{"id":1577,"date":"2025-10-17T17:21:22","date_gmt":"2025-10-17T17:21:22","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1577"},"modified":"2025-11-11T17:37:17","modified_gmt":"2025-11-11T17:37:17","slug":"applications-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/applications-learn-it-3\/","title":{"raw":"Applications: Learn It 3","rendered":"Applications: Learn It 3"},"content":{"raw":"<h2 data-type=\"title\">Forced Vibrations<\/h2>\r\n<p class=\"whitespace-normal break-words\">The last case we consider is when an external force acts on the system. In the case of the motorcycle suspension system, for example, the bumps in the road act as an external force acting on the system. Another example is a spring hanging from a support; if the support is set in motion, that motion would be considered an external force on the system.<\/p>\r\n<p class=\"whitespace-normal break-words\">We model these forced systems with the nonhomogeneous differential equation:<\/p>\r\n<p style=\"text-align: center;\">[latex]mx^{\\prime\\prime}+bx^\\prime+kx=f(t)[\/latex],<\/p>\r\n<p id=\"fs-id1170573582925\">where the external force is represented by the [latex]f(t)[\/latex] term. As we saw earlier, differential equations such as this have solutions of the form:<\/p>\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]x(t)=c_1x_1(t)+c_2x_2(t)+x_p(t)[\/latex],<\/span><\/p>\r\nwhere [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.\r\n\r\nIf the system is damped, [latex]\\underset{t\\to\\infty}{\\text{lim}}(c_1x_1(t)+c_2x_2(t))=0[\/latex]. Since these terms do not affect the long-term behavior of the system, we call this part of the solution the <strong>transient solution<\/strong>. The long-term behavior of the system is determined by [latex]x_p(t)[\/latex], so we call this part of the solution the <strong>steady-state solution<\/strong>.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>transient and steady-state solutions<\/h3>\r\n<p class=\"whitespace-normal break-words\">For a damped forced system with solution [latex]x(t)=c_1x_1(t)+c_2x_2(t)+x_p(t)[\/latex]:<\/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\">The <strong>transient solution<\/strong> is [latex]c_1x_1(t)+c_2x_2(t)[\/latex], which approaches zero as [latex]t\\to\\infty[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The <strong>steady-state solution<\/strong> is [latex]x_p(t)[\/latex], which determines the long-term behavior<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">A mass of\u00a0[latex]1[\/latex] slug stretches a spring\u00a0[latex]2[\/latex] ft and comes to rest at equilibrium. The system is attached to a dashpot that imparts a damping force equal to eight times the instantaneous velocity of the mass. Find the equation of motion if an external force equal to [latex]f(t)=8\\sin(4t)[\/latex] is applied to the system beginning at time [latex]t=0[\/latex]. What is the transient solution? What is the steady-state solution?[reveal-answer q=\"487492811\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"487492811\"]\r\n<p id=\"fs-id1170570978032\">We have [latex]mg=1(32)=2k[\/latex], so [latex]k=16[\/latex] and the differential equation is<\/p>\r\n<p style=\"text-align: center;\">[latex]x^{\\prime\\prime}+8^\\prime+16x=8\\sin(4t)[\/latex].<\/p>\r\n<p id=\"fs-id1170570995313\">The general solution to the complementary equation is<\/p>\r\n<p style=\"text-align: center;\">[latex]c_1e^{-4t}+c_2te^{-4t}[\/latex].<\/p>\r\n<p id=\"fs-id1170571227246\">Assuming a particular solution of the form [latex]x_p(t)=A\\cos(4t)+B\\sin(4t)[\/latex] and using the method of undetermined coefficients, we find [latex]x_p(t)=-\\frac14\\cos(4t)[\/latex], so<\/p>\r\n<p style=\"text-align: center;\">[latex]x(t)=c_1e^{-4t}+c_2te^{-4t}-\\frac14\\cos(4t)[\/latex].<\/p>\r\n<p id=\"fs-id1170571340436\">At [latex]t=0[\/latex], the mass is at rest in the equilibrium position, so [latex]x(0)=x^\\prime(0)=0[\/latex]. Applying these initial conditions to solve for [latex]c_1[\/latex] and [latex]c_2[\/latex], we get<\/p>\r\n<p style=\"text-align: center;\">[latex]x(t)=\\frac14e^{-4t}+te^{-4t}-\\frac14\\cos(4t)[\/latex].<\/p>\r\n<p id=\"fs-id1170573289060\">The transient solution is [latex]\\frac14e^{-4t}+te^{-4t}[\/latex]. The steady-state solution is [latex]-\\frac14\\cos(4t)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<h2 data-type=\"title\">Forced Vibrations<\/h2>\n<p class=\"whitespace-normal break-words\">The last case we consider is when an external force acts on the system. In the case of the motorcycle suspension system, for example, the bumps in the road act as an external force acting on the system. Another example is a spring hanging from a support; if the support is set in motion, that motion would be considered an external force on the system.