{"id":616,"date":"2017-09-18T18:05:30","date_gmt":"2017-09-18T22:05:30","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/chapter\/12-2-speed-of-sound-frequency-and-wavelength\/"},"modified":"2020-11-10T23:50:29","modified_gmt":"2020-11-11T04:50:29","slug":"12-2-speed-of-sound-frequency-and-wavelength","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/chapter\/12-2-speed-of-sound-frequency-and-wavelength\/","title":{"raw":"Speed of Sound, Frequency, and Wavelength","rendered":"Speed of Sound, Frequency, and Wavelength"},"content":{"raw":"<div>\r\n<div class=\"bcc-box bcc-highlight\">\r\n<h3>Summary<\/h3>\r\n<div>\r\n<ul>\r\n \t<li>Define pitch.<\/li>\r\n \t<li>Describe the relationship between the speed of sound, its frequency, and its wavelength.<\/li>\r\n \t<li>Describe the effects on the speed of sound as it travels through various media.<\/li>\r\n \t<li>Describe the effects of temperature on the speed of sound.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<figure>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"338\"]<img class=\"\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1107\/wp-content\/uploads\/sites\/1184\/2017\/09\/Figure_18_02_01a-1.jpg\" alt=\"A photograph of a fireworks display in the sky.\" width=\"338\" height=\"263\" \/> <strong>Figure 1.<\/strong> When a firework explodes, the light energy is perceived before the sound energy. Sound travels more slowly than light does. (credit: Dominic Alves, Flickr)[\/caption]<\/figure>\r\n<p id=\"import-auto-id3013063\">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 <strong><span id=\"import-auto-id1588064\">pitch<\/span><\/strong>. 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.<\/p>\r\n<p id=\"import-auto-id3154439\">The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves:<\/p>\r\n\r\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}} = f\\lambda},[\/latex]<\/div>\r\n<p id=\"import-auto-id2401892\">where[latex]\\boldsymbol{v_{\\textbf{w}}}[\/latex] is the speed of sound, [latex]\\boldsymbol{f}[\/latex] is its frequency, and [latex]\\boldsymbol{\\lambda}[\/latex] is its wavelength. The wavelength of a sound is the distance between adjacent identical parts of a wave\u2014for example, between adjacent compressions as illustrated in <a class=\"autogenerated-content\" href=\"#import-auto-id1538012\">Figure 2<\/a>. The frequency is the same as that of the source and is the number of waves that pass a point per unit time.<\/p>\r\n\r\n<figure id=\"import-auto-id1538012\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"350\"]<img src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1107\/wp-content\/uploads\/sites\/1184\/2017\/09\/Figure_18_02_02a-1.jpg\" alt=\"A picture of a vibrating tuning fork is shown. The sound wave compressions and rarefactions are shown to emanate from the fork on both the sides as semicircular arcs of alternate bold and dotted lines. The wavelength is marked as the distance between two successive bold arcs. The frequency of the vibrations is shown as f and velocity of the wave represented by v sub w.\" width=\"350\" height=\"546\" \/> <strong>Figure 2.<\/strong> A sound wave emanates from a source vibrating at a frequency<em><strong> f<\/strong><\/em>, propagates at<strong><em> v<\/em><sub>w<\/sub><\/strong>, and has a wavelength <em><strong>\u03bb<\/strong><\/em>.[\/caption]<\/figure>\r\n<a class=\"autogenerated-content\" href=\"#import-auto-id3177545\">Table 1<\/a> 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\u2019s rigidity (or compressibility in gases) and its density. The more rigid (or less compressible) the medium, the faster the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is directly proportional to the stiffness of the oscillating object. The greater the density of a medium, the slower the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is inversely proportional to the mass of the oscillating object. 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.\r\n<table id=\"import-auto-id3177545\" summary=\"Two-column table listing various media for sound in the first column and their speeds of sound in the second column. The list of media is divided into three groups: gases, liquids, and solids.\">\r\n<thead>\r\n<tr>\r\n<th>Medium<\/th>\r\n<th><em>v<\/em><sub>w <\/sub>(m\/s)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td colspan=\"2\"><strong><em>Gases at 0\u00baC<\/em><\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Air<\/td>\r\n<td>331<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Carbon dioxide<\/td>\r\n<td>259<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Oxygen<\/td>\r\n<td>316<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Helium<\/td>\r\n<td>965<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Hydrogen<\/td>\r\n<td>1290<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\"><strong><em>Liquids at 20\u00baC<\/em><\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Ethanol<\/td>\r\n<td>1160<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Mercury<\/td>\r\n<td>1450<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Water, fresh<\/td>\r\n<td>1480<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Sea water<\/td>\r\n<td>1540<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Human tissue<\/td>\r\n<td>1540<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\"><strong><em>Solids (longitudinal or bulk)<\/em><\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Vulcanized rubber<\/td>\r\n<td>54<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Polyethylene<\/td>\r\n<td>920<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Marble<\/td>\r\n<td>3810<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Glass, Pyrex<\/td>\r\n<td>5640<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Lead<\/td>\r\n<td>1960<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Aluminum<\/td>\r\n<td>5120<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Steel<\/td>\r\n<td>5960<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\"><strong>Table 1.