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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

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

where the temperature (denoted as T ) is in units of kelvin. While not negligible, this is not a strong dependence. At 0ºC , the speed of sound is 331 m/s, whereas at 20.0ºC it is 343 m/s, less than a 4% increase. [link] shows a use of the speed of sound by a bat to sense distances. Echoes are also used in medical imaging.

The picture is of a bat trying to catch its prey an insect using sound echoes. The incident sound and sound reflected from the bat are shown as semicircular arcs.
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

v w = fλ. size 12{v size 8{w}=fλ} {}

In a given medium under fixed conditions, v w is constant, so that there is a relationship between f size 12{f} {} and λ size 12{λ} {} ; the higher the frequency, the smaller the wavelength. See [link] and consider the following example.

Picture of a speaker having a woofer and a tweeter. High frequency sound coming out of the woofer shown as small circles closely spaced. Low frequency sound coming out of tweeter are shown as larger circles distantly spaced.
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.

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 air. (Assume that the frequency values are accurate to two significant figures.)

Strategy

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

Solution

  1. Identify knowns. The value for v w size 12{v size 8{w}} {} , is given by
    v w = 331 m/s T 273 K . size 12{v size 8{w}= left ("331"" m/s" right ) sqrt { { {T} over {"273"" K"} } } } {}
  2. Convert the temperature into kelvin and then enter the temperature into the equation
    v w = 331 m/s 303 K 273 K = 348 . 7 m/s .
  3. Solve the relationship between speed and wavelength for λ size 12{λ} {} :
    λ = v w f . size 12{λ= { {v size 8{w}} over {f} } } {}
  4. Enter the speed and the minimum frequency to give the maximum wavelength:
    λ max = 348 . 7 m/s 20 Hz = 17 m . size 12{λ size 8{"max"}= { {"348" "." 7" m/s"} over {"20 Hz"} } ="17"" m"} {}
  5. Enter the speed and the maximum frequency to give the minimum wavelength:
    λ min = 348 . 7 m/s 20 , 000 Hz = 0 . 017 m = 1 . 7 cm . size 12{λ size 8{"min"}= { {"348" "." 7" m/s"} over {"20" "." "000 Hz"} } =0 "." "0174"" m"=1 "." "74 cm"} {}

Discussion

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

Practice Key Terms 1

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Source:  OpenStax, Concepts of physics with linear momentum. OpenStax CNX. Aug 11, 2016 Download for free at http://legacy.cnx.org/content/col11960/1.9
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