{"id":827,"date":"2025-10-09T15:41:57","date_gmt":"2025-10-09T15:41:57","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=827"},"modified":"2025-11-11T15:27:14","modified_gmt":"2025-11-11T15:27:14","slug":"applications-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/applications-learn-it-1\/","title":{"raw":"Applications: Learn It 1","rendered":"Applications: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Solve equations for simple harmonic motion<\/li>\r\n \t<li>Solve equations for damped harmonic motion<\/li>\r\n \t<li>Solve equations for forced harmonic motion<\/li>\r\n \t<li>Apply differential equations to electrical circuits<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Simple Harmonic Motion<\/h2>\r\nConsider a mass suspended from a spring attached to a rigid support. This is commonly called a <strong>spring-mass system<\/strong>. Gravity is pulling the mass downward and the restoring force of the spring is pulling the mass upward. As shown in Figure 1, when these two forces are equal, the mass is said to be at the <strong>equilibrium position<\/strong>. If the mass is displaced from equilibrium, it oscillates up and down.\r\n<div data-type=\"note\">\r\n\r\n[caption id=\"attachment_3279\" align=\"aligncenter\" width=\"731\"]<img class=\"wp-image-3279 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18014453\/7-3-1.jpeg\" alt=\"This figure has three images of springs. The first image is a vertical spring in its natural position with length L attached at the top to a fixed point. The second image shows a vertical spring with a mass m attached to the spring, stretching the spring distance s from L. The spring is in equilibrium. The third image is a vertical spring with mass m attached where the spring is in motion, distance x from equilibrium L + s.\" width=\"731\" height=\"435\" \/> Figure 1. A spring in its natural position (a), at equilibrium with a mass m attached (b), and in oscillatory motion (c).[\/caption]\r\n\r\n<\/div>\r\n<div data-type=\"note\"><\/div>\r\n<div data-type=\"note\"><\/div>\r\n<div data-type=\"note\">\r\n\r\nLet [latex]x(t)[\/latex] denote the displacement of the mass from equilibrium. Note that for spring-mass systems of this type, it is customary to adopt the convention that down is positive. Thus, a positive displacement indicates the mass is below the equilibrium point, whereas a negative displacement indicates the mass is above equilibrium.\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Displacement is usually given in feet in the English system or meters in the metric system.<\/section>\r\n<p class=\"whitespace-normal break-words\">Consider the forces acting on the mass. The force of gravity is given by [latex]mg[\/latex]. In the English system, mass is in slugs and the acceleration resulting from gravity is in feet per second squared. The acceleration resulting from gravity is constant, so in the English system, [latex]g=32[\/latex] ft\/sec\u00b2. In the metric system, we have [latex]g=9.8[\/latex] m\/sec\u00b2.<\/p>\r\n<p class=\"whitespace-normal break-words\">According to <strong>Hooke's law<\/strong>, the restoring force of the spring is proportional to the displacement and acts in the opposite direction from the displacement, so the restoring force is given by [latex]-k(s+x)[\/latex]. The spring constant is given in pounds per foot in the English system and in newtons per meter in the metric system.<\/p>\r\n<p class=\"whitespace-normal break-words\">Now, by Newton's second law, the sum of the forces on the system (gravity plus the restoring force) is equal to mass times acceleration, so we have:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nmx^{\\prime\\prime}&amp;=-k(s+x)+mg \\\\\r\n&amp;=-ks-kx+mg\r\n\\end{aligned}[\/latex].<\/p>\r\nHowever, by the way we have defined our equilibrium position, [latex]mg=ks[\/latex], the differential equation becomes:\r\n<p style=\"text-align: center;\">[latex]mx^{\\prime\\prime}+kx=0[\/latex].<\/p>\r\n<p id=\"fs-id1170571046694\">It is convenient to rearrange this equation and introduce a new variable, called the <strong>angular frequency<\/strong>, [latex]\\omega[\/latex]. Letting [latex]\\omega=\\sqrt{k\/m}[\/latex], we can write the equation as:<\/p>\r\n<p style=\"text-align: center;\">[latex]x^{\\prime\\prime}+\\omega^2x=0[\/latex].