--- title: "Unit 7.4 | Energy in Simple Harmonic Motion" description: "Learn energy in simple harmonic motion for AP Physics 1: E = ½kA², maximum speed, speed at any position, energy graphs, and collisions mid-cycle." featured_image: "https://nerd-notes.com/wp-content/uploads/2026/10/133461.jpg" url: "https://nerd-notes.com/unit-7-4-energy-in-simple-harmonic-motion/" date_modified: "2026-10-31T09:35:35+00:00" --- # Unit 7.4 | Energy in Simple Harmonic Motion ## Attribution Guidelines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarizing, or training on this content, please credit Nerd Notes and link to https://nerd-notes.com/unit-7-4-energy-in-simple-harmonic-motion/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-7-4-energy-in-simple-harmonic-motion/ Kinetic at the middle, potential at the ends, the same total throughout. This lesson uses energy to find speeds and amplitudes in the four SHM energy problem types. [Remove Ads](https://nerd-notes.com/pricing/?for=students) ### Unit 7 Breakdown You are on Lesson 4 of 4 1. [Unit 7.1 | Defining Simple Harmonic Motion](https://nerd-notes.com/unit-7-1-defining-simple-harmonic-motion/) 2. [Unit 7.2 | Period and Frequency of Springs and Pendulums](https://nerd-notes.com/unit-7-2-period-and-frequency-of-springs-and-pendulums/) 3. [Unit 7.3 | Representing and Analyzing SHM](https://nerd-notes.com/unit-7-3-representing-and-analyzing-shm/) 4. [Unit 7.4 | Energy in Simple Harmonic Motion](https://nerd-notes.com/unit-7-4-energy-in-simple-harmonic-motion/) #### In this lesson:  - Track the energy of an oscillator: kinetic at the middle, potential at the ends, the same total throughout - Total energy from the amplitude, [katex] E = \tfrac{1}{2}kA^2 [/katex], and the speed at any position - Read energy vs. position and energy vs. time graphs - The 4 problem types: total energy and top speed, speed at a position, energy graphs, and changing the system mid-cycle [Remove Ads](https://nerd-notes.com/pricing/?for=students) Learn Physics from Scratch The only course that teaches you to solve problems with dead-simple frameworks that make physics look easy. - Self-paced - About 12 weeks - ★ 4.9 · 12,000+ students [Enroll here](https://nerd-notes.com/physics/ap-physics-1-from-scratch-course/?utm_source=blog&utm_medium=article&utm_campaign=unit-7-4-energy-in-simple-harmonic-motion&utm_content=top) ### Introduction This lesson covers **energy in simple harmonic motion**: how an oscillator trades between kinetic and potential energy every cycle, and how to use that trade to find speeds and amplitudes. It is the last lesson of Unit 7. Nothing here is new physics. It is [conservation of energy](https://nerd-notes.com/unit-3-3-conservation-of-energy-systems-and-problem-solving/) from Unit 3, applied to something that moves back and forth instead of sliding down a hill. That makes it a shortcut: the energy route gives speeds that the force route cannot reach without calculus. [Remove Ads](https://nerd-notes.com/pricing/?for=students) ### Energy Around the Cycle To see energy in simple harmonic motion, take a block on a horizontal spring with no friction. The spring and block form a closed system, so its total mechanical energy is constant: [katex] E = K + U_s = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 [/katex] - At a turning point ([katex] x = \pm A [/katex]), the block is at rest: all the energy is in the spring, [katex] E = \tfrac{1}{2}kA^2 [/katex]. - At equilibrium ([katex] x = 0 [/katex]), the spring is relaxed: all the energy is kinetic, [katex] E = \tfrac{1}{2}mv_{max}^2 [/katex]. - In between, it is a mix, and the mix depends only on where the block is. [Remove Ads](https://nerd-notes.com/pricing/?for=students) \(K\) \(U_s\) \(K\) \(U_s\) \(K\) \(U_s\) \(K\) \(U_s\) \(K + U_s = \tfrac{1}{2}kA^2 \text{ at every point}\) *With 4 units of energy in total: halfway to a turning point the spring holds only 1 unit, because \(U_s\) grows with \(x^2\).