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

148 7. Pipes, Waveguides, and ResonatorsIn conducting his experiments of complex musical tones, Helmholtz used aseries of resonators of the type illustrated in Figure 7.6(c), with small nipplesopposite the necks of the resonators. He used a graduated series of resonators, withdiffering volumes and neck sizes to achieve a wide range of resonant frequencies.When an incident sound wave contains a frequency component that correspondsto the resonant frequency of a Helmholtz device, a greatly amplified signal willbe generated within the cavity of the resonator, and it can be aurally detected byconnecting a short, flexible tube from the nipple to the ear. These phenomenaled to the definition of pressure amplification, which is the ratio of the acousticpressure amplitude P c inside the cavity to the external driving pressure amplitudeP. Equation (7.58) provides the pressure amplitude P c . At resonance|δ| =PAω 0 R rThen applying Equations (7.60 ) and (7.66), we find for the resonator with a flangedneck that√( )P cLP = 2π ′ 3V = QAThe gain, therefore, is the same as the value of the quality factor defined inEquation (7.67), and we see the Helmholtz resonator behaving as an amplifierof gain Q at resonance, a fact which explains why a loudspeaker mounted in aclosed chamber can be regarded as a Helmholtz resonator in which the air’s reactanceand the speaker cone mass constitute the effective mass of the system. The airresistance within the box and the cone’s stiffness combine to provide the effectivestiffness, while the effective resistance results from the sum of that attributable tothe radiation of acoustic energy and that arising from the mechanical resistance ofthe speaker cone.Example Problem 2A Helmholtz resonator is a sphere of 8 cm internal diameter. If it is to resonate at450 Hz in air, what is the hole diameter that should be drilled into the sphere?SolutionApplying Equation (7.66) we have√ √Aω 0 = 450 × 2π rad s −1 = cL ′ V = 341 m πa s−1 =21.5a × 4 π(0.04 m)33which results in a = 0.00865 m = 0.865 cm, or 1.73 cm diameter. Here we assumedan unflanged “pipe” with an effective value of L ′ = 1.5a.