College Physics • Oscillations & Simple Harmonic Motion

Representing and Analyzing SHM

Master the displacement, velocity, and acceleration functions that unify every oscillating system in physics.

An atom with an orbital path and traveling waves.
Inside this lesson 10 sections 5 questions 9 figures

SECTION 1

Historical Context & Motivation

Oscillatory motion — the rhythmic back-and-forth of a pendulum, the vibration of a guitar string, the pulsation of a quartz crystal — is among the most ubiquitous phenomena in nature. Long before physicists had the mathematical tools to describe it precisely, craftspeople and astronomers noticed that certain repetitive motions seemed to follow regular, predictable patterns. The quest to represent these patterns quantitatively gave rise to simple harmonic motion (SHM), a framework so powerful that it remains central to mechanics, electromagnetism, acoustics, and quantum mechanics alike. Understanding how SHM is represented and analyzed is therefore not merely an exercise in kinematics; it is a gateway to virtually every branch of modern physics.

1583 • Galileo's Pendulum Observation

Galileo Galilei, observing a swinging chandelier in the Cathedral of Pisa, noted that each oscillation appeared to take the same amount of time regardless of amplitude — the property of isochronism. This insight launched the scientific study of periodic motion.

1678 • Hooke's Law Published

Robert Hooke announced his law of elasticity — ut tensio, sic vis (as the extension, so the force) — providing the linear restoring force that underlies the spring-mass model of SHM.

1687 • Newton's Principia

Isaac Newton's second law, F = ma, combined with Hooke's linear restoring force, yielded the second-order differential equation whose solutions are sinusoidal functions — the mathematical heart of SHM.

1822 • Fourier's Analytical Theory of Heat

Joseph Fourier demonstrated that any periodic function can be decomposed into a sum of sinusoidal components, revealing SHM as the fundamental building block of all periodic phenomena.

From Galileo's qualitative observation to Fourier's sweeping generalization, a central question has persisted: how do we represent the instantaneous state of an oscillating system and extract measurable quantities — position, velocity, acceleration, energy — at any moment in time? This lesson develops the full representational toolkit: sinusoidal functions, phasor diagrams, phase-space plots, and energy graphs, equipping you to analyze SHM from every angle.

SECTION 2

Core Principles & Definitions

Before diving into equations and graphs, it is essential to establish the foundational ideas that define SHM and distinguish it from other types of oscillatory motion. Simple harmonic motion arises whenever the net restoring force on an object is directly proportional to its displacement from equilibrium and directed opposite to that displacement. This single condition — a linear restoring force — guarantees sinusoidal time dependence and makes the mathematics tractable. The following grid summarizes the key quantities and concepts you will work with throughout this lesson.

1

Amplitude (A)

The maximum displacement from the equilibrium position, measured in meters. It determines the energy scale of the oscillation: total mechanical energy is proportional to A².
2

Angular Frequency (ω)

The rate of change of the phase angle, measured in rad/s. Related to the period T by ω = 2π/T and to the frequency f by ω = 2πf. It is set entirely by system parameters (e.g., k/m for a spring).
3

Phase Constant (φ₀)

The initial phase angle at t = 0, encoding the starting conditions of the motion. Two oscillators with identical A and ω but different φ₀ are out of phase with each other.
4

Period (T) & Frequency (f)

The period is the time for one complete cycle (seconds), while the frequency is the number of cycles per second (Hz). They are reciprocals: f = 1/T. These are the most directly measurable SHM parameters.
5

Phase (ωt + φ₀)

The total argument of the sinusoidal function at time t. The phase determines the instantaneous displacement, velocity, and acceleration simultaneously, serving as the single independent variable that drives the entire motion.
✦

KEY TAKEAWAY

Think of simple harmonic motion as the shadow of uniform circular motion. Imagine a ball moving at constant speed around a vertical circle while a lamp casts the ball's shadow onto a wall. The shadow slides back and forth in perfect SHM — its position is the cosine of the ball's angle. The amplitude is the circle's radius, the angular frequency is the ball's angular speed, and the phase constant tells you where on the circle the ball starts. Every SHM equation is, in this sense, just a projection of something going around in a circle.

