }

Where γ=11+(kBcv)=cvcv+kB \gamma = {1 \over {1 + \left( {{{{k_B}} \over {{c_v}}}} \right)}} = {{{c_v}} \over {{c_v} + {k_B}}} . Suppose that Eq. (38) is differentiated twice with time, then it is obtained Eq. (39) and Eq. (40) (39) 1ρo∂2ρ∂t2=(1Po)γγ(γ−1)Pγ−2(∂P∂t)2+(1Po)γγPγ−1∂2P∂t2 \matrix{ {{1 \over {{\rho _o}}}{{{\partial ^2}\rho } \over {\partial {t^2}}}} \hfill & { = {{\left( {{1 \over {{P_o}}}} \right)}^\gamma }\gamma \left( {\gamma - 1} \right){P^{\gamma - 2}}{{\left( {{{\partial P} \over {\partial t}}} \right)}^2}} \hfill \cr {} \hfill & { + {{\left( {{1 \over {{P_o}}}} \right)}^\gamma }\gamma {P^{\gamma - 1}}{{{\partial ^2}P} \over {\partial {t^2}}}} \hfill \cr } (40) ∂2ρ∂t2=ρo[(1Po)γγ(γ−1)Pγ−2(∂P∂t)2+(1Po)γγPγ−1∂2P∂t2] {{{\partial ^2}\rho } \over {\partial {t^2}}} = {\rho _o}\left[ {{{\left( {{1 \over {{P_o}}}} \right)}^\gamma }\gamma \left( {\gamma - 1} \right){P^{\gamma - 2}}{{\left( {{{\partial P} \over {\partial t}}} \right)}^2}} \right.\left. { + {{\left( {{1 \over {{P_o}}}} \right)}^\gamma }\gamma {P^{\gamma - 1}}{{{\partial ^2}P} \over {\partial {t^2}}}} \right]

Substitute Eq. (40) to Eq. (25) then we get Eq. (41) (41) ∇2P=ρo[(1Po)γγ(γ−1)Pγ−2(∂P∂t)2+(1Po)γγPγ−1∂2P∂t2] {\nabla ^2}P = {\rho _o}\left[ {{{\left( {{1 \over {{P_o}}}} \right)}^\gamma }\gamma \left( {\gamma - 1} \right){P^{\gamma - 2}}{{\left( {{{\partial P} \over {\partial t}}} \right)}^2}} \right.\left. { + {{\left( {{1 \over {{P_o}}}} \right)}^\gamma }\gamma {P^{\gamma - 1}}{{{\partial ^2}P} \over {\partial {t^2}}}} \right]

Suppose that (∂P∂t)2 {\left( {{{\partial P} \over {\partial t}}} \right)^2} is very small (there is no sound absorption), then Eq. (42) to Eq. (44) (42) ∇2P=ρo[(1Po)γγPγ−1∂2P∂t2] {\nabla ^2}P = {\rho _o}\left[ {{{\left( {{1 \over {{P_o}}}} \right)}^\gamma }\gamma {P^{\gamma - 1}}{{{\partial ^2}P} \over {\partial {t^2}}}} \right] (43) ∇2P=ρoPo[(PPo)γ−1γ∂2P∂t2] {\nabla ^2}P = {{{\rho _o}} \over {{P_o}}}\left[ {{{\left( {{P \over {{P_o}}}} \right)}^{\gamma - 1}}\gamma {{{\partial ^2}P} \over {\partial {t^2}}}} \right] (44) ∇2P=ρoPo[(TTo)γ−1γ∂2P∂t2] {\nabla ^2}P = {{{\rho _o}} \over {{P_o}}}\left[ {{{\left( {{T \over {{T_o}}}} \right)}^{\gamma - 1}}\gamma {{{\partial ^2}P} \over {\partial {t^2}}}} \right]

Suppose that TTo≈1 {T \over {{T_o}}} \approx 1 (there is no significant temperature difference), then we get Eq. (45) (45) ∇2P=ρoPo[γ∂2P∂t2]=1v2d2Pdt2 {\nabla ^2}P = {{{\rho _o}} \over {{P_o}}}\left[ {\gamma {{{\partial ^2}P} \over {\partial {t^2}}}} \right] = {1 \over {{v^2}}}{{{d^2}P} \over {d{t^2}}}

