537_Physics ProblemsTechnical Physics

537_Physics ProblemsTechnical Physics - Chapter 18 P18.40...

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Chapter 18 539 P18.40 The air in the auditory canal, about 3 cm long, can vibrate with a node at the closed end and antinode at the open end, with d N to A cm == 3 4 λ so = 012 . m and f v 343 3 ms m kHz . A small-amplitude external excitation at this frequency can, over time, feed energy into a larger-amplitude resonance vibration of the air in the canal, making it audible. P18.41 For a closed box, the resonant frequencies will have nodes at both sides, so the permitted wavelengths will be Ln = 1 2 , n = 123 ,,, bg . i.e., L nn v f 22 and f nv L = 2 . Therefore, with L = 0860±m . and ′= L 210 . m, the resonant frequencies are fn n = 206 Hz a f for L = 0860 . m for each n from 1 to 9 and n 84 5 . H z af for L . m for each n from 2 to 23. P18.42 The wavelength of sound is = v f The distance between water levels at resonance is d v f = 2 ∴= = Rt r d rv f π 2 2 2 and t Rf = 2 2 . P18.43 For both open and closed pipes, resonant frequencies are equally spaced as numbers. The set of
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