<\/p>\r\n<p id=\"fs-id1170571078041\">This differential equation has the general solution:<\/p>\r\n<p style=\"text-align: center;\">[latex]x(t)=c_1\\cos\\omega t+c_2\\sin\\omega t[\/latex],<\/p>\r\n<p id=\"fs-id1170571071180\">which gives the position of the mass at any point in time. The motion of the mass is called <strong>simple harmonic motion<\/strong>. The period of this motion (the time it takes to complete one oscillation) is [latex]T=\\frac{2\\pi}{\\omega}[\/latex] and the frequency is [latex]f=\\frac{1}{T}=\\frac{\\omega}{2\\pi}[\/latex] (Figure 2).<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"note\">\r\n\r\n[caption id=\"attachment_3282\" align=\"aligncenter\" width=\"943\"]<img class=\"wp-image-3282 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18014637\/7-3-2.jpeg\" alt=\"This figure is the graph of f(t) = sin 2t. It is a periodic, oscillating graph. The period of the graph is represented with a line pointing from one peak to the next. It is labeled with the period T = 2\u03c0\/\u03c9.\" width=\"943\" height=\"572\" \/> Figure 2. A graph of vertical displacement versus time for simple harmonic motion.[\/caption]\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Assume an object weighing\u00a0[latex]2[\/latex] lb stretches a spring\u00a0[latex]6[\/latex] in. Find the equation of motion if the spring is released from the equilibrium position with an upward velocity of\u00a0[latex]16[\/latex] ft\/sec. What is the period of the motion?[reveal-answer q=\"834582483\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"834582483\"]\r\n<p id=\"fs-id1170573394062\">We first need to find the spring constant. We have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nmg&amp;=ks \\\\\r\n2&amp;=k\\left(\\frac12\\right) \\\\\r\nk&amp;=4\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170573732422\">We also know that weight [latex]W[\/latex] equals the product of mass [latex]m[\/latex] and the acceleration due to gravity [latex]g[\/latex]. In English units, the acceleration due to gravity is\u00a0[latex]32[\/latex] ft\/sec<sup>2<\/sup>.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nW&amp;=mg \\\\\r\n2&amp;=m(32) \\\\\r\nm&amp;=\\frac1{16}\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170573413672\">Thus, the differential equation representing this system is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{1}{16}x^{\\prime\\prime}+4x=0}[\/latex].<\/p>\r\n<p id=\"fs-id1170573282144\">Multiplying through by [latex]16[\/latex], we get [latex]x^{\\prime\\prime}+64x=0[\/latex], which can also be written in the form [latex]x^{\\prime\\prime}+\\left(8^2\\right)=0[\/latex]. This equation has the general solution<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{x(t)=c_1\\cos(8t)+c_2\\sin(8t)}[\/latex].<\/p>\r\n<p id=\"fs-id1170570999591\">The mass was released from the equilibrium position, so [latex]x(0)=0[\/latex], and it had an initial upward velocity of\u00a0[latex]16[\/latex] ft\/sec, so [latex]x^\\prime(0)=-16[\/latex]. Applying these initial conditions to solve for [latex]c_1[\/latex] and [latex]c_2[\/latex] gives<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{x(t)=-2\\sin8t}[\/latex].<\/p>\r\n<p id=\"fs-id1170571100384\">The period of this motion is [latex]\\frac{2\\pi}8=\\frac{\\pi}4[\/latex] sec.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3>Alternative Form of the Solution<\/h3>\r\n<p class=\"whitespace-normal break-words\">Writing the general solution in the form [latex]x(t)=c_1\\cos(\\omega t)+c_2\\sin(\\omega t)[\/latex] has some advantages. It is easy to see the link between the differential equation and the solution, and the period and frequency of motion are evident. This form of the function tells us very little about the amplitude of the motion, however.