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Chapter 16: Problem 18

(a) What is the sound intensity level in a car when the sound intensity is 0.500\(\mu \mathrm{W} / \mathrm{m}^{2} 7\) (b) What is the sound intensity level in the air near a jackhammer when the pressure amplitude of the sound is 0.150 \(\mathrm{Pa}\) and the temperature is \(20.0^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
(a) 57 dB, (b) 74 dB.

Step by step solution

01

Understanding the Sound Intensity Level Formula

The sound intensity level, measured in decibels (dB), is determined using the formula: \( L = 10 \log_{10}\left(\frac{I}{I_0}\right) \), where \( I \) is the sound intensity and \( I_0 = 1 \times 10^{-12} \text{ W/m}^2 \) is the reference intensity.
02

Calculate the Sound Intensity Level in the Car

Given \( I = 0.500 \times 10^{-6} \text{ W/m}^2 \), substitute into the formula: \[ L = 10 \log_{10}\left(\frac{0.500 \times 10^{-6}}{1 \times 10^{-12}}\right) \]. \[ L = 10 \log_{10}(0.500 \times 10^6) \]. Simplifying: \[ L = 10 \times (\log_{10}(0.500) + \log_{10}(10^6)) \]. Using \( \log_{10}(10^6) = 6 \) and \( \log_{10}(0.500) \approx -0.301 \), we get: \[ L = 10 \times (-0.301 + 6) \]. Hence, \[ L = 10 \times 5.699 \approx 57 \] dB.
03

Understanding the Relationship Between Pressure Amplitude and Intensity

The intensity \( I \) of a sound wave in terms of the pressure amplitude \( \Delta P \) is given by: \( I = \frac{(\Delta P)^2}{2 \rho v} \), where \( \rho \) is the density of the air and \( v \) is the speed of sound in the air. At \( 20^{\circ} \text{C} \), the speed of sound \( v \) is approximately \( 343 \text{ m/s} \), and the density \( \rho \) is approximately \( 1.2041 \text{ kg/m}^3 \).
04

Calculate the Intensity Near the Jackhammer

Given \( \Delta P = 0.150 \text{ Pa} \), use the formula: \[ I = \frac{0.150^2}{2 \times 1.2041 \times 343} \]. First calculate \( \Delta P^2 = 0.0225 \text{ Pa}^2 \). Then substitute values: \[ I = \frac{0.0225}{2 \times 1.2041 \times 343} \approx \frac{0.0225}{826.2026} \approx 2.723 \times 10^{-5} \text{ W/m}^2 \].
05

Calculate the Sound Intensity Level Near the Jackhammer

Using \( I = 2.723 \times 10^{-5} \text{ W/m}^2 \), substitute into the formula: \[ L = 10 \log_{10}\left(\frac{2.723 \times 10^{-5}}{1 \times 10^{-12}}\right) \]. Simplifying: \[ L = 10 \log_{10}(2.723 \times 10^7) \]. Evaluate: \( \log_{10}(2.723) \approx 0.435 \) so: \[ L = 10 \times (0.435 + 7) \approx 10 \times 7.435 = 74.35 \]. Thus, the sound intensity level is \( 74 \) dB.

