17.3: Speed of Sound, Frequency, and Wavelength
 Page ID
 1615
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 highpitch sounds, while large instruments, such as a tuba, typically make lowpitch 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 shortwavelength sounds. Similar arguments hold that a large instrument creates longwavelength sounds.
The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves:
\[v_w = f\lambda,\]
where \(v_w\) is the speed of sound, \(f\) is its frequency, and \(\lambda\) 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 \(\PageIndex{2}\). The frequency is the same as that of the source and is the number of waves that pass a point per unit time.
Figure \(\PageIndex{2}\): A sound wave emanates from a source vibrating at a frequency \(f\), propagates at \(v_w\), and has a wavelength \(\lambda\).
Table \(\PageIndex{1}\) 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.
Medium  \(v_w(m/s)\) 

Gases at \(0^oC\)  
Air  331 
Carbon dioxide  259 
Oxygen  316 
Helium  965 
Hydrogen  1290 
Liquids at \(20^oC\)  
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 
Lead  1960 
Aluminum  5120 
Steel  5960 
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 (Pwaves) in earthquakes in granite is significantly higher than the speed of transverse or shear waves (Swaves). Both components of earthquakes travel slower in less rigid material, such as sediments. Pwaves have speeds of 4 to 7 km/s, and Swaves correspondingly range in speed from 2 to 5 km/s, both being faster in more rigid material. The Pwave gets progressively farther ahead of the Swave as they travel through Earth’s crust. The time between the P and Swaves 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
\[v_w = (331 \, m/s)\sqrt{\dfrac{T}{273 \, K}},\]
where the temperature (denoted as \(T\)) is in units of kelvin. The speed of sound in gases is related to the average speed of particles in the gas, \(v_{rms}\), and that
\[v_{rms} = \sqrt{\dfrac{3 \, kT}{m}},\]
where \(k\) is the Boltzmann constant \((1.38 \times 10^{23} \, J/K)\) and \(m\) 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 \(0^oC\), the speed of sound is 331 m/s, whereas at \(20^oC\) it is 343 m/s, less than a 4% increase. Figure \(\PageIndex{3}\) shows a use of the speed of sound by a bat to sense distances. Echoes are also used in medical imaging.
Figure \(\PageIndex{3}\): 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 highfrequency sounds traveled faster—then the farther you were from the band, the more the sound from the lowpitch instruments would lag that from the highpitch 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
\[v_w = f\lambda.\]
In a given medium under fixed conditions, \(v_w\) is constant, so that there is a relationship between \(f\) and \(\lambda\); the higher the frequency, the smaller the wavelength. See Figure \(\PageIndex{4}\) and consider the following example.
Figure \(\PageIndex{4}\): Because they travel at the same speed in a given medium, lowfrequency sounds must have a greater wavelength than highfrequency sounds. Here, the lowerfrequency sounds are emitted by the large speaker, called a woofer, while the higherfrequency sounds are emitted by the small speaker, called a tweeter.
Example \(\PageIndex{1}\): 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 \(30.0^oC\) air. (Assume that the frequency values are accurate to two significant figures.)
Strategy
To find wavelength from frequency, we can use \(v_w = f\lambda\).
Solution
 Identify knowns. The value for \(v_w\), is given by \[v_w = (331 \, m/s)\sqrt{\dfrac{T}{273 \, K}}. \nonumber\]
 Convert the temperature into kelvin and then enter the temperature into the equation \[v_w = (331 \, m/s)\sqrt{\dfrac{303 \, K}{273 \, K}} = 348.7 \, m/s. \nonumber\]
 Solve the relationship between speed and wavelength for \(\lambda\): \[\lambda = \dfrac{v_w}{f}. \nonumber \]
 Enter the speed and the minimum frequency to give the maximum wavelength: \[\lambda_{max} = \dfrac{348.7 \, m/s}{20 \, Hz} = 17 \, m. \nonumber\]
 Enter the speed and the maximum frequency to give the minimum wavelength: \[\lambda_{min} = \dfrac{348.7 \, m/s}{20,000 \, Hz} = 0.017 \, m = 1.7 \, cm. \nonumber\]
Discussion
Because the product of \(f\) multiplied by \(\lambda\) equals a constant, the smaller \(f\) is, the larger \(\lambda\) 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 \(v_w\) changes and \(f\) remains the same, then the wavelength \(\lambda\) must change. That is, because \(v_w = f\lambda\), the higher the speed of a sound, the greater its wavelength for a given frequency.
MAKING CONNECTIONS: TAKEHOME 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.
Exercise \(\PageIndex{1A}\)
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.
 Answer

Sound and light both travel at definite speeds. The speed of sound is slower than the speed of light. The first firework is probably very close by, so the speed difference is not noticeable. The second firework is farther away, so the light arrives at your eyes noticeably sooner than the sound wave arrives at your ears.
Exercise \(\PageIndex{1B}\)
You observe two musical instruments that you cannot identify. One plays highpitch sounds and the other plays lowpitch sounds. How could you determine which is which without hearing either of them play?
 Answer

Compare their sizes. Highpitch instruments are generally smaller than lowpitch instruments because they generate a smaller wavelength.
Summary
 The relationship of the speed of sound \(v_w\), its frequency \(f\), and its wavelength \(\lambda\) is given by \(v_w = f\lambda,\) which is the same relationship given for all waves.
 In air, the speed of sound is related to air temperature \(T\) by \(v_w = (331 \, m/s) \sqrt{\dfrac{T}{273 \, K}}.\) \(v_w\) is the same for all frequencies and wavelengths.
Glossary
 pitch
 the perception of the frequency of a sound
Contributors
Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).