{"id":600,"date":"2020-01-13T15:13:26","date_gmt":"2020-01-13T20:13:26","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/chapter\/11-3-energy-in-waves-intensity\/"},"modified":"2020-11-10T23:50:04","modified_gmt":"2020-11-11T04:50:04","slug":"11-3-energy-in-waves-intensity","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/chapter\/11-3-energy-in-waves-intensity\/","title":{"raw":"Energy in Waves: Intensity","rendered":"Energy in Waves: Intensity"},"content":{"raw":"<div>\r\n<div class=\"bcc-box bcc-highlight\">\r\n<h3>Summary<\/h3>\r\n<ul>\r\n \t<li>Calculate the intensity and the power of rays and waves.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<figure id=\"import-auto-id1471656\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"395\"]<img class=\"\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1107\/wp-content\/uploads\/sites\/1184\/2020\/01\/Figure_17_11_01a-1.jpg\" alt=\"The destruction caused by an earthquake in Port-au-Prince, Haiti. Some buildings are shown on two sides of a street. Two buildings are completely destroyed. Rescue people are seen around.\" width=\"395\" height=\"282\" \/> <strong>Figure 1.<\/strong> The destructive effect of an earthquake is palpable evidence of the energy carried in these waves. The Richter scale rating of earthquakes is related to both their amplitude and the energy they carry. (credit: Petty Officer 2nd Class Candice Villarreal, U.S. Navy)[\/caption]<\/figure>\r\n<p id=\"import-auto-id2591664\">All waves carry energy. The energy of some waves can be directly observed. Earthquakes can shake whole cities to the ground, performing the work of thousands of wrecking balls. Loud sounds pulverize nerve cells in the inner ear, causing permanent hearing loss. Ultrasound is used for deep-heat treatment of muscle strains. A laser beam can burn away a malignancy. Water waves chew up beaches.<\/p>\r\n<p id=\"import-auto-id1236514\">The amount of energy in a wave is related to its amplitude. Large-amplitude earthquakes produce large ground displacements. Loud sounds have higher pressure amplitudes and come from larger-amplitude source vibrations than soft sounds. Large ocean breakers churn up the shore more than small ones. More quantitatively, a wave is a displacement that is resisted by a restoring force. The larger the displacement <em><strong>x<\/strong><\/em>, the larger the force <strong><em>F<\/em>=<em>kx<\/em><\/strong> needed to create it. Because work <em><strong>W<\/strong><\/em> is related to force multiplied by distance (<em><strong>Fx<\/strong><\/em>) and energy is put into the wave by the work done to create it, the energy in a wave is related to amplitude. In fact, a wave\u2019s energy is directly proportional to its amplitude squared because<\/p>\r\n\r\n<div id=\"eip-942\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{W\\:\\propto\\:Fx = kx^2}.[\/latex]<\/div>\r\n<p id=\"import-auto-id3034344\">The energy effects of a wave depend on time as well as amplitude. For example, the longer deep-heat ultrasound is applied, the more energy it transfers. Waves can also be concentrated or spread out. Sunlight, for example, can be focused to burn wood. Earthquakes spread out, so they do less damage the farther they get from the source. In both cases, changing the area the waves cover has important effects. All these pertinent factors are included in the definition of <strong><span id=\"import-auto-id969364\">intensity<\/span><\/strong> <em><strong>I<\/strong><\/em> as power per unit area:<\/p>\r\n\r\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I\\:=} [\/latex][latex size=\"2\"]\\boldsymbol{\\frac{P}{A}}[\/latex]<\/div>\r\n<p id=\"import-auto-id3073473\">where <em><strong>P<\/strong><\/em> is the power carried by the wave through area <em><strong>A<\/strong><\/em>. The definition of intensity is valid for any energy in transit, including that carried by waves. The SI unit for intensity is watts per square meter (<strong>W\/m<sup>2<\/sup><\/strong>). For example, infrared and visible energy from the Sun impinge on Earth at an intensity of\u00a0 <strong>1300 W\/m<sup>2<\/sup><\/strong> just above the atmosphere. There are other intensity-related units, the most common is the decibel (see <a href=\"\/douglasphys1108\/chapter\/12-3-sound-intensity-and-sound-level-2\/\">Chapter 12.3 Sound Intensity and Sound Level<\/a>).<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<div id=\"fs-id1420003\" class=\"example\">\r\n<h3 id=\"import-auto-id3017366\">Example 1: Calculating intensity and power: How much energy is in a ray of Sunlight?<\/h3>\r\nThe average intensity of sunlight on Earth\u2019s surface is about <strong>700 W\/m<sup>2<\/sup><\/strong>.\r\n<p id=\"import-auto-id3082715\">(a) Calculate the amount of energy that falls on a solar collector having an area of <strong>0.500 m<sup>2<\/sup><\/strong> in<strong> 4.00 h<\/strong>.