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Chapter 17: Problem 28

Show that the difference between decibel levels \(\beta_{1}\) and \(\beta_{2}\) of a sound is related to the ratio of the distances \(r_{1}\) and \(r_{2}\) from the sound source by $$\beta_{2}-\beta_{1}=20 \log \left(\frac{r_{1}}{r_{2}}\right)$$

Short Answer

Expert verified
Following the process, it has been confirmed that indeed \( \beta_{2} - \beta_{1} = 20 log \left( \frac{r_{1}}{r_{2}} \right) \).

Step by step solution

01

Express Intensities In Terms Of Distances

The intensity of sound decreases with distance as per the inverse square law. It can be expressed as \( I=\frac{P}{4 \pi r^{2}} \) where P is the power of the sound source and r is the distance from the source. Therefore \( I_{1}=\frac{P}{4 \pi r_{1}^{2}} \) and \( I_{2}=\frac{P}{4 \pi r_{2}^{2}} \).
02

Express Decibels In Terms Of Intensities

The sound level in decibels can be defined as \( \beta = 10 \log \left(\frac{I}{I_{0}}\right) \). By substituting \( I_{1} \) and \( I_{2} \) we found in the previous step, we get \( \beta_{1} = 10 \log \left(\frac{P/4 \pi r_{1}^{2}}{I_{0}}\right) \) and \( \beta_{2} = 10 \log \left(\frac{P/4 \pi r_{2}^{2}}{I_{0}}\right) \).
03

Find the Difference of Decibels

Find the difference of the decibel levels given by \( \beta_{2} - \beta_{1} = 10 \log \left(\frac{P/4 \pi r_{2}^{2}}{I_{0}}\right) - 10 \log \left(\frac{P/4 \pi r_{1}^{2}}{I_{0}}\right) \).
04

Simplify Using Logarithm Properties

The rules of logarithms state that \( log(a) - log(b) = log \left( \frac{a}{b} \right) \). Applying this: \( \beta_{2} - \beta_{1} = 10 log\left( \frac{P/4 \pi r_{2}^{2}}{I_{0} * P/4 \pi r_{1}^{2}} / I_{0} \right) = 10 log \left( \frac{r_{1}^{2}}{r_{2}^{2}} \right) \). Again applying the logarithm property \( log(a^{n}) = n log(a) \), this simplifies to \( \beta_{2} - \beta_{1} = 20 log \left( \frac{r_{1}}{r_{2}} \right) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Square Law
The inverse square law is a principle that helps us understand how the intensity of sound diminishes as it travels further from the source. When we talk about the intensity of sound, we're referring to the power per unit area, which is measured in watts per square meter. The key takeaway from the inverse square law is that the intensity of sound decreases proportionally to the square of the distance from the sound source.

Let's visualize it with an example. Imagine you're at a concert standing 1 meter away from a speaker. If you move to 2 meters away, the intensity isn't just halved, it is actually reduced to a quarter. This is because the energy from the speaker is spread over 4 times the area (since area increases with the square of the radius). The same principle applies to all kinds of waves spreading out from a point source, including light and radio waves. This law is crucial when we're trying to calculate changes in decibel levels over different distances.
Intensity of Sound
The intensity of sound is a measure that indicates how much energy passes through a certain area in a given time frame and is expressed in watts per square meter. When we measure the intensity of sound, we are essentially looking at how strong or loud a sound wave is as it reaches our ears or a measuring device.

Intensity is highly relevant for understanding auditory perception and is used in the formulation of the decibel scale. The decibel (dB) scale is a logarithmic one, which means that a small change in dB can represent a large change in intensity. This is because our ears perceive sound in a nonlinear fashion; we tend to notice differences in loudness more at lower intensities than at higher ones. In practice, doubling the sound intensity adds about 3 dB to its level, which is noticeable but not as dramatic as a doubling in loudness would imply. Understanding sound intensity is essential for evaluating noise pollution, designing soundproofing, and in various fields such as acoustics and audio engineering.
Logarithm Properties
Logarithms are mathematical operations that help us solve equations involving exponential relationships. They have properties that make working with exponential quantities more convenient. In the context of decibel calculations, two logarithm properties are particularly important:
  • The logarithm of a quotient is equal to the difference of the logarithms: \( \log(\frac{a}{b}) = \log(a) - \log(b) \).
  • The logarithm of a power is equal to the exponent times the logarithm of the base: \( \log(a^n) = n\log(a) \)

These properties allow us to simplify complex relationships into manageable calculations, such as determining the change in decibel levels when the distance from the sound source changes. By applying logarithm properties, we can express the change in decibel levels directly as a function of the distances (\( r_1 \) and \( r_2 \)), rather than working with the raw intensity values. This makes it easier to understand the relationship between distance and perceived loudness, and is a great example of how mathematical principles can provide practical answers to real-world problems.

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Most popular questions from this chapter

Find the speed of sound in mercury, which has a bulk modulus of approximately \(2.80 \times 10^{10} \mathrm{N} / \mathrm{m}^{2}\) and a density of \(13600 \mathrm{kg} / \mathrm{m}^{3}\)

A cowboy stands on horizontal ground between two parallel vertical cliffs. He is not midway between the cliffs. He fires a shot and hears its echoes. The second echo arrives \(1.92 \mathrm{s}\) after the first and \(1.47 \mathrm{s}\) before the third. Consider only the sound traveling parallel to the ground and reflecting from the cliffs. Take the speed of sound as \(340 \mathrm{m} / \mathrm{s}\) (a) What is the distance between the cliffs? (b) What If? If he can hear a fourth echo, how long after the third echo does it arrive?

This problem represents a possible (but not recommended) way to code instantaneous pressures in a sound wave into 16 -bit digital words. Example 17.2 mentions that the pressure amplitude of a \(120-\mathrm{dB}\) sound is \(28.7 \mathrm{N} / \mathrm{m}^{2}\) Let this pressure variation be represented by the digital -code \(65536 .\) Let zero pressure variation be represented on the recording by the digital word \(0 .\) Let other intermediate pressures be represented by digital words of intermediate -size, in direct proportion to the pressure. (a) What digital word would represent the maximum pressure in a \(40 \mathrm{dB}\) -sound? (b) Explain why this scheme works poorly for soft -sounds. (c) Explain how this coding scheme would clip off half of the waveform of any sound, ignoring the actual -shape of the wave and turning it into a string of zeros. By introducing sharp corners into every recorded waveform, this coding scheme would make everything sound like a buzzer or a kazoo.

Many artists sing very high notes in ad lib ornaments and cadenzas. The highest note written for a singer in a published score was F-sharp above high \(\mathrm{C}, 1.480 \mathrm{kHz}\), for Zerbinetta in the original version of Richard Strauss's opera Ariadne auf Naxos. (a) Find the wavelength of this sound in air. (b) Suppose people in the fourth row of seats hear this note with level \(81.0 \mathrm{dB} .\) Find the displacement amplitude of the sound. (c) What If? Because of complaints, Strauss later transposed the note down to \(\mathrm{F}\) above high \(\mathrm{C}, 1.397 \mathrm{kHz}\). By what increment did the wavelength change?

The tensile stress in a thick copper bar is \(99.5 \%\) of its elastic breaking point of \(13.0 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) If a \(500-\mathrm{Hz}\) sound wave is transmitted through the material, (a) what displacement amplitude will cause the bar to break? (b) What is the maximum speed of the elements of copper at this moment? (c) What is the sound intensity in the bar?
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