--- title: "Unit 7.1 | Defining Simple Harmonic Motion" description: "Learn simple harmonic motion for AP Physics 1: restoring force F = -kx, where force and speed peak, vertical springs, and pendulum forces, with practice." featured_image: "https://nerd-notes.com/wp-content/uploads/2023/06/12670.jpg" url: "https://nerd-notes.com/unit-7-1-defining-simple-harmonic-motion/" date_modified: "2026-10-31T09:33:51+00:00" --- # Unit 7.1 | Defining 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-7-1-defining-simple-harmonic-motion/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-7-1-defining-simple-harmonic-motion/ A restoring force proportional to the displacement is all it takes. This lesson shows what makes motion simple harmonic and solves the four SHM definition problem types. [Remove Ads](https://nerd-notes.com/pricing/?for=students) ### Unit 7 Breakdown You are on Lesson 1 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:  - What makes a motion **simple harmonic**: a restoring force proportional to the displacement, [katex] F = -kx [/katex] - Equilibrium, amplitude, and where the force, acceleration and speed are largest - Why a vertical spring and a small-angle pendulum are both simple harmonic oscillators - The 4 problem types: is it SHM, force and acceleration at a point, vertical springs, and pendulum forces [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-7-1-defining-simple-harmonic-motion&utm_content=top) ### Introduction This lesson covers **simple harmonic motion** (SHM): what it is, what causes it, and how to recognize it. It starts Unit 7, oscillations. A mass on a spring, a child on a swing, a guitar string and a bridge swaying in the wind all move back and forth around a resting point. You already have the tools to explain why: the [spring force](https://nerd-notes.com/unit-2-2-common-forces-in-physics/) from Unit 2 and [Newton’s second law](https://nerd-notes.com/unit-2-1-the-forces-that-move-us-unpacking-dynamics-in-ap-physics-1/). This lesson puts them together. [Remove Ads](https://nerd-notes.com/pricing/?for=students) ### Equilibrium, Displacement and Amplitude Every oscillator has an **equilibrium position**, the point where the net force on it is zero. Pull it away and let go, and it moves back and forth around that point. - **Displacement** [katex] x [/katex]: how far the object is from equilibrium right now, with a sign (one side positive, the other negative). - **Amplitude** [katex] A [/katex]: the largest displacement, the distance from equilibrium to either turning point. It is always positive. - **Period** [katex] T [/katex]: the time for one full cycle, out and back to where it started, moving the same way. [Remove Ads](https://nerd-notes.com/pricing/?for=students) A full cycle is *out and back*. Swinging from one side to the other is only half a cycle. ### The Restoring Force An oscillator always feels a force pointing back toward equilibrium. That is the **restoring force**. For a spring it is Hooke’s law: [katex] F_s = -kx [/katex] The minus sign is the whole idea. Stretch the spring ([katex] x > 0 [/katex]) and the force points the negative way. Compress it ([katex] x < 0 [/katex]) and the force points the positive way. Either way it pulls the object back toward [katex] x = 0 [/katex]. \(x = -A\) \(x = 0\) \(x = +A\) \(m\) \(F = -kA\) \(m\) \(v_{max}\) \(m\) \(F = +kA\) \(v = 0,\; |a| \text{ max}\) \(F = 0,\; a = 0,\; v \text{ max}\) \(v = 0,\; |a| \text{ max}\) *The force always points back to equilibrium, and it is largest at the turning points. At equilibrium the force is zero but the speed is largest.