<\/p>\r\n<p id=\"import-auto-id1578334\">(b) What intensity would such sunlight have if concentrated by a magnifying glass onto an area 200 times smaller than its own?<\/p>\r\n<p id=\"import-auto-id2599719\"><strong>Strategy a<\/strong><\/p>\r\n<p id=\"fs-id2437365\">Because power is energy per unit time or [latex]\\boldsymbol{P=\\frac{E}{t}},[\/latex] the definition of intensity can be written as [latex]\\boldsymbol{I=\\frac{P}{A}=\\frac{E\/t}{A}},[\/latex] and this equation can be solved for E with the given information.<\/p>\r\n<p id=\"import-auto-id3397936\"><strong>Solution a<\/strong><\/p>\r\n\r\n<ol id=\"fs-id3063523\">\r\n \t<li>Begin with the equation that states the definition of intensity:\r\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I\\:=}[\/latex][latex size=\"2\"]\\boldsymbol{\\frac{P}{A}}.[\/latex]<\/div><\/li>\r\n \t<li>Replace <em><strong>P<\/strong><\/em> with its equivalent <em><strong>E\/t<\/strong><\/em>:\r\n<div id=\"import-auto-id2437920\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I\\:=}[\/latex][latex size=\"2\"]\\boldsymbol{\\frac{E\/t}{A}}.[\/latex]<\/div><\/li>\r\n \t<li>Solve for <em><strong>E<\/strong><\/em>:\r\n<div id=\"import-auto-id1997176\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{E=IAt}.[\/latex]<\/div><\/li>\r\n \t<li>Substitute known values into the equation:\r\n<div id=\"import-auto-id3285558\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{E=(700\\textbf{ W\/m}^2)(0.500\\textbf{ m}^2)[(4.00\\textbf{ h})(3600\\textbf{ s\/h})]}.[\/latex]<\/div><\/li>\r\n \t<li>Calculate to find <em><strong>E<\/strong><\/em> and convert units:\r\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{5.04\\times10^6\\textbf{ J}},[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<p id=\"import-auto-id1860387\"><strong>Discussion a<\/strong><\/p>\r\nThe energy falling on the solar collector in 4 hours in part is enough to be useful\u2014for example, for heating a significant amount of water.\r\n<p id=\"import-auto-id1157183\"><strong>Strategy b<\/strong><\/p>\r\n<p id=\"fs-id3077951\">Taking a ratio of new intensity to old intensity and using primes for the new quantities, we will find that it depends on the ratio of the areas. All other quantities will cancel.<\/p>\r\n<p id=\"import-auto-id3022997\"><strong>Solution b<\/strong><\/p>\r\n\r\n<ol id=\"fs-id2684765\">\r\n \t<li>Take the ratio of intensities, which yields:\r\n<div id=\"eip-id1839375\" class=\"equation\" style=\"text-align: center\">[latex size=\"2\"]\\boldsymbol{\\frac{I^{\\prime}}{I}}[\/latex][latex]\\boldsymbol{=}[\/latex][latex size=\"2\"]\\boldsymbol{\\frac{P^{\\prime}\/A^{\\prime}}{P\/A}}[\/latex][latex]\\boldsymbol{=}[\/latex][latex size=\"2\"]\\boldsymbol{\\frac{A}{A^{\\prime}}}([\/latex][latex]\\boldsymbol{\\textbf{The powers cancel because }P^{\\prime}=P}[\/latex][latex size=\"2\"]).[\/latex]<\/div><\/li>\r\n \t<li>Identify the knowns:\r\n<div id=\"import-auto-id2692789\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{A=200A^{\\prime}},[\/latex]<\/div>\r\n<div id=\"import-auto-id2684269\" class=\"equation\" style=\"text-align: center\">[latex size=\"2\"]\\boldsymbol{\\frac{I^{\\prime}}{I}}[\/latex][latex]\\boldsymbol{=200}.[\/latex]<\/div><\/li>\r\n \t<li>Substitute known quantities:\r\n<div id=\"import-auto-id1318204\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I^{\\prime}=200I=200(700\\textbf{ W\/m}^2)}.[\/latex]<\/div><\/li>\r\n \t<li>Calculate to find <strong><em>I<\/em>\u2032<\/strong>:\r\n<div class=\"equation\" style=\"text-align: center\">\r\n\r\n[latex]\\boldsymbol{I^{\\prime}=1.40\\times10^5\\textbf{ W\/m}^2}.[\/latex]\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<p id=\"import-auto-id1177616\"><strong>Discussion b<\/strong><\/p>\r\n<p id=\"fs-id1607874\">Decreasing the area increases the intensity considerably. The intensity of the concentrated sunlight could even start a fire.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<div class=\"example\">\r\n<h3>Example 2: Determine the combined intensity of two waves: Perfect constructive interference<\/h3>\r\nIf two identical waves, each having an intensity of <strong>1.00 W\/m<sup>2<\/sup><\/strong>, interfere perfectly constructively, what is the intensity of the resulting wave?\r\n<p id=\"import-auto-id1859324\"><strong>Strategy<\/strong><\/p>\r\nWe know from <a href=\"\/douglasphys1108\/chapter\/11-2-superposition-and-interference\/\">Chapter 11.2 Superposition and Interference<\/a> that when two identical waves, which have equal amplitudes <em><strong>X<\/strong><\/em>, interfere perfectly constructively, the resulting wave has an amplitude of <strong>2<em>X<\/em><\/strong>.Because a wave\u2019s intensity is proportional to amplitude squared, the intensity of the resulting wave is four times as great as in the individual waves.