<\/strong>Speed of Sound in Various Media.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"import-auto-id1441562\">Earthquakes, essentially sound waves in Earth\u2019s 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\u2019s crust. The time between the P- and S-waves is routinely used to determine the distance to their source, the epicenter of the earthquake.<\/p>\r\n<p id=\"import-auto-id3246097\">The speed of sound is affected by temperature in a given medium. For air at sea level, the speed of sound is given by<\/p>\r\n\r\n<div id=\"eip-330\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}} = (331\\textbf{ m\/s})}[\/latex][latex size=\"2\"]\\boldsymbol{\\sqrt{\\frac{T}{273\\textbf{ K}}}},[\/latex]<\/div>\r\n<p id=\"import-auto-id1431577\">where the temperature (denoted as[latex]\\boldsymbol{T})[\/latex] is in units of kelvin. The speed of sound in gases is related to the average speed of particles in the gas, [latex]\\boldsymbol{v_{\\textbf{rms}}},[\/latex] and that<\/p>\r\n\r\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{rms}}\\:=}[\/latex][latex size=\"2\"]\\boldsymbol{\\sqrt{\\frac{3kT}{m}}},[\/latex]<\/div>\r\n<p id=\"import-auto-id2447377\">where [latex]\\boldsymbol{k}[\/latex] is the Boltzmann constant [latex](\\boldsymbol{1.38\\times10^{-23}\\textbf{ J\/K}})[\/latex] and [latex]\\boldsymbol{m}[\/latex] 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[latex]\\boldsymbol{0^{\\circ}\\textbf{C}},[\/latex] the speed of sound is 331 m\/s, whereas a t[latex]\\boldsymbol{20.0^{\\circ}\\textbf{C}}[\/latex] it is 343 m\/s, less than a 4% increase. <a href=\"#import-auto-id1578485\">Figure 3<\/a> shows a use of the speed of sound by a bat to sense distances. Echoes are also used in medical imaging.<\/p>\r\n\r\n<figure id=\"import-auto-id1578485\"><\/figure>\r\n[caption id=\"\" align=\"aligncenter\" width=\"250\"]<img src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1107\/wp-content\/uploads\/sites\/1184\/2017\/09\/Figure_18_02_03a-1.jpg\" alt=\"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.\" width=\"250\" height=\"593\" \/> <strong>Figure 3.<\/strong> 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.[\/caption]\r\n<p id=\"import-auto-id2444602\">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\u2014then 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<\/p>\r\n\r\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}} = f\\lambda}.[\/latex]<\/div>\r\nIn a given medium under fixed conditions, [latex]\\boldsymbol{v_{\\textbf{w}}}[\/latex] is constant, so that there is a relationship between [latex]\\boldsymbol{f}[\/latex] and [latex]\\boldsymbol{\\lambda};[\/latex] the higher the frequency, the smaller the wavelength. See <a class=\"autogenerated-content\" href=\"#import-auto-id1593942\">Figure 4<\/a> and consider the following example.\r\n<figure id=\"import-auto-id1593942\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"200\"]<img src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1107\/wp-content\/uploads\/sites\/1184\/2017\/09\/Figure_18_02_04a-1.jpg\" alt=\"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.\" width=\"200\" height=\"519\" \/> Figure 4. 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.[\/caption]<\/figure>\r\n<div class=\"textbox shaded\">\r\n<div id=\"fs-id2001560\" class=\"example\">\r\n<h3 class=\"title\">Example 1: Calculating Wavelengths: What Are the Wavelengths of Audible Sounds?<\/h3>\r\n<p id=\"import-auto-id1477887\">Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in [latex]\\boldsymbol{30.0^{\\circ}\\textbf{C}}[\/latex] air. (Assume that the frequency values are accurate to two significant figures.)<\/p>\r\n<strong>Strategy<\/strong>\r\n<p id=\"import-auto-id3396364\">To find wavelength from frequency, we can use [latex]\\boldsymbol{v_{\\textbf{w}}=f\\lambda}.[\/latex]<\/p>\r\n<p id=\"import-auto-id2381606\"><strong>Solution<\/strong><\/p>\r\n\r\n<ol id=\"fs-id2409442\">\r\n \t<li>Identify knowns. The value for[latex]\\boldsymbol{v_{\\textbf{w}}},[\/latex] is given by\r\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}}=(331\\textbf{ m\/s})}[\/latex][latex size=\"2\"]\\boldsymbol{\\sqrt{\\frac{T}{273\\textbf{ K}}}}.[\/latex]<\/div><\/li>\r\n \t<li id=\"import-auto-id1815176\">Convert the temperature into kelvin and then enter the temperature into the equation\r\n<div id=\"eip-575\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}}=(331\\textbf{ m\/s})}[\/latex][latex size=\"2\"]\\boldsymbol{\\sqrt{\\frac{303\\textbf{ K}}{273\\textbf{ K}}}}[\/latex][latex]\\boldsymbol{=348.7\\textbf{ m\/s}}.