By remembering that P=NkTV=ρkT P = {{NkT} \over V} = \rho kT , hence Pρ=kBT {P \over \rho } = {k_B}T we get Eq. (46) and Eq. (47) (46) ∇2P=1kBTo[γ∂2P∂t2] {\nabla ^2}P = {1 \over {{k_B}{T_o}}}\left[ {\gamma {{{\partial ^2}P} \over {\partial {t^2}}}} \right] (47) ∇2P−γkBTo∂2P∂t2=∇2P−1v2d2Pdt2=0 {\nabla ^2}P - {\gamma \over {{k_B}{T_o}}}{{{\partial ^2}P} \over {\partial {t^2}}} = {\nabla ^2}P - {1 \over {{v^2}}}{{{d^2}P} \over {d{t^2}}} = 0

Then the sound wave velocity is obtained as in Eq. (48) (48) v=kBToγ=1ρoϵ=Poρoγ v = \sqrt {{{{k_B}{T_o}} \over \gamma }} = {1 \over {\sqrt {{\rho _o}\epsilon } }} = \sqrt {{{{P_o}} \over {{\rho _o}\gamma }}}

With the solution of the wave equation is as follows Eq. (49) to Eq. (52) (49) 1v2∂2P(z,t)∂t2=∇2P(z,t) {1 \over {{v^2}}}{{{\partial ^2}P(z,t)} \over {\partial {t^2}}} = {\nabla ^2}P(z,t) (50) 1v21P(t)∂2P(t)∂t2=1P(z)∂2P(z)∂z2=−k2 {1 \over {{v^2}}}{1 \over {P(t)}}{{{\partial ^2}P(t)} \over {\partial {t^2}}} = {1 \over {P(z)}}{{{\partial ^2}P(z)} \over {\partial {z^2}}} = - {k^2} (51) 1v21P(t)∂2P(t)∂t2=−k2 {1 \over {{v^2}}}{1 \over {P(t)}}{{{\partial ^2}P(t)} \over {\partial {t^2}}} = - {k^2} (52a) 1P(t)∂2P(t)∂t2=−k2v2=−ω2 {1 \over {P(t)}}{{{\partial ^2}P(t)} \over {\partial {t^2}}} = - {k^2}{v^2} = - {\omega ^2}

With the wave number can be written as (52b) k2=ω2v2 {k^2} = {{{\omega ^2}} \over {{v^2}}}

The k value can be also called the propagation constant (wave propagation) and ω is the angular frequency of the wave with v called the velocity phase of the wave propagation. The sound pressure value P (z, t) can be described as follows Eq. (53) to Eq. (57): (53) ∂2P(z,t)∂z2=1v2d2P(z,t)dt2 {{{\partial ^2}P(z,t)} \over {\partial {z^2}}} = {1 \over {{v^2}}}{{{d^2}P(z,t)} \over {d{t^2}}} (54) P(t)∂2P(z)∂z2=1v2P(z)d2P(z)dt2 P(t){{{\partial ^2}P\left( z \right)} \over {\partial {z^2}}} = {1 \over {{v^2}}}P(z){{{d^2}P(z)} \over {d{t^2}}} (55) 1P(z)∂2P(z)∂z2=1P(t)1v2d2P(z)dt2 {1 \over {P(z)}}{{{\partial ^2}P\left( z \right)} \over {\partial {z^2}}} = {1 \over {P\left( t \right)}}{1 \over {{v^2}}}{{{d^2}P(z)} \over {d{t^2}}} (56) ∂2P(z)∂z2=−k2P(z) {{{\partial ^2}P(z)} \over {\partial {z^2}}} = - {k^2}P(z) (57) P(z)=Pocos(kz) P\left( z \right) = {P_o}\cos \left( {kz} \right)

Do the same for the function variable t, then after a little mathematical calculation, so we get Eq. (58) (58) 1P(t)∂2P(t)∂t2=−ω2 {1 \over {P(t)}}{{{\partial ^2}P(t)} \over {\partial {t^2}}} = - {\omega ^2}

Then obtained P (t) = P_o cos (−ωt), so we get the general function of the wave equation as follows Eq. (59) and Eq. (60) (59) P(x,t)=Pocos(kz−ωt). P\left( {x,t} \right) = {P_o}\cos (kz - \omega t). (60) P(z,t)=Pocos(2πλz−2πTt)=Pocos(2π(zλ−tT))=Pocos(2πλ(z−λtT))=Pocos(2πλ(z−vt)) \matrix{ {P\left( {z,t} \right)} \hfill & { = {P_o}\cos \left( {{{2\pi } \over \lambda }z - {{2\pi } \over T}t} \right)} \hfill \cr {} \hfill & { = {P_o}\cos \left( {2\pi \left( {{z \over \lambda } - {t \over T}} \right)} \right)} \hfill \cr {} \hfill & { = {P_o}\cos \left( {{{2\pi } \over \lambda }\left( {z - {{\lambda t} \over T}} \right)} \right)} \hfill \cr {} \hfill & { = {P_o}\cos \left( {{{2\pi } \over \lambda }\left( {z - vt} \right)} \right)} \hfill \cr }