SECTION 3

Visual Explanation — Displacement, Velocity & Acceleration Graphs

The most direct way to understand SHM is to see how displacement, velocity, and acceleration evolve simultaneously over time. The following diagram plots all three kinematic quantities for one complete cycle of a cosine-based SHM with phase constant φ₀ = 0. Notice three critical features: velocity leads displacement by a quarter-cycle (π/2 radians), acceleration leads velocity by another quarter-cycle, and acceleration is always proportional to and opposite in sign to displacement.

x(t), v(t), and a(t) for SHM (φ₀ = 0)0Time →Amplitude0T/4T/23T/4T+A−Ax(t) = A cos(ωt)v(t) = −Aω sin(ωt)a(t) = −Aω² cos(ωt)x = +Av = 0x = 0v = −Aω
Displacement x(t) (solid cyan) is a cosine function peaking at t = 0. Velocity v(t) (dashed violet) is the time derivative: it is zero when displacement is maximal and reaches its own extremum when x crosses zero. Acceleration a(t) (dotted pink) is always opposite to displacement, confirming the restoring-force nature of SHM.

A few patterns visible in the diagram deserve emphasis. At t = 0, displacement is at its positive maximum while velocity is zero — the object is momentarily at rest at the turning point. A quarter-period later, displacement has dropped to zero (the object crosses equilibrium) and velocity reaches its maximum negative value — the object is moving fastest through the center. Meanwhile, acceleration at t = 0 is at its negative extreme, pointing back toward equilibrium; it passes through zero at T/4 when the object is at the equilibrium point. These quarter-cycle phase relationships are the signature of SHM and arise directly from the fact that each kinematic quantity is the time derivative of the one before it.

SECTION 4

Mathematical Framework

The mathematical description of SHM flows naturally from Newton's second law applied to a linear restoring force. Consider a mass m attached to a spring of stiffness k displaced from equilibrium by x. The net force is F = −kx, so Newton's second law gives ma = −kx, or equivalently d²x/dt² = −(k/m)x. Defining ω² = k/m, the equation of motion becomes the SHM differential equation. Its general solution is a sinusoidal function characterized by three parameters: amplitude, angular frequency, and phase constant.

SHM DIFFERENTIAL EQUATION

d2x/dt2+ω2x=0d²x/dt² + ω²x = 0

x = displacement from equilibrium (m); ω = √(k/m) = angular frequency (rad/s); t = time (s). This second-order ODE has two linearly independent solutions: sin(ωt) and cos(ωt).

DISPLACEMENT

x(t)=Acos(ωt+φ0)x(t) = A cos(ωt + φ₀)

A = amplitude (m), the maximum |x|; ω = angular frequency (rad/s); φ₀ = phase constant (rad), determined by initial conditions. Equivalently, x(t) = A sin(ωt + φ₀ + π/2).

VELOCITY

v(t)=dx/dt=−Aωsin(ωt+φ0)v(t) = dx/dt = −Aω sin(ωt + φ₀)

Maximum speed vmax = Aω occurs at equilibrium (x = 0). Velocity leads displacement by π/2 radians.

ACCELERATION

a(t)=d2x/dt2=−Aω2cos(ωt+φ0)=−ω2x(t)a(t) = d²x/dt² = −Aω² cos(ωt + φ₀) = −ω²x(t)

Maximum acceleration |amax| = Aω² occurs at maximum displacement. The relation a = −ω²x is the defining kinematic signature of SHM.

💡

Connecting Initial Conditions to A and φ₀

Given initial position x₀ = x(0) and initial velocity v₀ = v(0), the amplitude and phase constant are found from A = √(x₀² + (v₀/ω)²) and tan(φ₀) = −v₀/(ωx₀). These two equations let you convert any set of initial conditions into the standard cosine representation.

Energy provides a complementary perspective. The kinetic energy K = ½mv² and the potential energy U = ½kx² both oscillate in time, but their sum — the total mechanical energy E = ½kA² — remains constant in the absence of dissipation. At maximum displacement, all energy is potential; at equilibrium, all energy is kinetic. This continuous exchange between kinetic and potential energy is what sustains the oscillation indefinitely in the ideal SHM model.

ENERGY CONSERVATION IN SHM

E=½kA2=½kx2+½mv2=constantE = ½kA² = ½kx² + ½mv² = constant

k = spring constant (N/m); m = mass (kg). At x = ±A, v = 0 and U = E. At x = 0, |v| = Aω and K = E.

