College Physics for AP® Courses

# 17.2Speed of Sound, Frequency, and Wavelength

College Physics for AP® Courses17.2 Speed of Sound, Frequency, and Wavelength

### Learning Objectives

By the end of this section, you will be able to:

• Define pitch.
• Describe the relationship between the speed of sound, its frequency, and its wavelength.
• Describe the effects on the speed of sound as it travels through various media.
• Describe the effects of temperature on the speed of sound.

The information presented in this section supports the following AP® learning objectives and science practices:

• 6.B.4.1 The student is able to design an experiment to determine the relationship between periodic wave speed, wavelength, and frequency, and relate these concepts to everyday examples. (S.P. 4.2, 5.1, 7.2)
Figure 17.8 When a firework explodes, the light energy is perceived before the sound energy. Sound travels more slowly than light does. (credit: Dominic Alves, Flickr)

Sound, like all waves, travels at a certain speed and has the properties of frequency and wavelength. You can observe direct evidence of the speed of sound while watching a fireworks display. The flash of an explosion is seen well before its sound is heard, implying both that sound travels at a finite speed and that it is much slower than light. You can also directly sense the frequency of a sound. Perception of frequency is called pitch. The wavelength of sound is not directly sensed, but indirect evidence is found in the correlation of the size of musical instruments with their pitch. Small instruments, such as a piccolo, typically make high-pitch sounds, while large instruments, such as a tuba, typically make low-pitch sounds. High pitch means small wavelength, and the size of a musical instrument is directly related to the wavelengths of sound it produces. So a small instrument creates short-wavelength sounds. Similar arguments hold that a large instrument creates long-wavelength sounds.

The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves:

$vw=fλ,vw=fλ, size 12{v size 8{w}=fλ} {}$
17.1

where $vwvw size 12{v size 8{w}} {}$ is the speed of sound, $ff size 12{f} {}$ is its frequency, and $λλ size 12{λ} {}$ is its wavelength. The wavelength of a sound is the distance between adjacent identical parts of a wave—for example, between adjacent compressions as illustrated in Figure 17.9. The frequency is the same as that of the source and is the number of waves that pass a point per unit time.

Figure 17.9 A sound wave emanates from a source vibrating at a frequency $f f$, propagates at $v w v w$, and has a wavelength $λλ size 12{λ} {}$.

Table 17.4 makes it apparent that the speed of sound varies greatly in different media. The speed of sound in a medium is determined by a combination of the medium’s rigidity (or compressibility in gases) and its density. The more rigid (or less compressible) the medium, the faster the speed of sound. For materials that have similar rigidities, sound will travel faster through the one with the lower density because the sound energy is more easily transferred from particle to particle. The speed of sound in air is low, because air is compressible. Because liquids and solids are relatively rigid and very difficult to compress, the speed of sound in such media is generally greater than in gases.

### Applying the Science Practices: Bottle Music

When liquid is poured into a small-necked container like a soda bottle, it can make for a fun musical experience! Find a small-necked bottle and pour water into it. When you blow across the surface of the bottle, a musical pitch should be created. This pitch, which corresponds to the resonant frequency of the air remaining in the bottle, can be determined using Equation 17.1. Your task is to design an experiment and collect data to confirm this relationship between the frequency created by blowing into the bottle and the depth of air remaining.

1. Use the explanation above to design an experiment that will yield data on depth of air column and frequency of pitch. Use the data table below to record your data.
Depth of air column (λ) Frequency of pitch generated (f)
Table 17.1
2. Construct a graph using the information collected above. The graph should include all five data points and should display frequency on the dependent axis.
3. What type of relationship is displayed on your graph? (direct, inverse, quadratic, etc.)
4. Does your graph align with equation 17.1, given earlier in this section? Explain.

Note: For an explanation of why a frequency is created when you blow across a small-necked container, explore Section 17.5 later in this chapter.