For the case ∂P∂t {{\partial P} \over {\partial t}} is not ignored (there is an absorption of sound waves in medium), then obtained Eq. (61) and Eq. (62) (61) ∇2P=ρo[(1Po)γγ(γ−1)Pγ−2(∂P∂t)2+(1Po)γγPγ−1∂2P∂t2] {\nabla ^2}{P} = {\rho _o}\left[ {{{\left( {{1 \over {{P_o}}}} \right)}^\gamma }\gamma \left( {\gamma - 1} \right){P^{\gamma - 2}}{{\left( {{{\partial P} \over {\partial t}}} \right)}^2}} \right.\left. { + {{\left( {{1 \over {{P_o}}}} \right)}^\gamma }\gamma {P^{\gamma - 1}}{{{\partial ^2}P} \over {\partial {t^2}}}} \right] (62) ∇2P=ρo[1Po2(PPo)γ−2γ(γ−1)(∂P∂t)2+(PPo)γ−1γPo∂2P∂t2] {\nabla ^2}{P} = {\rho _o}\left[ {{1 \over {{P_o}^2}}{{\left( {{P \over {{P_o}}}} \right)}^{\gamma - 2}}\gamma \left( {\gamma - 1} \right){{\left( {{{\partial P} \over {\partial t}}} \right)}^2}} \right.\left. { + {{\left( {{P \over {{P_o}}}} \right)}^{\gamma - 1}}{\gamma \over {{P_o}}}{{{\partial ^2}P} \over {\partial {t^2}}}} \right]

If the ratio PPo=TTo≈1 {P \over {{P_o}}} = {T \over {{T_o}}} \approx 1 , we get Eq. (63) and Eq. (64) (63) ∇2P=[ρoPoγ1Po(γ−1)(∂P∂t)2+ρoγPo∂2P∂t2] {\nabla ^2}{P} = \left[ {{{{\rho _o}} \over {{P_o}}}\gamma {1 \over {{P_o}}}\left( {\gamma - 1} \right){{\left( {{{\partial P} \over {\partial t}}} \right)}^2} + {{{\rho _o}\gamma } \over {{P_o}}}{{{\partial ^2}P} \over {\partial {t^2}}}} \right] (64) ∇2P=[γ(γ−1)kBToPo(∂P∂t)2+γkBTo∂2P∂t2]≈[γ(γ−1)kBTo1Po(∂P∂t)2+γkBTo∂2P∂t2] \matrix{ {{\nabla ^2}{P}} \hfill & { = \left[ {{{\gamma \left( {\gamma - 1} \right)} \over {{k_B}{T_o}{P_o}}}{{\left( {{{\partial P} \over {\partial t}}} \right)}^2} + {\gamma \over {{k_B}{T_o}}}{{{\partial ^2}P} \over {\partial {t^2}}}} \right]} \hfill \cr {} \hfill & { \approx \left[ {{{\gamma \left( {\gamma - 1} \right)} \over {{k_B}{T_o}}}{1 \over {{P_o}}}{{\left( {{{\partial P} \over {\partial t}}} \right)}^2} + {\gamma \over {{k_B}{T_o}}}{{{\partial ^2}P} \over {\partial {t^2}}}} \right]} \hfill \cr }

