{"id":3376,"date":"2026-08-23T14:53:41","date_gmt":"2026-08-23T14:53:41","guid":{"rendered":"https:\/\/us.allassignmentsupport.com\/blog\/?p=3376"},"modified":"2026-08-23T15:15:40","modified_gmt":"2026-08-23T15:15:40","slug":"simple-harmonic-motion-shm-concepts-formulas-and-worked-examples","status":"publish","type":"post","link":"https:\/\/us.allassignmentsupport.com\/blog\/simple-harmonic-motion-shm-concepts-formulas-and-worked-examples\/","title":{"rendered":"Simple Harmonic Motion (SHM): Concepts, Formulas, and Worked Examples"},"content":{"rendered":"<p dir=\"ltr\">Simple harmonic motion is where physics starts to feel less like &#8220;objects moving in straight lines or circles&#8221; and more like genuinely new territory \u2014 sine and cosine functions describing position, a restoring force that&#8217;s proportional to displacement, and energy sloshing back and forth between kinetic and potential forms. Students often understand the individual pieces but struggle to connect them into a single coherent picture, especially when a problem mixes concepts (asking for velocity at a specific position, say, rather than at a specific time). This guide builds that connected picture, with worked examples covering springs, pendulums, and energy in SHM.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_69_1 counter-hierarchy ez-toc-counter ez-toc-light-blue ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title \" >Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/simple-harmonic-motion-shm-concepts-formulas-and-worked-examples\/#What_Makes_Motion_%E2%80%9CSimple_Harmonic%E2%80%9D\" title=\"What Makes Motion &#8220;Simple Harmonic&#8221;\">What Makes Motion &#8220;Simple Harmonic&#8221;<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/simple-harmonic-motion-shm-concepts-formulas-and-worked-examples\/#The_Core_SHM_Equations\" title=\"The Core SHM Equations\">The Core SHM Equations<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/simple-harmonic-motion-shm-concepts-formulas-and-worked-examples\/#Worked_Example_1_Basic_Mass-Spring_System\" title=\"Worked Example 1: Basic Mass-Spring System\">Worked Example 1: Basic Mass-Spring System<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/simple-harmonic-motion-shm-concepts-formulas-and-worked-examples\/#Worked_Example_2_Finding_Velocity_at_a_Specific_Position_Not_Time\" title=\"Worked Example 2: Finding Velocity at a Specific Position (Not Time)\">Worked Example 2: Finding Velocity at a Specific Position (Not Time)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/simple-harmonic-motion-shm-concepts-formulas-and-worked-examples\/#Energy_in_Simple_Harmonic_Motion\" title=\"Energy in Simple Harmonic Motion\">Energy in Simple Harmonic Motion<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/simple-harmonic-motion-shm-concepts-formulas-and-worked-examples\/#The_Simple_Pendulum_A_Second_Common_SHM_System\" title=\"The Simple Pendulum: A Second Common SHM System\">The Simple Pendulum: A Second Common SHM System<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/simple-harmonic-motion-shm-concepts-formulas-and-worked-examples\/#A_Step-by-Step_Checklist_for_Students_Stuck_on_an_SHM_Problem\" title=\"A Step-by-Step Checklist for Students Stuck on an SHM Problem\">A Step-by-Step Checklist for Students Stuck on an SHM Problem<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/simple-harmonic-motion-shm-concepts-formulas-and-worked-examples\/#FAQs\" title=\"FAQs\">FAQs<\/a><\/li><\/ul><\/nav><\/div>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"What_Makes_Motion_%E2%80%9CSimple_Harmonic%E2%80%9D\"><\/span>What Makes Motion &#8220;Simple Harmonic&#8221;<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p dir=\"ltr\">Simple harmonic motion occurs whenever the restoring force on an object is directly proportional to its displacement from equilibrium, and points back toward equilibrium:<\/p>\n<p dir=\"ltr\"><strong>F = \u2212kx<\/strong><\/p>\n<p dir=\"ltr\">This is Hooke&#8217;s Law, and it&#8217;s the defining condition for SHM \u2014 a direct extension of the <strong><a href=\"https:\/\/us.allassignmentsupport.com\/blog\/newtons-laws-of-motion-a-problem-solving-guide-with-worked-examples\/\">force and acceleration relationship<\/a><\/strong> from Newton&#8217;s second law \u2014 any system obeying this relationship (not just springs) will oscillate sinusoidally. This is why a mass on a spring and a simple pendulum (for small angles) both qualify as SHM, even though they look like very different physical setups.