[\/latex]<\/div><\/li>\r\n \t<li id=\"import-auto-id2682073\">Solve the relationship between speed and wavelength for [latex]\\boldsymbol{\\lambda}:[\/latex]\r\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{\\lambda\\:=}[\/latex][latex size=\"2\"]\\boldsymbol{\\frac{v_{\\textbf{w}}}{f}}.[\/latex]<\/div><\/li>\r\n \t<li id=\"import-auto-id963388\">Enter the speed and the minimum frequency to give the maximum wavelength:\r\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{\\lambda_{\\textbf{max}}\\:=}[\/latex][latex size=\"2\"]\\boldsymbol{\\frac{348.7\\textbf{ m\/s}}{20\\textbf{ Hz}}}[\/latex][latex]\\boldsymbol{=17\\textbf{ m}}.[\/latex]<\/div><\/li>\r\n \t<li id=\"import-auto-id1816494\">Enter the speed and the maximum frequency to give the minimum wavelength:\r\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{\\lambda_{\\textbf{min}}=}[\/latex][latex size=\"2\"]\\boldsymbol{\\frac{348.7\\textbf{ m\/s}}{20,000\\textbf{ Hz}}}[\/latex][latex]\\boldsymbol{=0.017\\textbf{ m}=1.7\\textbf{ cm}}.[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<p id=\"import-auto-id2991817\"><strong>Discussion<\/strong><\/p>\r\n<p id=\"fs-id3250053\">Because the product of [latex]\\boldsymbol{f}[\/latex] multiplied by [latex]\\boldsymbol{\\lambda}[\/latex] equals a constant, the smaller [latex]\\boldsymbol{f}[\/latex] is, the larger [latex]\\boldsymbol{\\lambda}[\/latex] must be, and vice versa.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nThe 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. I f[latex]\\boldsymbol{v_{\\textbf{w}}}[\/latex] changes and [latex]\\boldsymbol{f}[\/latex]remains the same, then the wavelength [latex]\\boldsymbol{\\lambda}[\/latex] must change. That is, because [latex]\\boldsymbol{v_{\\textbf{w}}=f\\lambda}[\/latex]the higher the speed of a sound, the greater its wavelength for a given frequency.\r\n<div id=\"fs-id2383870\" class=\"note\">\r\n<div class=\"textbox shaded\">\r\n<div class=\"note\">\r\n<h3 class=\"title\">Making Connections: Take-Home Investigation\u2014Voice as a Sound Wave<\/h3>\r\n<p id=\"import-auto-id1427729\">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.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1437960\" class=\"exercise\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Check Your Understanding 1<\/h3>\r\n<div class=\"exercise\">\r\n<div id=\"fs-id2722326\" class=\"problem\">\r\n<p id=\"fs-id1622890\">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.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Check Your Understanding 2<\/h3>\r\n<div id=\"fs-id1526208\" class=\"exercise\">\r\n<div id=\"fs-id2737920\" class=\"problem\">\r\n<p id=\"fs-id2205803\">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?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1931189\" class=\"section-summary\">\r\n<h1>Section Summary<\/h1>\r\n<p id=\"import-auto-id2600539\">The relationship of the speed of sound [latex]\\boldsymbol{v_{\\textbf{w}}},[\/latex] its frequency [latex]\\boldsymbol{f},[\/latex] and its wavelength [latex]\\boldsymbol{\\lambda}[\/latex] is given by<\/p>\r\n\r\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}} = f\\lambda},[\/latex]<\/div>\r\n<p id=\"import-auto-id3021013\">which is the same relationship given for all waves.<\/p>\r\nIn air, the speed of sound is related to air temperature [latex]\\boldsymbol{T}[\/latex] by\r\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}}=(331\\textbf{ m\/s})}[\/latex][latex size=\"2\"]\\boldsymbol{\\sqrt{\\frac{T}{273\\textbf{ K}}}}.[\/latex]<\/div>\r\n<p id=\"import-auto-id2962616\">[latex]\\boldsymbol{v_{\\textbf{w}}}[\/latex] is the same for all frequencies and wavelengths.<\/p>\r\n\r\n<\/section><section id=\"fs-id1386961\" class=\"conceptual-questions\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Conceptual Questions<\/h3>\r\n<div id=\"fs-id1375143\" class=\"exercise\">\r\n<div id=\"fs-id1954277\" class=\"problem\">\r\n<p id=\"import-auto-id1410916\"><strong>1: <\/strong>How do sound vibrations of atoms differ from thermal motion?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id3008692\" class=\"exercise\">\r\n<div id=\"fs-id2992665\" class=\"problem\">\r\n<p id=\"import-auto-id3065090\"><strong>2: <\/strong>When sound passes from one medium to another where its propagation speed is different, does its frequency or wavelength change? Explain your answer briefly.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id2591417\" class=\"problems-exercises\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Exercises<\/h3>\r\n<div id=\"fs-id3451972\" class=\"exercise\">\r\n<div id=\"fs-id2931366\" class=\"problem\">\r\n<p id=\"import-auto-id2399660\"><strong>1: <\/strong>When poked by a spear, an operatic soprano lets out a 1200-Hz shriek. What is its wavelength if the speed of sound is 345 m\/s?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"exercise\">\r\n<div class=\"problem\">\r\n<p id=\"import-auto-id3110312\"><strong>2: <\/strong>What frequency sound has a 0.10-m wavelength when the speed of sound is 340 m\/s?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id3043771\" class=\"exercise\">\r\n<div id=\"fs-id1945477\" class=\"problem\">\r\n<p id=\"import-auto-id1870708\"><strong>3: <\/strong>Calculate the speed of sound on a day when a 1500 Hz frequency has a wavelength of 0.221 m.