The shape of the lesson
By the end of the lesson, learners can
- The energy swing: As an oscillator moves, energy sloshes between kinetic and potential.
- Total energy, fixed by the amplitude: The total energy of an undamped oscillation is E = ½mω²x₀².
- Defining condition: a = −ω²x, with ω = 2πf = 2π / T.
- Velocity: v = ±ω√(x₀² − x²), greatest (ωx₀) at the centre and zero at the extremes.
- Energy: total E = ½mω²x₀², shared between KE (greatest at the centre) and PE (greatest at the extremes).
One picture every idea returns to
In SHM kinetic and potential energy interchange while the total stays constant. KE is greatest at the centre, PE at the extremes, and the total is E = ½mω²x₀².
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Kinetic and potential, trading places.. Energy is the other way to read an oscillation. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: The energy swing | As an oscillator moves, energy sloshes between kinetic and potential. Project tab 2 of the lab (Energy in SHM) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Total energy, fixed by the amplitude | The total energy of an undamped oscillation is E = ½mω²x₀². | Anticipation / Reaction Guide |
| 30 to 43 min | Check and plenary | The quick-check questions from the student lesson, answered on mini whiteboards one at a time; re-teach the one most get wrong, then a random call on the last. | Individual, then whole class |
| 43 to 45 min | Exit and homework | Set the worksheet below (5 questions, 58 marks) from the topic practice set. Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Total energy, fixed by the amplitude, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
No examiner trap is recorded for this lesson. Use the quiz key below to spot the distractor most of the class picks.
Energy in simple harmonic motion: topic worksheet
5 questions, 58 marks. Written in the style of the papers; no past paper question is reproduced. These questions cover the whole topic, so sibling lessons share them.
A body of mass 0.20 kg oscillates with simple harmonic motion of amplitude 5.0 cm and frequency 2.0 Hz. Fig. 2.1 shows how the energy of the system varies with displacement.
(a) Describe how the kinetic and potential energy of the body change during one complete oscillation. [2]
(b) Calculate the angular frequency ω. [1]
(c) Calculate the total energy of the oscillation, E = ½mω²x₀². [3]
(d) State where in the motion the kinetic energy is greatest, and give its maximum value. [2]
(e) Calculate the potential energy of the body when its displacement is 3.0 cm. [3]
(f) State how the kinetic energy varies with displacement, with reference to Fig. 2.1. [2]
A 0.20 kg mass performs SHM with amplitude 6.0 cm and frequency 5.0 Hz. Parts (a) and (b) use the kinematics; parts (c) to (e) use the energy.
(a) Calculate the angular frequency. [1]
(b) Calculate the speed when the displacement is 3.0 cm. [3]
(c) Calculate the kinetic energy at this displacement. [2]
(d) Calculate the total energy of the oscillation. [2]
(e) Calculate the potential energy at x = 3.0 cm and confirm that KE + PE equals the total energy. [3]
A mass-spring system oscillates in SHM. Fig. 7.1 shows how the kinetic and potential energies vary with displacement.
(a) Write expressions for the kinetic and potential energies in terms of displacement x. [2]
(b) State where the kinetic energy and the potential energy are each maximum. [2]
(c) Show that the total energy is ½mω²x₀² and is constant. [2]
(d) Determine the displacement, as a fraction of the amplitude, at which KE = PE. [3]
(e) Describe how the kinetic and potential energies vary with displacement, as shown in Fig. 7.1. [2]
A 0.50 kg trolley on springs oscillates in SHM, completing 20 oscillations in 16 s with amplitude 4.0 cm. Part (a) is kinematics; the rest use the energy.
(a) Calculate the period and the angular frequency. [2]
(b) Calculate the maximum speed. [2]
(c) Calculate the maximum kinetic energy (the total energy). [2]
(d) Calculate the displacement at which the kinetic energy equals the potential energy. [2]
(e) Calculate the kinetic energy when the displacement is 2.0 cm. [3]
A mass on a spring is driven by an external periodic force of variable frequency. Fig. 10.1 shows the response. This question links the SHM, its energy and resonance.
