hw15 - center will be midway between two nodes, and there...

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PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS CLASS 15. 17.09, 33.03, 33.05; Q16.09, Q16.10 17.09. a) Hz 10 4.5 m/s 343 6 × = = f v air λ = 7.6 ×10 -5 m = 76 μ m b) Hz 10 4.5 m/s 500 1 6 × = = f v tissue = 3.3 ×10 -4 m = 330 m 33.03 a) The radius of Earth is R = 6.37 ×10 6 m. ) m 10 37 . 6 )( 10 0 . 1 ( m/s 10 0 . 3 6 4 8 × × × = = c f f = 4.7 ×10 -3 Hz b) Hz 10 7 . 4 1 1 3 - × = = f T T = 212 s = 3 min 32 s 33.05. The frequency of the oscillator must be equal to the frequency of the wave. Proceed by finding the frequency of the wave, and then set it equal to LC π 2 1 . Since C is given, you can solve for L . c f = , so LC c f 2 1 = = . Then, LC c 2 2 2 4 1 = , and 2 2 2 4 Cc L = 2 8 12 2 2 9 m) 10 0 . 3 ( F) 10 17 ( 4 m) 10 550 ( × × × = - - L L = 5.0 ×10 -21 H This is very small, much smaller than could be manufactured.
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Q16.09. a) For a string that is fixed at both ends, there are two nodes at the fundamental frequency (first harmonic), and a new node gets added for each higher harmonic. Therefore, the seventh harmonic would have 8 nodes . b) Since there are an even number of nodes there can be none in the center. Then, the
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Unformatted text preview: center will be midway between two nodes, and there will be an antinode at that point. c) The sixth harmonic will have only 7 nodes on the same string. Therefore, the distance between nodes will be greater, and the wavelength will be longer . d) Since the wavelength is longer, the frequency will be lower , since f = v / . Q16.10. Only in case (d) is it possible that the two strings oscillate at the same frequency. Since the tension is higher in string B , while the densities are the same, the wave speed is higher in string B than in string A . Therefore, waves in string B would have longer wavelengths than waves of the same frequency in string A . Case (d) is the only one to satisfy that condition....
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hw15 - center will be midway between two nodes, and there...

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