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.
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
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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
- 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 E_K + E_P = E_T, a constant.
| 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.
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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 E_T 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 E_T − E_K — 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.).