IB Physics HLTopic 3 — Oscillations & WavesPaper 1 & 2E = ½mω²x²~12 min read
Calculating Energy Changes in SHM
You already know the story: in SHM, energy trades between kinetic and potential while the total stays fixed. Now we put numbers on it. Three equations do the whole job — one for potential energy, one for kinetic, and one for the total — and they all share the same building blocks: mass, angular frequency, amplitude and displacement. Once you can find ω, the rest is substitution.
📘 What you need to know
Potential energy:EP = ½mω²x²
Total energy:ET = ½mω²x0²
Kinetic energy:EK = ½mω²(x0² − x²)
The total energy is fixed, so ET = EK + EP at every point
Total energy depends on amplitude squared — double the amplitude, quadruple the energy
Maximum KE (at the centre) equals maximum PE (at the extremes) equals the total energy
The three energy equations
Every energy question in SHM comes back to these three formulas. They use the mass m, the angular frequency ω, the amplitude x0, and the displacement x at the moment you care about.
Potential energy at displacement xEP = ½ mω²x²
Kinetic energy at displacement xEK = ½ mω²(x0² − x²)
Total energyET = ½ mω²x0²
Where m is mass (kg), ω is angular frequency (rad s−1), x0 is amplitude (m) and x is displacement (m). All three give energy in joules.
Look closely and you’ll see the three equations are really one idea. The total swaps x for x0 (energy when displacement is maximum). The PE uses the actual x. And the KE is just the total minus the PE: ½mω²x0² − ½mω²x² = ½mω²(x0² − x²). Learn one, and you can rebuild the others.
Where the equations come from
The kinetic energy formula is the key one — the others follow. Start from the ordinary kinetic energy EK = ½mv², then substitute the SHM velocity–displacement relation v = ±ω√(x0² − x²):
🔧 Building the energy equations
Start with EK = ½mv² and the SHM velocity v = ±ω√(x0² − x²).
Squaring v gives v² = ω²(x0² − x²), so EK = ½mω²(x0² − x²).
At the centre (x = 0), KE is maximum and PE is zero — so this is the total energy: ET = ½mω²x0².
Potential energy is whatever’s left over: EP = ET − EK = ½mω²x².
WE 1
A 0.30 kg mass oscillates in SHM with frequency 2.0 Hz and amplitude 0.05 m. Calculate the total energy of the oscillation.
Step 1 — find ωω = 2πf = 2π × 2.0 = 12.57 rad s⁻¹Step 2 — use E_T = ½mω²x₀²E_T = 0.5 × 0.30 × (12.57)² × (0.05)²E_T = 0.5 × 0.30 × 157.9 × 0.0025E_T = 0.059 J (59 mJ)Find ω first, then square it and square the amplitude. Both get squared — a common slip.
Total energy depends on amplitude squared
The total-energy formula hides an important rule: because x0 is squared, the energy grows with the square of the amplitude. Give an oscillator twice the amplitude and it carries four times the energy; triple the amplitude and it’s nine times the energy.
Energy scales with amplitude squaredET ∝ x0²
Handy consequence: the same rule applies to potential energy, EP ∝ x². That’s why, when an oscillator is only halfway out (x = x0/2), it holds just a quarter of its energy as PE — not half.
The energy–displacement picture
Plotting the three energies against displacement shows how the formulas fit together. PE is a parabola (from the x² term), KE is its upside-down partner, and the total is the flat line they always add up to.
PE (blue) is the parabola ½mω²x²; KE (red) is the total minus PE. They always add up to the flat total line ET.
WE 2
A 0.20 kg mass oscillates with angular frequency 10 rad s⁻¹ and amplitude 0.08 m. Find its kinetic and potential energies when the displacement is 0.04 m.
A 35 g ball is held between two springs and oscillates in SHM with frequency 5.0 Hz and amplitude 2.0 cm. Calculate the total energy of the oscillation.
Step 1 — convert units: m = 0.035 kg, x₀ = 0.020 mStep 2 — find ωω = 2πf = 2π × 5.0 = 31.42 rad s⁻¹Step 3 — use E_T = ½mω²x₀²E_T = 0.5 × 0.035 × (31.42)² × (0.020)²E_T = 0.5 × 0.035 × 987 × 0.0004E_T = 6.9 mJGrams to kg and cm to m first — skip that and the answer is out by a big factor.
Maximum energies
Because energy shuffles fully from one store to the other, all three of these are the same number:
All equal to the total energyET = EK(max) = EP(max) = ½mω²x0²
Maximum KE happens at the centre (all energy kinetic); maximum PE happens at the extremes (all energy potential). Each equals the fixed total, so if a question gives you the maximum of either, you already know the total.
At centre EK = max, EP = 0
EK + EP = ET (constant)
Halfway out ¾ KE, ¼ PE
EK + EP = ET (constant)
At extreme EP = max, EK = 0
The energy equations at a glance
Energy
Equation
Maximum where
Potential (EP)
½mω²x²
At the extremes (x = x0)
Kinetic (EK)
½mω²(x0² − x²)
At the centre (x = 0)
Total (ET)
½mω²x0²
Constant everywhere
💡 Top tips
Find ω first (from f or T) — all three energy equations need it.
Square bothω and the displacement/amplitude — the ² applies to each.
Total energy ∝ amplitude²: double the amplitude means four times the energy.
EK = ET − EP is often the quickest route — find the total, subtract the PE.
Convert grams to kg and cm to m before substituting.
⚠ Common mistakes
Squaring only ω and forgetting to square x or x0 (or vice versa)
Forgetting units — grams instead of kg, or cm instead of m
Thinking doubling the amplitude doubles the energy — it quadruples it
Using x instead of x0 in the total-energy formula (or the reverse)
Assuming PE is half the total at half the amplitude — it’s a quarter, because PE ∝ x²
Quick recap: Three equations cover all SHM energy calculations: EP = ½mω²x², EK = ½mω²(x0² − x²), and ET = ½mω²x0². They always satisfy ET = EK + EP, the maximum KE and PE both equal the total, and the total energy is proportional to amplitude squared.
You can now describe SHM, find its period for springs and pendulums, write its motion equations and calculate its energy. The last piece of the topic handles oscillators that don’t start neatly at the centre or the edge — they start partway through, described by a phase angle. That’s the next page: phase angles in SHM.
Want SHM energy sums to feel routine?
Book a free meeting and we’ll drill the three energy equations and past-paper questions together.