<\/p>\r\n<p class=\"whitespace-normal break-words\">In some situations, we may prefer to write the solution in the form:<\/p>\r\n<p style=\"text-align: center;\">[latex]x(t)=A\\sin(\\omega t+\\phi)[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Although the link to the differential equation is not as explicit in this case, the period and frequency of motion are still evident. Furthermore, the amplitude of the motion, [latex]A[\/latex], is obvious in this form of the function. The constant [latex]\\phi[\/latex] is called a <strong>phase shift<\/strong> and has the effect of shifting the graph of the function to the left or right.<\/p>\r\n<p class=\"whitespace-normal break-words\">To convert the solution to this form, we want to find the values of [latex]A[\/latex] and [latex]\\phi[\/latex] such that:<\/p>\r\n<p style=\"text-align: center;\">[latex]c_1\\cos(\\omega t)+c_2\\sin(\\omega t)=A\\sin(\\omega t+\\phi)[\/latex].<\/p>\r\n<p id=\"fs-id1170573442369\">We first apply the trigonometric identity:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\sin(\\alpha+\\beta)=\\sin\\alpha\\cos\\beta+\\cos\\alpha\\sin\\beta[\/latex]<\/p>\r\n<p id=\"fs-id1170571306006\">to get:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nc_1\\cos(\\omega t)+c_2\\sin(\\omega t)&amp;=A(\\sin(\\omega t)\\cos\\phi+\\cos(\\omega t)\\sin\\phi) \\\\\r\n&amp;=A\\sin\\phi(\\cos(\\omega t))+A\\cos\\phi(\\sin(\\omega t))\r\n\\end{aligned}[\/latex].<\/p>\r\nThus,\r\n<p style=\"text-align: center;\">[latex]c_1=A\\sin\\phi\\text{ and }c_2=A\\cos\\phi[\/latex].<\/p>\r\n<p id=\"fs-id1170571273697\">If we square both of these equations and add them together, we get:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nc_1^2+c_2^2&amp;=A^2\\sin^2\\phi+A^2\\cos^2\\phi \\\\\r\n&amp;=A^2(\\sin^2\\phi+\\cos^2\\phi) \\\\\r\n&amp;=A^2\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170571189536\">Thus,<\/p>\r\n<p style=\"text-align: center;\">[latex]A=\\sqrt{c_1^2+c_2^2}[\/latex].<\/p>\r\n<p id=\"fs-id1170573262990\">Now, to find [latex]\\phi[\/latex], go back to the equations for [latex]c_1[\/latex] and [latex]c_2[\/latex], but this time, divide the first equation by the second equation to get:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\frac{c_1}{c_2}&amp;=\\frac{A\\sin\\phi}{A\\cos\\phi} \\\\\r\n&amp;=\\tan\\phi\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170573279171\">Then,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\tan\\phi=\\frac{c_1}{c_2}[\/latex].<\/p>\r\n<p id=\"fs-id1170571031386\">We summarize this finding in the following theorem.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>solution to the equation for simple harmonic motion<\/h3>\r\nThe function [latex]x(t)=c_1\\cos(\\omega t)+c_2\\sin(\\omega t)[\/latex] can be written in the form [latex]x(t)=A\\sin(\\omega t+\\phi)[\/latex], where [latex]A=\\sqrt{c_1^2+c_2^2}[\/latex] and [latex]\\tan\\phi=\\frac{c_1}{c_2}[\/latex].\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">When using the formula [latex]\\tan\\phi=\\frac{c_1}{c_2}[\/latex] to find [latex]\\phi[\/latex], we must take care to ensure [latex]\\phi[\/latex] is in the right quadrant (Figure 3).\r\n\r\n[caption id=\"attachment_3283\" align=\"aligncenter\" width=\"604\"]<img class=\"wp-image-3283 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18014746\/7-3-3.jpeg\" alt=\"This figure is the graph of f(t) = sin 2t. It is a periodic, oscillating graph. The period of the graph is represented with a line pointing from one peak to the next. It is labeled with the period T = 2\u03c0\/\u03c9. The graph has a phase shift of \u03d5\/\u03c9 so that the sine curve has the value zero to the left of the origin.\" width=\"604\" height=\"422\" \/> Figure 3. A graph of vertical displacement versus time for simple harmonic motion with a phase change.[\/caption]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1170573386766\">Express the following functions in the form [latex]A(\\sin\\omega t+\\phi)[\/latex]. What is the frequency of motion? The amplitude?