SECTION 5

Multiple Representations — Phasors & Energy Diagrams

While x-vs-t graphs are the most intuitive representation of SHM, physicists routinely employ two additional visual frameworks. The phasor diagram represents the oscillation as a rotating vector (the phasor) of length A in the complex plane; the projection of this vector onto the real axis gives the instantaneous displacement. As time progresses, the phasor rotates counterclockwise at angular speed ω. This representation is especially powerful when you need to add two oscillations of the same frequency, because vector addition in the phasor plane replaces trigonometric identities. The energy diagram plots kinetic energy, potential energy, and total energy against displacement x, revealing how energy shuttles between forms as the object moves through the oscillation cycle.

Phasor DiagramEnergy vs. DisplacementReImAx(t)ωt+φ₀Phasor rotates CCW at ωxEnergy−A+AE = ½kA²K = ½mv²U = ½kx²K=E at x=0U=E
Left: The phasor (amber arrow of length A) rotates counterclockwise. Its real-axis projection (cyan dot) gives the instantaneous displacement x(t). Right: Energy versus displacement. The parabolic potential energy U (dashed violet) and its mirror-image kinetic energy K (solid pink) always sum to the constant total energy E (dashed green line). At x = 0 all energy is kinetic; at x = ±A all energy is potential.

The phasor diagram on the left reinforces the analogy between SHM and circular motion introduced earlier. The instantaneous displacement is simply the horizontal projection of the rotating phasor onto the real axis. When the phasor points to the right (angle = 0), x = +A; when it points straight up (angle = π/2), x = 0. This makes it visually obvious why displacement and velocity are π/2 out of phase — velocity corresponds to the projection onto the imaginary axis, which reaches its extreme precisely when the real-axis projection is zero.

The energy diagram on the right tells a complementary story. Notice that the potential energy parabola U = ½kx² opens upward, while the kinetic energy curve K = E − U is an inverted parabola that peaks at x = 0. The object's displacement can never exceed ±A because beyond that point the required potential energy would exceed the total energy — there is simply no kinetic energy left. This graphical constraint is the energy analog of the turning points in the x-vs-t plot.

SECTION 6

Worked Example

A 0.50 kg block is attached to a horizontal spring (k = 200 N/m) on a frictionless surface. At t = 0 the block is pulled 0.10 m from equilibrium and released from rest. Determine the angular frequency, the equations of motion x(t), v(t), and a(t), the maximum speed, the maximum acceleration, and the displacement and velocity at t = 0.040 s.

Spring-Mass Oscillator Analysis

  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 m/s. Since the block is released from rest at maximum displacement, the phase constant φ₀ = 0 when using the cosine form.
  2. 2

    Step 2 — Calculate Angular Frequency

    ω = √(k/m) = √(200/0.50) = √400 = 20 rad/s. The period is T = 2π/ω = 2π/20 ≈ 0.314 s, and the frequency is f = 1/T ≈ 3.18 Hz.
    ω = 20 rad/s
  3. 3

    Step 3 — Write Equations of Motion

    x(t) = A cos(ωt) = 0.10 cos(20t) m. Differentiating: v(t) = −Aω sin(ωt) = −(0.10)(20) sin(20t) = −2.0 sin(20t) m/s. Differentiating again: a(t) = −Aω² cos(ωt) = −(0.10)(400) cos(20t) = −40 cos(20t) m/s².
    x(t) = 0.10 cos(20t) m; v(t) = −2.0 sin(20t) m/s; a(t) = −40 cos(20t) m/s²
  4. 4

    Step 4 — Maximum Speed and Acceleration

    v_max = Aω = (0.10)(20) = 2.0 m/s, occurring at equilibrium (x = 0). a_max = Aω² = (0.10)(400) = 40 m/s², occurring at the turning points (x = ±A).
    v_max = 2.0 m/s; a_max = 40 m/s²
  5. 5

    Step 5 — Evaluate at t = 0.040 s

    Phase angle: ωt = (20)(0.040) = 0.80 rad ≈ 45.8°. Displacement: x = 0.10 cos(0.80) = 0.10 × 0.6967 ≈ 0.070 m. Velocity: v = −2.0 sin(0.80) = −2.0 × 0.7174 ≈ −1.43 m/s. The negative sign indicates the block is moving in the negative direction (toward equilibrium and beyond).
    x(0.040) ≈ 0.070 m; v(0.040) ≈ −1.43 m/s

SECTION 7

Strengths & Limitations of the SHM Model

The SHM model is one of the most widely applied idealizations in physics, but like all models, it has a domain of validity. Understanding where the model excels and where it breaks down is essential for applying it wisely in real-world contexts.

