P18_038 - the same frequency we solve this for the tension...

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38. (a) Using Eq. 18-39 with n = 1 (for the fundamental mode of vibration) and 343 m/s for the speed of sound, we obtain f = (1) v sound 4 L tube = 343 m / s 4(1 . 20 m) =71 . 5Hz . (b) For the wire (using Eq. 17-53) we have f 0 = nv wire 2 L wire = 1 2 L wire r τ µ where µ = m wire /L wire . Recognizing that f = f 0 (both the wire and the air in the tube vibrate at
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Unformatted text preview: the same frequency), we solve this for the tension τ : τ = (2 L wire f ) 2 µ m wire L wire ¶ = 4 f 2 m wire L wire = 4(71 . 5 Hz) 2 ( 9 . 60 × 10 − 3 kg ) (0 . 33 m) = 64 . 8 N ....
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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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