For case Poρo=kBTo {{{P_o}} \over {{\rho _o}}} = {k_B}{T_o} and γ(γ−1)Po2=σ {{\gamma \left( {\gamma - 1} \right)} \over {{P_o}^2}} = \sigma , we get Eq. (65) to Eq. (69) (65) ∇2P−ρoγ(γ−1)Po2(∂P∂t)2−ρoγPo∂2P∂t2=0 {\nabla ^2}{\bf P} - {{{\rho _o}\gamma \left( {\gamma - 1} \right)} \over {{P_o}^2}}{\left( {{{\partial {\boldsymbol {P}}} \over {\partial t}}} \right)^2} - {{{\rho _o}\gamma } \over {{P_o}}}{{{\partial ^2}{\boldsymbol {P}}} \over {\partial {t^2}}} = 0 (66) ∇2P−ρoσ(∂P∂t)2−ρoγPo∂2P∂t2=0 {\nabla ^2}{\bf P} - {\rho _o}\sigma {\left( {{{\partial {\boldsymbol {P}}} \over {\partial t}}} \right)^2} - {\rho _o}{\gamma \over {{P_o}}}{{{\partial ^2}{\boldsymbol {P}}} \over {\partial {t^2}}} = 0 (67) ∇2P−ρoσ(∂P∂t)2−ρoϵ∂2P∂t2≈∇2P−ρoσ(1+(∂P∂t−1))2−ρoϵ∂2P∂t2=∇2P−ρoσ(1+2(∂P∂t−1))−ρoϵ∂2P∂t2=0 \matrix{ {{\nabla ^2}{\bf P} - {\rho _o}\sigma {{\left( {{{\partial {\boldsymbol {P}}} \over {\partial t}}} \right)}^2} - {\rho _o}\epsilon {{{\partial ^2}{\boldsymbol {P}}} \over {\partial {t^2}}}} \hfill \cr { \approx {\nabla ^2}{\bf P} - {\rho _o}\sigma {{\left( {1 + \left( {{{\partial {\boldsymbol {P}}} \over {\partial t}} - 1} \right)} \right)}^2} - {\rho _o}\epsilon {{{\partial ^2}{\boldsymbol {P}}} \over {\partial {t^2}}}} \hfill \cr { = {\nabla ^2}{\bf P} - {\rho _o}\sigma \left( {1 + 2\left( {{{\partial {\boldsymbol {P}}} \over {\partial t}} - 1} \right)} \right) - {\rho _o}\epsilon {{{\partial ^2}{\boldsymbol {P}}} \over {\partial {t^2}}} = 0} \hfill \cr } (68) ∇2P−ρoσ(1+2(∂P∂t))−ρoϵ∂2P∂t2=ρoσ {\nabla ^2}{\bf P} - {\rho _o}\sigma \left( {1 + 2\left( {{{\partial {\boldsymbol {P}}} \over {\partial t}}} \right)} \right) - {\rho _o}\epsilon {{{\partial ^2}{\boldsymbol {P}}} \over {\partial {t^2}}} = {\rho _o}\sigma (69) ∇2P−2ρoσ∂P∂t−ρoϵ∂2P∂t2=ρoσ {\nabla ^2}{\bf P} - 2{\rho _o}\sigma {{\partial {\boldsymbol {P}}} \over {\partial t}} - {\rho _o}\epsilon {{{\partial ^2}{\boldsymbol {P}}} \over {\partial {t^2}}} = {\rho _o}\sigma

For case P_o is very highand ρ_o is very small, then σ=γ(γ−1)Po2≈0 \sigma = {{\gamma \left( {\gamma - 1} \right)} \over {{P_o}^2}} \approx 0 with the condition that σ ≤ 0, we get Eq. 70 (70) ∇2P−2ρoσ∂P∂t−ρoϵ∂2P∂t2=∇2P−2ρoσ∂P∂t−1v2∂2P∂t2≈0 \matrix{ {{\nabla ^2}{\bf P} - 2{\rho _o}\sigma {{\partial {\boldsymbol {P}}} \over {\partial t}} - {\rho _o}\epsilon {{{\partial ^2}{\boldsymbol {P}}} \over {\partial {t^2}}}} \hfill \cr { = {\nabla ^2}{\bf P} - 2{\rho _o}\sigma {{\partial {\boldsymbol {P}}} \over {\partial t}} - {1 \over {{v^2}}}{{{\partial ^2}{\boldsymbol {P}}} \over {\partial {t^2}}}} \hfill \cr { \approx 0} \hfill \cr }

For example the solution is P = P_oe^i(kz−ωt), then substitute to the wave equation ∇2P−2ρoσ∂P∂t−ρoϵ∂2P∂t2=0 {\nabla ^2}{\bf P} - 2{\rho _o}\sigma {{\partial {\boldsymbol {P}}} \over {\partial t}} - {\rho _o}\epsilon {{{\partial ^2}{\boldsymbol {P}}} \over {\partial {t^2}}} = 0 , so we get Eq. (71) and Eq. (72) (71) −k2−2iωρoσ+ω2ρoϵ=0 - {k^2} - 2i\omega {\rho _o}\sigma + {\omega ^2}{\rho _o}\epsilon = 0 (72) k2=ω2ρoϵ−2iωρoσ {k^2} = {\omega ^2}{\rho _o}\epsilon - 2i\omega {\rho _o}\sigma