\r\n\r\n<strong>Solution<\/strong>\r\n<ol id=\"fs-id1118590\">\r\n \t<li id=\"import-auto-id2445366\">Recall that intensity is proportional to amplitude squared.<\/li>\r\n \t<li id=\"import-auto-id3161645\">Calculate the new amplitude:\r\n<div id=\"import-auto-id969598\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I^{\\prime}\\propto(X^{\\prime})^2=(2X)^2=4X^2}.[\/latex]<\/div><\/li>\r\n \t<li>Recall that the intensity of the old amplitude was:\r\n<div id=\"import-auto-id1525319\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I\\propto{X}^2}.[\/latex]<\/div><\/li>\r\n \t<li>Take the ratio of new intensity to the old intensity. This gives:\r\n<div id=\"import-auto-id3181149\" class=\"equation\" style=\"text-align: center\">[latex size=\"2\"]\\boldsymbol{\\frac{I^{\\prime}}{I}}[\/latex][latex]\\boldsymbol{=4}.[\/latex]<\/div><\/li>\r\n \t<li>Calculate to find <strong><em>I<\/em>\u2032<\/strong>:\r\n<div id=\"import-auto-id1419064\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I^{\\prime}=4I=4.00\\textbf{ W\/m}^2}.[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<p id=\"import-auto-id1935435\"><strong>Discussion<\/strong><\/p>\r\nThe intensity goes up by a factor of 4 when the amplitude doubles. This answer is a little disquieting. The two individual waves each have intensities of <strong>1.00 W\/m<sup>2<\/sup><\/strong>, yet their sum has an intensity of <strong>4.00 W\/m<sup>2<\/sup><\/strong>, which may appear to violate conservation of energy. This violation, of course, cannot happen. What does happen is intriguing. The area over which the intensity is <strong>4.00 W\/m<sup>2<\/sup><\/strong> is much less than the area covered by the two waves before they interfered. There are other areas where the intensity is zero. The addition of waves is not as simple as our first look in <a href=\"\/douglasphys1108\/chapter\/11-2-superposition-and-interference\/\">Chapter 11.2 Superposition and Interference<\/a> suggested. We actually get a pattern of both constructive interference and destructive interference whenever two waves are added. For example, if we have two stereo speakers putting out <strong>1.00 W\/m<sup>2<\/sup><\/strong> each, there will be places in the room where the intensity is <strong>4.00 W\/m<sup>2<\/sup><\/strong>, other places where the intensity is zero, and others in between. <a class=\"autogenerated-content\" href=\"#import-auto-id1398723\">Figure 2<\/a> shows what this interference might look like. We will pursue interference patterns elsewhere in this text.\r\n<figure id=\"import-auto-id1398723\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1107\/wp-content\/uploads\/sites\/1184\/2020\/11\/Figure_17_11_02a-1.jpg\" alt=\"Two speakers are shown at the top of the figure at left and right side. Rarefactions are shown as dotted curves and compression as dark curves. The interference of the sound waves from these two speakers is shown. There are some red spots, showing constructive interference, are shown on the interfering waves.\" width=\"325\" height=\"252\" \/> <strong>Figure 2.<\/strong> These stereo speakers produce both constructive interference and destructive interference in the room, a property common to the superposition of all types of waves. The shading is proportional to intensity.[\/caption]<\/figure>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2445366\" class=\"exercise\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Check Your Understanding<\/h3>\r\n<div class=\"exercise\">\r\n<div id=\"fs-id3386450\" class=\"problem\">\r\n<p id=\"import-auto-id2594823\"><strong>1:<\/strong> Which measurement of a wave is most important when determining the wave's intensity?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id2459087\" class=\"section-summary\">\r\n<h1>Section Summary<\/h1>\r\n<p id=\"import-auto-id3176628\">Intensity is defined to be the power per unit area:<\/p>\r\n<p id=\"import-auto-id2611636\">[latex]\\boldsymbol{I=\\frac{P}{A}}[\/latex] and has units of <strong>W\/m<sup>2<\/sup><\/strong>.<\/p>\r\n\r\n<\/section><section id=\"fs-id2406988\" class=\"conceptual-questions\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Conceptual Questions<\/h3>\r\n<div class=\"exercise\">\r\n<div id=\"fs-id1931391\" class=\"problem\">\r\n<p id=\"import-auto-id2697452\"><strong>1: <\/strong>Two identical waves undergo pure constructive interference. Is the resultant intensity twice that of the individual waves? Explain your answer.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2391059\" class=\"exercise\">\r\n<div id=\"fs-id1389596\" class=\"problem\">\r\n\r\n<strong>2: <\/strong>Circular water waves decrease in amplitude as they move away from where a rock is dropped. Explain why.