<\/p>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"The_Core_SHM_Equations\"><\/span>The Core SHM Equations<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul dir=\"ltr\">\n<li><strong>Position:<\/strong> x(t) = A cos(\u03c9t + \u03c6), where A is amplitude, \u03c9 is angular frequency, and \u03c6 is the phase constant (determined by initial conditions)<\/li>\n<li><strong>Velocity:<\/strong> v(t) = \u2212A\u03c9 sin(\u03c9t + \u03c6)<\/li>\n<li><strong>Acceleration:<\/strong> a(t) = \u2212A\u03c9\u00b2 cos(\u03c9t + \u03c6) = \u2212\u03c9\u00b2x<\/li>\n<li><strong>Angular frequency (mass-spring system):<\/strong> \u03c9 = \u221a(k\/m)<\/li>\n<li><strong>Period:<\/strong> T = 2\u03c0\/\u03c9 = 2\u03c0\u221a(m\/k)<\/li>\n<li><strong>Frequency:<\/strong> f = 1\/T<\/li>\n<\/ul>\n<p dir=\"ltr\"><strong>A relationship worth memorizing directly, since it appears constantly in assignments:<\/strong> <strong>v_max = A\u03c9<\/strong> (maximum speed occurs at the equilibrium position, x = 0) <strong>a_max = A\u03c9\u00b2<\/strong> (maximum acceleration occurs at the extremes, x = \u00b1A)<\/p>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"Worked_Example_1_Basic_Mass-Spring_System\"><\/span>Worked Example 1: Basic Mass-Spring System<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p dir=\"ltr\"><strong>Problem:<\/strong> A 0.4 kg mass is attached to a spring with spring constant k = 100 N\/m. It is pulled 0.05 m from equilibrium and released from rest. Find (a) the angular frequency, (b) the period, and (c) the maximum speed.<\/p>\n<p dir=\"ltr\"><strong>Step 1 \u2014 Angular frequency:<\/strong> \u03c9 = \u221a(k\/m) = \u221a(100\/0.4) = \u221a250 = 15.8 rad\/s<\/p>\n<p dir=\"ltr\"><strong>Step 2 \u2014 Period:<\/strong> T = 2\u03c0\/\u03c9 = 2\u03c0\/15.8 = 0.398 s<\/p>\n<p dir=\"ltr\"><strong>Step 3 \u2014 Maximum speed<\/strong> (amplitude A = 0.05 m, since the mass is released from rest at its maximum displacement): v_max = A\u03c9 = (0.05)(15.8) = 0.79 m\/s<\/p>\n<p dir=\"ltr\"><strong>Answer:<\/strong> \u03c9 \u2248 15.8 rad\/s, T \u2248 0.398 s, v_max \u2248 0.79 m\/s.<\/p>\n<p dir=\"ltr\"><strong>Common mistake to avoid:<\/strong> Confusing the release displacement with something other than the amplitude. Since the mass is released <em>from rest<\/em>, the point of release is automatically the amplitude (the maximum displacement) \u2014 this wouldn&#8217;t be true if the mass were released with some initial velocity instead, which would require a slightly different approach to find A.<\/p>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"Worked_Example_2_Finding_Velocity_at_a_Specific_Position_Not_Time\"><\/span>Worked Example 2: Finding Velocity at a Specific Position (Not Time)<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p dir=\"ltr\">This is one of the most commonly assigned SHM problem types, and it&#8217;s where many students get stuck because they instinctively reach for the time-based velocity equation when the question doesn&#8217;t give a time at all.<\/p>\n<p dir=\"ltr\"><strong>Problem:<\/strong> Using the same spring system as above (A = 0.05 m, \u03c9 = 15.8 rad\/s), find the mass&#8217;s speed when it is 0.03 m from equilibrium.<\/p>\n<p dir=\"ltr\"><strong>Step 1 \u2014 Recognize this requires the position-velocity relationship<\/strong>, derived from conservation of energy (shown in the next section), rather than the time-based v(t) equation, since no time value is given:<\/p>\n<p dir=\"ltr\">v = \u03c9\u221a(A\u00b2 \u2212 x\u00b2)<\/p>\n<p dir=\"ltr\"><strong>Step 2 \u2014 Substitute values:<\/strong> v = 15.8\u221a(0.05\u00b2 \u2212 0.03\u00b2) = 15.8\u221a(0.0025 \u2212 0.0009) = 15.8\u221a0.0016 = 15.8(0.04) = 0.632 m\/s<\/p>\n<p dir=\"ltr\"><strong>Answer:<\/strong> The speed at x = 0.03 m is 0.632 m\/s.<\/p>\n<p dir=\"ltr\"><strong>Common mistake to avoid:<\/strong> Trying to first solve for t using x(t) = A cos(\u03c9t), then substituting that time into v(t) = \u2212A\u03c9 sin(\u03c9t). This works but is significantly more error-prone and time-consuming than using the direct position-velocity relationship shown above \u2014 recognizing when a problem gives position (not time) as the known variable is the key skill this example tests.