{"id":18808,"date":"2025-06-13T09:33:11","date_gmt":"2025-06-13T09:33:11","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=18808"},"modified":"2025-06-13T09:33:14","modified_gmt":"2025-06-13T09:33:14","slug":"a-harmonic-oscillator-is-described-by-the-function-xt-0-500-m-cos0-290t","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-harmonic-oscillator-is-described-by-the-function-xt-0-500-m-cos0-290t\/","title":{"rendered":"A harmonic oscillator is described by the function x(t) = (0.500 m) cos(0.290t)"},"content":{"rendered":"\n<p>A harmonic oscillator is described by the function x(t) = (0.500 m) cos(0.290t). Find the oscillator&#8217;s maximum velocity and maximum acceleration. Find the oscillator&#8217;s position, velocity, and acceleration when t = 1.50 s.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p>To solve this harmonic oscillator problem, we begin with the given displacement function: x(t)=(0.500\u2009m)cos\u2061(0.290t)x(t) = (0.500\\, \\text{m}) \\cos(0.290t)<\/p>\n\n\n\n<p>This function represents simple harmonic motion (SHM) of the form: x(t)=Acos\u2061(\u03c9t)x(t) = A \\cos(\\omega t)<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A=0.500\u2009mA = 0.500 \\, \\text{m} is the amplitude,<\/li>\n\n\n\n<li>\u03c9=0.290\u2009rad\/s\\omega = 0.290 \\, \\text{rad\/s} is the angular frequency.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udd39 <strong>Maximum Velocity and Acceleration<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Maximum velocity<\/strong> is given by:<\/li>\n<\/ol>\n\n\n\n<p>vmax=A\u03c9=(0.500)(0.290)=0.145\u2009m\/sv_{\\text{max}} = A\\omega = (0.500)(0.290) = 0.145 \\, \\text{m\/s}<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li><strong>Maximum acceleration<\/strong> is given by:<\/li>\n<\/ol>\n\n\n\n<p>amax=A\u03c92=(0.500)(0.290)2=0.500\u00d70.0841=0.04205\u2009m\/s2a_{\\text{max}} = A\\omega^2 = (0.500)(0.290)^2 = 0.500 \\times 0.0841 = 0.04205 \\, \\text{m\/s}^2<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udd39 <strong>At Time t=1.50\u2009st = 1.50\\, \\text{s}<\/strong><\/h3>\n\n\n\n<p>We differentiate x(t)x(t) to find velocity and acceleration:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Position<\/strong>:<\/li>\n<\/ul>\n\n\n\n<p>x(1.50)=0.500cos\u2061(0.290\u00d71.50)=0.500cos\u2061(0.435)\u22480.500\u00d70.906=0.453\u2009mx(1.50) = 0.500 \\cos(0.290 \\times 1.50) = 0.500 \\cos(0.435) \\approx 0.500 \\times 0.906 = 0.453\\, \\text{m}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Velocity<\/strong> (first derivative):<\/li>\n<\/ul>\n\n\n\n<p>v(t)=dxdt=\u2212A\u03c9sin\u2061(\u03c9t)v(t) = \\frac{dx}{dt} = -A\\omega \\sin(\\omega t) v(1.50)=\u22120.500\u00d70.290\u00d7sin\u2061(0.435)\u2248\u22120.145\u00d70.421=\u22120.061\u2009m\/sv(1.50) = -0.500 \\times 0.290 \\times \\sin(0.435) \\approx -0.145 \\times 0.421 = -0.061\\, \\text{m\/s}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Acceleration<\/strong> (second derivative):<\/li>\n<\/ul>\n\n\n\n<p>a(t)=d2xdt2=\u2212A\u03c92cos\u2061(\u03c9t)a(t) = \\frac{d^2x}{dt^2} = -A\\omega^2 \\cos(\\omega t) a(1.50)=\u22120.500\u00d7(0.290)2\u00d7cos\u2061(0.435)=\u22120.04205\u00d70.906=\u22120.0381\u2009m\/s2a(1.50) = -0.500 \\times (0.290)^2 \\times \\cos(0.435) = -0.04205 \\times 0.906 = -0.0381\\, \\text{m\/s}^2<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Final Answers<\/strong>:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Maximum velocity<\/strong>: 0.145\u2009m\/s0.145 \\, \\text{m\/s}<\/li>\n\n\n\n<li><strong>Maximum acceleration<\/strong>: 0.04205\u2009m\/s20.04205 \\, \\text{m\/s}^2<\/li>\n\n\n\n<li><strong>At t=1.50\u2009st = 1.50\\, \\text{s}<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Position: x=0.453\u2009mx = 0.453 \\, \\text{m}<\/li>\n\n\n\n<li>Velocity: v=\u22120.061\u2009m\/sv = -0.061 \\, \\text{m\/s}<\/li>\n\n\n\n<li>Acceleration: a=\u22120.0381\u2009m\/s2a = -0.0381 \\, \\text{m\/s}^2<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83e\udde0 <strong>Explanation (Conceptual)<\/strong><\/h3>\n\n\n\n<p>In SHM, the object oscillates back and forth around an equilibrium point. The cosine function implies it starts at maximum displacement. Velocity is greatest as the mass passes through equilibrium, and acceleration is greatest at maximum displacement. These values depend on both amplitude and angular frequency. By applying calculus to the position function, we obtained velocity and acceleration at any time, such as t=1.50\u2009st = 1.50\\, \\text{s}.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A harmonic oscillator is described by the function x(t) = (0.500 m) cos(0.290t). Find the oscillator&#8217;s maximum velocity and maximum acceleration. Find the oscillator&#8217;s position, velocity, and acceleration when t = 1.50 s. The correct answer and explanation is: To solve this harmonic oscillator problem, we begin with the given displacement function: x(t)=(0.500\u2009m)cos\u2061(0.290t)x(t) = (0.500\\, [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-18808","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18808","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=18808"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18808\/revisions"}],"predecessor-version":[{"id":18809,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18808\/revisions\/18809"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=18808"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=18808"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=18808"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}