Suppose that σ^* = 2σ, we get Eq. (73) and Eq. (74) (73) k2=ω2ρoϵ+2iωρoσ=ω2ρoϵ(1−2iσωϵ)=ω2ρoϵ(1−iσ*ωϵ) \matrix{ {{k^2}} \hfill & { = {\omega ^2}{\rho _o}\epsilon + 2i\omega {\rho _o}\sigma = {\omega ^2}{\rho _o}\epsilon \left( {1 - 2{{i\sigma } \over {\omega \epsilon }}} \right)} \hfill \cr {} \hfill & { = {\omega ^2}{\rho _o}\epsilon \left( {1 - {{i{\sigma ^*}} \over {\omega \epsilon }}} \right)} \hfill \cr } where wavenumber can be written as in Eq. (74) (74) k=ωρoϵ(1−iσ*ωϵ)1/2 k = \omega \sqrt {{\rho _o}\epsilon } {\left( {1 - {{i{\sigma ^*}} \over {\omega \epsilon }}} \right)^{1/2}}

Suppose that = α−iβ=ωρoϵ(1−iσ*ωϵ)12 \alpha - i\beta = \omega \sqrt {{\rho _o}\epsilon } {\left( {1 - {{i{\sigma ^*}} \over {\omega \epsilon }}} \right)^{{1 \over 2}}} , we get Eq. (75) (75) k=ωv(1−iσ*ωϵ)1/2=ωv(1−iσ*2πfϵ)1/2=ωv(1−i.0,16σ*fϵ)1/2 \matrix{ k \hfill & { = {\omega \over v}{{\left( {1 - {{i{\sigma ^*}} \over {\omega \epsilon }}} \right)}^{1/2}} = {\omega \over v}{{\left( {1 - {{i{\sigma ^*}} \over {2\pi f\epsilon }}} \right)}^{1/2}}} \hfill \cr {} \hfill & { = {\omega \over v}{{\left( {1 - i.0,16{{{\sigma ^*}} \over {f\epsilon }}} \right)}^{1/2}}} \hfill \cr }

Therefore we get Eq. (76) (76) k2=(α−iβ)(α−iβ)=α2−β2−2iαβ=ω2ρoϵ−iωρoσ* \matrix{ {{k^2}} \hfill & { = \left( {\alpha - i\beta } \right)\left( {\alpha - i\beta } \right) = {\alpha ^2} - {\beta ^2} - 2i\alpha \beta } \hfill \cr {} \hfill & { = {\omega ^2}{\rho _o}\epsilon - i\omega {\rho _o}{\sigma ^*}} \hfill \cr } where α² − β² = ω²ρ_oɛ and −2iαβ = −iωρ_oσ^* where is αβ=ωρoσ*2 \alpha \beta = {{\omega {\rho _o}{\sigma ^*}} \over 2} . The solution of the sound pressure can be obtained as in Eq. (77) (77) P=Poei(kz−ωt)=Poei((α−iβ)z−ωt)=Poe((iα−β)z−iωt)=Poe−βzei(αz−ωt) \matrix{ {\boldsymbol {P}} \hfill & { = {{\boldsymbol {P}}_o}{e^{i(kz - \omega t)}} = {{\boldsymbol {P}}_o}{e^{i(\left( {\alpha - i\beta } \right)z - \omega t)}} = {{\boldsymbol {P}}_o}{e^{(\left( {i\alpha - \beta } \right)z - i\omega t)}}} \hfill \cr {} \hfill & { = {{\boldsymbol {P}}_o}{e^{ - \beta z}}{e^{i(\alpha z - \omega t)}}} \hfill \cr }

For a case α = β, then β=ωρoσ*2 \beta = \sqrt {{{\omega {\rho _o}{\sigma ^*}} \over 2}} (78a) β=ωρoσ*2α=vρoσ*2 \beta = {{\omega {\rho _o}{\sigma ^*}} \over {2\alpha }} = {{v{\rho _o}{\sigma ^*}} \over 2}

