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94 CHAPTER 5. SOUND INSULATIONk = 2×1.4P 0D(5.31)The resonance frequency is equal to :f 0 = 1√k2π m = 1√ √2.8P0 12π mD ≈ 84 mD(5.32)If the sound has a frequency equal to f 0 both walls will resonate andthus transmit the sound. For common used double glazing the insulationat low frequencies (100-300 Hz) will therefore be low (doubleglazing with m = 10 kg/m 2 and a air layer thickness of 1 cm we foundf 0 = 266 Hz).7. The air layer between the walls has also an infinite range of eigenfrequenciesat which the system becomes transmissive for sound. Thefundamental frequency is found by taking the distance d as half of thewavelength :f 1 = c λ = c(5.33)2dThe harmonics are integer multiples : f 2 = 2f 1 , f 3 = 3f 1 , ... .These frequencies are usually high. One can damp all those resonancesby applying absorbing materials between the two panels, if possible.8. Each panel has its own critical coincidence frequency. It is preferredto select the thickness of both panels different such that the criticalfrequencies do not coincide.5.2.5 Insulation of a composite wallConsider a wall with total surface S composed of different materials (e.g.windows, doors, walls, etc.) In the following we will calculate the insulationvalue of composite walls. Suppose that the wall with surface S consists oftwo components with surfaces S 1 and S 2 and that these components have adifferent insulation value :R 1 = L z −L 1 +10log S 1AR 2 = L z −L 2 +10log S 2A
5.3. THE ACOUSTICAL BARRIER 95For the complete composed wall we have :R = L z −L+10log S A(5.34)Assume that the sounds transmitted by the separate components can beadded non-coherently :It then follows that :p 2 effp 2 0= (p2 1 ) effp 2 0+ (p2 2 ) effp 2 010 0.1(Lz−R+10log S A ) = 10 0.1(Lz−R 1+10log S 1A ) +10 0.1(Lz−R+2+10log S 2A )(5.35)After elimination of 10 0.1Lz−10logA in the left– and right hand side :10 −R10 +logS = 10 −R 110 +logS 1+10 −R 210 +logS 2(5.36)The insulation R of the composite wall can be written as :R = 10log10∑i S i∑i S i10 −R i10(5.37)Example : consider a wall of 1 m × 1 m with a good insulation value(R = 60 dB). If one would make in this wall an opening of 1 mm over thewhole length, one can calculate from equation 5.37 that()1R = 10log≈0.999∗10 −6 +0.001∗10 0()1= 30 dB (5.38)0.001∗10 0This shows that the sound insulation decreases drastically is a small hole ispresent. In general the sound insulation of a composite wall is determinedby the sound insulation of the worst sub wall.5.3 The acoustical barrierOne can often see along roads – and especially highways – walls that have thepurpose to form a sound barrier of the road noise to the inhabitants living inthe neighbourhood of the road. But sound barriers are also used frequentlyin offices and factories (under the form of a screen). Part of the sound is
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VRIJE UNIVERSITEITBRUSSELAcousticsB
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iiWhen sound becomes noise, it will
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ContentsI Introduction to acoustics
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CONTENTSvii4.3 Measuring the acoust
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Part IIntroduction to acoustics1
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4 CHAPTER 1. FUNDAMENTAL CONCEPTS O
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144 APPENDIX A. MATERIAL PROPERTIES
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152 BIBLIOGRAPHY[12] Bruel and Kjae
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