problem16_52

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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16.52: (a) The length of the string is , 10 L d = so its third harmonic has frequency . 3 2 1 string 3 μ F f d = The stopped pipe has length L , so its first harmonic has frequency . 4 pipe 1 L v f s = Equating these and using . 3600 1 gives 10 2 s μv F L d = = (b) If the tension is doubled, all the frequencies of the string will increase by a factor of 2 . In particular, the third harmonic of the string will no longer be in resonance with the first harmonic of the pipe because the frequencies will no longer match, so the sound
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Unformatted text preview: produced by the instrument will be diminished. (c) The string will be in resonance with a standing wave in the pipe when their frequencies are equal. Using string 1 pipe 1 3 f f = , the frequencies of the pipe are string 1 pipe 1 3 nf nf = , (where n =1, 3, 5…). Setting this equal to the frequencies of the string , string 1 f n ′ the harmonics of the string are ,... 15 , 9 , 3 3 = = ′ n n...
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