Track energy in simple harmonic motion
Explain the exchange between kinetic and potential energy and calculate the total energy from mass, angular frequency and amplitude.
Section 17 · Oscillations
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 17. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use displacement, velocity, acceleration, force, energy and trigonometric functions with SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
At an extreme position in ideal SHM, what is the kinetic energy?
What you will learn
- Describe kinetic–potential energy interchange during SHM
- Use E = ½mω²x₀² for total energy
- Relate energy changes to position in the cycle
Kinetic and potential energy exchange as the oscillator moves
As an oscillator moves from an extreme towards equilibrium, its speed increases: potential energy is transferred into kinetic energy. As it moves away from equilibrium, kinetic energy is transferred back into potential energy. In an ideal undamped oscillator, the total energy stays constant.
At an extreme, displacement magnitude is x₀ and velocity is zero, so the kinetic energy is zero and the oscillator's potential energy is greatest. At equilibrium, speed is greatest and the potential energy measured relative to equilibrium is minimum.
Amplitude sets the total energy
For a mass m undergoing SHM with angular frequency ω and amplitude x₀, the total energy is E = ½mω²x₀². It is proportional to mass, to ω² and to amplitude squared. Doubling amplitude therefore quadruples the total energy if m and ω stay fixed.
Use kilograms, rad s⁻¹ and metres to obtain joules. For an ideal oscillator, the total is shared between kinetic and potential energy; damping transfers mechanical energy to the surroundings, so the oscillation amplitude and total mechanical energy decrease.
Pause and explain
If the amplitude doubles while m and ω stay fixed, what happens to total energy?
Worked example
A 0.20 kg mass oscillates with ω = 10 rad s⁻¹ and amplitude 0.030 m. Calculate its total energy.
Show the worked solution
- Use E = ½mω²x₀².
- Substitute ½ × 0.20 × 10² × 0.030².
- Give energy in joules.
Answer E = 0.0090 J.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate E for a 0.50 kg oscillator with ω = 6.0 rad s⁻¹ and x₀ = 0.10 m.
Give me a hint
Substitute into E = ½mω²x₀².
Compare my reasoning
- E = ½ × 0.50 × 6.0² × 0.10².
- E = 0.25 × 36 × 0.010.
- Evaluate and state joules.
E = 0.090 J.
Look for these in your work
- Correct formula used
- Squares applied
- Joule unit stated
An oscillator has total energy 0.12 J at amplitude 0.040 m. Its amplitude is reduced to 0.020 m while m and ω remain fixed. Find its new energy.
Give me a hint
Energy scales with the square of amplitude.
Compare my reasoning
- The amplitude ratio is 0.020/0.040 = 1/2.
- The energy ratio is (1/2)² = 1/4.
- New energy = 0.12/4.
The new total energy is 0.030 J.
Look for these in your work
- Amplitude ratio found
- Square scaling applied
- New energy calculated
Describe the energy changes as an undamped oscillator moves from equilibrium to an extreme and back.
Give me a hint
Compare speed and displacement at the equilibrium and extreme positions.
Compare my reasoning
- At equilibrium, speed and kinetic energy are greatest.
- As it moves to an extreme, speed falls while potential energy rises.
- At the extreme speed is zero; on the return, potential energy converts back to kinetic energy.
Kinetic energy is greatest at equilibrium and transfers to potential energy towards the extreme, then back to kinetic energy on return.
Look for these in your work
- Equilibrium energy state identified
- Energy transfer on outward motion described
- Return motion described
Common mistakes
- Using the instantaneous displacement in place of amplitude in the total-energy formula
- Assuming kinetic energy is greatest at an extreme
- Claiming total mechanical energy stays constant when damping removes energy
- Forgetting that energy is proportional to amplitude squared
What can you explain now?
At an extreme position in ideal SHM, what is the kinetic energy?
Compare with the explanation
Zero
The oscillator is momentarily at rest before reversing direction.
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Make it stick
Where are kinetic energy and potential energy greatest during one ideal SHM cycle?
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