Physics I: Mechanics › Oscillations & gravitation › SHM energy

E = ½kA²Say it: “the total energy of a simple harmonic oscillator equals one half k A squared”

Energy in SHM

Energy sloshes between motion and spring — but the total never changes, and one number (the amplitude) fixes it.

Notation on this page: E is the total mechanical energy (joules), K = ½mv² kinetic, U = ½kx² spring potential, A the amplitude.

Before this lesson: Mass–spring, Simple pendulum

Where it comes from

Watch one swing of a mass on a spring and track the energy. At the turning points (x = ±A) the mass momentarily stops: all energy is spring potential, U = ½kA². At equilibrium (x = 0) the spring is relaxed but the mass flies: all energy is kinetic, K = ½mvmax². In between, it's a mixture — but the sum never changes:

E = ½mv² + ½kx² = ½kA²kinetic + potential at any instant = the constant total
Before reading on: where is the mass moving fastest — at the turning points or at equilibrium? And where is it instantaneously at rest? Answer from the energy trade-off before reading the verdict.

Fastest at equilibrium (all energy is kinetic there), at rest at the turning points (all energy is potential). The trade is exact: whatever K loses, U gains, and E = ½kA² sits above it all, unchanging.

Derivation

We prove E is constant by writing K + U with the SHM position and velocity, then watching the time dependence cancel. The identity sin² + cos² = 1 does the final magic.

E
=
½mv² + ½kx²
Step 1 — define the total. Kinetic plus spring potential, at an arbitrary instant. Nothing assumed about constancy yet.
=
½m[−Aω sin(ωt+φ)]² + ½k[A cos(ωt+φ)]²
Step 2 — insert SHM. x = A cos(ωt+φ) and v = −Aω sin(ωt+φ) from the SHM position page.
=
½mA²ω² sin²(ωt+φ) + ½kA² cos²(ωt+φ)
Step 3 — square it out. The minus sign dies in the square. Now use mω² = k (from ω = √(k/m)) on the first term.
E
=
½kA²[sin²(ωt+φ) + cos²(ωt+φ)] = ½kA²
Step 4 — the time cancels. sin² + cos² = 1 kills every t. E = ½kA² — a constant, set entirely by the amplitude. ∎

The shortcut version: evaluate E at a turning point, where v = 0 and x = A: E = 0 + ½kA². Since E is conserved, that one snapshot fixes it everywhere — the four-step proof above just confirms the conservation.

How to use it

The procedure, every time:

  1. Find E from the amplitude. E = ½kA² — amplitude in, energy out.
  2. Max speed from E. At equilibrium U = 0, so ½mvmax² = E → vmax = Aω = A√(k/m).
  3. Speed at any x. ½mv² = E − ½kx² → v = ±ω√(A² − x²).
  4. Amplitude from E. A = √(2E/k) — energy in, amplitude out.
  5. Pendulums too. Replace k with mg/L: E = ½(mg/L)smax².

The working forms

vmax = Aω = A√(k/m)top speed at equilibrium — bigger swing or stiffer spring, faster flybySay it: “v max equals A omega”
v(x) = ±ω√(A² − x²)speed anywhere — the ± is direction, the root is how much energy remains kinetic
Common mistake: writing E = ½kx² with the instantaneous x. ½kx² is the potential right now — only ½kA², with the constant amplitude, is the total.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: k = 100 N/m, A = 0.10 m, m = 0.25 kg

  1. Total energy. E = ½kA² = 0.5 × 100 × 0.10² = 0.50 J.
  2. Max speed. ω = √(k/m) = √(100/0.25) = 20 rad/s; vmax = Aω = 0.10 × 20 = 2.0 m/s.
  3. Cross-check via kinetic. ½mvmax² = 0.5 × 0.25 × 4.0 = 0.50 J ✓ — matches E, as it must at equilibrium.
Common mistake: E = ½ × 100 × 0.10 = 5 J — forgetting to square the amplitude. Energy goes as A²: double the swing, quadruple the energy.
Your turn — k = 200 N/m, A = 0.050 m, m = 0.10 kg. Find E and vmax.

Answer: E = 0.25 J, vmax ≈ 2.24 m/s. E = 0.5 × 200 × 0.0025 = 0.25 J. ω = √(200/0.10) = √2000 ≈ 44.7 rad/s; vmax = 0.050 × 44.7 ≈ 2.24 m/s. Check: 0.5 × 0.10 × 5.0 = 0.25 J ✓.