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section class=\"problems-exercises\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Exercises<\/h3>\r\n<div id=\"fs-id1818055\" class=\"exercise\">\r\n<div id=\"fs-id2422537\" class=\"problem\">\r\n<p id=\"import-auto-id2677330\"><strong>1: Medical Application<\/strong><\/p>\r\n<p id=\"eip-id1165652459328\">Ultrasound of intensity 1.50 \u00d7 10<sup>2<\/sup> W\/m<sup>2<\/sup> is produced by the rectangular head of a medical imaging device measuring 3.00 by 5.00 cm. What is its power output?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"exercise\">\r\n<div class=\"problem\">\r\n\r\n<strong>2: <\/strong>The low-frequency speaker of a stereo set has a surface area of 0.05 m<sup>2<\/sup> and produces 1W of acoustical power. What is the intensity at the speaker? If the speaker projects sound uniformly in all directions, at what distance from the speaker is the intensity 0.1\u00a0 W\/m<sup>2<\/sup>?\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1900500\" class=\"exercise\">\r\n<div id=\"fs-id1921665\" class=\"problem\">\r\n<p id=\"import-auto-id2684765\"><strong>3: <\/strong>To increase intensity of a wave by a factor of 50, by what factor should the amplitude be increased?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1908042\" class=\"exercise\">\r\n<div id=\"fs-id3098837\" class=\"problem\">\r\n<p id=\"import-auto-id1969714\"><strong>4: Engineering Application<\/strong><\/p>\r\nA device called an insolation meter is used to measure the intensity of sunlight has an area of 100 cm<sup>2<\/sup> and registers 6.50 W. What is the intensity in W\/m<sup>2<\/sup>?\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1060385\" class=\"exercise\">\r\n<div id=\"fs-id1942034\" class=\"problem\">\r\n<p id=\"import-auto-id2617344\"><strong>5: Astronomy Application<\/strong><\/p>\r\nEnergy from the Sun arrives at the top of the Earth\u2019s atmosphere with an intensity of 1.30 kW\/m<sup>2<\/sup>. How long does it take for 1.8 \u00d7 10<sup>9<\/sup> J to arrive on an area of 1.00\u00a0 m<sup>2<\/sup>?\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2055167\" class=\"exercise\">\r\n<div id=\"fs-id3398187\" class=\"problem\">\r\n\r\n<strong>6: <\/strong>Suppose you have a device that extracts energy from ocean breakers in direct proportion to their intensity. If the device produces 10.0 kW of power on a day when the breakers are 1.20 m high, how much will it produce when they are 0.600 m high?\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id3063311\" class=\"exercise\">\r\n<div class=\"problem\">\r\n<p id=\"import-auto-id3025485\"><strong>7: Engineering Application<\/strong><\/p>\r\n(a) A photovoltaic array of (solar cells) is 10.0% efficient in gathering solar energy and converting it to electricity. If the average intensity of sunlight on one day is 700 W\/m<sup>2<\/sup>, what area should your array have to gather energy at the rate of 100 W? (b) What is the maximum cost of the array if it must pay for itself in two years of operation averaging 10.0 hours per day? Assume that it earns money at the rate of 9.00 \u00a2 per kilowatt-hour.\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2655597\" class=\"exercise\">\r\n<div id=\"fs-id1969919\" class=\"problem\">\r\n\r\n<strong>8: <\/strong>A microphone receiving a pure sound tone feeds an oscilloscope, producing a wave on its screen. If the sound intensity is originally 2.00 \u00d7 10<sup>-5<\/sup> W\/m<sup>2<\/sup>, but is turned up until the amplitude increases by 30.0%, what is the new intensity?\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2626122\" class=\"exercise\">\r\n<div id=\"fs-id2931728\" class=\"problem\">\r\n<p id=\"import-auto-id956458\"><strong>9: Medical Application<\/strong><\/p>\r\n(a) What is the intensity in W\/m<sup>2<\/sup> of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of energy into a circular spot 2.00 mm in diameter in 4.00 s? (b) Discuss how this intensity compares to the average intensity of sunlight (about 700 W\/m<sup>2<\/sup>) and the implications that would have if the laser beam entered your eye. Note how your answer depends on the time duration of the exposure.\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-id1864544\" class=\"definition\">\r\n \t<dt>intensity<\/dt>\r\n \t<dd id=\"fs-id2962596\">power per unit area<\/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\r\n<\/strong>\r\n\r\n<strong>1:<\/strong> Amplitude, because a wave\u2019s energy is directly proportional to its amplitude squared.\r\n\r\n<strong>Problems &amp; Exercises<\/strong>\r\n\r\n<strong>1: <\/strong>0.225 W\r\n\r\n<strong>3: <\/strong>7.07\r\n\r\n<strong>5: <\/strong>16.0 days\r\n\r\n<strong>6: <\/strong>2.50 kW\r\n\r\n<strong>8: <\/strong>[latex]\\boldsymbol{3.38\\times10^{-5}\\textbf{ W\/m}^2}[\/latex]\r\n\r\n<\/div>","rendered":"<div>\n<div class=\"bcc-box bcc-highlight\">\n<h3>Summary<\/h3>\n<ul>\n<li>Calculate the intensity and the power of rays and waves.