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2443672\" class=\"exercise\">\r\n<div class=\"problem\">\r\n<p id=\"import-auto-id1549327\"><strong>4: <\/strong>(a) What is the speed of sound in a medium where a 100-kHz frequency produces a 5.96-cm wavelength? (b) Which substance in <a class=\"autogenerated-content\" href=\"#import-auto-id3177545\">Table 1<\/a> is this likely to be?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1587092\" class=\"exercise\">\r\n<div id=\"fs-id2681777\" class=\"problem\">\r\n<p id=\"import-auto-id2051396\"><strong>5: <\/strong>Show that the speed of sound in 20.0 <sup>o<\/sup>C air is 343 m\/s, as claimed in the text.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2666191\" class=\"exercise\">\r\n<div id=\"fs-id3116303\" class=\"problem\">\r\n<p id=\"import-auto-id3062563\"><strong>6: <\/strong>Air temperature in the Sahara Desert can reach [latex]\\boldsymbol{56.0^{\\circ}\\textbf{C}}[\/latex](about [latex]\\boldsymbol{134^{\\circ}\\textbf{F}}[\/latex]). What is the speed of sound in air at that temperature?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2032287\" class=\"exercise\">\r\n<div id=\"fs-id3055563\" class=\"problem\">\r\n\r\n<strong>7: <\/strong>Dolphins make sounds in air and water. What is the ratio of the wavelength of a sound in air to its wavelength in seawater? Assume air temperature is 20.0 <sup>0<\/sup>C.\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1560729\" class=\"exercise\">\r\n<div class=\"problem\">\r\n<p id=\"import-auto-id2597921\"><strong>8: <\/strong>A sonar echo returns to a submarine 1.20 s after being emitted. What is the distance to the object creating the echo? (Assume that the submarine is in the ocean, not in fresh water.)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1986379\" class=\"exercise\">\r\n<div id=\"fs-id3234387\" class=\"problem\">\r\n<p id=\"import-auto-id2667613\"><strong>9: <\/strong>(a) If a submarine\u2019s sonar can measure echo times with a precision of 0.0100 s, what is the smallest difference in distances it can detect? (Assume that the submarine is in the ocean, not in fresh water.)<\/p>\r\n<p id=\"eip-id1517636\">(b) Discuss the limits this time resolution imposes on the ability of the sonar system to detect the size and shape of the object creating the echo.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1381531\" class=\"exercise\">\r\n<div id=\"fs-id1849553\" class=\"problem\">\r\n<p id=\"import-auto-id2032403\"><strong>10: <\/strong>A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.400 s. (a) How far away is the explosion if air temperature is 24.0 <sup>o<\/sup>C and if you neglect the time taken for light to reach the physicist? (b) Calculate the distance to the explosion taking the speed of light into account. Note that this distance is negligibly greater.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1386574\" class=\"exercise\">\r\n<div id=\"fs-id2403356\" class=\"problem\">\r\n<p id=\"import-auto-id3455423\"><strong>11: <\/strong>Suppose a bat uses sound echoes to locate its insect prey, 3.00 m away. (See <a class=\"autogenerated-content\" href=\"#import-auto-id1578485\">Figure 3<\/a>.) (a) Calculate the echo times for temperatures of 5.00 <sup>o<\/sup>C and 35.0 <sup>o<\/sup>C. b) What percent uncertainty does this cause for the bat in locating the insect? (c) Discuss the significance of this uncertainty and whether it could cause difficulties for the bat. (In practice, the bat continues to use sound as it closes in, eliminating most of any difficulties imposed by this and other effects, such as motion of the prey.)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"import-auto-id2442201\" class=\"definition\">\r\n \t<dt>pitch<\/dt>\r\n \t<dd id=\"fs-id1414345\">the perception of the frequency of a sound<\/dd>\r\n<\/dl>\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Solutions<\/h3>\r\n<strong>Check Your Understanding 1\r\n<\/strong>\r\n\r\nSound 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.\r\n\r\n<strong>Check Your Understanding 2<\/strong>\r\n\r\nCompare their sizes. High-pitch instruments are generally smaller than low-pitch instruments because they generate a smaller wavelength.\r\n\r\n<strong>Problems &amp; Exercises<\/strong>\r\n\r\n<strong>1: <\/strong>0.288 m\r\n\r\n<strong>3: <\/strong>332 m\/s\r\n\r\n<strong>5: <\/strong>[latex]\\begin{array}{lcl} \\boldsymbol{v_{\\textbf{w}}} &amp; \\boldsymbol{=} &amp; \\boldsymbol{(331\\textbf{ m\/s})\\sqrt{\\frac{T}{273\\textbf{ K}}}=(331\\textbf{ m\/s})\\sqrt{\\frac{293\\textbf{ K}}{273\\textbf{ K}}}} \\\\ {} &amp; \\boldsymbol{=} &amp; \\boldsymbol{343\\textbf{ m\/s}} \\end{array}[\/latex]\r\n\r\n<strong>7: <\/strong>0.223\r\n\r\n<strong>9: <\/strong>(a) 7.70 m (b) This means that sonar is good for spotting and locating large objects, but it isn\u2019t able to resolve smaller objects, or detect the detailed shapes of objects. Objects like ships or large pieces of airplanes can be found by sonar, while smaller pieces must be found by other means.\r\n\r\n<strong>11: <\/strong>(a) 18.0 ms, 17.1 ms (b) 5.00% (c) This uncertainty could definitely cause difficulties for the bat, if it didn\u2019t continue to use sound as it closed in on its prey. A 5% uncertainty could be the difference between catching the prey around the neck or around the chest, which means that it could miss grabbing its prey.