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

310 12. Walls, Enclosures, and Barriersthe half-wavelengths of the generated noise. This can render the enclosure uselessfrom the viewpoint of noise attenuation unless the inside of the enclosure is linedwith sound-absorption material at least one-quarter wavelength thick. If there isno sound absorption material inside the enclosure, situations can occur where thenoise generated at resonant frequencies may actually be louder outside than wouldbe the case without the enclosure!12.16 Acoustic BarriersThe term acoustic barrier (or noise barrier) refers to an obstacle that interruptsthe line of sight between a noise source and receiver but does not enclose eithersource or receiver. An acoustic barrier may be in the form of a fence, a wall, a berm(a mound of earth), dense foliage, or a building between the noise source and thereceiver. Noise attenuation occurs from the fact that noise transmission through thebarrier is negligible in comparison with refracted noise, particularly if the barrieris solid, without holes or openings, and it is of sufficient mass. Because the soundreaches the receiver by an indirect path over the top of the barrier, the sound levelwill be less than the case would be if the sound had traveled the (shorter) directpath. The refraction phenomenon is highly dependent upon the frequency contentof the sound. The calculations for barrier attenuation are based in part on Fresnel’swork in optics.In Figure 12.16 consider a room prior to inserting a barrier in the position shown.The mean-square pressure at the sound receiver’s location is given by( Qp0 2 = Wρ 0c4πr + 4 )2 Rwhere p0 2 denotes the mean-square pressure without the barrier. The sound pressurelevel is expressed by( QL p0 = L w + 10 log4πr + 4 )2 RwhereL p0 = SPL without the barrierL w = power level of the sourceQ = directivity of the sourceR = room constant without the barrierr = distance between the source and the receiver.Now let us insert the barrier as shown in Figure 12.16. The mean-square pressurep2 2 , at the receiver with the barrier installed is given byp2 2 = p2 r2 + p2 d2 (12.50)