AP Physics C: Mechanics • Oscillations

Defining Simple Harmonic Motion (SHM)

Understanding the restoring-force criterion that governs oscillations from springs to pendulums to molecules.

A motion gauge with symbols for mass and movement.
Inside this lesson 10 sections 5 questions 9 figures

SECTION 1

Historical Context & Motivation

The study of oscillatory motion stretches back to antiquity, but the formal mathematical treatment of what we now call simple harmonic motion (SHM) crystallized over several centuries of observation, experiment, and theoretical insight. Galileo Galilei's legendary observation of a swinging chandelier in the Cathedral of Pisa—whether apocryphal or not—highlighted the remarkable regularity of a pendulum's period, independent of its amplitude for small swings. This insight planted the seed for precision timekeeping and, more broadly, for a rigorous theory of periodic motion. Subsequent work by Robert Hooke, Isaac Newton, and Leonhard Euler transformed these observations into a complete mathematical framework that remains one of the most widely applicable models in all of physics.

1583 • Galileo and the Pendulum

Galileo observes the isochronous nature of a swinging pendulum, noting that the period remains approximately constant for small angular displacements regardless of amplitude.

1660 • Hooke's Law Published

Robert Hooke publishes his anagram-encoded spring law, later decoded as ut tensio, sic vis ('as the extension, so the force'), establishing the linear restoring force central to SHM.

1687 • Newton's Principia

Newton's second law (F = ma) provides the framework to convert Hooke's linear force law into the differential equation of SHM, linking force, mass, and acceleration in oscillatory systems.

1736 • Euler's Analytical Methods

Leonhard Euler introduces the general sinusoidal solution x(t) = A cos(ωt + φ) for linear differential equations, giving SHM its elegant closed-form expression.

1900s • Quantum Harmonic Oscillator

The SHM framework is extended into quantum mechanics, where the quantum harmonic oscillator becomes foundational in molecular spectroscopy, quantum field theory, and solid-state physics.

What unifies all these developments is a single, powerful idea: whenever a system experiences a restoring force proportional to displacement, it oscillates sinusoidally about its equilibrium position. This criterion—the defining hallmark of SHM—appears across an astonishing range of physical contexts, from mass–spring systems and pendulums to LC circuits and vibrating molecules. Our task in this lesson is to rigorously define SHM, derive its governing differential equation, and understand the physical quantities that characterize the motion.

SECTION 2

Core Principles & Definitions

Simple harmonic motion is a specific type of periodic, oscillatory motion governed by a strict set of conditions. Before diving into the mathematics, it is essential to establish the foundational ideas that distinguish SHM from other forms of oscillation—such as damped, driven, or anharmonic oscillations. The following core principles collectively define SHM and provide the conceptual scaffolding for the differential equation we will encounter in Section 4.

1

Linear Restoring Force

The net force on the oscillating object is always directed toward the equilibrium position and is directly proportional to the displacement from equilibrium: Fnet = −kx. The negative sign ensures the force is restoring.
2

Sinusoidal Time Dependence

The position, velocity, and acceleration of the system are all sinusoidal functions of time. The position is described by x(t) = A cos(ωt + φ), with well-defined amplitude, angular frequency, and phase constant.
3

Constant Amplitude & Frequency

In ideal SHM (no friction or external driving forces), the amplitude A and angular frequency ω remain constant. Energy is conserved and continuously exchanged between kinetic and potential forms.
4

Isochronous Oscillation

The period T = 2π/ω is independent of amplitude—a property called isochronism. This is a direct consequence of the linearity of the restoring force and distinguishes SHM from anharmonic oscillations.
5

Equilibrium as Reference

All displacements, velocities, and forces are measured relative to a stable equilibrium position where the net force vanishes. Displacement x is defined as the deviation from this equilibrium.
✦

KEY TAKEAWAY

Think of SHM like a perfectly elastic rubber band attached to a ball on a frictionless surface: the farther you stretch it, the harder it pulls back—always proportionally. No matter how far you initially displace the ball, it always takes the same time to complete one full back-and-forth cycle. This proportional-restoring-force criterion is the single condition that defines SHM; every other property (sinusoidal motion, constant period, energy conservation) follows mathematically from it.

