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Unformatted text preview: Phys. 2212 Homework Chap. 21 21.7. Model: Reflections at the string boundaries cause a standing wave on the string. Solve: Figure EX21.7 indicates two full wavelengths on the string. Hence 1 2 (60 cm) 30 cm 0.30 m. λ = = = Thus ( )( ) 0.30 m 100 Hz 30 m/s v f λ = = = 21.52. Model: The openclosed tube forms standing waves. Visualize: Solve: When the air column length L is the proper length for a 580 Hz standing wave, a standing wave resonance will be created and the sound will be loud. From Equation 21.18, the standing wave frequencies of an openclosed tube are f m = m ( v /4 L ), where v is the speed of sound in air and m is an odd integer: m = 1, 3, 5, … The frequency is fixed at 580 Hz, but as the length L changes, 580 Hz standing waves will occur for different values of m . The length that causes the m th standing wave mode to be at 580 Hz is ( ) ( )( ) 343 m/s 4 580 Hz m L = We can place the values of L , and corresponding values of h = 1 m L , in a table: m L h = 1 m  L 1 0.148 m 0.852 m = 85.2 cm 3 0.444 m 0.556 m = 55.6 cm 5 0.739 m 0.261 m = 26.1 cm 7 1.035 m h can’t be negative So water heights of 26 cm, 56 cm, and 85 cm will cause a standing wave resonance at 580 Hz. The figure shows the m = 3 standing wave at h = 56 cm....
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This note was uploaded on 04/18/2010 for the course PHYS ? taught by Professor Zhou during the Spring '10 term at Georgia State.
 Spring '10
 ZHOU

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