College Physics (4th Edition)

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

Chapter 12 - Problems - Page 463: 15


(a) $P_0 = 28.7~Pa$ (b) $F = 1.58\times 10^{-3}~N$

Work Step by Step

(a) We can find the intensity of the sound: $\beta = 10~log\frac{I}{I_0}$ $120.0 = 10~log\frac{I}{I_0}$ $12.0 = log\frac{I}{I_0}$ $10^{12.0} = \frac{I}{I_0}$ $I = (10^{12.0})~I_0$ $I = (10^{12.0})~(1.0\times 10^{-12}~W/m^2)$ $I = 1.0~W/m^2$ 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: $P_0 = \sqrt{2I\rho v}$ $P_0 = \sqrt{(2)(1.0~W/m^2)(1.2~kg/m^3)(343~m/s)}$ $P_0 = 28.7~Pa$ (b) We can find the force exerted on the eardrum: $F = P~A$ $F = (28.7~Pa)(0.550\times 10^{-4}~m^2)$ $F = 1.58\times 10^{-3}~N$
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