<\/p>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"Energy_in_Simple_Harmonic_Motion\"><\/span>Energy in Simple Harmonic Motion<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p dir=\"ltr\">Total mechanical energy in SHM is constant and continuously exchanges between kinetic and potential energy:<\/p>\n<p dir=\"ltr\"><strong>E_total = \u00bdkA\u00b2 = \u00bdmv\u00b2 + \u00bdkx\u00b2<\/strong><\/p>\n<p dir=\"ltr\">This equation is exactly where the position-velocity relationship used above comes from \u2014 rearranging it for v gives v = \u03c9\u221a(A\u00b2 \u2212 x\u00b2), using the fact that \u03c9\u00b2 = k\/m.<\/p>\n<p dir=\"ltr\"><strong>Worked example 3: Using energy conservation directly<\/strong><\/p>\n<p dir=\"ltr\"><strong>Problem:<\/strong> A 0.25 kg mass on a spring (k = 60 N\/m) has an amplitude of 0.1 m. Find its kinetic energy and potential energy when it is at x = 0.06 m.<\/p>\n<p dir=\"ltr\"><strong>Step 1 \u2014 Total energy:<\/strong> E_total = \u00bdkA\u00b2 = \u00bd(60)(0.1\u00b2) = \u00bd(60)(0.01) = 0.3 J<\/p>\n<p dir=\"ltr\"><strong>Step 2 \u2014 Potential energy at x = 0.06 m:<\/strong> PE = \u00bdkx\u00b2 = \u00bd(60)(0.06\u00b2) = \u00bd(60)(0.0036) = 0.108 J<\/p>\n<p dir=\"ltr\"><strong>Step 3 \u2014 Kinetic energy (remaining energy):<\/strong> KE = E_total \u2212 PE = 0.3 \u2212 0.108 = 0.192 J<\/p>\n<p dir=\"ltr\"><strong>Answer:<\/strong> At x = 0.06 m, PE = 0.108 J and KE = 0.192 J.<\/p>\n<p dir=\"ltr\"><strong>Common mistake to avoid:<\/strong> Forgetting that total energy stays constant throughout the motion \u2014 a common error is recalculating total energy using the current position instead of the amplitude, which would incorrectly treat the current position as if it were the maximum displacement.<\/p>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"The_Simple_Pendulum_A_Second_Common_SHM_System\"><\/span>The Simple Pendulum: A Second Common SHM System<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p dir=\"ltr\">For small angles (typically under about 15\u00b0), a simple pendulum also exhibits SHM, with:<\/p>\n<p dir=\"ltr\"><strong>T = 2\u03c0\u221a(L\/g)<\/strong><\/p>\n<p dir=\"ltr\">where L is the pendulum&#8217;s length and g is gravitational acceleration. Notably, this period is independent of both mass and amplitude (for small angles) \u2014 a frequently tested conceptual point.<\/p>\n<p dir=\"ltr\"><strong>Worked example 4: Pendulum period<\/strong><\/p>\n<p dir=\"ltr\"><strong>Problem:<\/strong> Find the length of a simple pendulum that has a period of exactly 2 seconds on Earth (g = 9.8 m\/s\u00b2).<\/p>\n<p dir=\"ltr\"><strong>Step 1 \u2014 Rearrange the period formula for L:<\/strong> T = 2\u03c0\u221a(L\/g) T\u00b2 = 4\u03c0\u00b2(L\/g) L = gT\u00b2\/(4\u03c0\u00b2)<\/p>\n<p dir=\"ltr\"><strong>Step 2 \u2014 Substitute values:<\/strong> L = (9.8)(2\u00b2)\/(4\u03c0\u00b2) = (9.8)(4)\/(39.48) = 39.2\/39.48 = 0.993 m<\/p>\n<p dir=\"ltr\"><strong>Answer:<\/strong> The pendulum needs to be approximately 0.993 m (close to 1 m) long.<\/p>\n<p dir=\"ltr\"><strong>Common mistake to avoid:<\/strong> Assuming a heavier pendulum bob changes the period. Because mass doesn&#8217;t appear anywhere in the pendulum period formula, changing the bob&#8217;s mass alone has no effect on the period \u2014 only length (and, to a lesser degree at larger angles, amplitude) affects it, which is a common conceptual trap in exam-style questions.<\/p>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"A_Step-by-Step_Checklist_for_Students_Stuck_on_an_SHM_Problem\"><\/span>A Step-by-Step Checklist for Students Stuck on an SHM Problem<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ol dir=\"ltr\">\n<li>Identify what type of SHM system you&#8217;re dealing with (mass-spring or pendulum), since they use different formulas for angular frequency and period.<\/li>\n<li>Check whether the problem gives you a time value or a position value \u2014 this determines whether you need the time-based equations (x(t), v(t)) or the position-based energy relationship (v = \u03c9\u221a(A\u00b2 \u2212 x\u00b2)).<\/li>\n<li>Remember that amplitude A is the maximum displacement, and it only equals the release position if the object was released from rest.<\/li>\n<li>For energy problems, always calculate total energy using the amplitude (\u00bdkA\u00b2), since this stays constant throughout the motion, and use it to find KE or PE at any other position.<\/li>\n<li>For pendulum problems, double check whether the small-angle approximation applies, and remember mass does not affect the period.