/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Back in the days before computer... [FREE SOLUTION] | 91影视

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Chapter 23: Problem 15

Back in the days before computers, a single typist typing furiously could generate an average sound level nearby of \(60.0 \mathrm{~dB}\). What would be the decibel level in the vicinity if three equally noisy typists were working close to one another? If each typist emits the same amount of sound energy, then the final sound intensity \(I_{f}\) should be three times the initial intensity \(I_{i}\) We have $$ \begin{array}{l} \beta_{f}=10 \log _{10}\left(\frac{I_{f}}{I_{0}}\right)=10 \log _{10} I_{f}-10 \log _{10} I_{0} \\ \text { and } \beta_{i}=10 \log _{10} I_{i}-10 \log _{10} I_{0} \end{array} $$ Subtracting these yields the change in sound level in going from \(I_{i}\) $$ \begin{array}{l} \text { to } I_{f}=3 I_{i}, \\ \qquad \begin{aligned} \beta_{f}-\beta_{i}=10 \log _{10} I_{f}-10 \log _{10} I_{i} \\ \quad \beta_{f}=\beta_{i}+10 \log _{10}\left(\frac{I_{I}}{I_{i}}\right)=60.0 \mathrm{~dB}+10 \log _{10} 3=64.8 \mathrm{~dB} \end{aligned} \end{array} $$ The sound level, being a logarithmic measure, rises very slowly with the number of sources.

Short Answer

Expert verified
The sound level with three typists is 64.8 dB.

Step by step solution

01

Understand the Given Problem

We need to determine the new decibel level when three typists, each producing 60 dB, work together. Each typist is equally noisy, meaning their sound intensities are the same, and the final intensity is three times the initial intensity.
02

Recall the Formula for Sound Intensity Level

The sound intensity level in decibels (dB) is given by \( \beta = 10 \log_{10}\left(\frac{I}{I_0}\right) \), where \( \beta \) is the sound level, \( I \) is the sound intensity, and \( I_0 \) is the reference intensity.
03

Calculate Initial and Final Intensity Levels

Given \( \beta_i = 60 \text{ dB} \) for one typist, the final intensity (with three typists) is \( I_f = 3I_i \). The final sound level is calculated using the change in decibel formula.
04

Derive the Formula for Change in Decibel Level

Using the formulas: \( \beta_f = 10 \log_{10} I_f - 10 \log_{10} I_0 \) and \( \beta_i = 10 \log_{10} I_i - 10 \log_{10} I_0 \), subtract them to find \( \Delta \beta = \beta_f - \beta_i = 10 \log_{10}\left(\frac{I_f}{I_i}\right) \).
05

Calculate the Decibel Increase

Substitute \( I_f = 3I_i \) into the change formula: \( \Delta \beta = 10 \log_{10}(3) \). Calculate \( 10 \log_{10}(3) \approx 4.8 \text{ dB} \).
06

Calculate the Final Sound Level

Add the change to the initial sound level: \( \beta_f = \beta_i + \Delta \beta = 60 \text{ dB} + 4.8 \text{ dB} = 64.8 \text{ dB} \). The sound level when three typists are typing together is 64.8 dB.

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Key Concepts

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

Decibel Calculation
Decibels (dB) are used to measure sound intensity levels, which allows us to express how loud a sound is in a way that's easy to understand. The decibel scale is logarithmic, meaning that it measures the relative intensity between two sounds. A change in decibels corresponds to a multiplicative change in actual intensity, which is why increasing the number of sound sources doesn't lead to a direct, linear increase in dB levels.

In the given example, each of the three typists generates a sound level of 60 dB individually. When you have three equally loud sources, the total sound intensity increases. However, since decibels are a logarithmic measure, this increase is not linear. Instead, you add the change in intensity to the initial dB level, using log functions.

The calculation involves the formula for the change in sound level:
  • New sound level (\(eta_f\)): 64.8 dB
  • Initial sound level (\(eta_i\)): 60.0 dB
  • Change in dB: 10 log(\(3\))

This approach reveals that the final sound intensity level with three typists increases to 64.8 dB.
Logarithmic Measure
The concept of logarithmic measure is crucial for understanding sound levels expressed in decibels. The use of logarithms helps to accommodate the vast range of human hearing, from the faintest whisper to the roar of a jet engine, within a manageable scale. Since the ear perceives sound intensity on a logarithmic scale, the decibel scale reflects our hearing capabilities more accurately compared to a linear scale.

The formula used to calculate decibels is:
  • \( \beta = 10 \, \log_{10}\left( \frac{I}{I_0} \right) \)
where \( \beta \) denotes the sound intensity level in decibels, \( I \) is the sound intensity, and \( I_0 \) is the reference intensity, typically the threshold of human hearing. By taking the logarithm of the ratio of the intensity to the reference level and multiplying it by 10, we can convert the physical intensity of sound into a more perceivable scale.

