{"id":829,"date":"2025-10-09T15:42:03","date_gmt":"2025-10-09T15:42:03","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=829"},"modified":"2025-11-13T21:06:12","modified_gmt":"2025-11-13T21:06:12","slug":"applications-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/applications-fresh-take\/","title":{"raw":"Applications: Fresh Take","rendered":"Applications: Fresh Take"},"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\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">A spring-mass system oscillates when a mass is displaced from its <strong>equilibrium position<\/strong>\u2014the point where the spring's restoring force balances gravity. This oscillation is called <strong>simple harmonic motion<\/strong>.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Setting Up the Differential Equation<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Using Newton's second law and Hooke's law (restoring force proportional to displacement), the motion satisfies [latex]mx'' + kx = 0[\/latex], where [latex]m[\/latex] is mass, [latex]k[\/latex] is the spring constant, and [latex]x(t)[\/latex] is displacement from equilibrium. By convention, positive displacement means the mass is below equilibrium.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Introducing the <strong>angular frequency<\/strong> [latex]\\omega = \\sqrt{k\/m}[\/latex], we can rewrite this as [latex]x'' + \\omega^2 x = 0[\/latex]. The general solution is [latex]x(t) = c_1\\cos(\\omega t) + c_2\\sin(\\omega t)[\/latex].<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Key Physical Quantities<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The <strong>period<\/strong> [latex]T = \\frac{2\\pi}{\\omega}[\/latex] is the time for one complete oscillation. The <strong>frequency<\/strong> [latex]f = \\frac{1}{T} = \\frac{\\omega}{2\\pi}[\/latex] counts oscillations per unit time.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Finding the Spring Constant and Mass<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The spring constant comes from the equilibrium condition [latex]mg = ks[\/latex], where [latex]s[\/latex] is the stretch distance. In the English system, [latex]g = 32[\/latex] ft\/sec\u00b2; in metric, [latex]g = 9.8[\/latex] m\/sec\u00b2. Use [latex]W = mg[\/latex] to find mass from weight.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Alternative Form of the Solution<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The solution can be written as [latex]x(t) = A\\sin(\\omega t + \\phi)[\/latex], where [latex]A = \\sqrt{c_1^2 + c_2^2}[\/latex] is the <strong>amplitude<\/strong> (maximum displacement) and [latex]\\phi[\/latex] is the <strong>phase shift<\/strong>. To find [latex]\\phi[\/latex], use [latex]\\tan\\phi = \\frac{c_1}{c_2}[\/latex] and verify [latex]\\phi[\/latex] is in the correct quadrant based on the signs of [latex]c_1[\/latex] and [latex]c_2[\/latex].<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">A [latex]200[\/latex]-g mass stretches a spring\u00a0[latex]5[\/latex] cm. Find the equation of motion of the mass if it is released from rest from a position\u00a0[latex]10[\/latex] cm below the equilibrium position. What is the frequency of this motion?[reveal-answer q=\"753649205\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"753649205\"][latex]x(t)=0.1\\cos(14t)[\/latex] (in meters); frequency is [latex]\\frac{14}{2\\pi}[\/latex] Hz.[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Express the function [latex]x(t)=\\cos(4t)+4\\sin(4t)[\/latex] in the form [latex]A\\sin(\\omega t+\\phi)[\/latex]. What is the frequency of motion? The amplitude?[reveal-answer q=\"173489144\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"173489144\"][latex]x(t)=\\sqrt{17}\\sin(4t+0.245)[\/latex],\u00a0[latex]\\text{frequency}=\\frac4{2\\pi}\\approx0.637, \\ A=\\sqrt{17}[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250345&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=rIAHiAtDRAo&amp;video_target=tpm-plugin-l9eakoxz-rIAHiAtDRAo\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.16_transcript.html\">transcript for \u201cCP 7.16\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2 data-type=\"title\">Damped Vibrations<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">In real-world spring-mass systems, friction and air resistance cause oscillations to gradually die off\u2014an effect called <strong>damping<\/strong>. The damping force is proportional to velocity and acts opposite to motion, giving us the differential equation [latex]mx'' + bx' + kx = 0[\/latex], where [latex]b &gt; 0[\/latex] is the damping coefficient.