16.6

The Speed of Sound

| Gases

Sound travels through gases, liquids, and solids at considerably different speeds, as Table 16-1 reveals. Near room temperature, the speed of sound in air is 343 m/s (767 mi/h) and is markedly greater in liquids and solids. For example, sound travels more than four times faster in water and more than seventeen times faster in steel than it does in air. In general, sound travels slowest in gases, faster in liquids, and fastest in solids.

Table 16-1 Speed of Sound in Gases, Liquids, and Solids

Substance | Speed (m/s)
Gases
Air (0 °C) | 331
Air (20 °C) | 343
Carbon dioxide (0 °C) | 259
Oxygen (0 °C) | 316
Helium (0 °C) | 965
Liquids
Chloroform (20 °C) | 1004
Ethyl alcohol (20 °C) | 1162
Mercury (20 °C) | 1450
Fresh water (20 °C) | 1482
Seawater (20 °C) | 1522
Solids
Copper | 5010
Glass (Pyrex) | 5640
Lead | 1960
Steel | 5960

Like the speed of a wave on a guitar string, the speed of sound depends on the properties of the medium. In a gas, it is only when molecules collide that the condensations and rarefactions of a sound wave can move from place to place. It is reasonable, then, to expect the speed of sound in a gas to have the same order of magnitude as the average molecular speed between collisions. For an ideal gas this average speed is the translational rms speed given by Equation 14.6: , where T is the Kelvin temperature, m is the mass of a molecule, and k is Boltzmann’s constant. Although the expression for vrms overestimates the speed of sound, it does give the correct dependence on Kelvin temperature and particle mass. Careful analysis shows that the speed of sound in an ideal gas is given by

| (16.5)

where is the ratio of the specific heat capacity at constant pressure CP to the specific heat capacity at constant volume CV.
The factor is introduced in Section 15.5, where the adiabatic compression and expansion of an ideal gas is discussed. In Section 15.6 it is shown that has the value of for ideal monatomic gases and a higher value of for ideal diatomic gases. The value of appears in Equation 16.5 because the condensations and rarefactions of a sound wave are formed by adiabatic compressions and expansions of the gas. The regions that are compressed (the condensations) become slightly warmed, and the regions that are expanded (the rarefactions) become slightly cooled. However, no appreciable heat flows from a condensation to an adjacent rarefaction because the distance between the two (half a wavelength) is relatively large for most audible sound waves and a gas is a poor thermal conductor. Thus, the compression and expansion process is adiabatic. Example 4 illustrates the use of Equation 16.5.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 4 | | | An Ultrasonic Ruler

Figure 16-18 shows an ultrasonic ruler that is used to measure the distance to a target, such as a wall. To initiate the measurement, the ruler generates a pulse of ultrasonic sound that travels to the wall and, much like an echo, reflects from it. The reflected pulse returns to the ruler, which measures the time it takes for the round-trip. Using a preset value for the speed of sound, the unit determines the distance to the wall and displays it on a digital readout. Suppose the round-trip travel time is 20.0 ms on a day when the air temperature is 32 °C. Assuming that air is an ideal diatomic gas that the average molecular mass of air is 28.9 u, find the distance between the ultrasonic ruler and the wall.

Figure 16-18 An ultrasonic ruler uses sound with a frequency greater than 20 kHz to measure the distance x to the wall. The blue arcs and blue arrow denote the outgoing sound wave, and the red arcs and red arrow denote the wave reflected from the wall.

Reasoning The distance between the ruler and the wall is equal to the speed of sound multiplied by the time it takes for the sound pulse to reach the wall. The speed v of sound can be determined from a knowledge of the air temperature T and the average mass m of an air molecule by using the relation . The time can be deduced from the given data.
Knowns and Unknowns The data for this problem are listed below:

Description | Symbol | Value | Comment
Round-trip time of sound | tRT | 20.0 ms |
Air temperature | Tc | 32 °C |
Ratio of specific heats for air |  |  |
Average molecular mass of air | m | 28.9 u | Must convert “u” to kilograms.
Unknown Variable
Distance between ruler and wall | x | ? |

Modeling the Problem

| Step 1 Kinematics
| Since sound moves at a constant speed, the distance x it travels is the product of its speed v and the time t, or . The time for the sound to reach the wall is one-half the round-trip time tRT, SO . Thus, the distance to the wall is

The round-trip time tRT is known, but the speed of sound in air at 32 °C is not. We will find an expression for this speed in Step 2.

| (1)

| Step 2 Speed of Sound
| Since the air is assumed to be an ideal gas, the speed v of sound is related to the Kelvin temperature T and the average mass m of an air molecule by

| (16.5)

where is the ratio of the specific heat capacity of air at constant pressure to that at constant volume (see Section 15.5), and k is Boltzmann’s constant. The temperature in this expression must be the Kelvin temperature of the air, which is related to its Celsius perature Tc by (Equation 12.1). Thus, the speed of sound in air is

This expression for v can be substituted into Equation 1, as shown at the right.