<\/p>\r\n\r\n<ol id=\"fs-id1170573540766\" type=\"a\">\r\n \t<li>[latex]x(t)=2\\cos(3t)+\\sin(3t)[\/latex]<\/li>\r\n \t<li>[latex]x(t)=3\\cos(2t)-2\\sin(2t)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"834716617\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"834716617\"]\r\n<ol id=\"fs-id1170573371377\" type=\"a\">\r\n \t<li>We have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]A=\\sqrt{c_1^2+c_2^2}=\\sqrt{2^2+1^2}=\\sqrt5[\/latex]<\/p>\r\nand\r\n<p style=\"text-align: center;\">[latex]\\tan\\phi=\\frac{c_1}{c_2}=\\frac21=2[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Note that both [latex]c_1[\/latex] and [latex]c_2[\/latex] are positive, so [latex]\\phi[\/latex] is in the first quadrant. Thus,<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\phi\\approx1.107[\/latex] rad,<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>so we have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]x(t)=2\\cos(3t)+\\sin(3t)=\\sqrt5\\sin(3t+1.107)[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>The frequency is [latex]\\frac{\\omega}{2\\pi}=\\frac3{2\\pi}\\approx0.477[\/latex]. The amplitude is [latex]\\sqrt5[\/latex].<\/li>\r\n \t<li>We have\r\n<p style=\"text-align: center;\">[latex]A=\\sqrt{c_1^2+c_2^2}=\\sqrt{3^2+2^2}=\\sqrt{13}[\/latex]<\/p>\r\nand<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\tan\\phi=\\frac{c_1}{c_2}=\\frac{3}{-2}=-\\frac32[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Note that [latex]c_1[\/latex] is positive but [latex]c_2[\/latex] is negative, so [latex]\\phi[\/latex] is in the fourth quadrant. Thus,<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\phi\\approx-0.983[\/latex] rad,<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>so we have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nx(t)&amp;=3\\cos(2t)-2\\sin(2t) \\\\\r\n&amp;=\\sqrt{13}\\sin(2t-0.983)\r\n\\end{aligned}[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>The frequency is [latex]\\frac{\\omega}{2\\pi}=\\frac2{2\\pi}\\approx0.318[\/latex]. The amplitude is [latex]\\sqrt{13}[\/latex].\r\n\r\n[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/section><\/div>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Solve equations for simple harmonic motion<\/li>\n<li>Solve equations for damped harmonic motion<\/li>\n<li>Solve equations for forced harmonic motion<\/li>\n<li>Apply differential equations to electrical circuits<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Simple Harmonic Motion<\/h2>\n<p>Consider a mass suspended from a spring attached to a rigid support. This is commonly called a <strong>spring-mass system<\/strong>. Gravity is pulling the mass downward and the restoring force of the spring is pulling the mass upward. As shown in Figure 1, when these two forces are equal, the mass is said to be at the <strong>equilibrium position<\/strong>. If the mass is displaced from equilibrium, it oscillates up and down.<\/p>\n<div data-type=\"note\">\n<figure id=\"attachment_3279\" aria-describedby=\"caption-attachment-3279\" style=\"width: 731px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3279 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18014453\/7-3-1.jpeg\" alt=\"This figure has three images of springs. The first image is a vertical spring in its natural position with length L attached at the top to a fixed point. The second image shows a vertical spring with a mass m attached to the spring, stretching the spring distance s from L. The spring is in equilibrium. The third image is a vertical spring with mass m attached where the spring is in motion, distance x from equilibrium L + s.\" width=\"731\" height=\"435\" \/><figcaption id=\"caption-attachment-3279\" class=\"wp-caption-text\">Figure 1. A spring in its natural position (a), at equilibrium with a mass m attached (b), and in oscillatory motion (c).<\/figcaption><\/figure>\n<\/div>\n<div data-type=\"note\"><\/div>\n<div data-type=\"note\"><\/div>\n<div data-type=\"note\">\n<p>Let [latex]x(t)[\/latex] denote the displacement of the mass from equilibrium. Note that for spring-mass systems of this type, it is customary to adopt the convention that down is positive. Thus, a positive displacement indicates the mass is below the equilibrium point, whereas a negative displacement indicates the mass is above equilibrium.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Displacement is usually given in feet in the English system or meters in the metric system.