THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

292 12. Walls, Enclosures, and BarriersSolutionSurface density m = 8 in. × 1m/(39.37 in.) × 2406 kg/m 3 = 488.9 kg/m 2Using field incidence law Equation (12.13)TL − 20 log(fm) − 47 = 20 log(800 × 488.9) − 47 = 65 dBLaboratory tests indicate a TL closer to 50 dB. Actual installations may yield evenlower values because of flanking noise transmission paths.Example Problem 3Find the resonant frequency of a double-panel wall constructed of a 15-kg/m 2 anda 20-kg/m 2 gypsum board with 9-cm airspace.SolutionApplying Equation (12.23):√ √m1 + m 2 15 + 20f n = 60 = 60= 68 Hz.hm 1 m 2 0.09(15)(20)12.8 Measuring Transmission LossRecasting Equation (12.1) for transmission loss TL in terms of vector sound intensity,we write( )IITL = 10 logI Twhere I I = incident vector sound intensity and I T = transmitted vector soundintensity, For sound pressure level and particle velocity in phase, sound intensityis given byI = p2 rmsρcwith the result that transmission loss could be restated as( p2)TL = 10 log rmsI.p 2 rmsTBut the above equation is not usable for measurement purposes, since distinctioncannot be made between root-mean-square (rms) pressures attributable to incidentsound, transmitted sound and reflected sound.In order to determine the transmission loss, it is required to set up a sourcechamber and a receiving chamber, with a panel of the specimen material installedin a window between the two chambers. Such a setup is shown in Figure 12.8. Thechambers must be designed so as to minimize the sound transmission paths other