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Three Types of Damping<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The characteristic equation [latex]m\\lambda^2 + b\\lambda + k = 0[\/latex] has solutions [latex]\\lambda = \\frac{-b \\pm \\sqrt{b^2 - 4mk}}{2m}[\/latex]. The value of the discriminant [latex]b^2 - 4mk[\/latex] determines the system's behavior.<\/p>\r\n\r\n<ul>\r\n \t<li class=\"font-claude-response-body whitespace-normal break-words\"><strong>Overdamped<\/strong> ([latex]b^2 &gt; 4mk[\/latex]): The system has two distinct negative real roots. The general solution is [latex]x(t) = c_1e^{\\lambda_1 t} + c_2e^{\\lambda_2 t}[\/latex]. The mass returns to equilibrium without oscillating (at most one change of direction). The motion decays quickly.<\/li>\r\n \t<li class=\"font-claude-response-body whitespace-normal break-words\"><strong>Critically Damped<\/strong> ([latex]b^2 = 4mk[\/latex]): The system has one repeated negative real root. The general solution is [latex]x(t) = c_1e^{\\lambda t} + c_2te^{\\lambda t}[\/latex]. Like overdamped systems, critically damped systems don't oscillate. This represents the boundary between oscillatory and non-oscillatory behavior. In practice, it's nearly impossible to achieve exact critical damping.<\/li>\r\n \t<li class=\"font-claude-response-body whitespace-normal break-words\"><strong>Underdamped<\/strong> ([latex]b^2 &lt; 4mk[\/latex]): The system has complex conjugate roots [latex]\\alpha \\pm \\beta i[\/latex] where [latex]\\alpha &lt; 0[\/latex]. The general solution is [latex]x(t) = e^{\\alpha t}(c_1\\cos(\\beta t) + c_2\\sin(\\beta t))[\/latex]. The system oscillates, but the amplitude decreases exponentially over time due to the [latex]e^{\\alpha t}[\/latex] factor.<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Key Observation<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For all damped systems, [latex]\\lim_{t \\to \\infty} x(t) = 0[\/latex]. The system always returns to equilibrium eventually, regardless of damping type.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">A[latex]2[\/latex]-kg mass is attached to a spring with spring constant\u00a0[latex]24[\/latex] N\/m. The system is then immersed in a medium imparting a damping force equal to\u00a0[latex]16[\/latex] times the instantaneous velocity of the mass. Find the equation of motion if it is released from rest at a point\u00a0[latex]40[\/latex] cm below equilibrium.[reveal-answer q=\"482309319\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"482309319\"][latex]x(t)=0.6e^{-2t}-0.2e^{-6t}[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250344&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=66R4PCBUdV0&amp;video_target=tpm-plugin-rimxr4nx-66R4PCBUdV0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.17_transcript.html\">transcript for \u201cCP 7.17\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">A [latex]1[\/latex]-lb weight stretches a spring\u00a0[latex]6[\/latex] in., and the system is attached to a dashpot that imparts a damping force equal to half the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point\u00a0[latex]6[\/latex] in. below equilibrium.[reveal-answer q=\"983453984\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"983453984\"][latex]x(t)=\\frac12e^{-8t}+4te^{-8t}[\/latex][\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">A [latex]1[\/latex]-kg mass stretches a spring\u00a0[latex]49[\/latex] cm. The system is immersed in a medium that imparts a damping force equal to four times the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point\u00a0[latex]24[\/latex] cm above equilibrium.[reveal-answer q=\"904011382\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"904011382\"][latex]x(t)=-0.24e^{-2t}\\cos(4t)-0.12e^{-2t}\\sin(4t)[\/latex][\/hidden-answer]<\/section>\r\n<h2 data-type=\"title\">Forced Vibrations<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">When an external force acts on a spring-mass system (like bumps in a road or a moving support), we model it with the nonhomogeneous differential equation [latex]mx'' + bx' + kx = f(t)[\/latex], where [latex]f(t)[\/latex] represents the external force.