VRIJE UNIVERSITEIT BRUSSEL Acoustics - the Dept. of ...

5.2. AIRBORNE SOUND INSULATION OF A WALL 89with m the mass, k the wall stiffness and d the daming. For harmonic waveswe now that from x = Xexp(iωt) follows ẋ = iωx and ẍ = −ω 2 x. Equation5.16 can now be written in function of the amplitudes P and X (respectivelyof the sound pressure and particle displacement) :(−mω 2 +iωd+k)X = 2P i −iωρ 1 c 1 X −iωρ 2 c 2 X (5.17)or by introducing the velocity amplitude V = iωX :[i(ωm− k ω )+(d+ρ 1c 1 +ρ 2 c 2 )]V = 2P i (5.18)From the previous paragraph we know that P d = ρcV , and thus :Which gives us :P d = ρ 2 c 2 2P i [i(ωm− k ω )+(d+ρ 1c 1 +ρ 2 c 2 )] −1 (5.19)P iP d= 2(i(ωm−k) (ω d+ + ρ ) )1c 1+1 (5.20)ρ 2 c 2 ρ 2 c 2 ρ 2 c 2Three cases can now be distinguished depending on the frequency ω :1.ω ≪ ω 0 =√km⇒ R = 20logk −20logf −20log(4πρc) (5.21)For low frequencies the sound insulation of a wall is thus determinedby the wall stiffness.2.3.ω ≫ ω 0 =√km⇒ R = 20logm+20logf −20log(ρcπ ) (5.22)For high frequencies the mass of the wall is the determining factor forsound insulation (in this case the simple mass-frequency law is applicable).ω = ω 0 =√km ⇒ P ( )i d=P d ρc +2 ≈ 1 and R ≈ 0 (5.23)At the resonance frequency ω 0 the wall becomes transparant for sound.