1. As the depth of the air column increases, the frequency values must decrease. A sample set of data is displayed below.
Depth of air column (λ)Frequency of pitch generated (f)
24 cm689.6 Hz
22 cm752.3 Hz
20 cm827.5 Hz
18 cm919.4 Hz
16 cm1034.4 Hz
Table 17.2
2. The graph drawn should have frequency on the vertical axis, contain five data points, and trend downward and to the right. A graph using the sample data from above is displayed below.
Figure 17.10 A graph of the depth of air column versus the frequency of pitch generated.
3. Inverse relationship.
Depth of air column (λ)Frequency of pitch generated (f)Product of wavelength and frequency
24 cm689.6 Hz165.5
22 cm752.3 Hz165.5
20 cm827.5 Hz165.5
18 cm919.4 Hz165.5
16 cm1034.4 Hz165.5
Table 17.3
4. The graph does align with the equation v = f λ. As the wavelength decreases, the frequency of the pitch generated increases. This relationship is validated by both the sample data table and the sample graph. Additionally, as Table 17.1 demonstrates, the product of λ and f is constant across all five data points.

In addition to these explanations, the student may use the formula as given in the problem statement to show that the product f × air column height is consistently 165.5.

Medium vw(m/s)
Gases at $0ºC 0ºC$
Air 331
Carbon dioxide 259
Oxygen 316
Helium 965
Hydrogen 1290
Liquids at $20ºC 20ºC$
Ethanol 1160
Mercury 1450
Water, fresh 1480
Sea water 1540
Human tissue 1540
Solids (longitudinal or bulk)
Vulcanized rubber 54
Polyethylene 920
Marble 3810
Glass, Pyrex 5640
Aluminum 5120
Steel 5960
Table 17.4 Speed of Sound in Various Media

Earthquakes, essentially sound waves in Earth’s crust, are an interesting example of how the speed of sound depends on the rigidity of the medium. Earthquakes have both longitudinal and transverse components, and these travel at different speeds. The bulk modulus of granite is greater than its shear modulus. For that reason, the speed of longitudinal or pressure waves (P-waves) in earthquakes in granite is significantly higher than the speed of transverse or shear waves (S-waves). Both components of earthquakes travel slower in less rigid material, such as sediments. P-waves have speeds of 4 to 7 km/s, and S-waves correspondingly range in speed from 2 to 5 km/s, both being faster in more rigid material. The P-wave gets progressively farther ahead of the S-wave as they travel through Earth’s crust. The time between the P- and S-waves is routinely used to determine the distance to their source, the epicenter of the earthquake.

The speed of sound is affected by temperature in a given medium. For air at sea level, the speed of sound is given by

$vw=331 m/sT273K,vw=331 m/sT273K, size 12{v size 8{w}= left ("331"" m/s" right ) sqrt { { {T} over {"273"" K"} } } } {}$
17.2

where the temperature (denoted as $T T$) is in units of kelvin. The speed of sound in gases is related to the average speed of particles in the gas, $vrmsvrms size 12{v size 8{ ital "rms"}} {}$, and that

$vrms=3kTm,vrms=3kTm, size 12{v size 8{ ital "rms"}= sqrt { { {"3KT"} over {m} } } } {}$
17.3

where $k k$ is the Boltzmann constant ($1.38 × 10 −23 J/K 1.38 × 10 −23 J/K$) and $m m$ is the mass of each (identical) particle in the gas. So, it is reasonable that the speed of sound in air and other gases should depend on the square root of temperature. While not negligible, this is not a strong dependence. At $0ºC 0ºC$, the speed of sound is 331 m/s, whereas at $20.0ºC 20.0ºC$ it is 343 m/s, less than a 4% increase. Figure 17.11 shows a use of the speed of sound by a bat to sense distances. Echoes are also used in medical imaging.

Figure 17.11 A bat uses sound echoes to find its way about and to catch prey. The time for the echo to return is directly proportional to the distance.