SECTION 3

Visual Explanation — Anatomy of SHM

The following diagram illustrates a mass–spring system executing SHM alongside the corresponding position-versus-time graph. The visual connects the physical configuration of the oscillator at key instants (maximum displacement, equilibrium, and maximum displacement in the opposite direction) with the sinusoidal waveform that describes the position x(t). Pay particular attention to how the restoring force arrow reverses direction as the mass crosses equilibrium and grows in magnitude as the mass moves farther from center.

Mass–Spring System: Three Key Instants & Position Grapht = 0 (x = +A)mF = −kAx = 0t = T/4 (x = 0)mF = 0x = 0t = T/2 (x = −A)mF = +kAx = 0tx+A−At=0T/4T/23T/4Tx(t) = A cos(ωt + φ)
Top row: The mass–spring system at three key instants during one half-period. At t = 0 the spring is maximally stretched (x = +A) and the restoring force points left; at t = T/4 the mass passes through equilibrium with maximum speed and zero force; at t = T/2 the spring is maximally compressed (x = −A) and the force points right. Bottom: The corresponding cosine waveform, with colored dots matching each instant above.

Several critical features are visible in the diagram. First, notice that the restoring force arrows are longest at the turning points (x = ±A) and vanish at equilibrium—this is the graphical signature of the linear restoring force F = −kx. Second, at the equilibrium position the mass has its maximum velocity (all energy is kinetic), while at the turning points the velocity is momentarily zero (all energy is elastic potential). Finally, the position graph below is a smooth cosine curve, confirming that the motion is sinusoidal with constant amplitude A and period T. Every real SHM system—whether it involves a spring, a pendulum, or a vibrating tuning fork—shares precisely this qualitative structure.

A spring exerts a restoring force. For m = 0.50 kg and k = 200 N/m, displacement −0.10 m gives rightward force, zero displacement gives zero force, and +0.10 m gives leftward force. T = 0.314 s and maximum speed is 2 m/s.A spring exerts a restoring force. For m = 0.50 kg and k = 200 N/m, displacement −0.10 m gives rightward force, zero displacement gives zero force, and +0.10 m gives leftward force. T = 0.314 s and maximum speed is 2 m/s.

SECTION 4

Mathematical Framework

The mathematical definition of SHM emerges directly from Newton's second law applied to a linear restoring force. Let a particle of mass m be subject to a net force F = −kx, where x is the displacement from stable equilibrium and k is a positive constant (the spring constant for a Hookean spring). Newton's second law gives F = ma = m(d²x/dt²), and equating these expressions produces the defining differential equation of SHM.

DEFINING DIFFERENTIAL EQUATION OF SHM

d2x/dt2=−ω2xd²x/dt² = −ω²x

where ω = √(k/m) is the angular frequency (rad/s), x is displacement from equilibrium (m), and t is time (s). This equation states that the acceleration is always proportional to, and oppositely directed from, the displacement.

The general solution to this second-order linear ODE with constant coefficients is well known from differential equations. Because the characteristic equation r² + ω² = 0 has purely imaginary roots r = ±iω, the general real solution takes the form of a sinusoidal function with two constants of integration determined by initial conditions.

GENERAL SOLUTION — POSITION

x(t)=Acos(ωt+φ)x(t) = A cos(ωt + φ)

A is the amplitude—the maximum magnitude of displacement from equilibrium (m). φ is the phase constant (rad), determined by the initial position and velocity. An equivalent form is x(t) = C₁ cos(ωt) + C₂ sin(ωt), where A = √(C₁² + C₂²) and tan φ = −C₂/C₁.

Differentiating the position function once gives velocity; differentiating again gives acceleration. These expressions are essential for problem-solving on the AP exam.

VELOCITY IN SHM

v(t)=dx/dt=−Aωsin(ωt+φ)v(t) = dx/dt = −Aω sin(ωt + φ)

Maximum speed occurs at equilibrium (x = 0): vmax = Aω. The velocity leads the position by π/2 radians.

ACCELERATION IN SHM

a(t)=d2x/dt2=−Aω2cos(ωt+φ)=−ω2xa(t) = d²x/dt² = −Aω² cos(ωt + φ) = −ω²x

Maximum acceleration occurs at the turning points (x = ±A): amax = Aω². The acceleration is always anti-parallel to displacement—this is the mathematical restatement of the restoring force condition.

