Amplitude, Frequency & Phase — Interpreting Solutions (Amplitude, Frequency, Phase) - Differential Equations

SECTION 1

Historical Context & Motivation

People have been fascinated by repetitive motion — swinging pendulums, vibrating strings, ocean waves — for thousands of years. When mathematicians finally developed the tools to describe these motions precisely, they discovered that second-order linear differential equations were at the heart of it all. The solutions to these equations always involve sine and cosine functions, and buried inside those functions are three key quantities: how big the motion is, how fast it repeats, and where it starts. Understanding these quantities — amplitude, frequency, and phase — unlocks the physical meaning behind every oscillating system.

1583 • Galileo and the Pendulum

Galileo Galilei observed that a swinging chandelier took the same time to complete each swing regardless of its arc, noting the property of isochronism — one of the first quantitative studies of oscillation.

1687 • Newton's Principia

Isaac Newton published his laws of motion, providing the framework (F = ma) that would later be expressed as second-order differential equations for vibrating systems.

1739 • Euler's General Solutions

Leonhard Euler developed systematic techniques for solving second-order linear ODEs, showing that sine and cosine functions naturally emerge as solutions.

1822 • Fourier's Harmonic Analysis

Joseph Fourier demonstrated that any periodic signal can be decomposed into a sum of sinusoids, each described by its own amplitude, frequency, and phase.

So here is the central question: when you solve a second-order ODE and get an expression like y(t) = 3 sin(5t + π/4), what does each number actually mean? How do you connect the math back to the real-world motion it describes? That is exactly what this lesson will teach you.

SECTION 2

Core Principles & Definitions

When you solve a second-order linear homogeneous ODE with constant coefficients — for example, y″ + ω²y = 0 — the general solution takes the form y(t) = C₁ cos(ωt) + C₂ sin(ωt). This can always be rewritten in a single, more revealing form: y(t) = A sin(ωt + φ). Each piece of this expression carries distinct physical meaning.

Amplitude (A)

The amplitude is the maximum displacement from the equilibrium position. It tells you how big the oscillation is. A is always positive: A = √(C₁² + C₂²).

Angular Frequency (ω)

The angular frequency tells you how fast the oscillation repeats, measured in radians per second. The period T = 2π/ω, and the ordinary frequency f = ω/(2π) in cycles per second (hertz).

Phase (φ)

The phase (or phase shift) tells you where in its cycle the oscillation starts at t = 0. It is an angle, measured in radians: φ = arctan(C₁/C₂).

Period (T)

The period is the time for one complete oscillation cycle. It is the inverse of frequency: T = 1/f = 2π/ω. A shorter period means faster oscillation.

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KEY TAKEAWAY

Think of a singer holding a long note. The amplitude is the volume (how loud the note is), the frequency is the pitch (how high or low), and the phase is exactly when they started singing relative to the beat. The same three numbers completely describe any simple oscillation in physics.

SECTION 3

Visualizing Amplitude, Frequency & Phase

The diagram below shows three sine waves on the same time axis. Each wave has been modified in exactly one way — amplitude, frequency, or phase — so you can see the effect of each parameter in isolation. Pay attention to how each change reshapes the curve.

The cyan curve has a larger amplitude (taller peaks). The amber curve has a higher frequency (more cycles in the same time). The pink dashed curve has a phase shift (starts at a different point in its cycle). The violet curve is the unmodified reference.

Notice that changing the amplitude only stretches or compresses the wave vertically — it doesn't affect how quickly the wave oscillates. Changing the frequency compresses the wave horizontally so that more cycles fit in the same time window. Changing the phase slides the entire wave left or right along the time axis without altering its shape or speed.

SECTION 4

Mathematical Framework

Let's start from the ODE and trace the math all the way to the amplitude-phase form. Consider the simplest undamped oscillator equation, which models a mass on a spring with no friction.

SIMPLE HARMONIC OSCILLATOR ODE

y″ + ω²y = 0

Here y″ is the second derivative of y with respect to t (acceleration), and ω (omega) is a positive constant related to the stiffness of the system. The characteristic equation is r² + ω² = 0, giving roots r = ±ωi (purely imaginary).

GENERAL SOLUTION (SUM FORM)

y(t) = C₁ cos(ωt) + C₂ sin(ωt)

C₁ and C₂ are constants determined by the initial conditions y(0) and y′(0).

While the sum form is perfectly correct, it hides the physical story. We can convert it to a single sinusoid using a trigonometric identity. Define A and φ as follows.

AMPLITUDE FORMULA

A = √(C₁² + C₂²)

A is always non-negative. It represents the farthest the oscillating quantity ever gets from zero.