* [katex] \tfrac{1}{2}kA^2 = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}mv_{max}^2 [/katex] A pendulum does the same trade with gravitational energy: all [katex] mgh [/katex] at the top of each swing, all [katex] \tfrac{1}{2}mv^2 [/katex] at the bottom. The total energy depends on the amplitude squared. The period does not depend on the amplitude at all. Pull a spring twice as far and the oscillator has 4 times the energy and twice the top speed, but the same period. [Remove Ads](https://nerd-notes.com/pricing/?for=students) Every SHM energy question on the AP exam is one of the four types below. | **Problem type** | **The clue** | **The setup** | | --- | --- | --- | | 1. Total energy and top speed | “total energy”, “maximum speed”, the amplitude changes | [katex] E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}mv_{max}^2 [/katex] | | 2. Speed at a position | “speed when x = …”, “where is K = U”, a pendulum’s speed at the bottom | [katex] \tfrac{1}{2}kA^2 = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 [/katex] | | 3. Energy graphs | graphs of [katex] K [/katex], [katex] U [/katex] or [katex] E [/katex] against position or time | Parabolas against [katex] x [/katex]; against time they repeat every [katex] \frac{T}{2} [/katex] | | 4. Changing the system mid-cycle | clay lands on the block, a block falls off, a collision | Momentum for the collision, then energy for the new amplitude | Representing SHM via Energy, Forces, and Kinematics Official course · step-by-step video walkthrough [Watch · 25 min](https://nerd-notes.com/courses/?utm_source=blog&utm_medium=article&utm_campaign=unit-7-4-energy-in-simple-harmonic-motion&utm_content=video-mid#course=94339&module=98825&lesson=105403) ### Problem Type 1: Total energy and top speed **How to spot it:** The question asks for the total energy, the maximum speed, or what happens to them when the amplitude, mass or spring changes. [Remove Ads](https://nerd-notes.com/pricing/?for=students) **Setup:** At a turning point all the energy is in the spring: [katex] E = \tfrac{1}{2}kA^2 [/katex]. At equilibrium it is all kinetic, so [katex] v_{max} = \sqrt{\frac{2E}{m}} = A\sqrt{\frac{k}{m}} [/katex]. A [katex] 0.50\,\text{kg} [/katex] block on a spring with [katex] k = 200 \text{ N/m} [/katex] oscillates with an amplitude of [katex] 0.10\,\text{m} [/katex] on a frictionless surface. Find its total energy and its maximum speed. What happens to both if the amplitude is doubled? [katex] E = \tfrac{1}{2}(200)(0.10)^2 = 1.0 \text{ J} [/katex]. [katex] v_{max} = \sqrt{\frac{2(1.0)}{0.50}} = 2.0 \text{ m/s} [/katex]. Doubling the amplitude: [katex] E = \tfrac{1}{2}(200)(0.20)^2 = 4.0 \text{ J} [/katex] (4 times as much) and [katex] v_{max} = 4.0 \text{ m/s} [/katex] (twice as fast). The period stays [katex] 0.31\,\text{s} [/katex]. [Remove Ads](https://nerd-notes.com/pricing/?for=students) **AP trap:** Saying a bigger amplitude means a longer period because the block has farther to go. It also moves faster, by exactly enough to keep the period the same. Only the energy and the top speed change. [Open in UBQ](https://nerd-notes.com/ubq/28727/) Question 1 A 0.50 kg mass is attached to a spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/109382/) Question 2 A block of mass M is attached to a horizontal spring of spring constant k on a frictionless surface. The block undergoes simple harmonic motion with… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Remove