💡

Period and Frequency Relations

The angular frequency ω connects to the period T and ordinary frequency f via T = 2π/ω and f = 1/T = ω/(2π). For a mass–spring system, T = 2π√(m/k). Note that T depends on mass and spring constant but not on amplitude—this is the mathematical origin of isochronism.

Two hanging-spring setups and a coil close-up accompany a syntheticT²-versus-mass fit with a positive intercept.Measure displacement from equilibrium. For a linear restoring force, acceleration is proportional to that displacement and points in the opposite direction.

Displacement from loaded equilibrium. A 2 kg mass stretches the spring 0.050 m, giving k = 392 N/m and ω = 14 rad/s. With upward displacement measured from equilibrium and x(0) = −0.030 m, x(t) = −0.030 cos(14t) m.A 2 kg mass stretches the spring 0.050 m, giving k = 392 N/m and ω = 14 rad/s. With upward displacement measured from equilibrium and x(0) = −0.030 m, x(t) = −0.030 cos(14t) m.

SECTION 5

Energy in SHM & Graphical Relationships

A hallmark of SHM is the continuous, lossless exchange between kinetic energy and potential energy. For a mass on a spring, the elastic potential energy is U = ½kx² and the kinetic energy is K = ½mv². Because no non-conservative forces act in ideal SHM, the total mechanical energy E = K + U remains constant throughout the motion. Substituting the SHM solutions for x(t) and v(t) and using the identity sin²θ + cos²θ = 1 yields E = ½kA², confirming that the total energy depends only on the amplitude and the spring constant. The following diagram shows how x(t), v(t), K(t), and U(t) relate over one full period.

Position, Velocity & Energy vs. Time in SHM+A−Atx, vx(t)v(t)0T/4T/23T/4TtEnergyE½kA²UKT/4T/23T/4TK + U = E = ½kA² = constant (green dashed line)
Upper panel: Position x(t) (solid violet) and velocity v(t) (dashed cyan) are 90° out of phase. Lower panel: Kinetic energy K (pink) and potential energy U (amber) oscillate at twice the frequency of x(t), while the total energy E (dashed green) remains constant at ½kA².

Several important observations arise from this energy diagram. The kinetic and potential energy curves each oscillate between 0 and ½kA² at twice the frequency of the displacement—mathematically because sin²(ωt) and cos²(ωt) both have period π/ω = T/2. When the displacement is at a maximum (a turning point), the velocity is zero and all energy is potential. At equilibrium, the displacement is zero, the velocity is at its maximum, and all energy is kinetic. At every intermediate instant, K + U = ½kA², forming the horizontal dashed line in the diagram. This energy conservation result is routinely tested on the AP exam, both conceptually (interpreting energy bar charts) and quantitatively (solving for velocity at a given displacement using ½kA² = ½kx² + ½mv²).

ENERGY CONSERVATION IN SHM

E=½kA2=½kx2+½mv2E = ½kA² = ½kx² + ½mv²

Solving for speed at displacement x gives v = ω√(A² − x²). This is extremely useful for finding the speed at any position without needing to know the time.

SECTION 6

Worked Example — Mass–Spring SHM

A 0.50 kg block attached to a horizontal spring (k = 200 N/m) on a frictionless surface is pulled 0.10 m from its equilibrium position and released from rest. Find (a) the angular frequency and period, (b) the position as a function of time, (c) the maximum speed, and (d) the speed when the block is 0.060 m from equilibrium.

Block on a Horizontal Spring

  1. 1

    Step 1 — Identify Given Values

    Mass m = 0.50 kg, spring constant k = 200 N/m, initial displacement x₀ = 0.10 m, initial velocity v₀ = 0 (released from rest). The motion is SHM because the spring exerts a linear restoring force F = −kx.
  2. 2

    Step 2 — Angular Frequency and Period

    ω = √(k/m) = √(200/0.50) = √400 = 20 rad/s. The period is T = 2π/ω = 2π/20 = π/10 s ≈ 0.314 s.
    ω = 20 rad/s, T ≈ 0.314 s
  3. 3

    Step 3 — Position Function x(t)

    The general solution is x(t) = A cos(ωt + φ). At t = 0, x(0) = A cos φ = 0.10 m and v(0) = −Aω sin φ = 0. The velocity condition requires sin φ = 0, so φ = 0 (choosing the principal value). Then A cos 0 = A = 0.10 m.
    x(t) = 0.10 cos(20t) m
  4. 4