<\/li>\n<\/ol>\n<p>SHM problems can also appear alongside mechanics, energy, waves, and oscillations in broader university coursework. If you need support working through a complete physics assignment that combines these topics, you can explore our <a class=\"decorated-link\" href=\"https:\/\/us.allassignmentsupport.com\/physics-assignment-help\" target=\"_new\" rel=\"noopener\" data-start=\"520\" data-end=\"610\"><strong data-start=\"521\" data-end=\"548\">Physics Assignment Help<\/strong><\/a> service.<\/p>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"FAQs\"><\/span>FAQs<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p dir=\"ltr\"><strong>Q1: What&#8217;s the difference between period and angular frequency in SHM?<\/strong> Period (T) is the time for one complete oscillation, measured in seconds, while angular frequency (\u03c9) describes how quickly the phase of the oscillation advances, measured in radians per second. They&#8217;re related by \u03c9 = 2\u03c0\/T, and angular frequency is what appears directly inside the sine and cosine functions describing position, velocity, and acceleration.<\/p>\n<p dir=\"ltr\"><strong>Q2: Why does maximum speed occur at the equilibrium position, not at the extremes?<\/strong> At the extremes (x = \u00b1A), the object is momentarily at rest as it changes direction, so velocity is zero there. As the object moves back toward equilibrium, potential energy converts into kinetic energy, and speed increases continuously until reaching a maximum exactly at x = 0, where all the system&#8217;s energy is kinetic.<\/p>\n<p dir=\"ltr\"><strong>Q3: How do I know when to use v(t) versus v = \u03c9\u221a(A\u00b2 \u2212 x\u00b2)?<\/strong> Use v(t) = \u2212A\u03c9 sin(\u03c9t + \u03c6) when the problem gives you a specific time and asks for velocity at that time. Use v = \u03c9\u221a(A\u00b2 \u2212 x\u00b2) when the problem gives you a specific position and asks for velocity at that position \u2014 this second equation comes directly from energy conservation and avoids needing to solve for time first.<\/p>\n<p dir=\"ltr\"><strong>Q4: Does a pendulum&#8217;s amplitude affect its period?<\/strong> For the small-angle approximation used in most introductory SHM (typically under about 15\u00b0), amplitude does not significantly affect the period. At larger angles, this approximation breaks down and the period does begin to depend slightly on amplitude, though this more complex case is generally beyond the scope of introductory SHM coverage.<\/p>\n<p dir=\"ltr\"><strong>Q5: Is all oscillating motion simple harmonic motion?<\/strong> No \u2014 SHM specifically requires the restoring force to be directly proportional to displacement (F = \u2212kx). Many real oscillating systems only approximate this behavior under certain conditions (like a pendulum at small angles) and deviate from true SHM at larger displacements or under more complex force conditions, which is why the small-angle qualifier matters for pendulum problems specifically.<\/p>\n<p dir=\"ltr\"><strong>Q6: How does SHM connect to other topics you&#8217;ll study?<\/strong> The energy-exchange logic here is the exact same one introduced in <strong><a href=\"https:\/\/us.allassignmentsupport.com\/blog\/work-energy-and-power-in-physics-formulas-and-worked-examples\/\">Work, Energy, and Power in Physics<\/a><\/strong> \u2014 SHM just applies it to a system where kinetic and potential energy trade back and forth continuously rather than only once. If your course moves on to sound or light, the same sinusoidal position function x(t) = A cos(\u03c9t + \u03c6) reappears as the building block for <strong><a href=\"https:\/\/us.allassignmentsupport.com\/blog\/wave-interference-and-diffraction-understanding-youngs-double-slit-experiment\/\">Wave Interference and Diffraction<\/a><\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Simple harmonic motion is where physics starts to feel less like &#8220;objects moving in straight lines or circles&#8221; and more [&hellip;]<\/p>\n","protected":false},"author":2,"featured_media":3381,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_robots_primary_cat":"none","_seopress_titles_title":"Simple Harmonic Motion (SHM): Concepts, Formulas, and Worked Examples","_seopress_titles_desc":"A university-level guide to simple harmonic motion, covering springs, pendulums, energy in SHM, and fully worked examples for physics students struggling with oscillation problems.","_seopress_robots_index":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center 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