This logarithmic measure also means that the perception of sound intensity is not direct. For example, a doubling of sound power results in an increase of approximately 3 dB, and therefore, an additional sound source increases the dB level, but not linearly. Understanding this logarithmic relationship allows one to predict changes in sound intensity correctly.
Sound Intensity Formula
The sound intensity formula is fundamental to understanding how sound energy spreads in an environment. Sound intensity \( I \) refers to power per unit area and is measured in watts per square meter (W/m虏). When dealing with sound intensity levels, the formula helps to relate physical sound intensity to perceived loudness via decibels.

The basic formula for sound intensity is:
  • \( I = \frac{P}{A} \)
where \( I \) is the intensity, \( P \) is the power in watts, and \( A \) is the area in square meters over which the power is distributed. This formula highlights why increasing the number of sound sources increases the total sound intensity, which is a product of adding more sound energies in the vicinity.

In the classroom example, when calculating the new sound intensity level with three typists, the final sound intensity \( I_f \) becomes three times the initial intensity \( I_i \). By understanding how the formula works, you can calculate the resulting dB level accurately by using logarithmic measures to translate these into decibels.

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Most popular questions from this chapter

If the intensity of sound changes, the sound level will change. Suppose then that the sound level goes from \(\beta_{i}\) to \(\beta_{f}\) such that \(\beta_{f}-\) \(\beta_{i}=\Delta \beta .\) Show that $$ \Delta \beta=10 \log _{10}\left(I_{f} / I_{i}\right) $$ [ Hint: Remember that \(\log _{10} \mathrm{~A}-\log _{10} \mathrm{~B}=\log _{10} \mathrm{~A} / \mathrm{B} .\) ]

To determine the speed of a harmonic oscillator, a beam of sound is sent along the line of the oscillator's motion. The sound, which is emitted at a frequency of \(8000.0 \mathrm{~Hz}\), is reflected straight back by the oscillator to a detector system. The detector observes that the reflected beam varies in frequency between the limits of \(8003.1 \mathrm{~Hz}\) and \(7996.9 \mathrm{~Hz}\). What is the maximum speed of the oscillator? Take the speed of sound to be \(340 \mathrm{~m} / \mathrm{s}\).

An uncomfortably loud sound might have an intensity of \(0.54\) \(\mathrm{W} / \mathrm{m}^{2}\). Find the maximum displacement of the molecules of air in a sound wave if its frequency is \(800 \mathrm{~Hz}\). Take the density of air to be \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\) and the speed of sound to be \(340 \mathrm{~m} / \mathrm{s}\). We are given \(I, f, \rho\), and \(u\), and have to find \(a_{0}\). From \(I=2 \pi^{2} f^{2} \rho v a_{0}^{2}\) $$ a_{0}=\frac{1}{\pi f} \sqrt{\frac{1}{2 p v}}=\frac{1}{\left(800 \mathrm{~s}^{-1} \pi\right)} \sqrt{\frac{0.54 \mathrm{~W} / \mathrm{m}^{2}}{(2)\left(1.29 \mathrm{~kg} / \mathrm{m}^{3}\right)(340 \mathrm{~m} / \mathrm{s})}}=9.9 \times 10^{-6 \mathrm{~m}} \mathrm{~m}=9.9 \mu \mathrm{m} $$

An automobile moving at \(30.0 \mathrm{~m} / \mathrm{s}\) is approaching a factory whistle that has a frequency of \(500 \mathrm{~Hz}\). (a) If the speed of sound in air is \(340 \mathrm{~m} / \mathrm{s}\), what is the apparent frequency of the whistle as heard by the driver? (b) Repeat for the case of the car leaving the factory at the same speed. This is a Doppler shift problem. Draw an arrow from observer to source; this is the positive direction. Here in part \((a)\) the observer is moving in the positive direction, and \(u_{s}=0 .\) Hence, use \(+u_{o}\) and So (a) \(f_{f}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(500 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+30.0 \mathrm{~m} / \mathrm{s}}{340 \mathrm{~m} / \mathrm{s}-0}=544 \mathrm{~Hz}\) With the car leaving in the negative direction use \(-\mathrm{u}_{o}\) and (b) \(f_{n}=f_{0} \frac{v \pm v_{o}}{v \mp v_{3}}=(500 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}-30.0 \mathrm{~m} / \mathrm{s}}{340 \mathrm{~m} / \mathrm{s}-0}=456 \mathrm{~Hz}\)

Compute the speed of sound in neon gas at \(27.0{ }^{\circ} \mathrm{C}\). For neon, \(M=\) \(20.18 \mathrm{~kg} / \mathrm{kmol}\). Neon, being monatomic, has \(\gamma \approx 1.67\). Therefore, remembering that \(T\) is the absolute temperature, v=\sqrt{\frac{\gamma R T}{M}}=\sqrt{\frac{(1.67)(8314 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K})(300 \mathrm{~K})}{20.18 \mathrm{~kg} / \mathrm{kmol}}}=454 \mathrm{~m} / \mathrm{s}
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