    Step 4 — Maximum Speed

    v_max = Aω = (0.10)(20) = 2.0 m/s. This maximum speed occurs as the block passes through x = 0.
    v_max = 2.0 m/s
  5. 5

    Step 5 — Speed at x = 0.060 m

    Using energy conservation: v = ω√(A² − x²) = 20 × √(0.10² − 0.060²) = 20 × √(0.010 − 0.0036) = 20 × √0.0064 = 20 × 0.080 = 1.6 m/s.
    v = 1.6 m/s at x = 0.060 m

SECTION 7

SHM Compared to Other Types of Motion

SHM is an idealization—real oscillators inevitably encounter friction, external driving forces, or nonlinear restoring forces. Understanding how SHM relates to more general oscillatory motion helps clarify the assumptions and limitations of the model. The table below compares SHM with several closely related types of motion you may encounter on the AP exam or in more advanced coursework.

PropertySimple Harmonic MotionDamped OscillationDriven (Forced) Oscillation
Net restoring forceF = −kx (linear, proportional)F = −kx − bv (includes velocity-dependent damping)F = −kx − bv + F₀cos(ω_d t)
AmplitudeConstant — energy is conservedDecreases exponentially over timeReaches steady-state; can exhibit resonance
Period / FrequencyIndependent of amplitude (isochronous)Slightly altered by damping coefficientSteady-state freq equals driving frequency ω_d
EnergyE = ½kA² = constantDecreases due to dissipation by dampingExternal source replenishes energy; can grow at resonance
AP Exam relevancePrimary focus of Oscillations unitQualitative understanding expected; detailed math beyond scopeQualitative understanding of resonance expected

Comparison of SHM with damped and driven oscillatory motion

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KEY TAKEAWAY

SHM is to oscillatory motion what the ideal gas is to thermodynamics—a clean, exactly solvable reference model. Just as real gases deviate from ideal behavior at high pressures, real oscillators deviate from SHM when damping, driving forces, or nonlinear restoring forces become significant. On the AP exam, you will encounter problems set firmly in the SHM regime (frictionless surfaces, ideal springs, small-angle pendulums), but you should be prepared to qualitatively describe how damping or large amplitudes alter the motion.

A spring cart on a rail connected to a cutaway fluid dashpot with a visible piston assembly.With linear damping, the phase trajectory contracts and mechanical energy decreases at the rate −bv². The undamped comparison has constant total mechanical energy.

Damping contracts the phase trajectory. With linear damping, the phase trajectory contracts and mechanical energy decreases at the rate −bv². The undamped comparison has constant total mechanical energy.With linear damping, the phase trajectory contracts and mechanical energy decreases at the rate −bv². The undamped comparison has constant total mechanical energy.

SECTION 8

Connection to Advanced Theory — SHM Beyond the Spring

The power of SHM lies in its universality: any system whose potential energy has a local minimum can be approximated as a simple harmonic oscillator for small displacements about that minimum. If U(x) is the potential energy function for a one-dimensional system, a Taylor expansion about the equilibrium position x₀ gives U(x) ≈ U(x₀) + ½U″(x₀)(x − x₀)². The linear term vanishes because dU/dx = 0 at equilibrium, and the constant term merely shifts the energy reference. Identifying the effective spring constant as keff = U″(x₀), we recover the SHM equation d²x/dt² = −(keff/m)x. This technique is used extensively in classical mechanics, molecular physics, and even general relativity to analyze small oscillations about equilibria.

SystemRestoring 'Force' / TorqueEffective ωSHM Condition
Mass–springF = −kxω = √(k/m)Hooke's law valid (spring not overstretched)
Simple pendulumτ = −mgL sin θ ≈ −mgLθω = √(g/L)Small angle: sin θ ≈ θ
Physical pendulumτ = −mgh sin θ ≈ −mghθω = √(mgh/I)Small angle; I about pivot
Torsional oscillatorτ = −κθω = √(κ/I)Linear restoring torque (Hooke analog)
LC circuitV = −Q/C (analogous to F = −kx)ω = 1/√(LC)No resistance; ideal L and C

