Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

Chapter 20 - Traveling Waves - Exercises and Problems - Page 587: 31

Answer

The sound intensity level at a distance of 1.0 km is 109.5 dB

Work Step by Step

Let $I_2$ be the intensity at a distance of $r_2 = 1.0~km$. We can find an expression for $I_2$ in terms of $I_1$. $\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}$ $I_2 = \frac{r_1^2}{r_2^2}~I_1$ $I_2 = \frac{(30~m)^2}{(1000~m)^2}$ $I_2 = (9\times 10^{-4})~I_1$ We can find the sound intensity level $\beta_2$. $\beta_2 = 10~log(\frac{I_2}{I_0})$ $\beta_2 = 10~log(\frac{(9\times 10^{-4})~I_1}{I_0})$ $\beta_2 = 10~log(9\times 10^{-4})+10~log(\frac{I_1}{I_0})$ $\beta_2 = (-30.5~dB)+140~dB$ $\beta_2 = 109.5~dB$ The sound intensity level at a distance of 1.0 km is 109.5 dB.

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