The HL upgrade: You already know an oscillator keeps swapping kinetic and potential energy. At HL you get exact equations for each, written in terms of the mass m, the angular frequency ω, the amplitude x₀, and 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.
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
The kinetic energy depends on where the object is. At displacement x the booklet gives a formula that uses the difference x₀² − x²: it is largest at the centre (x = 0) and 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 whatever is left over, since the total is fixed. Subtracting gives a neat result: — zero at the centre and largest at the extremes, the mirror image of the kinetic energy.
- 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
Feeling unprepared for exams?
Get a clear study plan, practice with real questions, and know exactly where you stand before exam day. No more guessing.
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
- Use the given kinetic-energy formula:
- 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.
Memorize terms 3x faster
Smart flashcards show you cards right before you forget them. Perfect for definitions and key concepts.
Where it shows up: The SHM energy equations are HL only (C.1):
- 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 — speed 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
- Write the given kinetic-energy formula:
- 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.).