\r\n\r\n<\/div>","rendered":"<div>\n<div class=\"bcc-box bcc-highlight\">\n<h3>Summary<\/h3>\n<div>\n<ul>\n<li>Define pitch.<\/li>\n<li>Describe the relationship between the speed of sound, its frequency, and its wavelength.<\/li>\n<li>Describe the effects on the speed of sound as it travels through various media.<\/li>\n<li>Describe the effects of temperature on the speed of sound.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<figure>\n<figure style=\"width: 338px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1107\/wp-content\/uploads\/sites\/1184\/2017\/09\/Figure_18_02_01a-1.jpg\" alt=\"A photograph of a fireworks display in the sky.\" width=\"338\" height=\"263\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 1.<\/strong> When a firework explodes, the light energy is perceived before the sound energy. Sound travels more slowly than light does. (credit: Dominic Alves, Flickr)<\/figcaption><\/figure>\n<\/figure>\n<p id=\"import-auto-id3013063\">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 <strong><span id=\"import-auto-id1588064\">pitch<\/span><\/strong>. 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.<\/p>\n<p id=\"import-auto-id3154439\">The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves:<\/p>\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}} = f\\lambda},[\/latex]<\/div>\n<p id=\"import-auto-id2401892\">where[latex]\\boldsymbol{v_{\\textbf{w}}}[\/latex] is the speed of sound, [latex]\\boldsymbol{f}[\/latex] is its frequency, and [latex]\\boldsymbol{\\lambda}[\/latex] is its wavelength. The wavelength of a sound is the distance between adjacent identical parts of a wave\u2014for example, between adjacent compressions as illustrated in <a class=\"autogenerated-content\" href=\"#import-auto-id1538012\">Figure 2<\/a>. The frequency is the same as that of the source and is the number of waves that pass a point per unit time.<\/p>\n<figure id=\"import-auto-id1538012\"><figcaption><\/figcaption><figure style=\"width: 350px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1107\/wp-content\/uploads\/sites\/1184\/2017\/09\/Figure_18_02_02a-1.jpg\" alt=\"A picture of a vibrating tuning fork is shown. The sound wave compressions and rarefactions are shown to emanate from the fork on both the sides as semicircular arcs of alternate bold and dotted lines. The wavelength is marked as the distance between two successive bold arcs. The frequency of the vibrations is shown as f and velocity of the wave represented by v sub w.\" width=\"350\" height=\"546\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 2.<\/strong> A sound wave emanates from a source vibrating at a frequency<em><strong> f<\/strong><\/em>, propagates at<strong><em> v<\/em><sub>w<\/sub><\/strong>, and has a wavelength <em><strong>\u03bb<\/strong><\/em>.<\/figcaption><\/figure>\n<\/figure>\n<p><a class=\"autogenerated-content\" href=\"#import-auto-id3177545\">Table 1<\/a> 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\u2019s rigidity (or compressibility in gases) and its density. The more rigid (or less compressible) the medium, the faster the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is directly proportional to the stiffness of the oscillating object. The greater the density of a medium, the slower the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is inversely proportional to the mass of the oscillating object. 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.<\/p>\n<table id=\"import-auto-id3177545\" summary=\"Two-column table listing various media for sound in the first column and their speeds of sound in the second column. The list of media is divided into three groups: gases, liquids, and solids.\">\n<thead>\n<tr>\n<th>Medium<\/th>\n<th><em>v<\/em><sub>w <\/sub>(m\/s)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td colspan=\"2\"><strong><em>Gases at 0\u00baC<\/em><\/strong><\/td>\n<\/tr>\n<tr>\n<td>Air<\/td>\n<td>331<\/td>\n<\/tr>\n<tr>\n<td>Carbon dioxide<\/td>\n<td>259<\/td>\n<\/tr>\n<tr>\n<td>Oxygen<\/td>\n<td>316<\/td>\n<\/tr>\n<tr>\n<td>Helium<\/td>\n<td>965<\/td>\n<\/tr>\n<tr>\n<td>Hydrogen<\/td>\n<td>1290<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\"><strong><em>Liquids at 20\u00baC<\/em><\/strong><\/td>\n<\/tr>\n<tr>\n<td>Ethanol<\/td>\n<td>1160<\/td>\n<\/tr>\n<tr>\n<td>Mercury<\/td>\n<td>1450<\/td>\n<\/tr>\n<tr>\n<td>Water, fresh<\/td>\n<td>1480<\/td>\n<\/tr>\n<tr>\n<td>Sea water<\/td>\n<td>1540<\/td>\n<\/tr>\n<tr>\n<td>Human tissue<\/td>\n<td>1540<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\"><strong><em>Solids (longitudinal or bulk)<\/em><\/strong><\/td>\n<\/tr>\n<tr>\n<td>Vulcanized rubber<\/td>\n<td>54<\/td>\n<\/tr>\n<tr>\n<td>Polyethylene<\/td>\n<td>920<\/td>\n<\/tr>\n<tr>\n<td>Marble<\/td>\n<td>3810<\/td>\n<\/tr>\n<tr>\n<td>Glass, Pyrex<\/td>\n<td>5640<\/td>\n<\/tr>\n<tr>\n<td>Lead<\/td>\n<td>1960<\/td>\n<\/tr>\n<tr>\n<td>Aluminum<\/td>\n<td>5120<\/td>\n<\/tr>\n<tr>\n<td>Steel<\/td>\n<td>5960<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\"><strong>Table 1.<\/strong>Speed of Sound in Various Media.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"import-auto-id1441562\">Earthquakes, essentially sound waves in Earth\u2019s 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\u2019s crust. The time between the P- and S-waves is routinely used to determine the distance to their source, the epicenter of the earthquake.