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<figure id=\"import-auto-id1471656\">\n<figure style=\"width: 395px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1107\/wp-content\/uploads\/sites\/1184\/2020\/01\/Figure_17_11_01a-1.jpg\" alt=\"The destruction caused by an earthquake in Port-au-Prince, Haiti. Some buildings are shown on two sides of a street. Two buildings are completely destroyed. Rescue people are seen around.\" width=\"395\" height=\"282\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 1.<\/strong> The destructive effect of an earthquake is palpable evidence of the energy carried in these waves. The Richter scale rating of earthquakes is related to both their amplitude and the energy they carry. (credit: Petty Officer 2nd Class Candice Villarreal, U.S. Navy)<\/figcaption><\/figure>\n<\/figure>\n<p id=\"import-auto-id2591664\">All waves carry energy. The energy of some waves can be directly observed. Earthquakes can shake whole cities to the ground, performing the work of thousands of wrecking balls. Loud sounds pulverize nerve cells in the inner ear, causing permanent hearing loss. Ultrasound is used for deep-heat treatment of muscle strains. A laser beam can burn away a malignancy. Water waves chew up beaches.<\/p>\n<p id=\"import-auto-id1236514\">The amount of energy in a wave is related to its amplitude. Large-amplitude earthquakes produce large ground displacements. Loud sounds have higher pressure amplitudes and come from larger-amplitude source vibrations than soft sounds. Large ocean breakers churn up the shore more than small ones. More quantitatively, a wave is a displacement that is resisted by a restoring force. The larger the displacement <em><strong>x<\/strong><\/em>, the larger the force <strong><em>F<\/em>=<em>kx<\/em><\/strong> needed to create it. Because work <em><strong>W<\/strong><\/em> is related to force multiplied by distance (<em><strong>Fx<\/strong><\/em>) and energy is put into the wave by the work done to create it, the energy in a wave is related to amplitude. In fact, a wave\u2019s energy is directly proportional to its amplitude squared because<\/p>\n<div id=\"eip-942\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{W\\:\\propto\\:Fx = kx^2}.[\/latex]<\/div>\n<p id=\"import-auto-id3034344\">The energy effects of a wave depend on time as well as amplitude. For example, the longer deep-heat ultrasound is applied, the more energy it transfers. Waves can also be concentrated or spread out. Sunlight, for example, can be focused to burn wood. Earthquakes spread out, so they do less damage the farther they get from the source. In both cases, changing the area the waves cover has important effects. All these pertinent factors are included in the definition of <strong><span id=\"import-auto-id969364\">intensity<\/span><\/strong> <em><strong>I<\/strong><\/em> as power per unit area:<\/p>\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I\\:=}[\/latex][latex]\\boldsymbol{\\frac{P}{A}}[\/latex]<\/div>\n<p id=\"import-auto-id3073473\">where <em><strong>P<\/strong><\/em> is the power carried by the wave through area <em><strong>A<\/strong><\/em>. The definition of intensity is valid for any energy in transit, including that carried by waves. The SI unit for intensity is watts per square meter (<strong>W\/m<sup>2<\/sup><\/strong>). For example, infrared and visible energy from the Sun impinge on Earth at an intensity of\u00a0 <strong>1300 W\/m<sup>2<\/sup><\/strong> just above the atmosphere. There are other intensity-related units, the most common is the decibel (see <a href=\"\/douglasphys1108\/chapter\/12-3-sound-intensity-and-sound-level-2\/\">Chapter 12.3 Sound Intensity and Sound Level<\/a>).<\/p>\n<div class=\"textbox shaded\">\n<div id=\"fs-id1420003\" class=\"example\">\n<h3 id=\"import-auto-id3017366\">Example 1: Calculating intensity and power: How much energy is in a ray of Sunlight?<\/h3>\n<p>The average intensity of sunlight on Earth\u2019s surface is about <strong>700 W\/m<sup>2<\/sup><\/strong>.<\/p>\n<p id=\"import-auto-id3082715\">(a) Calculate the amount of energy that falls on a solar collector having an area of <strong>0.500 m<sup>2<\/sup><\/strong> in<strong> 4.00 h<\/strong>.<\/p>\n<p id=\"import-auto-id1578334\">(b) What intensity would such sunlight have if concentrated by a magnifying glass onto an area 200 times smaller than its own?<\/p>\n<p id=\"import-auto-id2599719\"><strong>Strategy a<\/strong><\/p>\n<p id=\"fs-id2437365\">Because power is energy per unit time or [latex]\\boldsymbol{P=\\frac{E}{t}},[\/latex] the definition of intensity can be written as [latex]\\boldsymbol{I=\\frac{P}{A}=\\frac{E\/t}{A}},[\/latex] and this equation can be solved for E with the given information.