Solution Combining the results of the modeling steps, we have

Since the average mass of an air molecule is given in atomic mass units (28.9 u), we must convert it to kilograms by using the conversion factor (see Section 14.1). Thus,

The distance from the ultrasonic ruler to the wall is

Problem solving insight When using equation to calculate the speed of sound in an ideal gas, be sure to express the temperature T in kelvins and not in degrees Celsius or Fahrenheit.

Related Homework: Problems 44 and 46

Check Your Understanding 2

Carbon monoxide (CO), hydrogen (H2), and nitrogen (N2) may be treated as ideal gases. Each has the same temperature and nearly the same value for the ratio of the specific heat capacities at constant pressure and constant volume. In which two of the three gases is the speed of sound approximately the same? (The answers are given at the end of the book.)
Background: This question deals with the factors that determine the speed of sound in an ideal gas. Consult the periodic table on the inside of the back cover, as needed.
For similar questions (including calculational counterparts), consult Self-Assessment Test 16.2. This test is described at the end of Section 16.9.

Sonar (sound navigation ranging) is a technique for determining water depth and locating underwater objects, such as reefs, submarines, and schools of fish. The core of a sonar unit consists of an ultrasonic transmitter and receiver mounted on the bottom of a ship. The transmitter emits a short pulse of ultrasonic sound, and at a later time the reflected pulse returns and is detected by the receiver. The water depth is determined from the electronically measured round-trip time of the pulse and a knowledge of the speed of sound in water; the depth registers automatically on an appropriate meter. Such a depth measurement is similar to the distance measurement discussed for the ultrasonic ruler in Example 4.
Conceptual Example 5 illustrates how the speed of sound in air can be used to estimate the distance to a thunderstorm, using a handy rule of thumb.

The physics of sonar.

BOX 16.1 |

| Conceptual Example 5 | | | Lightning, Thunder, and a Rule of Thumb

There is a rule of thumb for estimating how far away a thunderstorm is. After you see a flash of lightning, count off the seconds until the thunder is heard. Divide the number of seconds by five. The result gives the approximate distance (in miles) to the thunderstorm. Why does this rule work?
Reasoning and Solution Figure 16-19 shows a lightning bolt and a person who is standing one mile away. When the lightning occurs, light and sound (thunder) are produced very nearly at the same instant. Light travels so rapidly that it reaches the observer almost instantaneously. Its travel time is only . In comparison, sound travels very slowly . The time for the thunder to reach the person is . Thus, the time interval between seeing the flash and hearing the thunder is about 5 seconds for every mile of travel. This rule of thumb works because the speed of light is so much greater than the speed of sound that the time needed for the light to reach the observer is negligible compared to the time needed for the sound.

Figure 16-19 A lightning bolt from a thunderstorm generates a flash of light and sound (thunder). The speed of light is much greater than the speed of sound. Therefore, the light reaches the person first, followed about 5 seconds later by the sound.

| Liquids

In a liquid, the speed of sound depends on the density and the adiabatic bulk modulus Bad of the liquid:

| (16.6)

The bulk modulus is introduced in Section 10.7 in a discussion of the volume deformation of liquids and solids. There it is tacitly assumed that the temperature remains constant while the volume of the material changes; that is, the compression or expansion is isothermal. However, the condensations and rarefactions in a sound wave occur under adiabatic rather than isothermal conditions. Thus, the adiabatic bulk modulus Bad must be used when calculating the speed of sound in liquids. Values of Bad will be provided as needed in this text.
Table 16-1 gives some data for the speed of sound in liquids. In seawater, for instance, the speed is 1522 m/s, which is more than four times as great as the speed in air. The speed of sound is an important parameter in the measurement of distance, as discussed for the ultrasonic ruler in Example 4. Accurate distance measurements using ultrasonic sound play an important role in medicine, where the sound often travels through liquid-like materials in the body. A routine preoperative procedure in cataract surgery, for example, uses an ultrasonic probe called an A-scan to measure the length of the eyeball in front of the lens, the thickness of the lens, and the length of the eyeball between the lens and the retina (see Figure 16-20). The measurement is similar to that discussed in Example 4 and relies on the fact that the speed of sound in the material in front of and behind the lens of the eye is 1532 m/s, while that within the lens is 1641 m/s. In cataract surgery, the cataractous lens is removed and often replaced with an implanted artificial lens. Data provided by the A-scan facilitate the design of the size and the optical correction of the lens implant.

Figure 16-20 A cross-sectional view of the human eye.

The physics of cataract surgery.

| Solid Bars

When sound travels through a long slender solid bar, the speed of the sound depends on the properties of the medium according to

| (16.7)

where Y is Young’s modulus (defined in Section 10.7) and is the density.

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