<\/p>\n<p id=\"import-auto-id3246097\">The speed of sound is affected by temperature in a given medium. For air at sea level, the speed of sound is given by<\/p>\n<div id=\"eip-330\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}} = (331\\textbf{ m\/s})}[\/latex][latex]\\boldsymbol{\\sqrt{\\frac{T}{273\\textbf{ K}}}},[\/latex]<\/div>\n<p id=\"import-auto-id1431577\">where the temperature (denoted as[latex]\\boldsymbol{T})[\/latex] is in units of kelvin. The speed of sound in gases is related to the average speed of particles in the gas, [latex]\\boldsymbol{v_{\\textbf{rms}}},[\/latex] and that<\/p>\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{rms}}\\:=}[\/latex][latex]\\boldsymbol{\\sqrt{\\frac{3kT}{m}}},[\/latex]<\/div>\n<p id=\"import-auto-id2447377\">where [latex]\\boldsymbol{k}[\/latex] is the Boltzmann constant [latex](\\boldsymbol{1.38\\times10^{-23}\\textbf{ J\/K}})[\/latex] and [latex]\\boldsymbol{m}[\/latex] 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[latex]\\boldsymbol{0^{\\circ}\\textbf{C}},[\/latex] the speed of sound is 331 m\/s, whereas a t[latex]\\boldsymbol{20.0^{\\circ}\\textbf{C}}[\/latex] it is 343 m\/s, less than a 4% increase. <a href=\"#import-auto-id1578485\">Figure 3<\/a> shows a use of the speed of sound by a bat to sense distances. Echoes are also used in medical imaging.<\/p>\n<figure id=\"import-auto-id1578485\"><\/figure>\n<figure style=\"width: 250px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1107\/wp-content\/uploads\/sites\/1184\/2017\/09\/Figure_18_02_03a-1.jpg\" alt=\"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.\" width=\"250\" height=\"593\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 3.<\/strong> 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.<\/figcaption><\/figure>\n<p id=\"import-auto-id2444602\">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\u2014then 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<\/p>\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}} = f\\lambda}.[\/latex]<\/div>\n<p>In a given medium under fixed conditions, [latex]\\boldsymbol{v_{\\textbf{w}}}[\/latex] is constant, so that there is a relationship between [latex]\\boldsymbol{f}[\/latex] and [latex]\\boldsymbol{\\lambda};[\/latex] the higher the frequency, the smaller the wavelength. See <a class=\"autogenerated-content\" href=\"#import-auto-id1593942\">Figure 4<\/a> and consider the following example.<\/p>\n<figure id=\"import-auto-id1593942\">\n<figure style=\"width: 200px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1107\/wp-content\/uploads\/sites\/1184\/2017\/09\/Figure_18_02_04a-1.jpg\" alt=\"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.\" width=\"200\" height=\"519\" \/><figcaption class=\"wp-caption-text\">Figure 4. 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.<\/figcaption><\/figure>\n<\/figure>\n<div class=\"textbox shaded\">\n<div id=\"fs-id2001560\" class=\"example\">\n<h3 class=\"title\">Example 1: Calculating Wavelengths: What Are the Wavelengths of Audible Sounds?<\/h3>\n<p id=\"import-auto-id1477887\">Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in [latex]\\boldsymbol{30.0^{\\circ}\\textbf{C}}[\/latex] air. (Assume that the frequency values are accurate to two significant figures.)<\/p>\n<p><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id3396364\">To find wavelength from frequency, we can use [latex]\\boldsymbol{v_{\\textbf{w}}=f\\lambda}.[\/latex]<\/p>\n<p id=\"import-auto-id2381606\"><strong>Solution<\/strong><\/p>\n<ol id=\"fs-id2409442\">\n<li>Identify knowns. The value for[latex]\\boldsymbol{v_{\\textbf{w}}},[\/latex] is given by\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}}=(331\\textbf{ m\/s})}[\/latex][latex]\\boldsymbol{\\sqrt{\\frac{T}{273\\textbf{ K}}}}.[\/latex]<\/div>\n<\/li>\n<li id=\"import-auto-id1815176\">Convert the temperature into kelvin and then enter the temperature into the equation\n<div id=\"eip-575\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}}=(331\\textbf{ m\/s})}[\/latex][latex]\\boldsymbol{\\sqrt{\\frac{303\\textbf{ K}}{273\\textbf{ K}}}}[\/latex][latex]\\boldsymbol{=348.7\\textbf{ m\/s}}.[\/latex]<\/div>\n<\/li>\n<li id=\"import-auto-id2682073\">Solve the relationship between speed and wavelength for [latex]\\boldsymbol{\\lambda}:[\/latex]\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{\\lambda\\:=}[\/latex][latex]\\boldsymbol{\\frac{v_{\\textbf{w}}}{f}}.[\/latex]<\/div>\n<\/li>\n<li id=\"import-auto-id963388\">Enter the speed and the minimum frequency to give the maximum wavelength:\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{\\lambda_{\\textbf{max}}\\:=}[\/latex][latex]\\boldsymbol{\\frac{348.7\\textbf{ m\/s}}{20\\textbf{ Hz}}}[\/latex][latex]\\boldsymbol{=17\\textbf{ m}}.[\/latex]<\/div>\n<\/li>\n<li id=\"import-auto-id1816494\">Enter the speed and the maximum frequency to give the minimum wavelength:\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{\\lambda_{\\textbf{min}}=}[\/latex][latex]\\boldsymbol{\\frac{348.7\\textbf{ m\/s}}{20,000\\textbf{ Hz}}}[\/latex][latex]\\boldsymbol{=0.017\\textbf{ m}=1.7\\textbf{ cm}}.