Example 2 — speed mid-swing: same oscillator, at x = A/2

  1. Energy left for motion. U = ½kx² = 0.5 × 100 × 0.050² = 0.125 J; K = E − U = 0.50 − 0.125 = 0.375 J.
  2. Speed. v = √(2K/m) = √(0.75/0.25) = √3 ≈ 1.73 m/s.
  3. The pattern. v = ω√(A² − x²) = 20√(0.01 − 0.0025) = 20 × 0.0866 ≈ 1.73 m/s — that is (√3/2)vmax ≈ 0.866 × 2.0 ✓
Common mistake: “half the displacement, half the speed” — v = 1.0 m/s. Energy is quadratic: at x = A/2 three-quarters of E is still kinetic, so v = 0.866 vmax, not half.
Your turn — same oscillator: speed at x = A/√2?

Answer: v = vmax/√2 ≈ 1.41 m/s. U = ½k(A²/2) = E/2, so K = E/2 and v = vmax/√2. At x = A/√2 the energy is split exactly evenly: 0.25 J kinetic, 0.25 J potential.

Example 3 — working backwards: E = 2.0 J, k = 50 N/m. Find A.

  1. Rearrange first. E = ½kA² → A = √(2E/k).
  2. Substitute. A = √(4.0/50) = √0.08 ≈ 0.283 m.
  3. Sanity check. ½ × 50 × 0.08 = 2.0 J ✓ — and a soft 50 N/m spring needs a big 28 cm swing to hold 2 J. Plausible.
Common mistake: A = 2E/k = 0.08 m — forgetting the square root. A = 0.08 m would hold only ½ × 50 × 0.0064 = 0.16 J. Always verify by plugging back.
Your turn — E = 1.0 J, k = 800 N/m. Find A.

Answer: 0.050 m. A = √(2.0/800) = √0.0025 = 0.050 m. A stiff spring packs 1 J into just 5 cm of swing.

Before reading on: two oscillators hold the same 0.50 J, but one's spring is 4× stiffer. Which swings with the bigger amplitude — and what happens to each one's top speed?

Example 4 — judgment call: same energy, stiffer spring

  1. Amplitudes. A = √(2E/k): k = 100 → A = 0.10 m; k = 400 → A = √(1.0/400) = 0.050 m — halved.
  2. Top speeds. vmax = √(2E/m) — no k! Both give 2.0 m/s (m = 0.25 kg).
  3. The lesson. Same energy means same top speed regardless of stiffness; the stiffer spring just achieves it in a shorter, snappier swing (higher ω, smaller A).
Common mistake: “stiffer spring, faster top speed.” vmax = √(2E/m) doesn't contain k — stiffness changes how the energy is stored (small fast swing vs. big slow swing), not how fast the mass flies at equilibrium.
Your turn — same E, but the mass doubles (same k). What happens to A? To vmax?

Answer: A unchanged; vmax drops by 1/√2. A = √(2E/k) has no m. vmax = √(2E/m): double m → vmax/√2 ≈ 0.707×. Same swing size, lazier flyby.

Memorization tips

  • Chant it: “E equals one-half k A squared.” Amplitude in, energy out — and it never changes.
  • The A² rule: double the swing, quadruple the energy. Energy punishes big amplitudes quadratically.
  • A, not x: ½kA² is the constant total; ½kx² is the potential right now. Mixing them is the #1 error.
  • The two snapshots: turning point → E = ½kA² (all potential); equilibrium → E = ½mvmax² (all kinetic). Either snapshot rebuilds the formula.
  • vmax = Aω: the bridge between the energy picture and the motion picture.
  • Real-world asterisk: damping steals a little E each cycle, so A slowly shrinks — but the ½kA² bookkeeping still works swing to swing.

Final challenge

Five mixed questions — energy splits, scalings, and the traps, all in one. Score 5/5 and SHM energy is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

Is the total energy of an oscillator really constant?

For ideal SHM, yes: E = ½kA² never changes. Real oscillators lose a little to friction and air drag each cycle (damping), so the amplitude slowly decays — but over one swing the constant-E picture is excellent.

What is the energy of a pendulum in SHM terms?

The same idea with k replaced by mg/L: E = ½(mg/L)smax², or in angles E = ½mgL(θmax)² for small swings. Maximum kinetic energy at the bottom equals maximum gravitational potential at the turning points.

Why is it ½kA² and not ½kx²?

½kx² is the potential energy at one instant — it varies as the mass moves. ½kA² uses the amplitude A (a constant) and gives the total energy, which never varies.

Does the total energy depend on the mass?

Not directly: E = ½kA² has no m in it. But the mass sets how that energy splits into speed — vmax = A√(k/m) — so a heavier mass moves slower with the same energy.

Where is the kinetic energy maximum? The potential?

Kinetic is maximum at equilibrium (x = 0), where the speed peaks; potential ½kx² is maximum at the turning points (x = ±A), where the mass momentarily stops. They trade off so the sum stays ½kA².

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