College Physics (4th Edition)

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

Chapter 12 - Problems - Page 467: 77


The sound intensity level would be $~29.0~dB$

Work Step by Step

We can find the intensity when the intensity level is 38.0 dB: $10~log~\frac{I}{I_0} = \beta$ $10~log~\frac{I}{I_0} = 38.0$ $log~\frac{I}{I_0} = 3.80$ $\frac{I}{I_0} = 10^{3.80}$ $I = 10^{3.80}~I_0$ $I = 10^{3.80}~(1.0\times 10^{-12}~W/m^2)$ $I = 6.31\times 10^{-9}~W/m^2$ We can find the intensity of one violin: $I' = \frac{I}{8} = \frac{6.31\times 10^{-9}~W/m^2}{8} = 7.89\times 10^{-10}~W/m^2$ We can find the sound intensity level of one violin: $\beta = 10~log\frac{I'}{I_0}$ $\beta = 10~log\frac{7.89\times 10^{-10}~W/m^2}{1.0\times 10^{-12}~W/m^2}$ $\beta = 10~log~(789)$ $\beta = 29.0~dB$ The sound intensity level would be $~29.0~dB$.
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