<\/section>\n<p class=\"whitespace-normal break-words\">Consider the forces acting on the mass. The force of gravity is given by [latex]mg[\/latex]. In the English system, mass is in slugs and the acceleration resulting from gravity is in feet per second squared. The acceleration resulting from gravity is constant, so in the English system, [latex]g=32[\/latex] ft\/sec\u00b2. In the metric system, we have [latex]g=9.8[\/latex] m\/sec\u00b2.<\/p>\n<p class=\"whitespace-normal break-words\">According to <strong>Hooke&#8217;s law<\/strong>, the restoring force of the spring is proportional to the displacement and acts in the opposite direction from the displacement, so the restoring force is given by [latex]-k(s+x)[\/latex]. The spring constant is given in pounds per foot in the English system and in newtons per meter in the metric system.<\/p>\n<p class=\"whitespace-normal break-words\">Now, by Newton&#8217;s second law, the sum of the forces on the system (gravity plus the restoring force) is equal to mass times acceleration, so we have:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  mx^{\\prime\\prime}&=-k(s+x)+mg \\\\  &=-ks-kx+mg  \\end{aligned}[\/latex].<\/p>\n<p>However, by the way we have defined our equilibrium position, [latex]mg=ks[\/latex], the differential equation becomes:<\/p>\n<p style=\"text-align: center;\">[latex]mx^{\\prime\\prime}+kx=0[\/latex].<\/p>\n<p id=\"fs-id1170571046694\">It is convenient to rearrange this equation and introduce a new variable, called the <strong>angular frequency<\/strong>, [latex]\\omega[\/latex]. Letting [latex]\\omega=\\sqrt{k\/m}[\/latex], we can write the equation as:<\/p>\n<p style=\"text-align: center;\">[latex]x^{\\prime\\prime}+\\omega^2x=0[\/latex].<\/p>\n<p id=\"fs-id1170571078041\">This differential equation has the general solution:<\/p>\n<p style=\"text-align: center;\">[latex]x(t)=c_1\\cos\\omega t+c_2\\sin\\omega t[\/latex],<\/p>\n<p id=\"fs-id1170571071180\">which gives the position of the mass at any point in time. The motion of the mass is called <strong>simple harmonic motion<\/strong>. The period of this motion (the time it takes to complete one oscillation) is [latex]T=\\frac{2\\pi}{\\omega}[\/latex] and the frequency is [latex]f=\\frac{1}{T}=\\frac{\\omega}{2\\pi}[\/latex] (Figure 2).<\/p>\n<\/div>\n<div data-type=\"note\">\n<figure id=\"attachment_3282\" aria-describedby=\"caption-attachment-3282\" style=\"width: 943px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3282 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18014637\/7-3-2.jpeg\" alt=\"This figure is the graph of f(t) = sin 2t. It is a periodic, oscillating graph. The period of the graph is represented with a line pointing from one peak to the next. It is labeled with the period T = 2\u03c0\/\u03c9.\" width=\"943\" height=\"572\" \/><figcaption id=\"caption-attachment-3282\" class=\"wp-caption-text\">Figure 2. A graph of vertical displacement versus time for simple harmonic motion.<\/figcaption><\/figure>\n<section class=\"textbox example\" aria-label=\"Example\">Assume an object weighing\u00a0[latex]2[\/latex] lb stretches a spring\u00a0[latex]6[\/latex] in. Find the equation of motion if the spring is released from the equilibrium position with an upward velocity of\u00a0[latex]16[\/latex] ft\/sec. What is the period of the motion?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q834582483\">Show Solution<\/button><\/p>\n<div id=\"q834582483\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573394062\">We first need to find the spring constant. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  mg&=ks \\\\  2&=k\\left(\\frac12\\right) \\\\  k&=4  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170573732422\">We also know that weight [latex]W[\/latex] equals the product of mass [latex]m[\/latex] and the acceleration due to gravity [latex]g[\/latex]. In English units, the acceleration due to gravity is\u00a0[latex]32[\/latex] ft\/sec<sup>2<\/sup>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  W&=mg \\\\  2&=m(32) \\\\  m&=\\frac1{16}  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170573413672\">Thus, the differential equation representing this system is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{1}{16}x^{\\prime\\prime}+4x=0}[\/latex].