Lesson 6: Simple Harmonic Motion

Simple harmonic motion (SHM) is the motion of systems that are gently pushed away from equilibrium: pendulums, masses on springs, vibrating molecules, alternating currents. Theme C.1 defines SHM by a single condition, and every formula in this lesson follows from it.

The defining condition

SHM occurs when the acceleration is proportional to the displacement from equilibrium and directed towards it: a = −ω²x. The minus sign is essential (restoring direction) and ω is the angular frequency, ω = 2π/T = 2πf. From this definition the data booklet solutions follow: x = x₀ sin(ωt + φ) or x₀ cos(ωt + φ) depending on the starting point; v = ±ω√(x₀² − x²); maximum speed vmax = ωx₀ at the equilibrium position; maximum acceleration amax = ω²x₀ at the extremes. Kinetic and potential energy interchange continuously, with total energy E = ½mω²x₀² = ½kx₀² for a spring — constant in undamped motion and proportional to the square of the amplitude.

Periods of the two standard systems

• Mass on a spring: T = 2π√(m/k)
• Simple pendulum (small angles): T = 2π√(l/g)

Notice what is absent: the pendulum period does not depend on mass or (for small angles) amplitude — Galileo's observation that made accurate clocks possible. For the spring, T does not depend on g, so a mass–spring clock works in orbit while a pendulum clock does not.

Worked problem 1 — mass on a spring

A 0.35 kg mass hangs on a spring of constant k = 28 N m⁻¹ and oscillates with amplitude 0.060 m. Calculate (a) the period, (b) the maximum speed, (c) the total energy.

(a) T = 2π√(m/k) = 2π√(0.35/28) = 2π × 0.112 = 0.70 s
(b) ω = 2π/T = 8.95 rad s⁻¹ ;  vmax = ωx₀ = 8.95 × 0.060 = 0.54 m s⁻¹
(c) E = ½kx₀² = ½ × 28 × 0.060² = 0.050 J
Check: ½mv²max = ½ × 0.35 × 0.54² = 0.051 J ✓ (rounding)

Worked problem 2 — pendulum on the Moon

A pendulum of length 0.85 m is taken to the Moon (g = 1.62 m s⁻²). Find its period, and the length that would give a 2.0 s period (a "seconds pendulum") there.

T = 2π√(l/g) = 2π√(0.85/1.62) = 2π × 0.724 = 4.6 s
For T = 2.0 s: l = gT²/(4π²) = 1.62 × 4.0 / 39.48 = 0.16 m

The same pendulum on Earth has T = 1.85 s. A pendulum clock transported to the Moon would run slow by a factor √(9.81/1.62) = 2.46 — proportional reasoning again.

Practical worked example — finding g by timing oscillations

Timing single swings is dominated by reaction time, so we time many: 20 oscillations of a pendulum take 28.4 ± 0.2 s, giving T = 1.42 ± 0.01 s (the uncertainty is divided by 20 too — the main advantage of the method). With l = 0.512 ± 0.001 m (measured to the centre of the bob):

g = 4π²l / T² = 39.48 × 0.512 / 1.42² = 20.21 / 2.016 = 10.0 m s⁻²
% uncertainty: l contributes 0.2%; T² contributes 2 × 0.7% = 1.4%
Total ≈ 1.6% → g = 10.0 ± 0.2 m s⁻²

The accepted value 9.81 lies just inside the uncertainty range — a defensible conclusion. A systematic error to hunt: measuring l to the top of the bob underestimates l and hence g.

Nature of Science

The small-angle approximation sin θ ≈ θ converts the pendulum's true equation into SHM. The approximation is excellent below about 10° and quietly fails for large swings — models carry domains of validity, and stating them is a scientific habit the IB assesses in the IA evaluation.

Real-world application

Buildings in earthquake zones are fitted with tuned mass dampers — enormous pendulums (Taipei 101 has a 660 tonne steel sphere) whose oscillation, matched to the building's natural frequency, absorbs and dissipates seismic energy. Resonance, usually a hazard, is here turned into protection.

Common mistakes (examiner style)

• Using T = 2π√(m/k) for a pendulum or T = 2π√(l/g) for a spring.
• Confusing ω with f (a factor 2π error) or quoting ω in Hz.
• Stating the acceleration is maximum at equilibrium (it is zero there; velocity is maximum).
• Believing total energy depends on the position in the cycle, or that it is proportional to amplitude rather than amplitude squared.
• Forgetting the small-angle condition when applying the pendulum formula.

IB Exam Experience

Paper 1-style questions. 1. In SHM the acceleration is greatest when the: A. displacement is zero B. speed is greatest C. displacement is greatest D. kinetic energy is greatest. Key: C — a = −ω²x, maximum magnitude at maximum |x|. 2. The amplitude of an SHM oscillator is doubled. The total energy increases by a factor of: A. 2 B. 4 C. 8 D. √2. Key: B — E ∝ x₀². 3. A pendulum has period 2.0 s on Earth. On a planet where g is four times larger, its period is: A. 0.50 s B. 1.0 s C. 4.0 s D. 8.0 s. Key: B — T ∝ 1/√g, so T halves to 1.0 s.

