College Physics (4th Edition)

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

Chapter 12 - Problems - Page 463: 19


(a) The pressure amplitude is $0.051~Pa$ (b) The displacement amplitude is $1.5\times 10^{-7}~m$

Work Step by Step

(a) We can find the intensity of the sound: $\beta = 10~log\frac{I}{I_0}$ $65.0 = 10~log\frac{I}{I_0}$ $6.50 = log\frac{I}{I_0}$ $10^{6.50} = \frac{I}{I_0}$ $I = (10^{6.50})~I_0$ $I = (10^{6.50})~(1.0\times 10^{-12}~W/m^2)$ $I = 3.16\times 10^{-6}~W/m^2$ We can use $343~m/s$ as the speed of sound in air. We can use $\rho = 1.2~kg/m^3$ as the density of air. We can find the pressure amplitude: $P_0 = \sqrt{2I\rho v}$ $P_0 = \sqrt{(2)(3.16\times 10^{-6}~W/m^2)(1.2~kg/m^3)(343~m/s)}$ $P_0 = 0.051~Pa$ The pressure amplitude is $0.051~Pa$ (b) We can find the displacement amplitude: $s_0 = \sqrt{\frac{I}{2\pi^2 \rho f^2 v}}$ $s_0 = \sqrt{\frac{3.16\times 10^{-6}~W/m^2}{(2\pi^2)(1.2~kg/m^3)(131~Hz)^2(343~m/s)}}$ $s_0 = 1.5\times 10^{-7}~m$ The displacement amplitude is $1.5\times 10^{-7}~m$
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