(a) State what is meant by the natural frequency of the system. [1]
(b) The mass is 0.25 kg and the spring gives ω = 20 rad s⁻¹. Calculate the natural frequency. [2]
(c) Explain what is meant by resonance and state the driving frequency at which it occurs. [2]
(d) At resonance the amplitude is 6.0 cm. Calculate the total energy of the oscillation. [3]
(e) Describe and explain how increasing the damping changes the resonance peak, as in Fig. 10.1. [2]
(f) State one situation where resonance is useful and one where it must be avoided. [2]
Answer key and marking notes
- (a)KE is greatest at the centre and zero at the extremes; PE is greatest at the extremes and zero at the centre ✓; energy passes back and forth between the two while the total stays constant ✓
- (b)ω = 2πf = 2π × 2.0 = 12.6 rad s⁻¹ ✓
- (c)E = ½mω²x₀² = 0.5 × 0.20 × 12.57² × 0.050² ✓✓ = 3.9 × 10⁻² J ✓
- (d)Greatest at the centre (x = 0) ✓; there all the energy is kinetic, so KE_max = E = 3.9 × 10⁻² J ✓
- (e)PE = ½mω²x² = 0.5 × 0.20 × 12.57² × 0.030² ✓✓ = 1.4 × 10⁻² J ✓
- (f)KE is greatest at x = 0 and falls to zero at x = ±x₀, following the downward (inverted) curve in Fig. 2.1 ✓✓
- (a)ω = 2πf = 2π × 5.0 = 31.4 rad s⁻¹ ✓
- (b)v = ω√(x₀² − x²) = 31.4 × √(0.060² − 0.030²) ✓✓ = 1.6 m s⁻¹ ✓
- (c)KE = ½mv² = ½ × 0.20 × 1.63² ✓ = 0.27 J ✓
- (d)E = ½mω²x₀² = ½ × 0.20 × 31.4² × 0.060² ✓ = 0.36 J ✓
- (e)PE = ½mω²x² = ½ × 0.20 × 986 × 0.030² = 0.089 J ✓; KE + PE = 0.27 + 0.089 = 0.36 J = E ✓✓
- (a)KE = ½mω²(x₀² − x²) ✓; PE = ½mω²x² ✓
- (b)KE is maximum at x = 0; PE is maximum at x = ±x₀ ✓✓
- (c)KE + PE = ½mω²(x₀² − x²) + ½mω²x² = ½mω²x₀² ✓, which has no x dependence, so it is constant ✓
- (d)½mω²(x₀² − x²) = ½mω²x² ✓; x₀² = 2x² ✓; x = x₀/√2 ≈ 0.71 x₀ ✓
- (e)PE is a parabola (minimum at the centre, maximum at the extremes); KE is an inverted parabola (maximum at the centre); their sum is constant ✓✓
- (a)T = 16/20 = 0.80 s ✓; ω = 2π/0.80 = 7.9 rad s⁻¹ ✓
- (b)v_max = ωx₀ = 7.85 × 0.040 ✓ = 0.31 m s⁻¹ ✓
- (c)KE_max = ½mv_max² = ½ × 0.50 × 0.314² ✓ = 0.025 J ✓
- (d)x = x₀/√2 = 0.040/1.414 ✓ = 0.028 m (2.8 cm) ✓
- (e)KE = ½mω²(x₀² − x²) = 0.25 × 61.6 × (0.040² − 0.020²) ✓✓ = 0.018 J ✓
- (a)The frequency at which the system oscillates freely when displaced and released ✓
- (b)f₀ = ω/2π = 20/(2π) ✓ = 3.2 Hz ✓
- (c)The amplitude becomes very large when the driving frequency equals the natural frequency ✓, when energy transfer to the oscillator is greatest; this occurs at f₀ ✓
- (d)E = ½mω²x₀² = ½ × 0.25 × 20² × 0.060² ✓✓ = 0.18 J ✓
- (e)The peak becomes lower and broader ✓, with the maximum amplitude reduced (and shifted to a slightly lower frequency) ✓
- (f)Useful: tuning a radio or a musical instrument; to be avoided: bridges or buildings driven by wind or earthquakes ✓✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
For an undamped oscillator, the total energy during the motion is:
- Aconstant
- Bgreatest at the centre
- Cgreatest at the extremes
- Dsteadily decreasing
The kinetic energy of an oscillator is greatest:
- Aat the centre, where the speed is greatest
- Bat the extremes
- Ca quarter of the way out
- Deverywhere equally
The total energy of an oscillation is E = ½mω²x₀². Doubling the amplitude changes the total energy to:
- Afour times as large
- Btwice as large
- Chalf as large
- Dunchanged
At the extremes of the motion (x = ±x₀), the energy of the oscillator is:
- Aall potential
- Ball kinetic
- Chalf potential, half kinetic
- Dzero
Quiz key
Energy passes between kinetic and potential, but their sum, the total energy, stays constant for an undamped oscillation.
Speed is greatest at the centre, so the kinetic energy is greatest there and zero at the extremes.
E is proportional to x₀², so doubling the amplitude multiplies the total energy by four.
At the extremes the speed is zero, so the kinetic energy is zero and the energy is entirely potential.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.