PROBLEM 15 & 16 April 16 2010

PROBLEM 15 & 16 April 16 2010 - (like 14.40): What...

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SOLVING PROBLEM 11 April 16, 2010 Problem 15 (like 14.34): A mouse is hiding from a cat at the closed end of a piece of pipe 0.1 m long. What is the lowest resonant frequency above 1.5 kHz that the mouse could squeak to frighten the cat away from the open end? The velocity of sound in air is 343 m/s. a) 1715 Hz; b) 2572 Hz; c) 3430 Hz; d) 3087 Hz; e) 2744 Hz. What do you know? The resonator has a displacement node at one end and an antinode at the other end. Draw a picture. Now redraw the waves next to the resonator. It is clear that the resonant frequencies are odd multiples of the fundamental frequency, f = ( v sound / = [(343 m/s)/(0.4m)] 1, 3, 5, . .. or 857.5 Hz, 2572.5 Hz, 4287.5 Hz, … Problem 16
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Unformatted text preview: (like 14.40): What are the sum and difference frequencies heard between A, 440 Hz and C, 523.25 Hz? a) 440.25 and 463.25 Hz; b) 183.25 and 663.25 Hz; c) 83.25 and 963.25 Hz; d) 773.25 and 93.25 Hz; e) 220 and 1046.50 Hz. Answer: Sum = 963.25 Hz, Difference = 83.25 Hz What frequencies will a spectrum analyzer report when the two notes are played simultaneously? a) 83.25, 440 523.25, and 963.25 Hz; b) 440 and 523.25 Hz; c) 440 and 223.25 Hz; d) 445, 657, and 923.25 Hz; e) 220, 440, 523.25 and 1046.50 Hz. Answer: Although the nonlinear ear may hear the sum and difference frequencies, the spectrum analyzer only detects the Fourier components of the sound which are at 440 Hz and 523.25 Hz....
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This note was uploaded on 09/23/2010 for the course PH 02c taught by Professor Mile during the Spring '04 term at Riverside Community College.

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PROBLEM 15 & 16 April 16 2010 - (like 14.40): What...

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