[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p id=\"import-auto-id2991817\"><strong>Discussion<\/strong><\/p>\n<p id=\"fs-id3250053\">Because the product of [latex]\\boldsymbol{f}[\/latex] multiplied by [latex]\\boldsymbol{\\lambda}[\/latex] equals a constant, the smaller [latex]\\boldsymbol{f}[\/latex] is, the larger [latex]\\boldsymbol{\\lambda}[\/latex] must be, and vice versa.<\/p>\n<\/div>\n<\/div>\n<p>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. I f[latex]\\boldsymbol{v_{\\textbf{w}}}[\/latex] changes and [latex]\\boldsymbol{f}[\/latex]remains the same, then the wavelength [latex]\\boldsymbol{\\lambda}[\/latex] must change. That is, because [latex]\\boldsymbol{v_{\\textbf{w}}=f\\lambda}[\/latex]the higher the speed of a sound, the greater its wavelength for a given frequency.<\/p>\n<div id=\"fs-id2383870\" class=\"note\">\n<div class=\"textbox shaded\">\n<div class=\"note\">\n<h3 class=\"title\">Making Connections: Take-Home Investigation\u2014Voice as a Sound Wave<\/h3>\n<p id=\"import-auto-id1427729\">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.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1437960\" class=\"exercise\">\n<div class=\"bcc-box bcc-info\">\n<h3>Check Your Understanding 1<\/h3>\n<div class=\"exercise\">\n<div id=\"fs-id2722326\" class=\"problem\">\n<p id=\"fs-id1622890\">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.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Check Your Understanding 2<\/h3>\n<div id=\"fs-id1526208\" class=\"exercise\">\n<div id=\"fs-id2737920\" class=\"problem\">\n<p id=\"fs-id2205803\">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?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1931189\" class=\"section-summary\">\n<h1>Section Summary<\/h1>\n<p id=\"import-auto-id2600539\">The relationship of the speed of sound [latex]\\boldsymbol{v_{\\textbf{w}}},[\/latex] its frequency [latex]\\boldsymbol{f},[\/latex] and its wavelength [latex]\\boldsymbol{\\lambda}[\/latex] is given by<\/p>\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}} = f\\lambda},[\/latex]<\/div>\n<p id=\"import-auto-id3021013\">which is the same relationship given for all waves.<\/p>\n<p>In air, the speed of sound is related to air temperature [latex]\\boldsymbol{T}[\/latex] by<\/p>\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{v_{\\textbf{w}}=(331\\textbf{ m\/s})}[\/latex][latex]\\boldsymbol{\\sqrt{\\frac{T}{273\\textbf{ K}}}}.[\/latex]<\/div>\n<p id=\"import-auto-id2962616\">[latex]\\boldsymbol{v_{\\textbf{w}}}[\/latex] is the same for all frequencies and wavelengths.<\/p>\n<\/section>\n<section id=\"fs-id1386961\" class=\"conceptual-questions\">\n<div class=\"bcc-box bcc-info\">\n<h3>Conceptual Questions<\/h3>\n<div id=\"fs-id1375143\" class=\"exercise\">\n<div id=\"fs-id1954277\" class=\"problem\">\n<p id=\"import-auto-id1410916\"><strong>1: <\/strong>How do sound vibrations of atoms differ from thermal motion?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id3008692\" class=\"exercise\">\n<div id=\"fs-id2992665\" class=\"problem\">\n<p id=\"import-auto-id3065090\"><strong>2: <\/strong>When sound passes from one medium to another where its propagation speed is different, does its frequency or wavelength change? Explain your answer briefly.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id2591417\" class=\"problems-exercises\">\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<div id=\"fs-id3451972\" class=\"exercise\">\n<div id=\"fs-id2931366\" class=\"problem\">\n<p id=\"import-auto-id2399660\"><strong>1: <\/strong>When poked by a spear, an operatic soprano lets out a 1200-Hz shriek. What is its wavelength if the speed of sound is 345 m\/s?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\">\n<div class=\"problem\">\n<p id=\"import-auto-id3110312\"><strong>2: <\/strong>What frequency sound has a 0.10-m wavelength when the speed of sound is 340 m\/s?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id3043771\" class=\"exercise\">\n<div id=\"fs-id1945477\" class=\"problem\">\n<p id=\"import-auto-id1870708\"><strong>3: <\/strong>Calculate the speed of sound on a day when a 1500 Hz frequency has a wavelength of 0.221 m.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2443672\" class=\"exercise\">\n<div class=\"problem\">\n<p id=\"import-auto-id1549327\"><strong>4: <\/strong>(a) What is the speed of sound in a medium where a 100-kHz frequency produces a 5.96-cm wavelength? (b) Which substance in <a class=\"autogenerated-content\" href=\"#import-auto-id3177545\">Table 1<\/a> is this likely to be?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1587092\" class=\"exercise\">\n<div id=\"fs-id2681777\" class=\"problem\">\n<p id=\"import-auto-id2051396\"><strong>5: <\/strong>Show that the speed of sound in 20.0 <sup>o<\/sup>C air is 343 m\/s, as claimed in the text.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2666191\" class=\"exercise\">\n<div id=\"fs-id3116303\" class=\"problem\">\n<p id=\"import-auto-id3062563\"><strong>6: <\/strong>Air temperature in the Sahara Desert can reach [latex]\\boldsymbol{56.0^{\\circ}\\textbf{C}}[\/latex](about [latex]\\boldsymbol{134^{\\circ}\\textbf{F}}[\/latex]). What is the speed of sound in air at that temperature?