Simple harmonic motion (SHM) is the single most tested oscillation model in AP Physics 1, and the phrase defining simple harmonic motion on the exam almost always means a candidate must show that a system satisfies four specific conditions: equilibrium, a linear restoring force, energy interchange between kinetic and potential forms, and sinusoidal displacement in time. Unit 7 of the AP Physics 1 framework dedicates roughly 12–14% of the multiple-choice section to oscillation, and the topic resurfaces in qualitative-quantitative translation (QQT) FRQs and in the experimental-design FRQ whenever a spring or pendulum appears in the prompt. For a candidate targeting a 4 or 5, SHM is not an optional module — it is one of the four highest-leverage Unit 6–9 blocks the exam uses to differentiate score bands.
The four defining conditions of SHM that the AP Physics 1 exam tests explicitly
Most students enter Unit 7 thinking SHM is "anything that wiggles back and forth." That definition earns partial credit on a discussion question and zero on a calculation. The College Board expects students to recognise four non-negotiable conditions, and FRQ rubrics hand out the point for restoring force long before they credit the period formula. Walk through a typical QQT prompt and the grader's checklist becomes obvious: first, the system has a stable equilibrium position; second, the net force on the oscillating mass is directed toward that equilibrium; third, the magnitude of that force is proportional to displacement from equilibrium (linear, not quadratic or higher); and fourth, the resulting motion is sinusoidal in time, with constant amplitude, constant period, and zero net damping over the interval considered.
The first condition matters because AP Physics 1 question writers love to disguise a non-equilibrium setup as SHM. A bead sliding on a parabolic wire, for example, has a stable equilibrium at the bottom of the curve, but its restoring force for small displacements is linear — which makes it locally SHM. Push the bead too far, and the linear approximation breaks. The exam exploits this: a prompt will say "oscillates with small amplitude" precisely to signal that the small-angle or linear-restoring-force regime is in force. Candidates who skip that phrase and use the full nonlinear force expression will compute a period that does not match the answer choices, because the small-angle simplification is built into every formula the rubric awards credit for.
The second and third conditions are paired in most item stems. A horizontal mass-spring system delivers a clean F = −kx, so the linear-restoring-force requirement is automatic. A vertical mass-spring hangs at a new equilibrium where mg = kx₀, and SHM still holds about that shifted equilibrium — the exam will provide the unstretched length or the equilibrium position, then ask for the period. A pendulum is subtler: for small angles, the tangential restoring force is F = −mg sin θ ≈ −mgθ, and with s = Lθ the force becomes F = −(mg/L)s, which matches the linear form with an effective k of mg/L. A student who writes the pendulum period as T = 2π√(L/g) without justifying the small-angle assumption usually loses one point on a "justify your reasoning" stem.
Why the small-angle condition is itself an exam point
AP Physics 1 questions frequently distinguish between SHM and "approximate SHM." The reference to small amplitude appears in roughly one out of three Unit 7 stems, and the rubric explicitly awards the justification point to a student who notes that sin θ ≈ θ only for θ measured in radians and only when θ is small (a working threshold is θ < ~10° or θ < 0.17 rad). The graders do not expect a numerical cutoff, but they do expect a verbal caveat. In my experience, the candidates who write "assuming small-angle approximation, since sin θ ≈ θ" pick up the second point on a 4-point SHM FRQ when the more physics-confident candidate who silently uses the formula loses it.
The fourth condition — sinusoidal motion — is what closes the loop. A candidate should be able to write displacement as x(t) = A cos(ωt + φ) and identify the angular frequency ω = 2π/T = √(k/m) for a spring or ω = √(g/L) for a pendulum. Velocity and acceleration follow by differentiation: v(t) = −Aω sin(ωt + φ) and a(t) = −Aω² cos(ωt + φ) = −ω²x(t). The last identity is the exam's most elegant way to test the defining conditions: a candidate who can show that a(t) = −ω²x(t) from first principles has demonstrated linear restoring force and sinusoidal motion in two lines.
Period, frequency, and angular frequency: the three symbols the FRQs recycle
AP Physics 1 does not test Unit 7 in isolation. Period, frequency, and angular frequency appear across the entire Units 6–9 stretch, and the symbol conventions the exam uses are strict. Period T is measured in seconds, frequency f in hertz (cycles per second), and angular frequency ω in radians per second. The relationships are T = 1/f and ω = 2πf = 2π/T. A surprising number of otherwise strong candidates mix up the formulas on the exam by writing T = 2π√(m/k) as if it were already in angular form — the equation is correct, but the period inside the radical is the regular T in seconds, not ω. The two equations to memorise, in their cleanest form, are:
- Spring–mass: T = 2π√(m/k), ω = √(k/m), f = (1/2π)√(k/m)
- Simple pendulum (small angle): T = 2π√(L/g), ω = √(g/L), f = (1/2π)√(g/L)
The exam tests these formulas through substitution rather than derivation. A typical multiple-choice item gives a 0.50 kg block on a spring of k = 200 N/m and asks for the period; the answer is T = 2π√(0.50/200) ≈ 0.31 s. No calculus required. The harder item variant gives the period and asks for the spring constant, and the candidate must invert the formula correctly. Candidates who confuse T and ω lose 30–60 seconds of working time on each such item, which is the difference between finishing Module 1 of Section II and running out of clock on the last FRQ.