<\/p>\n<p id=\"import-auto-id3397936\"><strong>Solution a<\/strong><\/p>\n<ol id=\"fs-id3063523\">\n<li>Begin with the equation that states the definition of intensity:\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I\\:=}[\/latex][latex]\\boldsymbol{\\frac{P}{A}}.[\/latex]<\/div>\n<\/li>\n<li>Replace <em><strong>P<\/strong><\/em> with its equivalent <em><strong>E\/t<\/strong><\/em>:\n<div id=\"import-auto-id2437920\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I\\:=}[\/latex][latex]\\boldsymbol{\\frac{E\/t}{A}}.[\/latex]<\/div>\n<\/li>\n<li>Solve for <em><strong>E<\/strong><\/em>:\n<div id=\"import-auto-id1997176\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{E=IAt}.[\/latex]<\/div>\n<\/li>\n<li>Substitute known values into the equation:\n<div id=\"import-auto-id3285558\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{E=(700\\textbf{ W\/m}^2)(0.500\\textbf{ m}^2)[(4.00\\textbf{ h})(3600\\textbf{ s\/h})]}.[\/latex]<\/div>\n<\/li>\n<li>Calculate to find <em><strong>E<\/strong><\/em> and convert units:\n<div class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{5.04\\times10^6\\textbf{ J}},[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p id=\"import-auto-id1860387\"><strong>Discussion a<\/strong><\/p>\n<p>The energy falling on the solar collector in 4 hours in part is enough to be useful\u2014for example, for heating a significant amount of water.<\/p>\n<p id=\"import-auto-id1157183\"><strong>Strategy b<\/strong><\/p>\n<p id=\"fs-id3077951\">Taking a ratio of new intensity to old intensity and using primes for the new quantities, we will find that it depends on the ratio of the areas. All other quantities will cancel.<\/p>\n<p id=\"import-auto-id3022997\"><strong>Solution b<\/strong><\/p>\n<ol id=\"fs-id2684765\">\n<li>Take the ratio of intensities, which yields:\n<div id=\"eip-id1839375\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{\\frac{I^{\\prime}}{I}}[\/latex][latex]\\boldsymbol{=}[\/latex][latex]\\boldsymbol{\\frac{P^{\\prime}\/A^{\\prime}}{P\/A}}[\/latex][latex]\\boldsymbol{=}[\/latex][latex]\\boldsymbol{\\frac{A}{A^{\\prime}}}([\/latex][latex]\\boldsymbol{\\textbf{The powers cancel because }P^{\\prime}=P}[\/latex][latex]).[\/latex]<\/div>\n<\/li>\n<li>Identify the knowns:\n<div id=\"import-auto-id2692789\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{A=200A^{\\prime}},[\/latex]<\/div>\n<div id=\"import-auto-id2684269\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{\\frac{I^{\\prime}}{I}}[\/latex][latex]\\boldsymbol{=200}.[\/latex]<\/div>\n<\/li>\n<li>Substitute known quantities:\n<div id=\"import-auto-id1318204\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I^{\\prime}=200I=200(700\\textbf{ W\/m}^2)}.[\/latex]<\/div>\n<\/li>\n<li>Calculate to find <strong><em>I<\/em>\u2032<\/strong>:\n<div class=\"equation\" style=\"text-align: center\">\n<p>[latex]\\boldsymbol{I^{\\prime}=1.40\\times10^5\\textbf{ W\/m}^2}.[\/latex]<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<p id=\"import-auto-id1177616\"><strong>Discussion b<\/strong><\/p>\n<p id=\"fs-id1607874\">Decreasing the area increases the intensity considerably. The intensity of the concentrated sunlight could even start a fire.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"example\">\n<h3>Example 2: Determine the combined intensity of two waves: Perfect constructive interference<\/h3>\n<p>If two identical waves, each having an intensity of <strong>1.00 W\/m<sup>2<\/sup><\/strong>, interfere perfectly constructively, what is the intensity of the resulting wave?<\/p>\n<p id=\"import-auto-id1859324\"><strong>Strategy<\/strong><\/p>\n<p>We know from <a href=\"\/douglasphys1108\/chapter\/11-2-superposition-and-interference\/\">Chapter 11.2 Superposition and Interference<\/a> that when two identical waves, which have equal amplitudes <em><strong>X<\/strong><\/em>, interfere perfectly constructively, the resulting wave has an amplitude of <strong>2<em>X<\/em><\/strong>.Because a wave\u2019s intensity is proportional to amplitude squared, the intensity of the resulting wave is four times as great as in the individual waves.<\/p>\n<p><strong>Solution<\/strong><\/p>\n<ol id=\"fs-id1118590\">\n<li id=\"import-auto-id2445366\">Recall that intensity is proportional to amplitude squared.<\/li>\n<li id=\"import-auto-id3161645\">Calculate the new amplitude:\n<div id=\"import-auto-id969598\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I^{\\prime}\\propto(X^{\\prime})^2=(2X)^2=4X^2}.[\/latex]<\/div>\n<\/li>\n<li>Recall that the intensity of the old amplitude was:\n<div id=\"import-auto-id1525319\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I\\propto{X}^2}.[\/latex]<\/div>\n<\/li>\n<li>Take the ratio of new intensity to the old intensity. This gives:\n<div id=\"import-auto-id3181149\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{\\frac{I^{\\prime}}{I}}[\/latex][latex]\\boldsymbol{=4}.