The HL upgrade: You already know an oscillator keeps swapping kinetic and potential energy. At HL you get exact equations for each.
The equations use
- the mass m
- the angular frequency ω
- the amplitude x₀
- the displacement x
The headline result: The total energy never changes: it only ever moves between the two stores.
- total energy of the oscillation (J)
- mass of the oscillating object (kg)
- angular frequency (rad s⁻¹)
- amplitude — the maximum displacement (m)
What the total energy depends on: Because x₀ and ω are squared, the total energy is proportional to x₀² and proportional to ω².
- Double the amplitude → four times the energy. - Double the angular frequency → four times the energy.
The words to know
| Word | What it means |
|---|---|
| Total energy ET | Kinetic + potential: constant, ½mω²x₀² |
| Potential energy EP | Stored energy at displacement x: ½mω²x² |
| Kinetic energy EK | Energy of motion: ET − EP |
| Amplitude x₀ | The largest displacement from equilibrium |
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The energies depend on where the object is. At displacement x the booklet gives the potential energy, EP = ½mω²x²; the kinetic energy is what is left of the total, so it uses the difference x₀² − x².
How the kinetic energy changes
- It is largest at the centre (x = 0).
- It falls to zero at the extremes (x = ±x₀).
- kinetic energy at displacement x (J)
- mass (kg)
- angular frequency (rad s⁻¹)
- amplitude (m)
- displacement from the centre (m)
The potential energy is given: zero at the centre and largest at the extremes, the mirror image of the kinetic energy. Since the total is fixed, too.
- potential energy at displacement x (J)
- total energy (J)
- kinetic energy (J)
- displacement from the centre (m)
At the centre (x = 0)
- x = 0, so x₀² − x² = x₀²
- Kinetic energy is maximum — equal to the total Eₜ
- Potential energy is zero
- The object is moving fastest here
At the extremes (x = ±x₀)
- x = ±x₀, so x₀² − x² = 0
- Kinetic energy is zero — the object is momentarily at rest
- Potential energy is maximum — equal to the total Eₜ
- The object turns around here
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The plan: (1) Use for the total energy. (2) Use at the displacement asked for. (3) Get the potential energy by subtracting: . Keep every answer in joules.
IB-style question — energy of a mass on a spring
A 0.20 kg mass oscillates in SHM with angular frequency ω = 5.0 rad s⁻¹ and amplitude x₀ = 0.10 m. Determine the total energy of the oscillation.
Solution
- Write the given formula for total energy first:
- Substitute m = 0.20, ω = 5.0, x₀ = 0.10:
- Evaluate the squares, then multiply — keep the unit:
Final answer
ET = 0.025 J. This value stays the same all through the motion.
Worked example — KE and PE at a point
For the same oscillator (m = 0.20 kg, ω = 5.0 rad s⁻¹, x₀ = 0.10 m, ET = 0.025 J), find the kinetic and potential energies at displacement x = 0.060 m.
Solution
- Kinetic energy = total − potential, ET − EP:
- Substitute, using x₀² − x² = 0.10² − 0.060² = 0.0064:
- Evaluate — keep the unit:
- Potential energy is the total minus the kinetic:
Final answer
EK = 0.016 J and EP = 0.009 J. They add to 0.025 J = ET, as they must.
Energy is conserved — it just moves: As the object swings from the centre out to an extreme, kinetic energy turns into potential energy; on the way back, potential turns back into kinetic. At every instant EK + EP = ET, a constant.
Animated graph
Watch the graph build step by step in study mode.
| Position | Kinetic energy EK | Potential energy EP | Speed |
|---|---|---|---|
| Centre (x = 0) | Maximum (= ET) | Zero | Fastest |
| Halfway-ish (0 < x < x₀) | Some | Some | Slowing |
| Extreme (x = ±x₀) | Zero | Maximum (= ET) | Momentarily at rest |
Watch the squares: Both energy formulas use x², not x. So an object at x = +0.06 m and one at x = −0.06 m have the same kinetic and potential energy — the sign of the displacement does not matter to the energy.
Energy at half the amplitude
Amplitude from the spring energies: A vertical spring stays stretched, so its elastic PE never reaches zero. From the largest and smallest elastic PE, find the largest and smallest extensions with ½kx²; the amplitude is half their difference.
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Where it shows up: The SHM energy equations are HL only (C.1). They come up in Paper 1A and Paper 2.
| Paper | What they ask |
|---|---|
| Paper 1A | A one-step 'find the total energy', or 'where is the KE a maximum?', or 'how does ET change if the amplitude doubles?' |
| Paper 2 | Determine the KE or speed at a stated displacement, or show the KE and PE add to the total |
Three easy marks: (1) Square ω and the displacements before multiplying. (2) Find EP fast as ET − EK — don't recompute from scratch. (3) Quote energies in joules (J).
IB-style question — kinetic energy at a displacement
A 0.50 kg trolley oscillates in SHM with ω = 4.0 rad s⁻¹ and amplitude x₀ = 0.080 m. Determine the kinetic energy of the trolley when its displacement is x = 0.048 m.
Solution
- Kinetic energy = total − potential, ET − EP:
- Work out x₀² − x² = 0.080² − 0.048² = 0.0064 − 0.002304 = 0.004096:
- Evaluate ½ × 0.50 × 16 × 0.004096 — keep the unit:
Final answer
EK = 0.016 J (to 2 s.f.).