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Two Parts of the Solution<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The general solution has the form [latex]x(t) = c_1x_1(t) + c_2x_2(t) + x_p(t)[\/latex], where [latex]c_1x_1(t) + c_2x_2(t)[\/latex] is the general solution to the complementary equation and [latex]x_p(t)[\/latex] is a particular solution to the nonhomogeneous equation.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Transient vs. Steady-State Behavior<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For damped systems, the complementary solution [latex]c_1x_1(t) + c_2x_2(t)[\/latex] approaches zero as [latex]t \\to \\infty[\/latex]. This portion is called the <strong>transient solution<\/strong> because it only affects the system's short-term behavior and eventually dies out.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The particular solution [latex]x_p(t)[\/latex] is called the <strong>steady-state solution<\/strong> because it determines the long-term behavior of the system. After enough time passes, the transient effects disappear and the system's motion is dominated by the steady-state response to the external force.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Finding the Solutions<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Solve the complementary equation [latex]mx'' + bx' + kx = 0[\/latex] to get the transient solution. Use the method of undetermined coefficients or variation of parameters to find the particular solution [latex]x_p(t)[\/latex]. The steady-state solution typically oscillates at the same frequency as the external forcing function [latex]f(t)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox interact\" aria-label=\"Interact\">This\u00a0<a href=\"https:\/\/demonstrations.wolfram.com\/ForcedOscillations\/\" target=\"_blank\" rel=\"noopener nofollow\">website<\/a>\u00a0shows a simulation of forced vibrations.<\/section><section class=\"textbox example\" aria-label=\"Example\">A mass of\u00a0[latex]2[\/latex] kg is attached to a spring with constant\u00a0[latex]32[\/latex] N\/m and comes to rest in the equilibrium position. Beginning at time [latex]t=0[\/latex], an external force equal to [latex]f(t)=68e^{-2t}\\cos(4t)[\/latex] is applied to the system. Find the equation of motion if there is no damping. What is the transient solution? What is the steady-state solution?[reveal-answer q=\"093470388\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"093470388\"][latex]x(t)=-\\frac12\\cos(4t)+\\frac94\\sin(4t)+\\frac12e^{-2t}\\cos(4t)-2e^{-2t}\\sin(4t)[\/latex]Transient solution: [latex]\\frac12e^{-2t}\\cos(4t)-2e^{-2t}\\sin(4t)[\/latex]Steady-state solution: [latex]-\\frac12\\cos(4t)+\\frac94\\sin(4t)[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250343&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=gAX6Y3QGW-0&amp;video_target=tpm-plugin-pmpfgtlb-gAX6Y3QGW-0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.20_transcript.html\">transcript for \u201cCP 7.20\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2>The\u00a0RLC\u00a0Series Circuit<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">An <strong>RLC series circuit<\/strong> contains a resistor, inductor, and capacitor connected in series. These circuits appear in many applications, including radio tuners. The circuit can be modeled using a second-order differential equation that closely parallels the spring-mass system.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Building the Differential Equation<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Using Kirchhoff's voltage rule (the sum of voltage drops around a closed loop equals zero), we account for voltage drops across each component:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Inductor: [latex]E_L = L\\frac{dI}{dt}[\/latex] (proportional to the rate of change of current)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Resistor: [latex]E_R = RI[\/latex] (proportional to current, by Ohm's law)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Capacitor: [latex]E_C = \\frac{1}{C}q[\/latex] (proportional to charge)<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Adding these with the voltage source [latex]E(t)[\/latex] and using [latex]I = \\frac{dq}{dt}[\/latex], we get:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]L\\frac{d^2q}{dt^2} + R\\frac{dq}{dt} + \\frac{1}{C}q = E(t)[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Spring-Mass Analogy<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">This equation has the same mathematical form as the forced spring-mass