Systems that exhibit SHM under appropriate conditions

For the AP Physics C: Mechanics exam, you are expected to recognize and analyze SHM in mass–spring systems and simple pendulums quantitatively. Physical pendulums and torsional oscillators also appear, requiring you to set up the rotational analog of the SHM differential equation (Iα = −κθ or equivalent). The LC circuit analogy belongs to AP Physics C: E&M, but appreciating the structural similarity deepens your understanding of why SHM is such a universal paradigm. In more advanced mechanics courses, the small-oscillation approximation via Taylor expansion becomes a standard tool for analyzing motion near any stable equilibrium in multi-degree-of-freedom systems.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A particle moves along the x-axis under the influence of a single force. At any displacement x from equilibrium, the acceleration of the particle is given by a = −9x. Which of the following statements about this motion is correct?
  • A. The motion is SHM with angular frequency ω = 3 rad/s.
  • B. The motion is SHM with angular frequency ω = 9 rad/s.
  • C. The motion is SHM with angular frequency ω = 81 rad/s.
  • D. The motion is not SHM because no spring constant is given.
Show answer (A)
The defining equation of SHM is a = −ω²x. Comparing a = −9x with a = −ω²x gives ω² = 9, so ω = 3 rad/s. Choice A is correct. Choice B incorrectly identifies ω = 9 (this would require ω² = 81). Choice C confuses ω² with ω⁴ or simply squares unnecessarily. Choice D is wrong because the SHM criterion is purely about the mathematical form of the restoring force (or equivalently the acceleration), not whether a physical spring is present.

Question 2: BASIC CALCULATION

A 0.25 kg mass on a horizontal spring oscillates with a period of 0.50 s. What is the spring constant k?
  • A. k ≈ 9.9 N/m
  • B. k ≈ 19.7 N/m
  • C. k ≈ 39.5 N/m
  • D. k ≈ 78.9 N/m
Show answer (C)
From T = 2π√(m/k), solving for k: k = m(2π/T)² = 0.25 × (2π/0.50)² = 0.25 × (4π)² = 0.25 × 157.9 ≈ 39.5 N/m. Choice C is correct. Choice A results from using ω = 2π/T but forgetting to square it. Choice B results from an algebraic error (dividing by 2 instead of squaring properly). Choice D would result from doubling the correct answer, possibly by mishandling the mass value. Always check units: [kg][rad/s]² = [kg][s⁻²] = [N/m].

Question 3: INTERMEDIATE

A block oscillating on a spring has amplitude A = 0.20 m and angular frequency ω = 10 rad/s. At what displacement from equilibrium is the speed of the block equal to half its maximum speed?
  • A. x = 0.10 m
  • B. x ≈ 0.15 m
  • C. x ≈ 0.17 m
  • D. x ≈ 0.19 m
Show answer (C)
The speed at displacement x is v = ω√(A² − x²), and v_max = Aω. Setting v = v_max/2: ω√(A² − x²) = Aω/2, so A² − x² = A²/4, giving x² = 3A²/4, thus x = A√3/2 = 0.20 × (√3/2) ≈ 0.20 × 0.866 ≈ 0.173 m ≈ 0.17 m. Choice C is correct. Choice A (x = A/2) is a common error from assuming v = v_max/2 when x = A/2—this confuses linear proportion with the quadratic energy relation. Choice B is close but corresponds to no clean algebraic result. Choice D is nearly A itself, where the speed would be close to zero, not half-maximum.

Question 4: APPLIED

10 pts
A 2.0 kg block is attached to a vertical spring and allowed to come to rest at the new equilibrium. The spring stretches 0.050 m from its natural length under the weight of the block. The block is then pulled 0.030 m below this new equilibrium and released. (a) Show that the subsequent motion is SHM and find the angular frequency ω. (b) Write x(t), defining the positive direction as upward from the new equilibrium. (c) Find the maximum speed of the block. (d) Find the maximum net force on the block during the oscillation.
Show answer
Rubric (10 points): (a) Showing SHM and finding ω (4 pts): • (1 pt) At the new equilibrium, mg = kΔy₀, so k = mg/Δy₀ = (2.0)(9.8)/0.050 = 392 N/m. • (1 pt) Define displacement x from the new equilibrium. The net force is F_net = −k(Δy₀ + x) + mg = −kx (since kΔy₀ = mg). The gravitational and equilibrium-stretch terms cancel. • (1 pt) Therefore F_net = −kx, which is a linear restoring force → SHM. • (1 pt) ω = √(k/m) = √(392/2.0) = √196 = 14 rad/s. (b) Writing x(t) (2 pts): • (1 pt) At t = 0, x(0) = −0.030 m (below equilibrium, so negative if up is positive) and v(0) = 0. • (1 pt) x(t) = −0.030 cos(14t) m. [Equivalently, x(t) = 0.030 cos(14t + π) m.] (c) Maximum speed (2 pts): • (1 pt) v_max = Aω = (0.030)(14) = 0.42 m/s. • (1 pt) This occurs as the block passes through x = 0 (the equilibrium position). (d) Maximum net force (2 pts): • (1 pt) F_max = kA = (392)(0.030) = 11.8 N. • (1 pt) Equivalently, F_max = ma_max = m(Aω²) = 2.0 × 0.030 × 196 = 11.8 N. This occurs at the turning points x = ±A.