One of the more important properties of sound is that its speed is nearly independent of frequency. This independence is certainly true in open air for sounds in the audible range of 20 to 20,000 Hz. If this independence were not true, you would certainly notice it for music played by a marching band in a football stadium, for example. Suppose that high-frequency sounds traveled faster—then the farther you were from the band, the more the sound from the low-pitch instruments would lag that from the high-pitch ones. But the music from all instruments arrives in cadence independent of distance, and so all frequencies must travel at nearly the same speed. Recall that

$vw=fλ.vw=fλ. size 12{v size 8{w}=fλ} {}$
17.4

In a given medium under fixed conditions, $vwvw$ is constant, so that there is a relationship between $ff size 12{f} {}$ and $λλ size 12{λ} {}$; the higher the frequency, the smaller the wavelength. See Figure 17.12 and consider the following example.

Figure 17.12 Because they travel at the same speed in a given medium, low-frequency sounds must have a greater wavelength than high-frequency sounds. Here, the lower-frequency sounds are emitted by the large speaker, called a woofer, while the higher-frequency sounds are emitted by the small speaker, called a tweeter.

### Example 17.1

#### Calculating Wavelengths: What Are the Wavelengths of Audible Sounds?

Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in $30.0ºC 30.0ºC$ air. (Assume that the frequency values are accurate to two significant figures.)

#### Strategy

To find wavelength from frequency, we can use $vw=fλvw=fλ size 12{v size 8{w}=fλ} {}$.

#### Solution

1. Identify knowns. The value for $vwvw size 12{v size 8{w}} {}$, is given by
$vw=331m/sT273K.vw=331m/sT273K. size 12{v size 8{w}= left ("331"" m/s" right ) sqrt { { {T} over {"273"" K"} } } } {}$
17.5
2. Convert the temperature into kelvin and then enter the temperature into the equation
$vw=331m/s303 K273K=348.7m/s.vw=331m/s303 K273K=348.7m/s.$
17.6
3. Solve the relationship between speed and wavelength for $λλ size 12{λ} {}$:
$λ=vwf.λ=vwf. size 12{λ= { {v size 8{w}} over {f} } } {}$
17.7
4. Enter the speed and the minimum frequency to give the maximum wavelength:
$λmax=348.7m/s20 Hz=17m.λmax=348.7m/s20 Hz=17m. size 12{λ size 8{"max"}= { {"348" "." 7" m/s"} over {"20 Hz"} } ="17"" m"} {}$
17.8
5. Enter the speed and the maximum frequency to give the minimum wavelength:
$λmin=348.7m/s20,000 Hz=0.017m=1.7 cm.λmin=348.7m/s20,000 Hz=0.017m=1.7 cm. size 12{λ size 8{"min"}= { {"348" "." 7" m/s"} over {"20" "." "000 Hz"} } =0 "." "0174"" m"=1 "." "74 cm"} {}$
17.9

#### Discussion

Because the product of $ff size 12{f} {}$ multiplied by $λλ size 12{λ} {}$ equals a constant, the smaller $ff size 12{f} {}$ is, the larger $λλ size 12{λ} {}$ must be, and vice versa.

The speed of sound can change when sound travels from one medium to another. However, the frequency usually remains the same because it is like a driven oscillation and has the frequency of the original source. If $vwvw size 12{v size 8{w}} {}$ changes and $ff size 12{f} {}$ remains the same, then the wavelength $λλ size 12{λ} {}$ must change. That is, because $vw=fλvw=fλ$, the higher the speed of a sound, the greater its wavelength for a given frequency.

### Making Connections: Take-Home Investigation—Voice as a Sound Wave

Suspend a sheet of paper so that the top edge of the paper is fixed and the bottom edge is free to move. You could tape the top edge of the paper to the edge of a table. Gently blow near the edge of the bottom of the sheet and note how the sheet moves. Speak softly and then louder such that the sounds hit the edge of the bottom of the paper, and note how the sheet moves. Explain the effects.

Imagine you observe two fireworks explode. You hear the explosion of one as soon as you see it. However, you see the other firework for several milliseconds before you hear the explosion. Explain why this is so.

You observe two musical instruments that you cannot identify. One plays high-pitch sounds and the other plays low-pitch sounds. How could you determine which is which without hearing either of them play?