system [latex]mx'' + bx' + kx = f(t)[\/latex]. The correspondences are:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Inductance [latex]L[\/latex] \u2194 mass [latex]m[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Resistance [latex]R[\/latex] \u2194 damping coefficient [latex]b[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Reciprocal of capacitance [latex]1\/C[\/latex] \u2194 spring constant [latex]k[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Voltage source [latex]E(t)[\/latex] \u2194 external force [latex]f(t)[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">This means all the solution techniques and concepts from spring-mass systems (overdamped, critically damped, underdamped, transient and steady-state solutions) apply directly to RLC circuits. The charge [latex]q(t)[\/latex] on the capacitor behaves like the displacement [latex]x(t)[\/latex] in a mechanical system.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the charge on the capacitor in an\u00a0<em data-effect=\"italics\">RLC<\/em>\u00a0series circuit where [latex]L=1\/5 [\/latex] H, [latex]R=2\/5\\Omega[\/latex], [latex]C=1\/2[\/latex] F, and [latex]E(t)=50[\/latex] V. Assume the initial charge on the capacitor is 0 [latex]C[\/latex] and the initial current is [latex]4 \\ A[\/latex].[reveal-answer q=\"981734122\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"981734122\"][latex]q(t)=-25e^{-t}\\cos(3t)-7e^{-t}\\sin(3t)+25[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250346&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=7FbR492dU2g&amp;video_target=tpm-plugin-9g88d5wl-7FbR492dU2g\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.21_transcript.html\">transcript for \u201cCP 7.21\u201d here (opens in new window).<\/a><\/center><\/section>","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<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">A spring-mass system oscillates when a mass is displaced from its <strong>equilibrium position<\/strong>\u2014the point where the spring&#8217;s restoring force balances gravity. This oscillation is called <strong>simple harmonic motion<\/strong>.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Setting Up the Differential Equation<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Using Newton&#8217;s second law and Hooke&#8217;s law (restoring force proportional to displacement), the motion satisfies [latex]mx'' + kx = 0[\/latex], where [latex]m[\/latex] is mass, [latex]k[\/latex] is the spring constant, and [latex]x(t)[\/latex] is displacement from equilibrium. By convention, positive displacement means the mass is below equilibrium.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Introducing the <strong>angular frequency<\/strong> [latex]\\omega = \\sqrt{k\/m}[\/latex], we can rewrite this as [latex]x'' + \\omega^2 x = 0[\/latex]. The general solution is [latex]x(t) = c_1\\cos(\\omega t) + c_2\\sin(\\omega t)[\/latex].<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Key Physical Quantities<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The <strong>period<\/strong> [latex]T = \\frac{2\\pi}{\\omega}[\/latex] is the time for one complete oscillation. The <strong>frequency<\/strong> [latex]f = \\frac{1}{T} = \\frac{\\omega}{2\\pi}[\/latex] counts oscillations per unit time.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Finding the Spring Constant and Mass<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The spring constant comes from the equilibrium condition [latex]mg = ks[\/latex], where [latex]s[\/latex] is the stretch distance. In the English system, [latex]g = 32[\/latex] ft\/sec\u00b2; in metric, [latex]g = 9.8[\/latex] m\/sec\u00b2. Use [latex]W = mg[\/latex] to find mass from weight.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Alternative Form of the Solution<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The solution can be written as [latex]x(t) = A\\sin(\\omega t + \\phi)[\/latex], where [latex]A = \\sqrt{c_1^2 + c_2^2}[\/latex] is the <strong>amplitude<\/strong> (maximum displacement) and [latex]\\phi[\/latex] is the <strong>phase shift<\/strong>. To find [latex]\\phi[\/latex], use [latex]\\tan\\phi = \\frac{c_1}{c_2}[\/latex] and verify [latex]\\phi[\/latex] is in the correct quadrant based on the signs of [latex]c_1[\/latex] and [latex]c_2[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">A [latex]200[\/latex]-g mass stretches a spring\u00a0[latex]5[\/latex] cm. Find the equation of motion of the mass if it is released from rest from a position\u00a0[latex]10[\/latex] cm below the equilibrium position. What is the frequency of this motion?