College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 12 - Problems - Page 465: 52


(a) The depth of the water is $521~m$ (b) $\lambda = 4.0~cm$ (c) $\lambda = 9.2~mm$

Work Step by Step

(a) We can use $v = 1533~m/s$ as the speed of sound in the water. We can find the total distance the sound traveled in the water: $d = v~t = (1533~m/s)(0.68~s) = 1042~m$ Since the sound reached the bottom and returned to the ship, the depth of the water is $\frac{1042~m}{2} = 521~m$ (b) We can find the wavelength of the wave in the water: $\lambda = \frac{v}{f}$ $\lambda = \frac{1533~m/s}{38,000~Hz}$ $\lambda = 0.040~m$ $\lambda = 4.0~cm$ (c) When a wave moves from one medium into another medium, the frequency does not change. We can find the wavelength of the wave in the air: $\lambda = \frac{v}{f}$ $\lambda = \frac{350~m/s}{38,000~Hz}$ $\lambda = 0.0092~m$ $\lambda = 9.2~mm$
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