* [Remove Ads](https://nerd-notes.com/pricing/?for=students) #### The definition of SHM Put the restoring force into Newton’s second law, [katex] ma = -kx [/katex]: [katex] a = -\frac{k}{m}\,x [/katex] **Simple harmonic motion** is motion where the restoring force, and so the acceleration, is proportional to the displacement and points the opposite way. If doubling the displacement doubles the restoring force, the motion is SHM. This one equation tells you a lot without any numbers: - At the turning points ([katex] x = \pm A [/katex]), the object stops for an instant: [katex] v = 0 [/katex]. But the force and acceleration are at their **largest**, [katex] kA [/katex] and [katex] \frac{k}{m}A [/katex]. - At equilibrium ([katex] x = 0 [/katex]), the force and acceleration are **zero**, and the speed is at its **largest**. The object coasts through. - The acceleration always points toward equilibrium. So the object speeds up as it moves toward the middle and slows down as it moves away. [Remove Ads](https://nerd-notes.com/pricing/?for=students) Every SHM definition question on the AP exam is one of the four types below. | **Problem type** | **The clue** | **The setup** | | --- | --- | --- | | 1. Is it SHM? | a force law, a force vs. position graph, or “which of these is simple harmonic” | Restoring and proportional to [katex] x [/katex]: [katex] F = -kx [/katex] | | 2. Force and acceleration at a point | “at the turning point”, “at x = …”, “where is the speed greatest” | [katex] F = -kx [/katex], [katex] a = -\frac{k}{m}x [/katex] | | 3. Vertical springs | a mass hangs from a spring and bobs up and down | Equilibrium stretch [katex] d = \frac{mg}{k} [/katex], then [katex] x [/katex] from there | | 4. Pendulum forces | a pendulum bob at an angle, “restoring force”, “small angles” | Restoring force [katex] mg\sin\theta [/katex], SHM only for small angles | Simple Harmonic Motion Official course · step-by-step video walkthrough [Watch · 19 min](https://nerd-notes.com/courses/?utm_source=blog&utm_medium=article&utm_campaign=unit-7-1-defining-simple-harmonic-motion&utm_content=video-mid#course=94339&module=98825&lesson=105401) ### Problem Type 1: Is it simple harmonic motion? **How to spot it:** The question gives a force law, a graph of force against position, or several motions and asks which one is simple harmonic. [Remove Ads](https://nerd-notes.com/pricing/?for=students) **Setup:** Check two things. Does the force point back toward equilibrium (restoring)? Is it proportional to the displacement? On a force vs. position graph, SHM is a straight line through the origin with a negative slope, and the slope is [katex] -k [/katex]. Two restoring forces nerd-notes.com *Both pull back toward x = 0, but only the straight line is proportional to x. That one gives simple harmonic motion.* An object moves under one of three forces: (a) [katex] F = -20x [/katex], (b) [katex] F = -5x^3 [/katex], (c) [katex] F = +8x [/katex], with [katex] F [/katex] in newtons and [katex] x [/katex] in meters. Which give simple harmonic motion? Only (a). It points back toward equilibrium and is proportional to [katex] x [/katex], like a spring with [katex] k = 20 \text{ N/m} [/katex]. Force (b) is restoring, so the object still oscillates, but doubling [katex] x [/katex] makes the force 8 times bigger, not 2 times, so it is not SHM. Force (c) points *away* from equilibrium: the object never comes back. [Remove Ads](https://nerd-notes.com/pricing/?for=students) Plenty of motions repeat without being SHM. A ball bouncing on the floor repeats, but while it is in the air the only force is its constant weight, which does not depend on how high it is. **AP trap:** Calling every repeating motion simple harmonic. Periodic is not enough: the restoring force has to be proportional to the displacement. [Open in UBQ](https://nerd-notes.com/ubq/105734/) Question 1 What is the defining characteristic of the restoring force that causes an object to undergo Simple Harmonic Motion (SHM)? 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/122765/) Question 2 An object of mass m on a horizontal frictionless surface is attached to a non-linear spring that exerts a restoring force of magnitude F = bx³ on the… 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-7-1-defining-simple-harmonic-motion&utm_content=mid) ### Problem Type 2: Force and acceleration at a point in the cycle **How to spot it:** The question names a point (“at the turning point”, “at x = [katex] 0.05\,\text{m} [/katex]”, “halfway to equilibrium”) and asks for the force, acceleration or speed there, or asks where one of them is largest or zero. **Setup:** Use [katex] F = -kx [/katex] and [katex] a = -\frac{k}{m}x [/katex] with the displacement from equilibrium. Both are largest at [katex] x = \pm A [/katex] and zero at [katex] x = 0 [/katex]. Speed does the opposite. A [katex] 0.50\,\text{kg} [/katex] block on a frictionless table is attached to a spring with [katex] k = 200 \text{ N/m} [/katex]. It is pulled [katex] 0.10\,\text{m} [/katex] to the right and released. Find the largest force and acceleration on the block, and the force and acceleration when it is [katex] 0.05\,\text{m} [/katex] to the left of equilibrium. [Remove Ads](https://nerd-notes.com/pricing/?for=students) The largest force is at the turning points: [katex] F_{max} = kA = (200)(0.10) = 20 \text{ N} [/katex], so [katex] a_{max} = \frac{20}{0.50} = 40 \text{ m/s}^2 [/katex]. At [katex] x = -0.05 \text{ m} [/katex]: [katex] F = -(200)(-0.05) = +10 \text{ N} [/katex] and [katex] a = \frac{10}{0.50} = +20 \text{ m/s}^2 [/katex], pointing right, back toward equilibrium. Half the displacement, half the force. **AP trap:** Saying the acceleration is zero at the turning points because the block stops there. The velocity is zero for an instant, but the spring is stretched the most, so the force and acceleration are at their largest. Zero velocity does not mean zero acceleration. [Open in UBQ](https://nerd-notes.com/ubq/39141/) Question 3 A block attached to a spring undergoes simple harmonic motion. The acceleration of the block has its maximum magnitude at the point where: 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/105739/) Question 4 A mass–spring system is oscillating in simple harmonic motion. At the exact moment the mass passes through its equilibrium position, which of the… 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) ### Problem Type 3: Vertical springs **How to spot it:** A mass hangs from a spring, or sits on top of one, and bobs up and down. **Setup:** Gravity only moves the equilibrium. Find the new equilibrium from [katex] kd = mg [/katex], then measure [katex] x [/katex] from there: the net force is still [katex] -kx [/katex], with the same [katex] k [/katex]. \(m\) \(d\) \(kd = mg\) \(m\) \(A\) \(A\) \(x = 0\) \(\text{Measure } x \text{ from the new equilibrium: } F_{net} = -kx\) *Hanging the mass stretches the spring by \(d = \tfrac{mg}{k}\). Pulled further and released, it oscillates around that new equilibrium, not around the natural length.* A [katex] 0.40\,\text{kg} [/katex] mass hangs from a spring with [katex] k = 98 \text{ N/m} [/katex]. How far does it stretch the spring at equilibrium? The mass is then pulled down another [katex] 3.0\,\text{cm} [/katex] and released. Find the amplitude, and the net force on the mass at its lowest point. [Remove Ads](https://nerd-notes.com/pricing/?for=students) Equilibrium: [katex] d = \frac{mg}{k} = \frac{(0.40)(9.8)}{98} = 0.040 \text{ m} [/katex], or [katex] 4.0\,\text{cm} [/katex]. It is released from rest [katex] 3.0\,\text{cm} [/katex] below that point, so the amplitude is [katex] 3.0\,\text{cm} [/katex]: it moves between [katex] 1.0\,\text{cm} [/katex] and [katex] 7.0\,\text{cm} [/katex] below the natural length. At the lowest point the spring is stretched [katex] 7.0\,\text{cm} [/katex], so it pulls up with [katex] (98)(0.070) = 6.86 \text{ N} [/katex], while gravity pulls down with [katex] (0.40)(9.8) = 3.92 \text{ N} [/katex]. Net force: [katex] 2.94 \text{ N} [/katex] upward. The shortcut gives the same answer: [katex] kA = (98)(0.030) = 2.94 \text{ N} [/katex]. ▶ **Watch:** a [katex] 1.0\,\text{kg} [/katex] block on a [katex] 50\,\text{N/m} [/katex] spring that settles [katex] 0.20\,\text{m} [/katex] lower, right at its equilibrium, and the energy change of the block-spring-Earth system (starts at 14:48). [Remove Ads](https://nerd-notes.com/pricing/?for=students) [Embedded content](https://www.youtube-nocookie.com/embed/6ItRgoQqVNM?start=888&rel=0) **AP trap:** Measuring the amplitude from the spring’s natural length. A hanging mass oscillates around its new equilibrium, so the amplitude is how far it was pulled from *there*. [Open in UBQ](https://nerd-notes.com/ubq/109389/) Question 5 A block of mass m is suspended from an ideal vertical