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q753649205\">Show Solution<\/button><\/p>\n<div id=\"q753649205\" class=\"hidden-answer\" style=\"display: none\">[latex]x(t)=0.1\\cos(14t)[\/latex] (in meters); frequency is [latex]\\frac{14}{2\\pi}[\/latex] Hz.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Express the function [latex]x(t)=\\cos(4t)+4\\sin(4t)[\/latex] in the form [latex]A\\sin(\\omega t+\\phi)[\/latex]. What is the frequency of motion? The amplitude?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q173489144\">Show Solution<\/button><\/p>\n<div id=\"q173489144\" class=\"hidden-answer\" style=\"display: none\">[latex]x(t)=\\sqrt{17}\\sin(4t+0.245)[\/latex],\u00a0[latex]\\text{frequency}=\\frac4{2\\pi}\\approx0.637, \\ A=\\sqrt{17}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250345&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=rIAHiAtDRAo&amp;video_target=tpm-plugin-l9eakoxz-rIAHiAtDRAo\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.16_transcript.html\">transcript for \u201cCP 7.16\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2 data-type=\"title\">Damped Vibrations<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">In real-world spring-mass systems, friction and air resistance cause oscillations to gradually die off\u2014an effect called <strong>damping<\/strong>. The damping force is proportional to velocity and acts opposite to motion, giving us the differential equation [latex]mx'' + bx' + kx = 0[\/latex], where [latex]b > 0[\/latex] is the damping coefficient.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Three Types of Damping<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The characteristic equation [latex]m\\lambda^2 + b\\lambda + k = 0[\/latex] has solutions [latex]\\lambda = \\frac{-b \\pm \\sqrt{b^2 - 4mk}}{2m}[\/latex]. The value of the discriminant [latex]b^2 - 4mk[\/latex] determines the system&#8217;s behavior.<\/p>\n<ul>\n<li class=\"font-claude-response-body whitespace-normal break-words\"><strong>Overdamped<\/strong> ([latex]b^2 > 4mk[\/latex]): The system has two distinct negative real roots. The general solution is [latex]x(t) = c_1e^{\\lambda_1 t} + c_2e^{\\lambda_2 t}[\/latex]. The mass returns to equilibrium without oscillating (at most one change of direction). The motion decays quickly.<\/li>\n<li class=\"font-claude-response-body whitespace-normal break-words\"><strong>Critically Damped<\/strong> ([latex]b^2 = 4mk[\/latex]): The system has one repeated negative real root. The general solution is [latex]x(t) = c_1e^{\\lambda t} + c_2te^{\\lambda t}[\/latex]. Like overdamped systems, critically damped systems don&#8217;t oscillate. This represents the boundary between oscillatory and non-oscillatory behavior. In practice, it&#8217;s nearly impossible to achieve exact critical damping.<\/li>\n<li class=\"font-claude-response-body whitespace-normal break-words\"><strong>Underdamped<\/strong> ([latex]b^2 < 4mk[\/latex]): The system has complex conjugate roots [latex]\\alpha \\pm \\beta i[\/latex] where [latex]\\alpha < 0[\/latex]. The general solution is [latex]x(t) = e^{\\alpha t}(c_1\\cos(\\beta t) + c_2\\sin(\\beta t))[\/latex]. The system oscillates, but the amplitude decreases exponentially over time due to the [latex]e^{\\alpha t}[\/latex] factor.<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Key Observation<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For all damped systems, [latex]\\lim_{t \\to \\infty} x(t) = 0[\/latex]. The system always returns to equilibrium eventually, regardless of damping type.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">A[latex]2[\/latex]-kg mass is attached to a spring with spring constant\u00a0[latex]24[\/latex] N\/m. The system is then immersed in a medium imparting a damping force equal to\u00a0[latex]16[\/latex] times the instantaneous velocity of the mass. Find the equation of motion if it is released from rest at a point\u00a0[latex]40[\/latex] cm below equilibrium.