Explanation Given Info : The speed of the sound in air is 343 m/s and the frequency of the hearing report in first resonance is 52.0 Hz Formula to calculate the depth of the well under the resonance condition L = ( 2 n − 1 ) λ 1 4 (1) L is the depth of the well. n is the integer. λ 1 is the wavelength of the first resonance condition. The relation between the wavelength and frequency is, λ 1 = v f 1 (2) v is the speed of the sound in air. f 1 is the frequency of the hearing report in first resonance. Use equation (2) to rewrite equation (1) we get, L = ( 2 n − 1 ) v 4 f 1 Substitute 343 m/s for v and 52.0 Hz for f 1 . L = ( 2 n − 1 ) 343 m/s 4 ( 52.0 Hz ) = 1.65 m ( 2 n − 1 ) (3) Thus, student measure an audio oscillator under the depth of the water well for first successive resonance is 1.65 m ( 2 n − 1 ) . Formula to calculate the depth of the well under the resonance condition L = ( 2 n + 1 ) v 4 f 2 (4) f 2 is the frequency of the hearing report in second resonance. Substitute 343 m/s for v and 60.0 Hz for f 2 . L = ( 2 n + 1 ) 343 m/s 4 ( 60.0 Hz ) = 1.43 m ( 2 n + 1 ) (5) Under the resonance condition, the student measure an audio oscillator under the depth of the water well for first successive resonance is equal to the second resonance ( L = L )

11th Edition

ISBN: 9781337741644

Author: SERWAY

Publisher: CENGAGE L

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Chapter 14, Problem 74AP

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The depth of a water well.

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Chapter 14, Problem 73AP

Chapter 14, Problem 75AP

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Chapter 14 Solutions

COLLEGE PHYSICS (LL W/WEBASSIGN)

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The depth of a water well.

Solution Summary: The author explains how to determine the depth of a water well. The speed of sound in air is 343 m/s and the frequency of the hearing report in first resonance.

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Chapter 14 Solutions