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Key Concepts

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

Sound Intensity
Sound intensity is the power per unit area carried by a sound wave. It is measured in watts per square meter (W/m虏). Sound intensity reflects how much acoustic power flows through a particular area, and it decreases with distance from the sound source. The intensity of a sound can tell us how loud it will be perceived by an ear located at a certain point. For example, in a car or near a loud machine, knowing the sound intensity helps in assessing how much sound energy is affecting that spot. This measurement plays a crucial role in evaluating environments where hearing protection might be required. To calculate sound intensity, it can be understood as the square of the sound pressure divided by the density of the medium and the speed of sound in that medium.
Pressure Amplitude
Pressure amplitude is the variation in pressure caused by the propagation of sound waves. It is the peak value of the pressure change during a sound wave and is measured in Pascals (Pa). This quantity tells us how strong the sound is at its peak moment. In simple terms, higher pressure amplitude means a louder sound to the human ear. The pressure amplitude is related to the loudness and power of the sound, with higher amplitudes corresponding to louder sounds. In the example of a jackhammer, a pressure amplitude of 0.150 Pa means the pressure changes significantly at the wave's peak, resulting in a loud sound.
Decibels
Decibels (dB) are the units used to measure sound intensity levels. The decibel scale is logarithmic, meaning each increment represents a tenfold change in intensity. This scale allows us to handle the vast range of human hearing within a manageable framework.
  • A change of 10 dB represents a sound intensity change by a factor of 10.
  • A 20 dB increase means the sound intensity is 100 times greater.
This is why a slight increase in dB reflects a significant increase in sound intensity or loudness. Common sounds have various decibel levels, such as a quiet car interior which might be around 57 dB, while a jackhammer near your ear can be around 74 dB, demonstrating much greater intensity.
Speed of Sound
The speed of sound is the rate at which sound waves travel through a medium. In air at room temperature, this speed is approximately 343 meters per second (m/s). This speed can vary based on temperature and the medium through which it travels. For instance, sound travels faster in warmer air because the molecules are moving more rapidly and can transmit the wave more quickly. Understanding the speed of sound is crucial for determining how quickly sound reaches our ears or measuring the intensity of sounds in different environments.
Density of Air
Density of air refers to the mass of air particles in a given volume, typically expressed in kilograms per cubic meter (kg/m鲁). At 20掳C, the average density of air is approximately 1.2041 kg/m鲁. This value is essential when calculating sound intensity because it acts along with speed and pressure amplitude to influence how sound travels and disperses in the air. The density affects the sound wave's ability to carry through the medium, and changes in density can result from factors like altitude or temperature. Even small variations in air density can significantly affect sound propagation.

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Most popular questions from this chapter

Horseshoe bats (genus Rhinolophus) emit sounds from their nostrils and then listen to the frequency of the sound reflected from their prey to determine the prey's speed. (The "horseshoe" that gives the hat its name is a depression around the nostrils that acts like a focusing mirror, so that the hat emits sound in a narrow beam like a flashlight.) A Rhinolophus flying at speed \(v_{\text { tot }}\) emits sound of fre-quency \(f_{\text { but }}\) ; the sound it hears reflected from an insect flying toward it has a higher frequency \(f_{\text { rent }}(\text { a) Show that the speed of the insect is }\) where \(v\) is the speed of sound. (b) If \(f_{\mathrm{bat}}=80.7 \mathrm{kHz}, \quad f_{\mathrm{rell}}=\) \(83.5 \mathrm{kHz},\) and \(v_{\mathrm{bat}}=3.9 \mathrm{m} / \mathrm{s},\) calculate the speed of the insect.

How fast (as a percentage of light speed) would a star have to be moving so that the frequency of the light we recelve from it is 10.0\(\%\) higher than the frequency of the light it is emitting? Would it be moving away from us or toward us? (Assume it is moving either directly away from us or directly toward us.)

A person is playing a small fute 10.75 \(\mathrm{cm}\) long, open at one end and closed at the other, near a taut string having a fundamental frequency of 600.0 \(\mathrm{Hz}\) . If the speed of sound is 344.0 \(\mathrm{m} / \mathrm{s}\) , for which harmonics of the flute will the string resonate? In each case, which harmonic of the string is in resonance?

A small sphere of radius \(R\) is arranged to pulsate so that its radius varies in simple harmonic motion between a minimum of \(R-\Delta R\) and a maximum of \(R+\Delta R\) with frequency \(f .\) This produces sound waves in the surrounding air of density \(\rho\) and bulk modulus \(B\) (a) Find the intensity of sound waves at the surface of the sphere. (The amplitude of oscillation of the sphere is the same as that of the air at the surface of the sphere.) (b) Find the total acoustic powcr radiated by the sphcrc. (c) At a distance \(d \gg R\) from the center of the sphere, find the amplitude, pressure amplitude, and intensity of the sound wave.

Doppler Radar. A giant thunderstorm is moving toward a weather station at \(45.0 \mathrm{mi} / \mathrm{h}(20.1 \mathrm{m} / \mathrm{s}) .\) If the station sends a radar beam of frequency 200.0 \(\mathrm{MHz}\) toward the storm, what is the difference in frequency between the emitted beam and the beam reflected back from the storm? Be careful to carry plenty of significant figures! (Hint The storm reflects the same frequency that it receives.)
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