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Chapter 16: Problem 26

Compare the intensity of sound at the pain level, \(120 \mathrm{~dB}\), with that at the whisper level, \(20 \mathrm{~dB}\).

Short Answer

Expert verified
Answer: The intensity of sound at the pain level is 10^10 times greater than that at the whisper level.

Step by step solution

01

Know the decibel scale formula

The decibel scale formula is: \(I_{dB} = 10 \times \log_{10}(\frac{I}{I_0})\) where \(I_{dB}\) is the intensity level in decibels, \(I\) is the intensity of the sound, and \(I_0\) is the reference intensity (\(10^{-12}\) W/m虏).
02

Find the intensity of sound at the pain level

We are given that the intensity level at the pain level is 120 dB. We will plug in the values to the decibel scale formula and solve for the intensity \(I\): \(120 = 10 \times \log_{10}(\frac{I}{10^{-12}})\) First, divide both sides by 10: \(12 = \log_{10}(\frac{I}{10^{-12}})\) Now, calculate the inverse logarithm to find the intensity: \(I = 10^{12} \times 10^{-12}\) \(I = 1\,\text{W/m}^2\)
03

Find the intensity of sound at the whisper level

Similarly, we will find the intensity at the whisper level, which is 20 dB: \(20 = 10 \times \log_{10}(\frac{I}{10^{-12}})\) Divide both sides by 10: \(2 = \log_{10}(\frac{I}{10^{-12}})\) Now, calculate the inverse logarithm to determine the intensity: \(I = 10^{2} \times 10^{-12}\) \(I = 10^{-10}\,\text{W/m}^2\)
04

Compare the intensities

Now that we have the intensities of both the pain level and the whisper level sounds, we can compare them: Intensity at pain level: \(1\,\text{W/m}^2\) Intensity at whisper level: \(10^{-10}\,\text{W/m}^2\) To show the difference, we can calculate the ratio of the intensities: \(\text{Intensity ratio} = \frac{1\,\text{W/m}^2}{10^{-10}\,\text{W/m}^2} = 10^{10}\) The intensity of sound at the pain level is \(10^{10}\) times greater than that at the whisper level.

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