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2032287\" class=\"exercise\">\n<div id=\"fs-id3055563\" class=\"problem\">\n<p><strong>7: <\/strong>Dolphins make sounds in air and water. What is the ratio of the wavelength of a sound in air to its wavelength in seawater? Assume air temperature is 20.0 <sup>0<\/sup>C.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1560729\" class=\"exercise\">\n<div class=\"problem\">\n<p id=\"import-auto-id2597921\"><strong>8: <\/strong>A sonar echo returns to a submarine 1.20 s after being emitted. What is the distance to the object creating the echo? (Assume that the submarine is in the ocean, not in fresh water.)<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1986379\" class=\"exercise\">\n<div id=\"fs-id3234387\" class=\"problem\">\n<p id=\"import-auto-id2667613\"><strong>9: <\/strong>(a) If a submarine\u2019s sonar can measure echo times with a precision of 0.0100 s, what is the smallest difference in distances it can detect? (Assume that the submarine is in the ocean, not in fresh water.)<\/p>\n<p id=\"eip-id1517636\">(b) Discuss the limits this time resolution imposes on the ability of the sonar system to detect the size and shape of the object creating the echo.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1381531\" class=\"exercise\">\n<div id=\"fs-id1849553\" class=\"problem\">\n<p id=\"import-auto-id2032403\"><strong>10: <\/strong>A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.400 s. (a) How far away is the explosion if air temperature is 24.0 <sup>o<\/sup>C and if you neglect the time taken for light to reach the physicist? (b) Calculate the distance to the explosion taking the speed of light into account. Note that this distance is negligibly greater.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1386574\" class=\"exercise\">\n<div id=\"fs-id2403356\" class=\"problem\">\n<p id=\"import-auto-id3455423\"><strong>11: <\/strong>Suppose a bat uses sound echoes to locate its insect prey, 3.00 m away. (See <a class=\"autogenerated-content\" href=\"#import-auto-id1578485\">Figure 3<\/a>.) (a) Calculate the echo times for temperatures of 5.00 <sup>o<\/sup>C and 35.0 <sup>o<\/sup>C. b) What percent uncertainty does this cause for the bat in locating the insect? (c) Discuss the significance of this uncertainty and whether it could cause difficulties for the bat. (In practice, the bat continues to use sound as it closes in, eliminating most of any difficulties imposed by this and other effects, such as motion of the prey.)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"import-auto-id2442201\" class=\"definition\">\n<dt>pitch<\/dt>\n<dd id=\"fs-id1414345\">the perception of the frequency of a sound<\/dd>\n<\/dl>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Solutions<\/h3>\n<p><strong>Check Your Understanding 1<br \/>\n<\/strong><\/p>\n<p>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.<\/p>\n<p><strong>Check Your Understanding 2<\/strong><\/p>\n<p>Compare their sizes. High-pitch instruments are generally smaller than low-pitch instruments because they generate a smaller wavelength.<\/p>\n<p><strong>Problems &amp; Exercises<\/strong><\/p>\n<p><strong>1: <\/strong>0.288 m<\/p>\n<p><strong>3: <\/strong>332 m\/s<\/p>\n<p><strong>5: <\/strong>[latex]\\begin{array}{lcl} \\boldsymbol{v_{\\textbf{w}}} & \\boldsymbol{=} & \\boldsymbol{(331\\textbf{ m\/s})\\sqrt{\\frac{T}{273\\textbf{ K}}}=(331\\textbf{ m\/s})\\sqrt{\\frac{293\\textbf{ K}}{273\\textbf{ K}}}} \\\\ {} & \\boldsymbol{=} & \\boldsymbol{343\\textbf{ m\/s}} \\end{array}[\/latex]<\/p>\n<p><strong>7: <\/strong>0.223<\/p>\n<p><strong>9: <\/strong>(a) 7.70 m (b) This means that sonar is good for spotting and locating large objects, but it isn\u2019t able to resolve smaller objects, or detect the detailed shapes of objects. Objects like ships or large pieces of airplanes can be found by sonar, while smaller pieces must be found by other means.<\/p>\n<p><strong>11: <\/strong>(a) 18.0 ms, 17.1 ms (b) 5.00% (c) This uncertainty could definitely cause difficulties for the bat, if it didn\u2019t continue to use sound as it closed in on its prey. A 5% uncertainty could be the difference between catching the prey around the neck or around the chest, which means that it could miss grabbing its prey.<\/p>\n<\/div>\n","protected":false},"author":9,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-616","chapter","type-chapter","status-publish","hentry"],"part":603,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/pressbooks\/v2\/chapters\/616","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/wp\/v2\/users\/9"}],"version-history":[{"count":2,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/pressbooks\/v2\/chapters\/616\/revisions"}],"predecessor-version":[{"id":1116,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/pressbooks\/v2\/chapters\/616\/revisions\/1116"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/pressbooks\/v2\/parts\/603"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/pressbooks\/v2\/chapters\/616\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/wp\/v2\/media?parent=616"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/pressbooks\/v2\/chapter-type?post=616"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/wp\/v2\/contributor?post=616"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/wp\/v2\/license?post=616"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}