Lesson 15: Simple Harmonic MotionAP

Swing a pendulum, bounce a mass on a spring, or pluck a guitar string and you meet the same pattern: simple harmonic motion (SHM, CED Unit 8) — oscillation about a stable equilibrium under a restoring influence proportional to displacement. SHM matters far beyond this unit: it is the mathematics of everything from quartz clocks to molecular vibrations to alternating current, and the College Board tests it as a system of interchangeable descriptions — equations, graphs, and energy.

The defining condition and its consequences

SHM occurs whenever the net force (or torque) on a system is F = −k_eff x: directed back toward equilibrium and proportional to displacement. The constant's form varies (a literal spring constant for a mass on a spring), but the consequences are universal:

• The motion is sinusoidal: x(t) = A cos(ωt + φ), with amplitude A and angular frequency ω = 2πf = 2π/T.
• Acceleration is a = −ω²x — largest at the extremes, zero at equilibrium, always opposite in sign to x.
• Velocity is maximal at equilibrium (v_max = Aω) and zero at the extremes.
• The period is independent of amplitude (for true SHM) — the deep reason pendulum clocks keep time regardless of swing size (small swings, at least).

For the mass–spring system, ω = √(k/m) and T = 2π√(m/k): heavier mass, slower oscillation; stiffer spring, faster. These proportionalities — T ∝ √m, T ∝ 1/√k — are tested constantly, often as graph-slope questions.

Energy in SHM

With no friction, mechanical energy is constant and trades between kinetic and elastic potential: E = ½kA². At displacement x, U = ½kx² and K = ½k(A² − x²), so the speed at any position is v = ω√(A² − x²). At the extremes all energy is potential; at equilibrium all is kinetic; at x = A/√2 the two are exactly equal. Bar charts at three positions (extreme, middle, equilibrium) are the fastest route through energy questions in this unit.

Reading SHM graphs

Given an x–t graph, read A from the peak, T from the repeating pattern, and the phase from where the curve starts. The v–t graph is the same sinusoid shifted by a quarter period (velocity leads displacement); the a–t graph is the displacement graph inverted and scaled by ω². If a problem gives you one graph, sketching the other two takes seconds and prevents most sign errors.

Worked example 1 — the oscillator's vital statistics

A 0.50 kg mass on a spring with k = 200 N/m oscillates. Find (a) the angular frequency, (b) the period and frequency. Given: m = 0.50 kg, k = 200 N/m.

(a) ω = √(k/m) = √(200/0.50) = √400 = 20 rad/s
(b) T = 2π/ω = 2π/20 ≈ 0.31 s;   f = 1/T ≈ 3.2 Hz

Check: a stiff spring and a light mass should oscillate fast — three cycles per second is plausible for these values; quadrupling the mass would only double T (square-root dependence), a useful sanity pattern.

Worked example 2 — speed at a given displacement

The mass above oscillates with amplitude 0.10 m. Find (a) the maximum speed, (b) the speed at x = 0.060 m, (c) the maximum acceleration. Given: A = 0.10 m, ω = 20 rad/s.

(a) v_max = Aω = 0.10 × 20 = 2.0 m/s (at equilibrium)
(b) v = ω√(A² − x²) = 20 × √(0.0100 − 0.0036) = 20 × 0.080 = 1.6 m/s
(c) a_max = Aω² = 0.10 × 400 = 40 m/s² (at the extremes)

Check: at x = 0.6A the speed is 0.8 of maximum — matching √(1 − 0.36) = 0.8 from energy, a second route to the same number ✓. The maximum acceleration (~4 g) occurs exactly where the speed is zero — extremes trade speed for acceleration.

Real-world application: engineering with oscillators

Car suspension is tuned SHM: the spring constant and the car's mass set ω, and engineers choose it near 1–1.5 Hz because humans find that frequency comfortable; shock absorbers then add damping so the oscillation dies instead of continuing. Seismometers invert the idea — a mass that stays nearly still while the ground oscillates around it — and quartz watches exploit a crystal whose SHM frequency (32,768 Hz) is so stable it can count seconds for years.

Experimental design note

Investigate T = 2π√(m/k) honestly: time 20 oscillations (not 1) to shrink reaction-time error, vary the mass, and plot T² against m. The straight line's slope is 4π²/k, yielding k from the graph; a nonzero intercept quietly reveals the spring's own effective mass. Amplitude should be varied once, deliberately, to demonstrate the period's independence from it — that control test is what separates an AP design from a school lab.

Common mistakes (from College Board scoring commentary)

• Placing maximum speed at the extremes (it is zero there) or maximum acceleration at equilibrium (zero there).
• Using degrees in x = A cos(ωt) or confusing ω (rad/s) with f (Hz) — a factor of 2π error.
• Believing period depends on amplitude for a spring or (small-angle) pendulum.
• Guessing T ∝ m rather than √m, and misreading T²-graph slopes as k itself rather than 4π²/k.

