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How does energy move between kinetic and potential forms in an oscillator, and how does the total energy depend on amplitude?

Topic 7.4 Energy of Simple Harmonic Oscillators: analyze the interchange of kinetic and elastic potential energy in an oscillator and relate the total energy to the amplitude.

A focused answer to AP Physics 1 Topic 7.4, covering the continuous interchange of kinetic and elastic potential energy in SHM, the conservation of total mechanical energy, the relation E = half k A squared, and how the total energy scales with the square of the amplitude, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

In a frictionless oscillator, kinetic and elastic potential energy continuously interchange while the total mechanical energy stays constant: $E = \tfrac{1}{2}kx^2 + \tfrac{1}{2}mv^2 = \text{constant}$ . At the turning points ( $x = \pm A$ ) the energy is entirely potential, $E = \tfrac{1}{2}kA^2$ ; at equilibrium ( $x = 0$ ) it is entirely kinetic, $E = \tfrac{1}{2}mv_{max}^2$ , where the speed is greatest. Equating these gives the maximum speed. The total energy depends on the square of the amplitude: $E = \tfrac{1}{2}kA^2$ , so doubling the amplitude quadruples the energy. The speed at any displacement follows from splitting the fixed total between the potential at that $x$ and the remaining kinetic energy.

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What this topic is asking
Energy interchange in an oscillator
The two extremes set the energy
Energy scales with amplitude squared
Try this

What this topic is asking

The College Board (Topic 7.4) wants you to analyze the energy of a simple harmonic oscillator: the continuous interchange between kinetic and elastic potential energy, the conservation of the total mechanical energy, and the relation $E = \tfrac{1}{2}kA^2$ that ties the total energy to the amplitude. This applies the energy methods of Unit 3 to oscillation.

Energy interchange in an oscillator

This is the conservation-of-energy idea of Topic 3.4 applied to oscillation. As the block moves outward, it slows and stores energy in the spring (kinetic to potential); as it moves back in, the spring releases energy and the block speeds up (potential to kinetic). The total stays fixed throughout, so knowing it at any one point fixes it everywhere.

The two extremes set the energy

These two snapshots are the workhorses of energy problems in SHM. The total energy is most easily found at a turning point, where it is purely potential, $\tfrac{1}{2}kA^2$ . The maximum speed is most easily found at equilibrium, where that same total is purely kinetic. Equating the two values links amplitude to maximum speed without any reference to time.

Energy scales with amplitude squared

The relation $E = \tfrac{1}{2}kA^2$ shows that the total energy depends on the square of the amplitude. Doubling the amplitude quadruples the energy; tripling it multiplies the energy by nine. This is the most-tested feature of the topic and mirrors the way elastic potential energy and kinetic energy both depend on squared quantities. To find the speed at an intermediate displacement, split the fixed total: the potential energy there is $\tfrac{1}{2}kx^2$ , so the kinetic energy is the remainder, $E - \tfrac{1}{2}kx^2$ , and the speed follows from $\tfrac{1}{2}mv^2$ . The strategic insight is that energy methods let you find speeds at any position without solving the time-dependent equations of Topic 7.3: you simply track how the conserved total is partitioned. This connects the oscillator back to the broader energy framework, where the spring stores potential energy (Topic 3.3), the moving mass carries kinetic energy (Topic 3.1), and the total is conserved when no friction is present (Topic 3.4). Energy is the most efficient lens for "how fast is it moving here" questions in oscillation.

Maximum speed and speed at a point

A $0.25$ kg block oscillates on a spring of constant $100$ N/m with amplitude $0.080$ m on a frictionless surface. Find the total energy, the maximum speed, and the speed at $x = 0.040$ m.

step 1 Total mechanical energy

At the amplitude all the energy is elastic potential: $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(100)(0.080)^2 = \tfrac{1}{2}(100)(0.0064) = 0.32$ J.

step 2 Maximum speed at equilibrium

There all $0.32$ J is kinetic: $\tfrac{1}{2}(0.25)v_{max}^2 = 0.32$ , so $v_{max}^2 = 2(0.32)/0.25 = 2.56$ and $v_{max} = 1.6$ m/s.

step 3 Speed at x = 0.040 m

Potential there: $\tfrac{1}{2}(100)(0.040)^2 = \tfrac{1}{2}(100)(0.0016) = 0.08$ J. Kinetic: $0.32 - 0.08 = 0.24$ J. So $\tfrac{1}{2}(0.25)v^2 = 0.24$ , giving $v^2 = 1.92$ and $v = 1.39$ m/s.

step 4 Interpret

At half the amplitude the block still has most of its kinetic energy, because potential energy grows with $x^2$ : a quarter of the energy is potential at $x = A/2$ , leaving three-quarters kinetic. The speed there is well above half the maximum.

Try this

Q1. A spring of constant $200$ N/m oscillates with amplitude $0.050$ m. Calculate the total mechanical energy. [2 points]

Cue. $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(200)(0.050)^2 = \tfrac{1}{2}(200)(0.0025) = 0.25$ J.

Q2. An oscillator's amplitude is tripled. State how its total energy changes. [1 point]

Cue. $E \propto A^2$ , so the energy increases by a factor of $9$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)7 marksSection II (long FRQ). A

0.40

kg block on a frictionless surface oscillates on a spring of constant

160

N/m with amplitude

0.10

m. (a) Calculate the total mechanical energy of the oscillator. (b) Calculate the maximum speed of the block. (c) Calculate the speed when the block is at

x = 0.060

m. (d) State how the total energy changes if the amplitude is doubled, and justify your answer.

Show worked answer →

A 7-point FRQ on energy in simple harmonic motion.

(a) Total energy (2 points): at the amplitude all the energy is elastic potential, $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(160)(0.10)^2 = \tfrac{1}{2}(160)(0.010) = 0.80$ J.
(b) Maximum speed (2 points): at equilibrium all the energy is kinetic, $\tfrac{1}{2}mv_{max}^2 = 0.80$ J. So $v_{max}^2 = 2(0.80)/0.40 = 4.0$ and $v_{max} = 2.0$ m/s.
(c) Speed at x = 0.060 m (2 points): $E = \tfrac{1}{2}kx^2 + \tfrac{1}{2}mv^2$ . Potential there is $\tfrac{1}{2}(160)(0.060)^2 = 0.288$ J, so kinetic is $0.80 - 0.288 = 0.512$ J. Then $v^2 = 2(0.512)/0.40 = 2.56$ and $v = 1.6$ m/s.
(d) Doubled amplitude (1 point): $E = \tfrac{1}{2}kA^2 \propto A^2$ , so doubling the amplitude quadruples the total energy, to $3.2$ J.

Markers reward $E = \tfrac{1}{2}kA^2$ , converting it all to kinetic at equilibrium, the energy split at an intermediate point, and the squared dependence on amplitude.

AP 2023 (style)1 marksSection I (multiple choice). At what point in its motion does an SHM oscillator have its maximum kinetic energy? (A) at the turning points (B) at the equilibrium position (C) halfway between (D) it is constant. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the energy interchange. The answer is (B).

Total mechanical energy is conserved and shared between kinetic and potential. At equilibrium the potential energy is zero (displacement zero), so all the energy is kinetic and the speed is greatest. At the turning points it is all potential. The trap is (A): the turning points are where kinetic energy is least, not greatest.

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Sources & how we know this

AP Physics 1: Algebra-Based Course and Exam Description — College Board (2024)