Paper 2-style structured question (7 marks). A block of mass 0.20 kg oscillates on a spring with displacement x = 0.040 cos(15t), x in metres, t in seconds. (a) State the amplitude and the period of the motion. [2] (b) Calculate the maximum speed of the block. [2] (c) Calculate the spring constant. [3] *Marking notes:* (a) x₀ = 0.040 m; ω = 15 rad s⁻¹ so T = 2π/15 = 0.42 s. (b) vmax = ωx₀ = 15 × 0.040 = 0.60 m s⁻¹. (c) ω² = k/m → k = mω² = 0.20 × 225 = 45 N m⁻¹. Candidates who quote T = 1/15 s have confused ω with f — the most common error on this standard question.

IA idea: Investigate how the period of a pendulum depends on amplitude beyond the small-angle range, and find the angle at which the SHM model breaks down by more than 2%.

Energy in SHM, position by position

At displacement x, an oscillator with amplitude x₀ has potential energy fraction (x/x₀)² of the total and kinetic energy fraction 1 − (x/x₀)². At x = x₀/2, exactly one quarter of the energy is potential and three quarters kinetic, so the speed there is (√3/2)vmax. These ratios need no numbers and answer most "fraction" questions immediately. The average kinetic energy over a full cycle is half the total (the other half is the average potential energy) — equipartition in miniature.

Damping and resonance

Real oscillators lose energy: light damping leaves a slowly decaying sinusoid (the envelope decays exponentially, with the same mathematics as Lessons 12 and 13); critical damping returns the system to equilibrium in the shortest time without oscillation (car suspension, door closers aim slightly under critical); heavy (over-) damping crawls back without crossing. Resonance occurs when a system is driven at its natural frequency: amplitude grows until input power balances losses. Bridges and buildings are designed so environmental driving (wind, footsteps, earthquakes) avoids their natural frequencies — the Millennium Bridge's lateral wobble in 2000, caused by pedestrians synchronising with a 0.48 Hz sway mode, was fixed by adding dampers: this lesson's physics, at civil-engineering scale.

Worked extra — velocity at a given displacement

An oscillator has amplitude 0.12 m and period 0.90 s. Find its speed at x = 0.072 m from equilibrium. ω = 2π/0.90 = 6.98 rad s⁻¹; v = ω√(x₀² − x²) = 6.98 × √(0.0144 − 0.00518) = 6.98 × 0.0960 = 0.67 m s⁻¹. Check: vmax = ωx₀ = 0.84 m s⁻¹, and (0.072/0.12)² = 0.36 of the energy is potential, leaving 64% kinetic → v = 0.80 × vmax = 0.67 ✓.

Phase and the starting condition

Whether x = x₀ cos ωt or x = x₀ sin ωt describes a given oscillator depends entirely on the state at t = 0: released from rest at maximum displacement → cosine; passing through equilibrium at t = 0 → sine (with the sign setting the direction). Velocity is always a quarter-cycle ahead of displacement and acceleration a quarter-cycle ahead of velocity (or equivalently, in exact antiphase with displacement). A robust routine: sketch all three curves on shared axes before answering any phase question — "at what displacement is the speed half its maximum?" then reads off the energy relation of the X-section, and "is the acceleration positive when x is negative?" answers itself from a = −ω²x. Phase errors are the single largest source of wrong answers in SHM multiple choice, and they are entirely eliminated by the 20-second sketch.

The seconds pendulum and precision timekeeping

A pendulum with T = 2.0 s needs l = gT²/(4π²) = 0.994 m on Earth — conveniently near a metre, which is no accident historically: the metre was nearly defined as the length of the seconds pendulum before the meridian definition won. Temperature affects precision clocks through thermal expansion of the rod: a steel rod (α ≈ 1.2 × 10⁻⁵ K⁻¹) warmed by 10 K lengthens by 0.012%, lengthening T by half that fraction — about 5 seconds per day of drift. Gridiron and mercury-compensated pendulums were the engineering answer; quartz and atomic clocks are the modern one.

Beats and coupled oscillators: SHM in conversation

Two oscillators of nearly equal frequency, sounded together, produce beats: the amplitude swells and fades at the difference frequency |f₁ − f₂| — piano tuners literally count beats to zero. Two pendulums joined by a weak spring exchange energy back and forth completely, each alternately stopping: energy sloshes between them at the difference of their normal-mode frequencies. Both phenomena are pure superposition of SHM solutions, and both reappear in quantum mechanics (tunnelling oscillations) and electrical engineering (coupled resonators) — one more reason this topic repays deep study.