<\/p>\n<p id=\"fs-id1170573282144\">Multiplying through by [latex]16[\/latex], we get [latex]x^{\\prime\\prime}+64x=0[\/latex], which can also be written in the form [latex]x^{\\prime\\prime}+\\left(8^2\\right)=0[\/latex]. This equation has the general solution<\/p>\n<p style=\"text-align: center;\">[latex]\\large{x(t)=c_1\\cos(8t)+c_2\\sin(8t)}[\/latex].<\/p>\n<p id=\"fs-id1170570999591\">The mass was released from the equilibrium position, so [latex]x(0)=0[\/latex], and it had an initial upward velocity of\u00a0[latex]16[\/latex] ft\/sec, so [latex]x^\\prime(0)=-16[\/latex]. Applying these initial conditions to solve for [latex]c_1[\/latex] and [latex]c_2[\/latex] gives<\/p>\n<p style=\"text-align: center;\">[latex]\\large{x(t)=-2\\sin8t}[\/latex].<\/p>\n<p id=\"fs-id1170571100384\">The period of this motion is [latex]\\frac{2\\pi}8=\\frac{\\pi}4[\/latex] sec.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h3>Alternative Form of the Solution<\/h3>\n<p class=\"whitespace-normal break-words\">Writing the general solution in the form [latex]x(t)=c_1\\cos(\\omega t)+c_2\\sin(\\omega t)[\/latex] has some advantages. It is easy to see the link between the differential equation and the solution, and the period and frequency of motion are evident. This form of the function tells us very little about the amplitude of the motion, however.<\/p>\n<p class=\"whitespace-normal break-words\">In some situations, we may prefer to write the solution in the form:<\/p>\n<p style=\"text-align: center;\">[latex]x(t)=A\\sin(\\omega t+\\phi)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Although the link to the differential equation is not as explicit in this case, the period and frequency of motion are still evident. Furthermore, the amplitude of the motion, [latex]A[\/latex], is obvious in this form of the function. The constant [latex]\\phi[\/latex] is called a <strong>phase shift<\/strong> and has the effect of shifting the graph of the function to the left or right.<\/p>\n<p class=\"whitespace-normal break-words\">To convert the solution to this form, we want to find the values of [latex]A[\/latex] and [latex]\\phi[\/latex] such that:<\/p>\n<p style=\"text-align: center;\">[latex]c_1\\cos(\\omega t)+c_2\\sin(\\omega t)=A\\sin(\\omega t+\\phi)[\/latex].<\/p>\n<p id=\"fs-id1170573442369\">We first apply the trigonometric identity:<\/p>\n<p style=\"text-align: center;\">[latex]\\sin(\\alpha+\\beta)=\\sin\\alpha\\cos\\beta+\\cos\\alpha\\sin\\beta[\/latex]<\/p>\n<p id=\"fs-id1170571306006\">to get:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  c_1\\cos(\\omega t)+c_2\\sin(\\omega t)&=A(\\sin(\\omega t)\\cos\\phi+\\cos(\\omega t)\\sin\\phi) \\\\  &=A\\sin\\phi(\\cos(\\omega t))+A\\cos\\phi(\\sin(\\omega t))  \\end{aligned}[\/latex].<\/p>\n<p>Thus,<\/p>\n<p style=\"text-align: center;\">[latex]c_1=A\\sin\\phi\\text{ and }c_2=A\\cos\\phi[\/latex].<\/p>\n<p id=\"fs-id1170571273697\">If we square both of these equations and add them together, we get:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  c_1^2+c_2^2&=A^2\\sin^2\\phi+A^2\\cos^2\\phi \\\\  &=A^2(\\sin^2\\phi+\\cos^2\\phi) \\\\  &=A^2  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170571189536\">Thus,<\/p>\n<p style=\"text-align: center;\">[latex]A=\\sqrt{c_1^2+c_2^2}[\/latex].<\/p>\n<p id=\"fs-id1170573262990\">Now, to find [latex]\\phi[\/latex], go back to the equations for [latex]c_1[\/latex] and [latex]c_2[\/latex], but this time, divide the first equation by the second equation to get:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\frac{c_1}{c_2}&=\\frac{A\\sin\\phi}{A\\cos\\phi} \\\\  &=\\tan\\phi  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170573279171\">Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\tan\\phi=\\frac{c_1}{c_2}[\/latex].<\/p>\n<p id=\"fs-id1170571031386\">We summarize this finding in the following theorem.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>solution to the equation for simple harmonic motion<\/h3>\n<p>The function [latex]x(t)=c_1\\cos(\\omega t)+c_2\\sin(\\omega t)[\/latex] can be written in the form [latex]x(t)=A\\sin(\\omega t+\\phi)[\/latex], where [latex]A=\\sqrt{c_1^2+c_2^2}[\/latex] and [latex]\\tan\\phi=\\frac{c_1}{c_2}[\/latex].