AP Exam Experience

MCQ 1. A mass on a spring oscillates with period T. If the mass is quadrupled (same spring), the new period is:

• (A) T/2
• (B) T
• (C) 2T
• (D) 4T

Answer: (C). T = 2π√(m/k) ∝ √m; ×4 in mass gives ×2 in period.

MCQ 2. An oscillator's displacement is x = (0.20 m) cos(5.0t). Its maximum speed is:

• (A) 0.20 m/s
• (B) 0.50 m/s
• (C) 1.0 m/s
• (D) 5.0 m/s

Answer: (C). v_max = Aω = 0.20 × 5.0 = 1.0 m/s. (D) reads ω alone; (A) reads A alone.

MCQ 3. A mass in SHM passes through equilibrium. At that instant, which is true?

• (A) acceleration and speed are both maximal
• (B) acceleration is zero and speed is maximal
• (C) acceleration is maximal and speed is zero
• (D) acceleration and speed are both zero

Answer: (B). At equilibrium x = 0, so a = −ω²x = 0, while all the energy is kinetic so speed is at its maximum.

FRQ. A 0.30 kg block attached to a horizontal spring (k = 60 N/m) on a frictionless surface is pulled 0.12 m from equilibrium and released. (a) Calculate the period of oscillation. (b) Calculate the block's maximum speed. (c) Calculate the block's speed when it is 0.060 m from equilibrium. (d) On axes provided, sketch the block's kinetic energy and elastic potential energy as functions of time over one full period, labelling which is which. Scoring guideline. (a) 1 point: T = 2π√(m/k) = 2π√(0.30/60) = 2π × 0.0707 ≈ 0.44 s. (b) 1 point: ω = √(k/m) ≈ 14.1 rad/s; v_max = Aω = 0.12 × 14.1 ≈ 1.7 m/s (or via ½kA² = ½mv²). (c) 1 point: v = ω√(A² − x²) = 14.1 × √(0.0144 − 0.0036) = 14.1 × 0.104 ≈ 1.5 m/s. (d) 2 points: two sinusoidal curves exchanging twice per period, K starting at zero (released at extreme) and peaking at T/4, U starting at maximum (½kA² = 0.43 J) and dipping to zero at T/4, with K + U constant (1 point curves in antiphase, 1 point correct start values/constant sum).

The reference circle: where the cosine comes from

Why a cosine? Imagine a point moving in a circle at constant angular speed ω, radius A. Its shadow on a diameter — its projection — moves back and forth, fastest at the centre, pausing at the edges. That projection is SHM: x = A cos(ωt) for a point starting at the extreme. The restoring acceleration follows because the projection of the centripetal acceleration (ω²A toward the centre) onto the diameter is −ω²x. This is not a metaphor: it is why ω in SHM is called angular frequency, why the same 2π appears, and why phase differences are quoted in radians. A mass on a spring and a pendulum bob (small angles) both behave as projections of an invisible uniform rotation.

Worked example 3 — reading an equation like a graph

An oscillator obeys x = 0.050 cos(4π t) (metres, seconds). Find (a) A, ω, T and f; (b) v_max and a_max; (c) the first time the mass passes through equilibrium.

(a) A = 0.050 m; ω = 4π ≈ 12.6 rad/s; T = 2π/ω = 0.50 s; f = 2.0 Hz
(b) v_max = Aω = 0.050 × 4π ≈ 0.63 m/s; a_max = Aω² = 0.050 × (4π)² ≈ 7.9 m/s²
(c) Equilibrium first reached at T/4 = 0.125 s (cos falls from 1 to 0 in a quarter cycle)

Check: at t = 0 the mass sits at +A with v = 0 — consistent with the cosine form and with being released from rest at the extreme. A sine-form equation would describe release from equilibrium with a push; matching the formula to the story is the skill being tested.

Is it really SHM? The diagnostic test

Not every oscillation is simple harmonic. Test: is the acceleration proportional to displacement and oppositely directed, a = −(constant)x? For a bouncing ball, the acceleration is just ±g (constant, set by which phase of flight) — not proportional to x — so a bouncing ball is not SHM, however regular it looks; its period depends on drop height, unlike a true oscillator's. For a mass bobbing in water, buoyancy gives a restoring force proportional to submerged depth change — that one is SHM (within limits). Learn to run the test before reaching for SHM formulas: check proportionality, not rhythm.

Worked example 4 — from graph to equation

An x–t graph for a spring oscillator shows peaks at +0.080 m, with successive peaks 0.80 s apart. Write the motion's equation and find the speed as the mass passes x = +0.040 m moving toward equilibrium.

A = 0.080 m; T = 0.80 s; ω = 2π/T ≈ 7.85 rad/s → x = 0.080 cos(7.85t)
v = ω√(A² − x²) = 7.85 × √(0.0064 − 0.0016) = 7.85 × 0.069 ≈ 0.54 m/s

Check: v_max = Aω ≈ 0.63 m/s, and at x = A/2 the speed should be (√3/2) of that ≈ 0.54 ✓ (the quarter rule of Lesson 16 arriving early). Graph-to-equation translation is a named AP skill ("representations"), and this two-step — read A and T off the graph, then compute — is its complete method.