538_Physics ProblemsTechnical Physics

538_Physics ProblemsTechnical Physics - 540 P18.45...

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540 Superposition and Standing Waves P18.45 For resonance in a narrow tube open at one end, fn v L n == 4 135 ,,, bg . (a) Assuming n = 1 and n = 3, 384 40228 = v . a f and 384 3 40683 = v . a f . In either case, v = 350 m s . (b) For the next resonance n = 5, and L v f = 5 4 5350 4384 114 1 ms s m ej .. 22.8 cm 68.3 cm f = 384 Hz warm air FIG. P18.45 P18.46 The length corresponding to the fundamental satisfies f v L = 4 : L v f 1 4 34 4512 0167 = a f . m . Since L > 20 0 . cm, the next two modes will be observed, corresponding to f v L = 3 4 2 and f v L = 5 4 3 . or L v f 2 3 4 0502 . m and L v f 3 5 4 0837 m . P18.47 We suppose these are the lowest resonances of the enclosed air columns. For one, λ = v f 343 256 134 1 s m . length
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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