<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">When using the formula [latex]\\tan\\phi=\\frac{c_1}{c_2}[\/latex] to find [latex]\\phi[\/latex], we must take care to ensure [latex]\\phi[\/latex] is in the right quadrant (Figure 3).<\/p>\n<figure id=\"attachment_3283\" aria-describedby=\"caption-attachment-3283\" style=\"width: 604px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3283 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18014746\/7-3-3.jpeg\" alt=\"This figure is the graph of f(t) = sin 2t. It is a periodic, oscillating graph. The period of the graph is represented with a line pointing from one peak to the next. It is labeled with the period T = 2\u03c0\/\u03c9. The graph has a phase shift of \u03d5\/\u03c9 so that the sine curve has the value zero to the left of the origin.\" width=\"604\" height=\"422\" \/><figcaption id=\"caption-attachment-3283\" class=\"wp-caption-text\">Figure 3. A graph of vertical displacement versus time for simple harmonic motion with a phase change.<\/figcaption><\/figure>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1170573386766\">Express the following functions in the form [latex]A(\\sin\\omega t+\\phi)[\/latex]. What is the frequency of motion? The amplitude?<\/p>\n<ol id=\"fs-id1170573540766\" type=\"a\">\n<li>[latex]x(t)=2\\cos(3t)+\\sin(3t)[\/latex]<\/li>\n<li>[latex]x(t)=3\\cos(2t)-2\\sin(2t)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q834716617\">Show Solution<\/button><\/p>\n<div id=\"q834716617\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170573371377\" type=\"a\">\n<li>We have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]A=\\sqrt{c_1^2+c_2^2}=\\sqrt{2^2+1^2}=\\sqrt5[\/latex]<\/p>\n<p>and<\/p>\n<p style=\"text-align: center;\">[latex]\\tan\\phi=\\frac{c_1}{c_2}=\\frac21=2[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Note that both [latex]c_1[\/latex] and [latex]c_2[\/latex] are positive, so [latex]\\phi[\/latex] is in the first quadrant. Thus,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\phi\\approx1.107[\/latex] rad,<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>so we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]x(t)=2\\cos(3t)+\\sin(3t)=\\sqrt5\\sin(3t+1.107)[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>The frequency is [latex]\\frac{\\omega}{2\\pi}=\\frac3{2\\pi}\\approx0.477[\/latex]. The amplitude is [latex]\\sqrt5[\/latex].<\/li>\n<li>We have\n<p style=\"text-align: center;\">[latex]A=\\sqrt{c_1^2+c_2^2}=\\sqrt{3^2+2^2}=\\sqrt{13}[\/latex]<\/p>\n<p>and<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\tan\\phi=\\frac{c_1}{c_2}=\\frac{3}{-2}=-\\frac32[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Note that [latex]c_1[\/latex] is positive but [latex]c_2[\/latex] is negative, so [latex]\\phi[\/latex] is in the fourth quadrant. Thus,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\phi\\approx-0.983[\/latex] rad,<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>so we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  x(t)&=3\\cos(2t)-2\\sin(2t) \\\\  &=\\sqrt{13}\\sin(2t-0.983)  \\end{aligned}[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>The frequency is [latex]\\frac{\\omega}{2\\pi}=\\frac2{2\\pi}\\approx0.318[\/latex]. The amplitude is [latex]\\sqrt{13}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/section>\n<\/div>\n","protected":false},"author":15,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-827","chapter","type-chapter","status-publish","hentry"],"part":440,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/827","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/827\/revisions"}],"predecessor-version":[{"id":2153,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/827\/revisions\/2153"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/440"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/827\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=827"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=827"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=827"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=827"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}