For the case for α = β, then with further mathematical calculations in Eq. (76), Eq. (78b) is obtained. (78b) k=α−iβ=ωρoϵ2(1+(1Q)2+1)1/2−iωρoϵ2(1+(1Q)2−1)1/2 \matrix{ k \hfill & { = \alpha - i\beta = \omega \sqrt {{{{\rho _o}\epsilon } \over 2}} {{\left( {\sqrt {1 + {{\left( {{1 \over Q}} \right)}^2}} + 1} \right)}^{1/2}}} \hfill \cr {} \hfill & { - i\omega \sqrt {{{{\rho _o}\epsilon } \over 2}} {{\left( {\sqrt {1 + {{\left( {{1 \over Q}} \right)}^2}} - 1} \right)}^{1/2}}} \hfill \cr } where Q=ωϵσ* Q = {{\omega \epsilon } \over {{\sigma ^*}}} . If it is defined that the thickness of the absorption δ=1β=2vρoσ* \delta = {1 \over \beta } = {2 \over {v{\rho _o}{\sigma ^*}}} which is the thickness when the sound waves disappear, then for ω=2πT \omega = {{2\pi } \over T} and v=λT v = {\lambda \over T} , we get Eq. (79) (79) δ=1β=2vρoσ*=2λTρo(2σ)=2(2π)λωρo(2σ)=2πλωρoσ=1λfρoσ \matrix{ \delta \hfill & { = {1 \over \beta } = {2 \over {v{\rho _o}{\sigma ^*}}} = {2 \over {{\lambda \over T}{\rho _o}\left( {2\sigma } \right)}} = {{2\left( {2\pi } \right)} \over {\lambda \omega {\rho _o}(2\sigma )}} = {{2\pi } \over {\lambda \omega {\rho _o}\sigma }}} \hfill \cr {} \hfill & { = {1 \over {\lambda f{\rho _o}\sigma }}} \hfill \cr } the impedance can be written as Z=ρv=ρωk Z = \rho v = \rho {\omega \over k} . By knowing the wavenumber, k, then the impedance value Z. Large β can be determined related to the energy dissipation (loss of energy) in the sound wave into the form of heat energy. Large impedance values show that in the real part there is an acoustic resistance, while in the imaginary part there is an acoustic reactant or a phase change or energy storage mechanism. The value of reflection and transmission without any absorption event can be described as follows for incident wave pressure, reflection pressure and transmittance pressure as shown in Eq. (80) to Eq. (82) (80) Pi(z,t)=Picos(k1z−ωt) {{\boldsymbol {P}}_{\boldsymbol {i}}}\left( {z,t} \right) = {P_i}\cos ({k_1}z - \omega t) (81) PR(z,t)=PRcos(−k1z−ωt) {{\boldsymbol {P}}_{\boldsymbol {R}}}\left( {z,t} \right) = {P_R}\cos ( - {k_1}z - \omega t) (82) PT(z,t)=PTcos(k2z−ωt) {{\boldsymbol {P}}_{\boldsymbol {T}}}\left( {z,t} \right) = {P_T}\cos ({k_2}z - \omega t)

To get the P_R and P_T values, it can be determined from the boundary conditions, which are as follows, for when z = 0, we get Eq. (83) to Eq. (85) (83) Pi(0,t)+PR(0,t)=PT(0,t) {{\boldsymbol {P}}_{\boldsymbol {i}}}\left( {0,t} \right) + {{\boldsymbol {P}}_{\boldsymbol {R}}}\left( {0,t} \right) = {{\boldsymbol {P}}_{\boldsymbol {T}}}\left( {0,t} \right) (84) Picos(k10−ωt)+PRcos(−k10−ωt)=PTcos(k20−ωt) {P_i}\cos ({k_1}0 - \omega t) + {P_R}\cos ( - {k_1}0 - \omega t) = {P_T}\cos ({k_2}0 - \omega t) (85) Pi+PR=PT {P_i} + {P_R} = {P_T} differentiated with z, then Eq. (83) can be written as shown in Eq. (86) to Eq. (88) (86) ∂∂z(Pi(0,t)+PR(0,t))=∂PT(0,t)∂z {\partial \over {\partial z}}\left( {{{\boldsymbol {P}}_{\boldsymbol {i}}}\left( {0,t} \right) + {{\boldsymbol {P}}_{\boldsymbol {R}}}\left( {0,t} \right)} \right) = {{\partial {{\boldsymbol {P}}_{\boldsymbol {T}}}\left( {0,t} \right)} \over {\partial z}} (87) k1Pi−k1PR=k2PT {k_1}{P_i} - {k_1}{P_R} = {k_2}{P_T} (88) Pi−PR=k2k1PT {P_i} - {P_R} = {{{k_2}} \over {{k_1}}}{P_T}

Can be eliminated P_i + P_R = P_T and Pi−PR=k2k1PT {P_i} - {P_R} = {{{k_2}} \over {{k_1}}}{P_T} , we get Eq. (89) (89) PT=2k1k1+k2Pi {P_T} = {{2{k_1}} \over {{k_1} + {k_2}}}{P_i}