spring. The block is pulled downward from its equilibrium position and released from rest… 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/111604/) Question 6 A block of mass m is attached to the lower end of a vertical ideal spring with spring constant k. The upper end of the spring is fixed to a ceiling… 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) ### Problem Type 4: Pendulum forces **How to spot it:** A pendulum bob is pulled to an angle. The question asks for the restoring force, the force at the bottom, or when a pendulum counts as SHM. **Setup:** The restoring force is the part of the weight along the arc, [katex] mg\sin\theta [/katex]. For small angles [katex] \sin\theta \approx \theta [/katex] (in radians), so the restoring force is proportional to the displacement and the motion is close to SHM. \(\theta\) \(L\) \(mg\) \(mg\cos\theta\) \(mg\sin\theta\) \(T\) \(\text{restoring force} = mg\sin\theta\) \(\text{small angle: } \sin\theta \approx \theta = \tfrac{s}{L}\) \(F \approx -\tfrac{mg}{L}\,s \;\Rightarrow\; \text{SHM}\) *Tension points along the string and does not pull the bob back. The part of the weight along the arc, \(mg\sin\theta\), does.* [Remove Ads](https://nerd-notes.com/pricing/?for=students) The tension and the part of the weight along the string, [katex] mg\cos\theta [/katex], point along the string, so they cannot push the bob along its arc. Only [katex] mg\sin\theta [/katex] does. With [katex] \theta = \frac{s}{L} [/katex] for an arc length [katex] s [/katex]: [katex] F \approx -mg\,\theta = -\frac{mg}{L}\,s [/katex] That is Hooke’s law with [katex] \frac{mg}{L} [/katex] in place of [katex] k [/katex]. A pendulum at small angles is a simple harmonic oscillator. A [katex] 2.0\,\text{kg} [/katex] bob hangs on a [katex] 1.0\,\text{m} [/katex] string. Find the restoring force when it is pulled to [katex] 10^\circ [/katex], and compare it with the small-angle estimate. Then do the same at [katex] 60^\circ [/katex]. At [katex] 10^\circ [/katex]: [katex] mg\sin\theta = (2.0)(9.8)\sin 10^\circ = 3.40 \text{ N} [/katex]. The estimate uses the angle in radians, [katex] 10^\circ = 0.1745 \text{ rad} [/katex]: [katex] (2.0)(9.8)(0.1745) = 3.42 \text{ N} [/katex], within 1%. At [katex] 60^\circ [/katex]: [katex] (19.6)\sin 60^\circ = 17.0 \text{ N} [/katex], but [katex] (19.6)(1.047) = 20.5 \text{ N} [/katex], about 21% too big. At [katex] 60^\circ [/katex] the pendulum still swings, but its motion is not simple harmonic. [Remove Ads](https://nerd-notes.com/pricing/?for=students) **AP trap:** Saying the net force on the bob is zero at the bottom of the swing. The restoring force is zero there, but the bob is moving in a circle, so the tension is bigger than [katex] mg [/katex] and the net force points up, toward the pivot. That is the [centripetal force](https://nerd-notes.com/unit-3-1-understanding-circular-motion-centripetal-forces/). [Open in UBQ](https://nerd-notes.com/ubq/105742/) Question 7 When can the motion of a pendulum be modeled as simple harmonic motion? 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/111610/) Question 8 A small sphere of mass m is attached to a string of length L and displaced to an angle θ = 30° from the vertical, as shown in the figure. The sphere… 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-7-1-defining-simple-harmonic-motion&utm_content=practice-practice#course=94339&module=98825&lesson=105752) Quiz Lab Simple Harmonic Motion What makes motion simple harmonic, and how to read its graphs. 15 questions · Free [Take the quiz](https://nerd-notes.com/quiz/?id=187) 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-7-1-defining-simple-harmonic-motion&utm_content=end) ### Lesson 7.2 Preview Now you know what SHM is. In the [next lesson](https://nerd-notes.com/unit-7-2-period-and-frequency-of-springs-and-pendulums/) we time it: the **period** of a mass on a spring and of a pendulum, and the surprising list of things that do not change it. ## 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-7-1-defining-simple-harmonic-motion/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-7-1-defining-simple-harmonic-motion/