P18_030 - during its vibration. Consequently, the...

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30. (a) From Eq. 17-53, we have f = nv 2 L = (1)(250 m / s) 2(0 . 150 m) = 833 Hz . (b) The frequency of the wave on the string is the same as the frequency of the sound wave it produces
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Unformatted text preview: during its vibration. Consequently, the wavelength in air is = v sound f = 348 m / s 833 Hz = 0 . 418 m ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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