With v=ωk=zρ v = {\omega \over k} = {z \over \rho } , for a medium that has more pivot, it can be assumed that ρ₁ ≈ ρ₂, then we get Eq. (90) and Eq. (91) (90) PT=2k1k1+k2Pi {P_T} = {{2{k_1}} \over {{k_1} + {k_2}}}{P_i} (91) PT=2v2v1+v2Pi {P_T} = {{2{v_2}} \over {{v_1} + {v_2}}}{P_i}

Due to Z = ρv, we get Eq. (92) (92) PT=2Z2Z1+Z2Pi {P_T} = {{2{Z_2}} \over {{Z_1} + {Z_2}}}{P_i} do the same calculation to find the value of P_R, then Eq. (93) to Eq. (94) are obtained (93) PR=v2−v1v2+v1PI {P_R} = {{{v_2} - {v_1}} \over {{v_2} + {v_1}}}{P_I} (94) PR=Z2−Z1Z2+Z1PI {P_R} = {{{Z_2} - {Z_1}} \over {{Z_2} + {Z_1}}}{P_I}

It is defined that the wave intensity is as in Eq. (95) [25] (95) I=PpowerA(kgs3) I = {{{P_{power}}} \over A}\left( {{{kg} \over {{s^3}}}} \right) or it can be written as shown in Eq. (96) [25] (96) I=Ppressure2Z(kgs3) I = {{{P_{pressure}}^2} \over Z}\left( {{{kg} \over {{s^3}}}} \right)

It is defined that the reflection coefficient, transmission coefficient and absorption coefficient are as follows as Eq. (96) (97) aR=IRIi=PR2Z1Pi2Z1=PR2Pi2=(Z2−Z1Z2+Z1)2=R2 {a_R} = {{{I_R}} \over {{I_i}}} = {{{{{P_R}^2} \over {{Z_1}}}} \over {{{{P_i}^2} \over {{Z_1}}}}} = {{{P_R}^2} \over {{P_i}^2}} = {\left( {{{{Z_2} - {Z_1}} \over {{Z_2} + {Z_1}}}} \right)^2} = {R^2} if Z₁ is the impedance in ambient air, then Z₁ = ρ_ov, we get Eq. (98) (98) aR=(Z2Z1−1Z2Z1+1)2=(Z2ρov−1Z2ρov+1)2=R2 {a_R} = {\left( {{{{{{Z_2}} \over {{Z_1}}} - 1} \over {{{{Z_2}} \over {{Z_1}}} + 1}}} \right)^2} = {\left( {{{{{{Z_2}} \over {{\rho _o}v}} - 1} \over {{{{Z_2}} \over {{\rho _o}v}} + 1}}} \right)^2} = {R^2}

We get that Z2ρov=1+R1−R {{{Z_2}} \over {{\rho _o}v}} = {{1 + R} \over {1 - R}} and also we can found Eq. (99) (99) aT=ITIi=PT2Z2Pi2Z1=PT2Pi2Z1Z2=(2Z2Z1+Z2)2Z1Z2=4Z2Z1(Z1+Z2)2=T2 \matrix{ {{a_T}} \hfill & { = {{{I_T}} \over {{I_i}}} = {{{{{P_T}^2} \over {{Z_2}}}} \over {{{{P_i}^2} \over {{Z_1}}}}} = {{{P_T}^2} \over {{P_i}^2}}{{{Z_1}} \over {{Z_2}}} = {{\left( {{{2{Z_2}} \over {{Z_1} + {Z_2}}}} \right)}^2}{{{Z_1}} \over {{Z_2}}}} \hfill \cr {} \hfill & { = {{4{Z_2}{Z_1}} \over {{{\left( {{Z_1} + {Z_2}} \right)}^2}}} = {T^2}} \hfill \cr }

a_T is called the reflection factor or the pressure reflection coefficient. Sound absorption coefficient can be defined as Eq. (100) (100) aTaa=IaIi=PA2Pi2Z1Z2=A2 {a_T}{a_a} = {{{I_a}} \over {{I_i}}} = {{{P_A}^2} \over {{P_i}^2}}{{{Z_1}} \over {{Z_2}}} = {A^2}