One frequently tested detail: the period of a spring depends only on mass and stiffness, not on amplitude or on gravitational field. This is the "isochronous" property of SHM, and the exam presents it as a check on conceptual understanding. A stem will describe two identical springs with a 1 kg and a 2 kg mass, then ask which has the longer period; the correct answer is the heavier mass, with a ratio of √2. The amplitude is a distractor. A stem will then place the system on the Moon and ask if the period changes; for a horizontal spring, the answer is no; for a vertical spring, the answer is also no, because the new equilibrium absorbs the weight without changing ω. Candidates who say "the period on the Moon is longer because gravity is weaker" are applying pendulum logic to a spring system and lose the point.
Energy in SHM: the second FRQ workhorse
Unit 7's second scoring pillar is energy. The total mechanical energy of an SHM oscillator is conserved in the absence of damping, and the rubric awards two points for setting up E_total = (1/2)kA² = (1/2)mv² + (1/2)kx². The first equality links amplitude to total energy; the second partitions that total into kinetic energy (function of velocity) and elastic potential energy (function of displacement). On a 4-point QQT FRQ, the energy equation usually shows up in parts (b) and (c): part (b) might give the amplitude and ask for the maximum speed; part (c) gives a specific position x and asks for the speed at that position.
The work pattern is mechanical. From v_max = ωA = A√(k/m), the candidate computes maximum kinetic energy as (1/2)m(ωA)² = (1/2)kA², which equals the total energy. At an arbitrary position x, the elastic potential energy is (1/2)kx², and the kinetic energy is E_total − (1/2)kx² = (1/2)k(A² − x²). Setting this equal to (1/2)mv² gives v = ±ω√(A² − x²). The exam sometimes phrases this as "at what position is the speed equal to half the maximum speed?" — the algebra resolves to x = A√(3)/2, and the rubric wants both the magnitude and a sketch showing that position between equilibrium and amplitude.
A second energy variant is the energy bar chart, which the AP Physics 1 redesign introduced around 2021 and which now appears at least once per exam. The candidate draws three vertical bars — kinetic, potential, total — at three labelled positions: equilibrium, amplitude, and an intermediate point. The correct chart has total energy constant (flat bar), kinetic energy maximum at equilibrium and zero at amplitude, and potential energy zero at equilibrium and maximum at amplitude. A bar at the intermediate point should show a 1:3 split (or whatever the position demands) with the total unchanged. Candidates who draw a sloping total-energy bar lose the conceptual point, even if the kinetic and potential bars are right. In my experience, students who rehearse the chart on graph paper once per week for the last month of preparation internalise the bar height ratios in a way that raw equation-memorising never achieves.
Common pitfalls and how to avoid them on energy FRQs
Three errors account for the majority of lost points on Unit 7 energy questions. The first is mixing up A and x: amplitude A is a constant of the motion, while displacement x varies. Substituting A for x in the elastic potential energy formula gives the total energy, not the instantaneous potential energy, and the candidate then computes a constant KE that does not match the answer. The fix is to write the formula with explicit symbols: E_p = (1/2)k x(t)², where x(t) = A cos(ωt + φ). The second error is forgetting that the period formula for a vertical spring is identical to the horizontal case once equilibrium is established. The third error is treating gravitational potential energy as part of the SHM budget when the equilibrium shift already accounts for it. On a vertical spring, the gravitational PE is linear in displacement; combining it with the quadratic elastic PE produces a shifted parabola, and SHM occurs about the new minimum. Candidates who write the total energy as (1/2)kA² + mgA double-count the work done by gravity and overshoot the answer.
The reference circle: how AP Physics 1 connects rotation to SHM
One of the most elegant shortcuts on the AP Physics 1 exam is the relationship between uniform circular motion and SHM. Project the position of an object moving in a circle of radius A at angular speed ω onto a diameter, and the projection traces out exactly x(t) = A cos(ωt + φ). The vertical projection gives y(t) = A sin(ωt + φ). This is the reference-circle model, and the exam uses it in two ways: first, to derive velocity and acceleration components without calculus, and second, to explain the phase relationship between displacement, velocity, and acceleration in SHM.
The velocity of the circular-motion object is tangent to the circle, with magnitude v = Aω. Project that velocity onto the diameter perpendicular to the displacement projection, and the SHM velocity is v(t) = −Aω sin(ωt + φ). The negative sign and the sine reflect the 90° phase lead of velocity over displacement. Similarly, the centripetal acceleration of the circular-motion object has magnitude a = Aω², directed toward the centre; project that onto the displacement axis and you get a(t) = −Aω² cos(ωt + φ) = −ω²x(t). The acceleration leads or lags displacement by 180° — they are always in opposite directions, which is precisely the linear-restoring-force condition. Candidates who can draw the reference circle and label the projections in roughly 90 seconds on scratch paper have a reliable way to recover the phase relationships even if they forget the calculus derivation under time pressure.
The exam also uses the reference circle to derive the period of a pendulum in a single line. A pendulum bob moves on an arc of length s = Lθ, and the restoring force is the tangential component of gravity: F = −mg sin θ. For small θ, sin θ ≈ θ, so F ≈ −(mg/L)s, which is linear in s with effective k = mg/L. Substituting into T = 2π√(m/k) gives T = 2π√(L/g). The mass cancels, which is itself a non-trivial conceptual result the exam likes to test. A stem will sometimes give a 0.20 kg and a 0.40 kg pendulum bob on identical strings and ask whether their periods differ; the answer is no, which surprises most students on first encounter.