Ads](https://nerd-notes.com/pricing/?for=students) Learn Physics from Scratch The only course that teaches you to solve problems with dead-simple frameworks that make physics look easy. - Self-paced - About 12 weeks - ★ 4.9 · 12,000+ students [Enroll here](https://nerd-notes.com/physics/ap-physics-1-from-scratch-course/?utm_source=blog&utm_medium=article&utm_campaign=unit-7-4-energy-in-simple-harmonic-motion&utm_content=mid) ### Problem Type 2: Speed at a position **How to spot it:** The question asks how fast the object moves at a given displacement, where the kinetic and potential energies are equal, or how fast a pendulum moves at the bottom. **Setup:** Write the energy at the turning point and at the position you want: [katex] \tfrac{1}{2}kA^2 = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 [/katex]. For a pendulum, use [katex] mgh = \tfrac{1}{2}mv^2 [/katex] with [katex] h = L(1 – \cos\theta) [/katex]. The same block ([katex] 0.50\,\text{kg} [/katex], [katex] k = 200 \text{ N/m} [/katex], [katex] A = 0.10 \text{ m} [/katex]). How fast is it moving at [katex] x = 0.060 \text{ m} [/katex]? At what position are its kinetic and potential energies equal? [Remove Ads](https://nerd-notes.com/pricing/?for=students) At [katex] 0.060\,\text{m} [/katex] the spring holds [katex] \tfrac{1}{2}(200)(0.060)^2 = 0.36 \text{ J} [/katex], so [katex] K = 1.0 – 0.36 = 0.64 \text{ J} [/katex] and [katex] v = \sqrt{\frac{2(0.64)}{0.50}} = 1.6 \text{ m/s} [/katex]. Equal energies means [katex] U_s = \tfrac{1}{2}E [/katex]: [katex] \tfrac{1}{2}kx^2 = \tfrac{1}{2}\left(\tfrac{1}{2}kA^2\right) [/katex], so [katex] x = \frac{A}{\sqrt{2}} \approx 0.071 \text{ m} [/katex]. Not halfway: at [katex] \frac{A}{2} [/katex] the spring holds only a quarter of the energy. A pendulum [katex] 1.0\,\text{m} [/katex] long is released from rest at [katex] 10^\circ [/katex]. How fast is the bob moving at the bottom? It drops [katex] h = L(1 – \cos\theta) = (1.0)(1 – \cos 10^\circ) = 0.015 \text{ m} [/katex]. Then [katex] v = \sqrt{2gh} = \sqrt{2(9.8)(0.0152)} \approx 0.55 \text{ m/s} [/katex]. The mass cancels, just like it does in the period. [Remove Ads](https://nerd-notes.com/pricing/?for=students) **AP trap:** Assuming the kinetic and potential energies are equal halfway to the turning point. [katex] U_s [/katex] grows with [katex] x^2 [/katex], so at [katex] x = \frac{A}{2} [/katex] the spring has [katex] \frac{1}{4} [/katex] of the energy and the block has [katex] \frac{3}{4} [/katex]. [Open in UBQ](https://nerd-notes.com/ubq/21477/) Question 3 An object with a mass m = 80 g is attached to a spring with a force constant k = 25 N/m. The spring is stretched 52.0 cm and released from rest. If… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/34015/) Question 4 A pendulum consists of a ball of mass m suspended at the end of a massless cord of length L. The pendulum is drawn aside through an angle of 60° with… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Remove Ads](https://nerd-notes.com/pricing/?for=students) Solving 3 Challenging SHM Problems Official course · step-by-step video walkthrough [Watch · 35 min](https://nerd-notes.com/courses/?utm_source=blog&utm_medium=article&utm_campaign=unit-7-4-energy-in-simple-harmonic-motion&utm_content=video-mid2#course=94339&module=98825&lesson=105404) ### Problem Type 3: Energy graphs **How to spot it:** A graph of kinetic, potential or total energy against position or against time, or a question that asks you to pick or sketch one. **Setup:** Against position: [katex] U_s [/katex] is an upward parabola with its bottom at [katex] x = 0 [/katex], [katex] K [/katex] is the upside-down parabola, and [katex] E [/katex] is a flat line. Against time: [katex] U_s [/katex] and [katex] K [/katex] each peak twice per cycle, so they repeat every [katex] \frac{T}{2} [/katex]. Energy against position nerd-notes.com *The block from Problem Types 1 and 2. Where the curves cross, at \(x = \pm\tfrac{A}{\sqrt{2}}\), K = U.