In this study, there are two types of equations in the event of the propagation of sound waves in a medium, namely the absorption and the absence of absorption in the medium. in the absence of absorption, it can be written that a_R + a_T = 1, while in the event of absorption by the medium, it can be written a_R + a_T + a_A = 1, so the sound absorption coefficient can be written as follows a_a = 1 − a_R − a_T, then obtained Eq. (101) and Eq. (102) (101) aa+aT=1−aR=1−IiIR=1−Pi2Z1PR2Z1=1−PR2Pi2=1−(Z2−Z1Z2+Z1)2 \matrix{ {{a_a} + {a_T}} \hfill & { = 1 - {a_R} = 1 - {{{I_i}} \over {{I_R}}} = 1 - {{{{{P_i}^2} \over {{Z_1}}}} \over {{{{P_R}^2} \over {{Z_1}}}}} = 1 - {{{P_R}^2} \over {{P_i}^2}}} \hfill \cr {} \hfill & { = 1 - {{\left( {{{{Z_2} - {Z_1}} \over {{Z_2} + {Z_1}}}} \right)}^2}} \hfill \cr } (102) aa+aT=4Z2Z1(Z1+Z2)2=4Z2Z1(Z1+Z2)2 {a_a} + {a_T} = {{4{Z_2}{Z_1}} \over {{{\left( {{Z_1} + {Z_2}} \right)}^2}}} = {{4{Z_2}{Z_1}} \over {{{\left( {{Z_1} + {Z_2}} \right)}^2}}}

With Z = ρv, which is the speed on the material has the relationship as written v=Kρ=Kρ v = \sqrt {{K \over \rho }} = \sqrt {K\rho } with K is the modulus of elasticity in the unit (N/m²) and ρ (kg/m³) is the volume density. The relationship of the parameters can be written as v_Solid > v_liquid > v_gas, then Z_Solid > Z_liquid > Z_gas. if Z₂ > Z₁ and Z1Z2=n {{{Z_1}} \over {{Z_2}}} = n , then we get Eq. (103) (103) aa+aT=4Z2Z1(Z1+Z2)2=4Z2Z1(Z1+Z2)2=4Z2Z1(Z2(1+Z1Z2))2=4Z2Z1Z12(1+Z1Z2)2=4Z1Z2(1+Z1Z2)2=4n(1+n)2 \matrix{ {{a_a} + {a_T}} \hfill & { = {{4{Z_2}{Z_1}} \over {{{\left( {{Z_1} + {Z_2}} \right)}^2}}} = {{4{Z_2}{Z_1}} \over {{{\left( {{Z_1} + {Z_2}} \right)}^2}}}} \hfill \cr {} \hfill & { = {{4{Z_2}{Z_1}} \over {{{\left( {{Z_2}\left( {1 + {{{Z_1}} \over {{Z_2}}}} \right)} \right)}^2}}} = {{4{Z_2}{Z_1}} \over {{Z_1}^2{{\left( {1 + {{{Z_1}} \over {{Z_2}}}} \right)}^2}}}} \hfill \cr {} \hfill & { = {{4{Z_1}} \over {{Z_2}{{\left( {1 + {{{Z_1}} \over {{Z_2}}}} \right)}^2}}} = {{4n} \over {{{\left( {1 + n} \right)}^2}}}} \hfill \cr }

Suppose that Z₁ = 1, then a_a depends on Z₂, by remembering that Z = ρv, then we get Eq. (104) to Eq. (106) (104) aa+aT=4Z2(1+Z2)2=4v2(1+v2)2 {a_a} + {a_T} = {{4{Z_2}} \over {{{\left( {1 + {Z_2}} \right)}^2}}} = {{4{v_2}} \over {{{\left( {1 + {v_2}} \right)}^2}}} (105) aa+aT=4ω2/k2(1+ω2k2)2=42πf2/k2(1+2πf2k2)2=4f2/Jnm(1+f2/Jnm)2=4m(1+m)2 \matrix{ {{a_a} + {a_T}} \hfill & { = {{4{\omega _2}/{k_2}} \over {{{\left( {1 + {{{\omega _2}} \over {{k_2}}}} \right)}^2}}} = {{42\pi {f_2}/{k_2}} \over {{{\left( {1 + {{2\pi {f_2}} \over {{k_2}}}} \right)}^2}}}} \hfill \cr {} \hfill & { = {{4{f_2}/{J_{nm}}} \over {{{\left( {1 + {f_2}/{J_{nm}}} \right)}^2}}} = {{4m} \over {{{\left( {1 + m} \right)}^2}}}} \hfill \cr } (106) aa=4m(1+m)2−aT {a_a} = {{4m} \over {{{\left( {1 + m} \right)}^2}}} - {a_T}

Which requires that the value is m=f2Jnm m = {{{f_2}} \over {{J_{nm}}}} . with J_nm is a constant depending on the material, f₂ is the frequency