[\/latex]<\/div>\n<\/li>\n<li>Calculate to find <strong><em>I<\/em>\u2032<\/strong>:\n<div id=\"import-auto-id1419064\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{I^{\\prime}=4I=4.00\\textbf{ W\/m}^2}.[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p id=\"import-auto-id1935435\"><strong>Discussion<\/strong><\/p>\n<p>The intensity goes up by a factor of 4 when the amplitude doubles. This answer is a little disquieting. The two individual waves each have intensities of <strong>1.00 W\/m<sup>2<\/sup><\/strong>, yet their sum has an intensity of <strong>4.00 W\/m<sup>2<\/sup><\/strong>, which may appear to violate conservation of energy. This violation, of course, cannot happen. What does happen is intriguing. The area over which the intensity is <strong>4.00 W\/m<sup>2<\/sup><\/strong> is much less than the area covered by the two waves before they interfered. There are other areas where the intensity is zero. The addition of waves is not as simple as our first look in <a href=\"\/douglasphys1108\/chapter\/11-2-superposition-and-interference\/\">Chapter 11.2 Superposition and Interference<\/a> suggested. We actually get a pattern of both constructive interference and destructive interference whenever two waves are added. For example, if we have two stereo speakers putting out <strong>1.00 W\/m<sup>2<\/sup><\/strong> each, there will be places in the room where the intensity is <strong>4.00 W\/m<sup>2<\/sup><\/strong>, other places where the intensity is zero, and others in between. <a class=\"autogenerated-content\" href=\"#import-auto-id1398723\">Figure 2<\/a> shows what this interference might look like. We will pursue interference patterns elsewhere in this text.<\/p>\n<figure id=\"import-auto-id1398723\">\n<figure style=\"width: 325px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1107\/wp-content\/uploads\/sites\/1184\/2020\/11\/Figure_17_11_02a-1.jpg\" alt=\"Two speakers are shown at the top of the figure at left and right side. Rarefactions are shown as dotted curves and compression as dark curves. The interference of the sound waves from these two speakers is shown. There are some red spots, showing constructive interference, are shown on the interfering waves.\" width=\"325\" height=\"252\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 2.<\/strong> These stereo speakers produce both constructive interference and destructive interference in the room, a property common to the superposition of all types of waves. The shading is proportional to intensity.<\/figcaption><\/figure>\n<\/figure>\n<\/div>\n<\/div>\n<div id=\"fs-id2445366\" class=\"exercise\">\n<div class=\"bcc-box bcc-info\">\n<h3>Check Your Understanding<\/h3>\n<div class=\"exercise\">\n<div id=\"fs-id3386450\" class=\"problem\">\n<p id=\"import-auto-id2594823\"><strong>1:<\/strong> Which measurement of a wave is most important when determining the wave&#8217;s intensity?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id2459087\" class=\"section-summary\">\n<h1>Section Summary<\/h1>\n<p id=\"import-auto-id3176628\">Intensity is defined to be the power per unit area:<\/p>\n<p id=\"import-auto-id2611636\">[latex]\\boldsymbol{I=\\frac{P}{A}}[\/latex] and has units of <strong>W\/m<sup>2<\/sup><\/strong>.<\/p>\n<\/section>\n<section id=\"fs-id2406988\" class=\"conceptual-questions\">\n<div class=\"bcc-box bcc-info\">\n<h3>Conceptual Questions<\/h3>\n<div class=\"exercise\">\n<div id=\"fs-id1931391\" class=\"problem\">\n<p id=\"import-auto-id2697452\"><strong>1: <\/strong>Two identical waves undergo pure constructive interference. Is the resultant intensity twice that of the individual waves? Explain your answer.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2391059\" class=\"exercise\">\n<div id=\"fs-id1389596\" class=\"problem\">\n<p><strong>2: <\/strong>Circular water waves decrease in amplitude as they move away from where a rock is dropped. Explain why.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"problems-exercises\">\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<div id=\"fs-id1818055\" class=\"exercise\">\n<div id=\"fs-id2422537\" class=\"problem\">\n<p id=\"import-auto-id2677330\"><strong>1: Medical Application<\/strong><\/p>\n<p id=\"eip-id1165652459328\">Ultrasound of intensity 1.50 \u00d7 10<sup>2<\/sup> W\/m<sup>2<\/sup> is produced by the rectangular head of a medical imaging device measuring 3.00 by 5.00 cm. What is its power output?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\">\n<div class=\"problem\">\n<p><strong>2: <\/strong>The low-frequency speaker of a stereo set has a surface area of 0.05 m<sup>2<\/sup> and produces 1W of acoustical power. What is the intensity at the speaker? If the speaker projects sound uniformly in all directions, at what distance from the speaker is the intensity 0.1\u00a0 W\/m<sup>2<\/sup>?