FeatureStrengthLimitation
Restoring forceExact for any system with a strictly linear restoring force (ideal springs, LC circuits)Most real restoring forces are only approximately linear for small displacements; large amplitudes require anharmonic corrections
DampingCaptures the essential frequency and phase relationships even in lightly damped systemsIgnores energy loss entirely; real oscillators dissipate energy through friction, air resistance, or radiation
Amplitude independencePeriod is independent of amplitude — a powerful simplification enabling precision timekeepingIn real pendulums, period increases slightly with amplitude; in stiffening or softening springs, the nonlinearity alters the period
SuperpositionSHM solutions can be linearly superposed (Fourier analysis), enabling decomposition of complex periodic signalsNonlinear oscillators do not obey superposition; mode coupling and chaos can arise
Energy descriptionClean quadratic potential energy gives exact analytical solutions and elegant energy graphsReal potentials include cubic and higher-order terms, and energy is not conserved in dissipative systems

Strengths and limitations of the ideal SHM model

✦

KEY TAKEAWAY

SHM is to oscillatory motion what the ideal gas law is to thermodynamics: a powerful first-order model that captures the essential physics and serves as the baseline from which corrections (damping, nonlinearity, driving forces) are measured. In engineering, the standard practice is to begin every vibration analysis with an SHM model and then add complexity only as measurements demand. This is possible because near any stable equilibrium, the restoring force is approximately linear — making SHM not just a simplification, but a genuinely excellent approximation for small oscillations.

SECTION 8

Connection to Damped, Driven, and Coupled Oscillations

Ideal SHM is the foundation, but real physical systems introduce complications that require extensions of the basic model. The three most important extensions are damped oscillations (energy dissipation reduces amplitude over time), driven (forced) oscillations (an external periodic force sustains or modifies the motion), and coupled oscillations (two or more oscillators interact, giving rise to normal modes). Each of these builds directly on the SHM framework you have learned in this lesson.

PropertyIdeal SHMDamped / Driven / Coupled
Equation of motiond²x/dt² + ω²x = 0d²x/dt² + 2γ(dx/dt) + ω₀²x = (F₀/m) cos(ω_d t)
AmplitudeConstant for all timeDecays exponentially (damped) or reaches a steady-state value set by driving frequency and damping (driven)
Frequencyω = √(k/m), fixedShifts to ω' = √(ω₀² − γ²) for underdamped; resonance peak near ω₀ for driven systems
EnergyTotal mechanical energy E = ½kA² = constantEnergy decays (damped) or reaches dynamic equilibrium where power input equals dissipation (driven)
Key new phenomenonNone — pure sinusoidal motionResonance: amplitude peaks dramatically when driving frequency matches natural frequency, with applications from MRI to bridge engineering

Ideal SHM compared with its real-world extensions

The transition from ideal to damped SHM is mathematically elegant: one simply adds a velocity-dependent damping term 2γ(dx/dt) to the differential equation. When γ < ω₀ (the underdamped regime), the solution is still sinusoidal but with an exponentially decaying envelope: x(t) = Ae−γt cos(ω't + φ₀). In the driven case, adding a periodic external force introduces the spectacular phenomenon of resonance, where the steady-state amplitude reaches a maximum when the driving frequency matches the system's natural frequency. Both extensions rely on the SHM kinematic quantities — amplitude, frequency, and phase — developed in this lesson, which is why mastering the representation of ideal SHM is the essential first step.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A mass on a spring passes through the equilibrium position. At this instant, which of the following is true about its speed, acceleration, and potential energy? Explain your reasoning using the phase relationships among x(t), v(t), and a(t).
Show answer
At equilibrium (x = 0), the displacement is zero so the restoring force and therefore the acceleration are also zero (a = −ω²x = 0). Potential energy U = ½kx² = 0 as well. By contrast, the velocity reaches its maximum magnitude v_max = Aω because all energy is kinetic: K = ½mv² = ½kA² = E. This follows directly from the quarter-cycle phase lead of velocity over displacement: velocity is at an extremum precisely when displacement passes through zero.

Question 2: BASIC CALCULATION

A 0.25 kg mass oscillates on a spring with k = 100 N/m. If the amplitude is 0.08 m and the phase constant φ₀ = 0, find the period, the maximum speed, and the displacement at t = T/8.
Show answer
Angular frequency: ω = √(k/m) = √(100/0.25) = 20 rad/s. Period: T = 2π/ω = 2π/20 ≈ 0.314 s. Maximum speed: v_max = Aω = (0.08)(20) = 1.6 m/s. At t = T/8: ωt = (20)(0.314/8) = π/4 rad. x = A cos(π/4) = 0.08 × (√2/2) ≈ 0.057 m. So at one-eighth of the period, the mass has moved roughly 57% of the amplitude toward equilibrium.