Question 5: CRITICAL THINKING

7 pts
A small marble of mass m rolls without slipping inside a hemispherical bowl of radius R. For small displacements from the bottom of the bowl, the marble's translational coordinate along the bowl follows x(t) approximately as SHM. (a) Using energy methods or torque analysis, derive the angular frequency of oscillation for the marble. (Recall that the moment of inertia of a solid sphere about its center is I = ²⁄₅mr².) Clearly state any approximations. (b) Compare the resulting period to that of a simple pendulum of length R and explain the physical reason for any difference. (c) Does the period depend on the mass of the marble? Justify your answer using the derived expression. (d) A student replaces the solid marble with a thin hollow spherical shell of the same radius and mass. Predict qualitatively how the period changes and explain why.
Show answer
Rubric (7 points): (a) Derivation of ω (3 pts): • (1 pt) For small angular displacement θ from the bottom, the restoring torque about the contact point is τ = −mg(R − r)sinθ ≈ −mg(R − r)θ (small angle approximation). For r ≪ R this simplifies to τ ≈ −mgRθ. • (1 pt) The total moment of inertia about the contact point (using parallel axis theorem) is I_contact = I_cm + mr² = ²⁄₅mr² + mr² = ⁷⁄₅mr². Setting Iα = τ: (⁷⁄₅mr²)α = −mg(R − r)θ. For r ≪ R: α = −(5g)/(7R) × θ. • (1 pt) This has the SHM form α = −ω²θ, so ω = √(5g/(7R)). More precisely ω = √(5g/[7(R − r)]). (b) Comparison with simple pendulum (1 pt): • (1 pt) A simple pendulum of length R has ω_p = √(g/R). Since 5/7 < 1, ω_marble < ω_pendulum, so T_marble > T_pendulum. The marble oscillates more slowly because rotational kinetic energy takes up part of the total energy, effectively adding inertia and reducing the angular frequency. (c) Mass independence (1 pt): • (1 pt) The mass m cancels completely in ω = √(5g/(7R)). Both the restoring torque (∝ m) and the rotational inertia (∝ m) are proportional to mass, so T is independent of m, just as for a simple pendulum. (d) Hollow shell prediction (2 pts): • (1 pt) A thin hollow spherical shell has I_cm = ²⁄₃mr², which is larger than ²⁄₅mr² for the solid sphere. The total moment of inertia becomes ²⁄₃mr² + mr² = ⁵⁄₃mr², giving ω = √(3g/(5R)). • (1 pt) Since 3/5 < 5/7, ω_shell < ω_solid, so the period increases. Physically, a hollow shell concentrates mass farther from its center, increasing rotational inertia and slowing the oscillation.

SUMMARY

Lesson Summary

Summary

Simple harmonic motion is defined by a linear restoring force proportional to displacement: F = −kx. Applying Newton's second law yields the defining differential equation d²x/dt² = −ω²x, whose general solution is the sinusoidal function x(t) = A cos(ωt + φ). The angular frequency ω = √(k/m) determines the period T = 2π/ω, which is independent of amplitude (isochronism). Velocity v(t) = −Aω sin(ωt + φ) leads position by 90°, and acceleration a(t) = −ω²x is always anti-parallel to displacement.

Energy in SHM is conserved: E = ½kA² is constant, with kinetic and potential energy exchanging at twice the oscillation frequency. The SHM framework applies to any system with a quadratic potential energy minimum—mass–spring systems, small-angle pendulums, torsional oscillators, and more. Master the differential equation, the energy conservation relation v = ω√(A² − x²), and the ability to identify effective spring constants in novel contexts, and you will be well-prepared for the AP exam's Oscillations questions.

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