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1900500\" class=\"exercise\">\n<div id=\"fs-id1921665\" class=\"problem\">\n<p id=\"import-auto-id2684765\"><strong>3: <\/strong>To increase intensity of a wave by a factor of 50, by what factor should the amplitude be increased?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1908042\" class=\"exercise\">\n<div id=\"fs-id3098837\" class=\"problem\">\n<p id=\"import-auto-id1969714\"><strong>4: Engineering Application<\/strong><\/p>\n<p>A device called an insolation meter is used to measure the intensity of sunlight has an area of 100 cm<sup>2<\/sup> and registers 6.50 W. What is the intensity in W\/m<sup>2<\/sup>?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1060385\" class=\"exercise\">\n<div id=\"fs-id1942034\" class=\"problem\">\n<p id=\"import-auto-id2617344\"><strong>5: Astronomy Application<\/strong><\/p>\n<p>Energy from the Sun arrives at the top of the Earth\u2019s atmosphere with an intensity of 1.30 kW\/m<sup>2<\/sup>. How long does it take for 1.8 \u00d7 10<sup>9<\/sup> J to arrive on an area of 1.00\u00a0 m<sup>2<\/sup>?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2055167\" class=\"exercise\">\n<div id=\"fs-id3398187\" class=\"problem\">\n<p><strong>6: <\/strong>Suppose you have a device that extracts energy from ocean breakers in direct proportion to their intensity. If the device produces 10.0 kW of power on a day when the breakers are 1.20 m high, how much will it produce when they are 0.600 m high?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id3063311\" class=\"exercise\">\n<div class=\"problem\">\n<p id=\"import-auto-id3025485\"><strong>7: Engineering Application<\/strong><\/p>\n<p>(a) A photovoltaic array of (solar cells) is 10.0% efficient in gathering solar energy and converting it to electricity. If the average intensity of sunlight on one day is 700 W\/m<sup>2<\/sup>, what area should your array have to gather energy at the rate of 100 W? (b) What is the maximum cost of the array if it must pay for itself in two years of operation averaging 10.0 hours per day? Assume that it earns money at the rate of 9.00 \u00a2 per kilowatt-hour.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2655597\" class=\"exercise\">\n<div id=\"fs-id1969919\" class=\"problem\">\n<p><strong>8: <\/strong>A microphone receiving a pure sound tone feeds an oscilloscope, producing a wave on its screen. If the sound intensity is originally 2.00 \u00d7 10<sup>-5<\/sup> W\/m<sup>2<\/sup>, but is turned up until the amplitude increases by 30.0%, what is the new intensity?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2626122\" class=\"exercise\">\n<div id=\"fs-id2931728\" class=\"problem\">\n<p id=\"import-auto-id956458\"><strong>9: Medical Application<\/strong><\/p>\n<p>(a) What is the intensity in W\/m<sup>2<\/sup> of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of energy into a circular spot 2.00 mm in diameter in 4.00 s? (b) Discuss how this intensity compares to the average intensity of sunlight (about 700 W\/m<sup>2<\/sup>) and the implications that would have if the laser beam entered your eye. Note how your answer depends on the time duration of the exposure.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"import-auto-id1864544\" class=\"definition\">\n<dt>intensity<\/dt>\n<dd id=\"fs-id2962596\">power per unit area<\/dd>\n<\/dl>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Solutions<\/h3>\n<p><strong>Check Your Understanding<br \/>\n<\/strong><\/p>\n<p><strong>1:<\/strong> Amplitude, because a wave\u2019s energy is directly proportional to its amplitude squared.<\/p>\n<p><strong>Problems &amp; Exercises<\/strong><\/p>\n<p><strong>1: <\/strong>0.225 W<\/p>\n<p><strong>3: <\/strong>7.07<\/p>\n<p><strong>5: <\/strong>16.0 days<\/p>\n<p><strong>6: <\/strong>2.50 kW<\/p>\n<p><strong>8: <\/strong>[latex]\\boldsymbol{3.38\\times10^{-5}\\textbf{ W\/m}^2}[\/latex]<\/p>\n<\/div>\n","protected":false},"author":9,"menu_order":3,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-600","chapter","type-chapter","status-publish","hentry"],"part":577,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/pressbooks\/v2\/chapters\/600","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\/600\/revisions"}],"predecessor-version":[{"id":1113,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/pressbooks\/v2\/chapters\/600\/revisions\/1113"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/pressbooks\/v2\/parts\/577"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/pressbooks\/v2\/chapters\/600\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/wp\/v2\/media?parent=600"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/pressbooks\/v2\/chapter-type?post=600"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/wp\/v2\/contributor?post=600"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1108\/wp-json\/wp\/v2\/license?post=600"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}