Question 3: INTERMEDIATE

An object in SHM has a displacement of x = 0.030 m and a velocity of v = −0.40 m/s when ω = 10 rad/s. Determine the amplitude A and the phase constant φ₀ (using the cosine form x = A cos(ωt + φ₀)), and state whether the object is moving toward or away from equilibrium at this instant.
Show answer
Using A = √(x₀² + (v₀/ω)²) = √((0.030)² + (−0.40/10)²) = √(0.0009 + 0.0016) = √0.0025 = 0.050 m. The phase constant satisfies tan(φ₀) = −v₀/(ωx₀) = −(−0.40)/((10)(0.030)) = 0.40/0.30 = 4/3, so φ₀ = arctan(4/3) ≈ 0.927 rad ≈ 53.1°. Since the velocity is negative and the displacement is positive, the object is moving in the negative direction — that is, back toward and through equilibrium. The positive phase constant indicates that at t = 0 the phasor has already rotated 53.1° past the positive real axis, consistent with having moved away from maximum displacement.

Question 4: APPLIED

The piston of a car engine moves in approximately simple harmonic motion with a stroke (peak-to-peak displacement) of 0.090 m. If the engine runs at 3000 rpm, calculate the maximum speed and maximum acceleration of the piston. Express the maximum acceleration as a multiple of g = 9.8 m/s².
Show answer
The stroke equals 2A, so A = 0.045 m. The engine speed is 3000 rpm = 3000/60 = 50 rev/s, giving ω = 2π(50) = 100π ≈ 314.2 rad/s. Maximum speed: v_max = Aω = (0.045)(314.2) ≈ 14.1 m/s. Maximum acceleration: a_max = Aω² = (0.045)(314.2)² ≈ (0.045)(98,700) ≈ 4440 m/s². In multiples of g: 4440/9.8 ≈ 453g. This enormous acceleration explains why engine components must withstand extreme inertial forces and why material strength is critical in high-rpm engine design.

Question 5: CRITICAL THINKING

A horizontal spring-mass system oscillates with amplitude A on a frictionless surface. At the instant the mass is at x = A/2, what fraction of the total energy is kinetic and what fraction is potential? Derive your answer from the energy expressions and comment on what this tells you about where the oscillator spends most of its time.
Show answer
At x = A/2: U = ½k(A/2)² = ½k(A²/4) = (1/4)(½kA²) = E/4. Therefore K = E − U = E − E/4 = 3E/4. So at the midpoint of the displacement range, 75% of the energy is kinetic and only 25% is potential. This means the object is moving quickly through this region. By contrast, near the turning points (x ≈ ±A) almost all energy is potential and the speed is near zero, so the object moves slowly and lingers. This analysis reveals a general principle: an SHM oscillator spends more time near its turning points than near equilibrium, a fact that can be formalized by computing the probability distribution of displacement — which diverges at x = ±A. This result has deep connections to the quantum harmonic oscillator, where the classical turning-point concentration reappears as high probability density at the edges of the allowed region.

SUMMARY

Lesson Summary

Summary

Simple harmonic motion arises whenever a linear restoring force acts on a system displaced from stable equilibrium. The resulting motion is completely described by three parameters — amplitude A, angular frequency ω, and phase constant φ₀ — and represented by the sinusoidal function x(t) = A cos(ωt + φ₀). Velocity and acceleration follow by successive differentiation, each leading the previous quantity by π/2 radians. The defining kinematic relationship a = −ω²x confirms that acceleration is always proportional to and opposite in sign to displacement.

Multiple representations deepen understanding: x-vs-t graphs show temporal evolution, phasor diagrams reveal phase relationships via rotating vectors, and energy diagrams illustrate the continuous exchange between kinetic and potential energy, whose sum E = ½kA² remains constant. The total mechanical energy scales with the square of the amplitude, and the period is independent of amplitude — both hallmarks of ideal SHM. This framework serves as the essential foundation for the study of damped, driven, and coupled oscillations, as well as wave phenomena and quantum mechanics.

Varsity Tutors • College Physics • Representing and Analyzing SHM