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q482309319\">Show Solution<\/button><\/p>\n<div id=\"q482309319\" class=\"hidden-answer\" style=\"display: none\">[latex]x(t)=0.6e^{-2t}-0.2e^{-6t}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250344&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=66R4PCBUdV0&amp;video_target=tpm-plugin-rimxr4nx-66R4PCBUdV0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.17_transcript.html\">transcript for \u201cCP 7.17\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">A [latex]1[\/latex]-lb weight stretches a spring\u00a0[latex]6[\/latex] in., and the system is attached to a dashpot that imparts a damping force equal to half the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point\u00a0[latex]6[\/latex] in. below equilibrium.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q983453984\">Show Solution<\/button><\/p>\n<div id=\"q983453984\" class=\"hidden-answer\" style=\"display: none\">[latex]x(t)=\\frac12e^{-8t}+4te^{-8t}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">A [latex]1[\/latex]-kg mass stretches a spring\u00a0[latex]49[\/latex] cm. The system is immersed in a medium that imparts a damping force equal to four times the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point\u00a0[latex]24[\/latex] cm above equilibrium.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q904011382\">Show Solution<\/button><\/p>\n<div id=\"q904011382\" class=\"hidden-answer\" style=\"display: none\">[latex]x(t)=-0.24e^{-2t}\\cos(4t)-0.12e^{-2t}\\sin(4t)[\/latex]<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Forced Vibrations<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">When an external force acts on a spring-mass system (like bumps in a road or a moving support), we model it with the nonhomogeneous differential equation [latex]mx'' + bx' + kx = f(t)[\/latex], where [latex]f(t)[\/latex] represents the external force.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Two Parts of the Solution<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The general solution has the form [latex]x(t) = c_1x_1(t) + c_2x_2(t) + x_p(t)[\/latex], where [latex]c_1x_1(t) + c_2x_2(t)[\/latex] is the general solution to the complementary equation and [latex]x_p(t)[\/latex] is a particular solution to the nonhomogeneous equation.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Transient vs. Steady-State Behavior<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For damped systems, the complementary solution [latex]c_1x_1(t) + c_2x_2(t)[\/latex] approaches zero as [latex]t \\to \\infty[\/latex]. This portion is called the <strong>transient solution<\/strong> because it only affects the system&#8217;s short-term behavior and eventually dies out.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The particular solution [latex]x_p(t)[\/latex] is called the <strong>steady-state solution<\/strong> because it determines the long-term behavior of the system. After enough time passes, the transient effects disappear and the system&#8217;s motion is dominated by the steady-state response to the external force.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Finding the Solutions<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Solve the complementary equation [latex]mx'' + bx' + kx = 0[\/latex] to get the transient solution. Use the method of undetermined coefficients or variation of parameters to find the particular solution [latex]x_p(t)[\/latex]. The steady-state solution typically oscillates at the same frequency as the external forcing function [latex]f(t)[\/latex].<\/p>\n<\/div>\n<section class=\"textbox interact\" aria-label=\"Interact\">This\u00a0<a href=\"https:\/\/demonstrations.wolfram.com\/ForcedOscillations\/\" target=\"_blank\" rel=\"noopener nofollow\">website<\/a>\u00a0shows a simulation of forced vibrations.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">A mass of\u00a0[latex]2[\/latex] kg is attached to a spring with constant\u00a0[latex]32[\/latex] N\/m and comes to rest in the equilibrium position. Beginning at time [latex]t=0[\/latex], an external force equal to [latex]f(t)=68e^{-2t}\\cos(4t)[\/latex] is applied to the system. Find the equation of motion if there is no damping. What is the transient solution? What is the steady-state solution?