IB Physics SL Topic C.1 — Oscillations & SHM Paper 1 & 2 E = EK + EP ~7 min read

Energy in Simple Harmonic Motion

Every oscillator is really just an energy see-saw. As it swings and bounces, energy keeps pouring back and forth between kinetic and potential — but the two always add up to the same total.

📘 What you need to know

The Great Energy Swap

Picture a mass on a spring pulled out to one side and released. At that far point it’s not moving, so all its energy is stored as potential energy. As the spring pulls it back, it speeds up — potential energy is being converted into kinetic energy. It rockets through the middle at top speed (all kinetic, no potential), then climbs out the other side, slowing down as kinetic turns back into potential. And then the whole thing runs in reverse, over and over.

At the amplitude
PE max, KE = 0
speeds up →
PE → KE
At equilibrium
KE max, PE = 0
slows down →
KE → PE
Other amplitude
PE max, KE = 0

A swinging pendulum tells the same story, just with gravitational potential energy: it has the most PE at the top of its swing (highest point) and the most KE at the bottom (lowest and fastest).

Total Energy Is Conserved

Here’s the key idea: as one energy store grows, the other shrinks by exactly the same amount. Add them together at any instant and you always get the same number — the total energy never changes (as long as there’s no friction or air resistance draining it away).

Total energy of an SHM system E = EK + EP

To put numbers to it, you’ll lean on the three energy equations you already know:

The energy stores EK = ½mv2  •  EP (elastic) = ½kx2  •  EP (grav) = mgh

A neat consequence: since the object is momentarily still at the amplitude, all the energy is potential there — so the total energy equals the maximum potential energy. And since PE is zero at equilibrium, the total also equals the maximum kinetic energy. Both give you the total.

Energy Against Displacement

Plot the two energies against displacement and you get a pair of matching parabolas. Potential energy is smallest (zero) in the middle and largest at the ends — a U-shape. Kinetic energy is the exact opposite: largest in the middle, zero at the ends — an upside-down n-shape. Stack them and their heights always add up to the flat total energy line on top.

Energy x total E PE KEx0 x0 0
Potential energy (blue, U-shape) and kinetic energy (teal, n-shape) are mirror images; at every displacement they add up to the constant total energy (orange).

Energy Against Time

Now watch the same energies tick along in time. Both rise and fall as smooth periodic curves, always in opposite step — when KE peaks, PE is at zero, and vice versa. Notice they never dip below the axis: energy is always positive. And because the object reaches maximum displacement twice in each full oscillation, each energy curve completes two cycles per period.

T 2T Energy time total E KE PE
Kinetic (teal) and potential (blue) energy swap places twice every period, always summing to the constant total (orange) — and both stay above zero.

🧭 Tackling an SHM energy problem

  1. Find the total — it equals the max KE (read at equilibrium) or the max PE (read at the amplitude)
  2. Use E = EK + EP to get whichever store you’re missing at a given point
  3. For a speed, put the kinetic energy into EK = ½mv2 and rearrange to v = √(2EK/m)
  4. On an energy–displacement graph, remember PE is the U-shape and KE is the n-shape
  5. Watch your units — millijoules to joules, centimetres to metres — before substituting
Quick recap: KE and PE trade places every quarter-swing, but E = EK + EP stays fixed — max KE at the middle, max PE at the ends, and the total is always a flat line.
WE 1

A 0.50 kg object oscillates with SHM. Its kinetic energy is 60 mJ at the equilibrium position and falls to zero at a displacement of 2.0 cm.

(a) State the total energy. (b) State the amplitude. (c) Find the maximum speed. (d) The graph shows the KE has dropped to 45 mJ at a displacement of 1.0 cm — find the potential energy there.

Part (a) At equilibrium PE = 0, so the total equals the max KE E = 60 mJ Part (b) KE is zero at the amplitude, so x0 = the displacement where KE = 0 x0 = 2.0 cm Part (c) EK = ½mv² so v = √(2EK/m), with EK = 60 mJ = 0.060 J v = √(2 × 0.060 ÷ 0.50) = √0.24 v ≈ 0.49 m s⁻¹ Part (d) EP = E − EK = 60 − 45 EP = 15 mJ
WE 2

A mass on a spring oscillates horizontally with a total energy of 0.80 J and an amplitude of 0.10 m.

(a) Calculate the spring constant. (b) Find the kinetic energy when the displacement is 5.0 cm.

Part (a) At the amplitude all the energy is elastic PE: E = ½kx0² k = 2E ÷ x0² = (2 × 0.80) ÷ 0.10² k = 160 N m⁻¹ Part (b) PE at x = 0.050 m: EP = ½kx² = ½ × 160 × 0.050² EP = 0.20 J, so EK = E − EP = 0.80 − 0.20 EK = 0.60 J Halfway out in displacement, but most of the energy is still kinetic — the PE only catches up near the very ends.

💡 Top tips

⚠ Common mistakes

That wraps up Simple Harmonic Motion — you can now describe an oscillator, apply a = −ω2x, find the period of springs and pendulums, and follow the energy all the way round. Next we build on these oscillations to explore how they travel as waves.

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