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Chapter 19: Problem 14

Find the intensity ratio of two sounds whose sound levels differ by \(1.00 \mathrm{~dB}\).

Short Answer

Expert verified
The intensity ratio of the two sounds is 1.259.

Step by step solution

01

Understanding the Concept

The Decibel (dB) is a logarithmic unit used to express the ratio of two quantities, usually power or intensity. In the case of sound, it expresses the ratio between the intensity of the sound under consideration and a reference intensity, typically the lowest intensity sound that the average human ear can hear.
02

Formulate the Given Information and Apply the Formula

We know from the problem that the difference in sound levels is 1.00 dB. This means that \( dB_2 - dB_1 =1.00 \). Applying the formula \( dB =10 \log_{10}(\frac{I}{I_0}) \) separately for the two sounds we get \( dB_1 =10 \log_{10}(\frac{I_1}{I_0}) \) and \( dB_2 =10 \log_{10}(\frac{I_2}{I_0}) \). If we subtract these two equations we obtain that \( dB_2 - dB_1 =10 \log_{10}(\frac{I_2}{I_0}) - 10 \log_{10}(\frac{I_1}{I_0}) = 10 \log_{10}(\frac{I_2}{I_1}) \). We can therefore solve for \( \frac{I_2}{I_1} \) by substituting \( dB_2 - dB_1 =1.00 \) and using that \( 10 \log_{10}(x) \) is the inverse function of \( 10^x \). We therefore obtain \( \frac{I_2}{I_1} =10^{(dB_2 - dB_1) /10} =10^{1.00 /10} =1.259 \).
03

Interpret the Result

The intensity ratio \( \frac{I_2}{I_1} = 1.259 \) means that the second sound is roughly 1.259 times as intense as the first sound. This is the result when their decibel levels differ by 1.00 dB.

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Most popular questions from this chapter

A well with vertical sides and water at the bottom resonates at \(7.20 \mathrm{~Hz}\) and at no lower frequency. The air in the well has a density of \(1.21 \mathrm{~kg} / \mathrm{m}^{3}\) and a bulk modulus of \(1.41 \times 10^{5} \mathrm{~Pa}\). How deep is the well?

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