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q093470388\">Show Solution<\/button><\/p>\n<div id=\"q093470388\" class=\"hidden-answer\" style=\"display: none\">[latex]x(t)=-\\frac12\\cos(4t)+\\frac94\\sin(4t)+\\frac12e^{-2t}\\cos(4t)-2e^{-2t}\\sin(4t)[\/latex]Transient solution: [latex]\\frac12e^{-2t}\\cos(4t)-2e^{-2t}\\sin(4t)[\/latex]Steady-state solution: [latex]-\\frac12\\cos(4t)+\\frac94\\sin(4t)[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250343&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=gAX6Y3QGW-0&amp;video_target=tpm-plugin-pmpfgtlb-gAX6Y3QGW-0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.20_transcript.html\">transcript for \u201cCP 7.20\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2>The\u00a0RLC\u00a0Series Circuit<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">An <strong>RLC series circuit<\/strong> contains a resistor, inductor, and capacitor connected in series. These circuits appear in many applications, including radio tuners. The circuit can be modeled using a second-order differential equation that closely parallels the spring-mass system.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Building the Differential Equation<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Using Kirchhoff&#8217;s voltage rule (the sum of voltage drops around a closed loop equals zero), we account for voltage drops across each component:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Inductor: [latex]E_L = L\\frac{dI}{dt}[\/latex] (proportional to the rate of change of current)<\/li>\n<li class=\"whitespace-normal break-words\">Resistor: [latex]E_R = RI[\/latex] (proportional to current, by Ohm&#8217;s law)<\/li>\n<li class=\"whitespace-normal break-words\">Capacitor: [latex]E_C = \\frac{1}{C}q[\/latex] (proportional to charge)<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Adding these with the voltage source [latex]E(t)[\/latex] and using [latex]I = \\frac{dq}{dt}[\/latex], we get:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]L\\frac{d^2q}{dt^2} + R\\frac{dq}{dt} + \\frac{1}{C}q = E(t)[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Spring-Mass Analogy<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">This equation has the same mathematical form as the forced spring-mass system [latex]mx'' + bx' + kx = f(t)[\/latex]. The correspondences are:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Inductance [latex]L[\/latex] \u2194 mass [latex]m[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Resistance [latex]R[\/latex] \u2194 damping coefficient [latex]b[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Reciprocal of capacitance [latex]1\/C[\/latex] \u2194 spring constant [latex]k[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Voltage source [latex]E(t)[\/latex] \u2194 external force [latex]f(t)[\/latex]<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\">This means all the solution techniques and concepts from spring-mass systems (overdamped, critically damped, underdamped, transient and steady-state solutions) apply directly to RLC circuits. The charge [latex]q(t)[\/latex] on the capacitor behaves like the displacement [latex]x(t)[\/latex] in a mechanical system.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find the charge on the capacitor in an\u00a0<em data-effect=\"italics\">RLC<\/em>\u00a0series circuit where [latex]L=1\/5[\/latex] H, [latex]R=2\/5\\Omega[\/latex], [latex]C=1\/2[\/latex] F, and [latex]E(t)=50[\/latex] V. Assume the initial charge on the capacitor is 0 [latex]C[\/latex] and the initial current is [latex]4 \\ A[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q981734122\">Show Solution<\/button><\/p>\n<div id=\"q981734122\" class=\"hidden-answer\" style=\"display: none\">[latex]q(t)=-25e^{-t}\\cos(3t)-7e^{-t}\\sin(3t)+25[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250346&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=7FbR492dU2g&amp;video_target=tpm-plugin-9g88d5wl-7FbR492dU2g\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.21_transcript.html\">transcript for \u201cCP 7.21\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":22,"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-829","chapter","type-chapter","status-publish","hentry"],"part":440,"module-header":"fresh_take","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\/829","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":9,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/829\/revisions"}],"predecessor-version":[{"id":2270,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/829\/revisions\/2270"}],"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\/829\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=829"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=829"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=829"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=829"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}