ACOUSTIC POWER AND ENERGYACOUSTIC INTENSITYL23-1ACOUSTIC WAVES IN GASESBasic Differences with EM Waves:Electromagnetic Waves Acoustic WavesE, H are vectors SLinear physics⊥ U(velocity) // S, P(pressure) is scalarNon-linear physics, use perturbationsAcoustic Non-linearities:Compression heats the gas; cooling by conduction and radiation(adiabatic assumption—no heat transfer)Compression and advection introduce position shifts in waveWave velocity depends on pressure, varies along wave(loud sounds form shock waves)Pzshock frontsAcoustic Variables:[][]()2o1oo3o1Pressure: P Nm P pVelocity: U ms U u u set U 0 hereDensity: kg m−−=+=+= =⎡⎤ρ=ρ+ρ⎣⎦use perturbationsL23-2ACOUSTIC EQUATIONSConstitutive Equation:Fractional changes in gas density and pressure are proportional:3 Equations, 3 Unknowns:Reduce to 2 unknowns()p,u"adiabatic exponent" γ = 5/3 monotomic gas, ~1.4 air, 1-2 elseo1o(d1dP )pPP=ρρ=⇒ρργ γMass Conservation Equation:()()()()∂ρ∇• =∇•ρ =−∂∂ρ∇•ρ =−∂∂ρ +ρ∂ρ∇• ρ +ρ + ≅− =−∂∂ρeeo11oo1nd1Recall: J u Conservation of chargetAcoustics: u Conservation of masstLinearize: U uttDrop 2 order term uLinearized Conservation of Mass:Linearized Force Equation (f = ma):0∂ρρ∇• ≅−∂1out∂∇=−ρ∂ouptL23-3()()2pt,r p t k r [Nm ]−=ω−•Differential Equations:Acoustic Wave Equation:Solution:ACOUSTIC EQUATIONS3-22oup [Nm ] [kg m s ]t−−∂∇=−ρ∂Newton’s Law (f = ma):Conservation of Mass:"Acoustic Wave Equation"∇•∇p ⇒2ndspatial derivative = 2ndderivative in time22o2opp0Ptρ∂∇− =γ∂−∂∇• ≅−γ∂1op1u [s ]PtL23-4Example: assume p(t,r) = cos(ωt-kz):()ooou1kˆˆp u z p dt z cos t kzt∂∇=−ρ ⇒ =− ∇ = ω−∂ρρω∫Acoustic Impedance of Gas:(ηs≅ 425 Nsm-3[≠Ω] for air 20°C)Substituting solution into wave equation⇒ “Acoustic Dispersion Relation”:UNIFORM PLANE WAVESVelocity of Sound:Velocity of Sound in Liquids and Solids:Phase velocity:Group velocity:cs= (K/ρo)0.5≅ 1,500 ms-1in water, ≅ 1,500 – 13,000 in solids“Bulk modulus”ηs-1ooskPcρω=ω =γosooPkωρη= = ργ−γω== =ργ∂⎛⎞===⎜⎟∂ω ρ⎝⎠opso1ogsoPvckPkvcExample:Air at 0oC, Surface at Po⇒ cs≅ 330 m/s(G = 1.4, ρo= 1.29 kg/m3Po= 1.01×105N/m2)L23-5ACOUSTIC POWER AND ENERGYPoynting Theorem, differential form:∂∂∇• = •∇ + ∇• =−ρ • −∂γ ∂∂⎡⎤∇• =− ρ −⎢⎥∂γ⎣⎦oo223oopu1up u p p u u p tP t11up u p [W/m ]2t PTry:AVˆnda22oAVopupu nda dvt22P⎡⎤∂•=− ρ +⎢⎥∂γ⎣⎦∫∫Acoustic intensity Kinetic Potential [Watts]-13-22oo-2 2 1pu1Recall: p [Nm ] [kg m s ] u [s ]tPtNote: Wave intensity [Wm ] pu [(Nm )(ms )]−−−−∂∂∇=−ρ ∇•≅−∂γ∂=Integral form:Wk [Jm-3] WpI[Wm-2]Plane Wave Intensity I = ˆ⎡⎤⎣⎦-2pu•n Wm :==η = = =ηηηη= ργ ≅2222oo-2sosss-3soopupIntensity: I(t) u [Wm ], I I(t)22Where: P { 425 Nsm in surface air}Acoustic Poynting TheoremL23-6ACOUSTIC INTENSITYIthresh≅ 0 dB (acoustic scale) = 10-12[Wm-2]uo= po/ηs= 3×10-5/425 ≅ 7 × 10-8[ms-1]Plane Wave Intensity I = ˆ⎡⎤⎣⎦-2pu•n Wm :Example: small radio at beach: Example: Threshold of hearing: Io= 1 [Wm-2] at 1 kHz⇒uo= po/ηs= 0.07 [ms-1]; Δz = 2uo/ω = 10 microns2222oo-2so sss-3sooppuIntensity: I(t) u [Wm ], I I(t)22Where: P { 425 Nsm in surface air}==η = = =ηηηη= ργ ≅2osop 2 I 850 ~ 30 [N/m ]=η= =12 5 2osop 2 I 850 10 ~ 3 10 [N/m ]−−=η= × = ×-8o3-11u7×10z 2 2 2×10 [m] = 0.2 Å (< atom)7×10Δ≅ ≅ =ωL23-7BOUNDARY CONDITIONSInterfaces between gases or liquids:Pressure: Δp = 0 (otherwise ∞ acceleration of zero-mass boundary)Velocity: Δu⊥= 0 (otherwise ∞ mass-density accumulation)Rigid Boundaries:Pressure: Unconstrained Velocity: Δu⊥= 0 (rigid body is motionless)koxzktθtθrθikzReflection at Non-Rigid Boundaries:Matching Phases:−−−======→o,ro,too,ro,tocos sinjk riooicos sinjk rrrr rocos sinjk rttt tor,Incident wave: p p e pReflected wave: p p e pTransmitted wave: p p e pVelocities: Same, but p u , p+jk θ x- jk θ zoioi-jk θ x- jk θ zoror+jk θ x- jk θ ztttteee→r,ttuSnell's Law: sinθi/sinθt= kt/ki=θr= θiCritical angle:⇒ookPρ=ωγ− −ργ⎛⎞θ= =⎜⎟ργ⎝⎠1c1tiititcsin sinctiitργργL23-8REFLECTIONS AT BOUNDARIESEvanescent Waves (θi> θc):Normal Incidence, two gases:warmcoolzxθitp“lunch!”0zReflections from Solid Surface()ˆn•u=0 :()()()−+ −⊥+Γ = →+Γ= = Δ=η− Γη= η →−Γ= η η = Δ ==η η+η η=ωρ =ργjk z jk z jk zoo tii iio i o i t ottt o o o oop: pe p e pTe 1 T at z 0 p 0u: p p pT 1 T at z 0 u 0Solving: T 2 where k PuRecall:Where: cos θt= and sin θt= sin θi> 1So: ktcos θt= ±jα(k cosθ )(ksinθ )z (k sinθ )ztt tt tttto topp p==-j x- j -αx- jeeitit/ργ ργ−⎛⎞θ=⎜⎟⎝⎠1citcsinc−2t1sinθMIT OpenCourseWare http://ocw.mit.edu 6.013 Electromagnetics and Applications Spring 2009 For information about citing these materials or our Terms of Use, visit:
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