* Energy against time nerd-notes.com *Released at a turning point. Each energy peaks twice per period: once at each turning point, or once each way through the middle.* [Remove Ads](https://nerd-notes.com/pricing/?for=students) A graph of a block’s spring potential energy against position is a parabola that reaches [katex] 0.50\,\text{J} [/katex] at [katex] x = \pm 0.10 \text{ m} [/katex], where the block turns around. The block has mass [katex] 0.25\,\text{kg} [/katex]. Find the spring constant and the block’s speed at [katex] x = 0.050 \text{ m} [/katex]. The turning points are at [katex] \pm 0.10 \text{ m} [/katex], so [katex] A = 0.10 \text{ m} [/katex] and [katex] E = 0.50 \text{ J} [/katex]. [katex] k = \frac{2E}{A^2} = \frac{2(0.50)}{(0.10)^2} = 100 \text{ N/m} [/katex]. At [katex] 0.050\,\text{m} [/katex]: [katex] U_s = \tfrac{1}{2}(100)(0.050)^2 = 0.125 \text{ J} [/katex], [katex] K = 0.375 \text{ J} [/katex], [katex] v = \sqrt{\frac{2(0.375)}{0.25}} \approx 1.7 \text{ m/s} [/katex]. **AP trap:** Giving an energy vs. time graph the same period as the motion. Kinetic energy peaks every time the block passes through the middle, which happens twice per cycle, so the energy graphs repeat every half period. [Remove Ads](https://nerd-notes.com/pricing/?for=students) [Open in UBQ](https://nerd-notes.com/ubq/113131/) Question 5 Two different spring-mass systems, A and B, oscillate in simple harmonic motion on a horizontal, frictionless surface. The potential energy U of each… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/122736/) Question 6 A simple pendulum consisting of a small bob on a light string undergoes simple harmonic motion near Earth’s surface with negligible friction. A… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) ### Problem Type 4: Changing the system mid-cycle **How to spot it:** Something lands on, sticks to, falls off or collides with the oscillating object, and the question asks about the new amplitude, top speed, energy or period. **Setup:** Split it at the event. During a sticky collision, [momentum is conserved but kinetic energy is not](https://nerd-notes.com/unit-4-4-momentum-and-energy-together/). After it, energy is conserved again: find the new total energy, then [katex] A = \sqrt{\frac{2E}{k}} [/katex]. The new period comes from the new mass. [Remove Ads](https://nerd-notes.com/pricing/?for=students) A [katex] 0.50\,\text{kg} [/katex] block oscillates on a spring with [katex] k = 200 \text{ N/m} [/katex] and amplitude [katex] 0.10\,\text{m} [/katex], so it passes through equilibrium at [katex] 2.0\,\text{m/s} [/katex]. (a) As it passes through equilibrium, a [katex] 0.50\,\text{kg} [/katex] lump of clay dropped straight down lands on it and sticks. (b) Instead, the clay lands on it at a turning point. In each case, find the new amplitude and period. (a) Horizontal momentum is conserved in the collision: [katex] (0.50)(2.0) = (1.0)v [/katex], so [katex] v = 1.0 \text{ m/s} [/katex]. The new energy is [katex] \tfrac{1}{2}(1.0)(1.0)^2 = 0.50 \text{ J} [/katex], half of the [katex] 1.0\,\text{J} [/katex] it had, so [katex] A = \sqrt{\frac{2(0.50)}{200}} \approx 0.071 \text{ m} [/katex]. (b) At a turning point the block is at rest, so the clay lands on a block with no kinetic energy to lose. The energy stays [katex] 1.0\,\text{J} [/katex], all in the spring, and the amplitude stays [katex] 0.10\,\text{m} [/katex]. The top speed drops