## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 16 - Traveling Waves - Exercises and Problems: 72

#### Answer

When the sound level increases by 1 dB, the sound intensity increases by a factor of 1.26

#### Work Step by Step

Let $\beta_2 = 61~dB$ when the intensity is $I_2$. Let $\beta_1 = 60~dB$ when the intensity is $I_1$. Therefore; $\beta_2-\beta_1 = 10~log(\frac{I_2}{I_0})-10~log(\frac{I_1}{I_0})$ $61~dB-60~dB = 10~[log(\frac{I_2}{I_0})-log(\frac{I_1}{I_0})]$ $1~dB = 10~log\frac{(\frac{I_2}{I_0})}{(\frac{I_1}{I_0})}$ $0.1 = log\frac{I_2}{I_1}$ $10^{0.1} = \frac{I_2}{I_1}$ $I_2 = 10^{0.1}~I_1$ $I_2 = 1.26~I_1$ When the sound level increases by 1 dB, the sound intensity increases by a factor of 1.26.

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