<\/p>\n<p class=\"whitespace-normal break-words\">We model these forced systems with the nonhomogeneous differential equation:<\/p>\n<p style=\"text-align: center;\">[latex]mx^{\\prime\\prime}+bx^\\prime+kx=f(t)[\/latex],<\/p>\n<p id=\"fs-id1170573582925\">where the external force is represented by the [latex]f(t)[\/latex] term. As we saw earlier, differential equations such as this have solutions of the form:<\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]x(t)=c_1x_1(t)+c_2x_2(t)+x_p(t)[\/latex],<\/span><\/p>\n<p>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>If the system is damped, [latex]\\underset{t\\to\\infty}{\\text{lim}}(c_1x_1(t)+c_2x_2(t))=0[\/latex]. Since these terms do not affect the long-term behavior of the system, we call this part of the solution the <strong>transient solution<\/strong>. The long-term behavior of the system is determined by [latex]x_p(t)[\/latex], so we call this part of the solution the <strong>steady-state solution<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>transient and steady-state solutions<\/h3>\n<p class=\"whitespace-normal break-words\">For a damped forced system with solution [latex]x(t)=c_1x_1(t)+c_2x_2(t)+x_p(t)[\/latex]:<\/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\">The <strong>transient solution<\/strong> is [latex]c_1x_1(t)+c_2x_2(t)[\/latex], which approaches zero as [latex]t\\to\\infty[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">The <strong>steady-state solution<\/strong> is [latex]x_p(t)[\/latex], which determines the long-term behavior<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">A mass of\u00a0[latex]1[\/latex] slug stretches a spring\u00a0[latex]2[\/latex] ft and comes to rest at equilibrium. The system is attached to a dashpot that imparts a damping force equal to eight times the instantaneous velocity of the mass. Find the equation of motion if an external force equal to [latex]f(t)=8\\sin(4t)[\/latex] is applied to the system beginning at time [latex]t=0[\/latex]. 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=\"q487492811\">Show Solution<\/button><\/p>\n<div id=\"q487492811\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170570978032\">We have [latex]mg=1(32)=2k[\/latex], so [latex]k=16[\/latex] and the differential equation is<\/p>\n<p style=\"text-align: center;\">[latex]x^{\\prime\\prime}+8^\\prime+16x=8\\sin(4t)[\/latex].<\/p>\n<p id=\"fs-id1170570995313\">The general solution to the complementary equation is<\/p>\n<p style=\"text-align: center;\">[latex]c_1e^{-4t}+c_2te^{-4t}[\/latex].<\/p>\n<p id=\"fs-id1170571227246\">Assuming a particular solution of the form [latex]x_p(t)=A\\cos(4t)+B\\sin(4t)[\/latex] and using the method of undetermined coefficients, we find [latex]x_p(t)=-\\frac14\\cos(4t)[\/latex], so<\/p>\n<p style=\"text-align: center;\">[latex]x(t)=c_1e^{-4t}+c_2te^{-4t}-\\frac14\\cos(4t)[\/latex].<\/p>\n<p id=\"fs-id1170571340436\">At [latex]t=0[\/latex], the mass is at rest in the equilibrium position, so [latex]x(0)=x^\\prime(0)=0[\/latex]. Applying these initial conditions to solve for [latex]c_1[\/latex] and [latex]c_2[\/latex], we get<\/p>\n<p style=\"text-align: center;\">[latex]x(t)=\\frac14e^{-4t}+te^{-4t}-\\frac14\\cos(4t)[\/latex].<\/p>\n<p id=\"fs-id1170573289060\">The transient solution is [latex]\\frac14e^{-4t}+te^{-4t}[\/latex]. The steady-state solution is [latex]-\\frac14\\cos(4t)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":19,"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-1577","chapter","type-chapter","status-publish","hentry"],"part":440,"module-header":"- Select Header -","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\/1577","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":7,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1577\/revisions"}],"predecessor-version":[{"id":2158,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1577\/revisions\/2158"}],"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\/1577\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1577"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1577"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1577"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1577"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}