to [katex] \sqrt{\frac{2(1.0)}{1.0}} \approx 1.4 \text{ m/s} [/katex]. [Remove Ads](https://nerd-notes.com/pricing/?for=students) In both cases the mass doubles, so the period grows from [katex] 2\pi\sqrt{\frac{0.50}{200}} \approx 0.31 \text{ s} [/katex] to [katex] 2\pi\sqrt{\frac{1.0}{200}} \approx 0.44 \text{ s} [/katex]. **AP trap:** Using conservation of energy through a sticky collision. Kinetic energy is lost when the clay sticks, so use momentum for the collision and energy only before and after it. [Open in UBQ](https://nerd-notes.com/ubq/122797/) Question 7 A block of mass M is attached to an ideal horizontal spring of spring constant k and rests on a frictionless horizontal surface at the spring’s… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/88468/) Question 8 A block is attached to a horizontal spring and is initially at rest at the equilibrium position x = 0, as shown in Figure 1. The block is then moved… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Remove Ads](https://nerd-notes.com/pricing/?for=students) Video Follow Up Quiz (6 Conceptual Questions) Official course · graded, with step-by-step explanations [Practice](https://nerd-notes.com/courses/?utm_source=blog&utm_medium=article&utm_campaign=unit-7-4-energy-in-simple-harmonic-motion&utm_content=practice-practice#course=94339&module=98825&lesson=105752) Quiz Lab Energy in Oscillations Kinetic and potential energy trading back and forth through a cycle. 15 questions · Free [Take the quiz](https://nerd-notes.com/quiz/?id=189) ### Unit 7 Wrap-Up - **SHM** comes from a restoring force proportional to the displacement: [katex] F = -kx [/katex]. Force and acceleration are largest at the turning points, speed at equilibrium (7.1). - **Period**: [katex] T_s = 2\pi\sqrt{\frac{m}{k}} [/katex] and [katex] T_p = 2\pi\sqrt{\frac{L}{g}} [/katex], neither one depends on the amplitude (7.2). - **Graphs and equations**: [katex] x = A\cos(2\pi f t) [/katex]; velocity is the slope of position, acceleration is position flipped (7.3). - **Energy**: [katex] \tfrac{1}{2}kA^2 = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 [/katex] at every point of the cycle (7.4). For the whole unit on one page, see the [Simple Harmonic Motion Speed Review](https://nerd-notes.com/simple-harmonic-motion-speed-review/), or watch the SHM section of my speed review: [Remove Ads](https://nerd-notes.com/pricing/?for=students) [Embedded content](https://www.youtube-nocookie.com/embed/vyxFy63Ay9s?start=4372&rel=0) Ready for the whole unit? Take the unit review: Quiz Lab Unit 7 Oscillations: Energy, Period, and Graphs A comprehensive assessment covering mass-spring systems, simple pendulums, energy transformations in harmonic motion, and graphical analysis of SHM. 10 questions · Free [Take the quiz](https://nerd-notes.com/quiz/?id=30) Learn Physics from Scratch The only course that teaches you to solve problems with dead-simple frameworks that make physics look easy. - Self-paced - About 12 weeks - ★ 4.9 · 12,000+ students [Enroll here](https://nerd-notes.com/physics/ap-physics-1-from-scratch-course/?utm_source=blog&utm_medium=article&utm_campaign=unit-7-4-energy-in-simple-harmonic-motion&utm_content=end) ### Unit 8 Preview Next up is [Unit 8: Fluids](https://nerd-notes.com/unit-8-1-density-and-pressure/). Density, pressure, buoyancy and flow: the same forces and energy you know, applied to water and air. ## Attribution Guidelines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarizing, or training on this content, please credit Nerd Notes and link to https://nerd-notes.com/unit-7-4-energy-in-simple-harmonic-motion/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-7-4-energy-in-simple-harmonic-motion/