PHASE FORMULA

φ = arctan(C₁ / C₂) (adjusted for quadrant)

Use the signs of C₁ and C₂ to determine the correct quadrant. In many textbooks you will see tan(φ) = C₁/C₂ when the solution is written as A sin(ωt + φ).

AMPLITUDE-PHASE FORM

y(t) = A sin(ωt + φ)

This single expression tells the full story: A = maximum displacement, ω = angular frequency, φ = phase shift.

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Cosine vs. Sine Convention

Some textbooks write the solution as y(t) = A cos(ωt − φ) instead of A sin(ωt + φ). Both are correct — they just define the phase angle differently. The formulas for A and ω stay the same either way. Always check which convention your course uses.

SECTION 5

Detailed Breakdown — Reading the Solution

Once you have the amplitude-phase form, you can extract a wealth of information without any further calculation. The table below summarizes what each parameter tells you about the motion, how to find it, and what its units are.

Parameter	Symbol	What It Tells You	How to Find It
Amplitude	A	Maximum displacement from equilibrium	A = √(C₁² + C₂²)
Angular Frequency	ω	Rate of oscillation in rad/s	Coefficient of t inside the sine/cosine
Ordinary Frequency	f	Cycles per second (Hz)	f = ω / (2π)
Period	T	Time for one full cycle (seconds)	T = 2π / ω = 1 / f
Phase	φ	Starting position in the cycle at t = 0	φ = arctan(C₁ / C₂)

Key parameters extracted from the amplitude-phase solution

This annotated diagram shows a single sinusoidal solution. The amplitude A is the vertical distance from the center line to the peak. The period T = 2π/ω is the horizontal distance for one complete cycle. The phase shift φ/ω is the horizontal offset from the origin.

One subtle but important point: the phase φ is an angle in radians, but it produces a time shift equal to φ/ω. A positive φ shifts the wave to the left (it starts earlier), and a negative φ shifts it to the right (it starts later). This makes intuitive sense — if you are already partway through your cycle at t = 0, the wave appears shifted backward in time.

SECTION 6

Worked Example

A spring-mass system is modeled by the ODE y″ + 9y = 0, with initial conditions y(0) = 4 and y′(0) = −6. Find the general solution, convert it to amplitude-phase form, and identify A, ω, φ, the period T, and the ordinary frequency f.

Finding Amplitude, Frequency & Phase from an IVP

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Step 1 — Identify ω from the ODE

The ODE is y″ + 9y = 0. Comparing with y″ + ω²y = 0, we see that ω² = 9, so ω = 3 rad/s.

ω = 3 rad/s
2
Step 2 — Write the general solution

With purely imaginary roots r = ±3i, the general solution is y(t) = C₁ cos(3t) + C₂ sin(3t).
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Step 3 — Apply initial conditions

From y(0) = 4: C₁ cos(0) + C₂ sin(0) = C₁ = 4. Next, y′(t) = −3C₁ sin(3t) + 3C₂ cos(3t), so y′(0) = 3C₂ = −6, giving C₂ = −2.

C₁ = 4, C₂ = −2 → y(t) = 4 cos(3t) − 2 sin(3t)
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Step 4 — Compute the amplitude A

A = √(C₁² + C₂²) = √(16 + 4) = √20 = 2√5 ≈ 4.47.

A = 2√5 ≈ 4.47
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Step 5 — Compute the phase φ

To write the solution as y(t) = A sin(3t + φ), we use the identity A sin(3t + φ) = A cos(φ) sin(3t) + A sin(φ) cos(3t). Matching coefficients: A sin(φ) = C₁ = 4 and A cos(φ) = C₂ = −2. So tan(φ) = C₁/C₂ = 4/(−2) = −2. Since sin(φ) > 0 and cos(φ) < 0, φ is in the second quadrant: φ = π − arctan(2) ≈ π − 1.107 ≈ 2.034 rad.

φ ≈ 2.034 rad (about 116.6°)
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Step 6 — Find the period and frequency

T = 2π/ω = 2π/3 ≈ 2.094 seconds. The ordinary frequency is f = 1/T = 3/(2π) ≈ 0.477 Hz.

T ≈ 2.094 s, f ≈ 0.477 Hz
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Step 7 — Write the final amplitude-phase form

Combining everything, the solution is:

y(t) = 2√5 sin(3t + 2.034)

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Quick Check

You can verify by plugging t = 0 into the amplitude-phase form: y(0) = 2√5 sin(2.034) ≈ 4.47 × 0.894 ≈ 4.0 ✓. Always check your answer against the original initial conditions!

SECTION 7

Strengths, Limitations & Comparisons

The amplitude-phase form is incredibly useful, but it has a specific domain of applicability. Here we compare the two common ways of writing the solution and discuss when each shines.

Feature	Sum Form: C₁cos + C₂sin	Amplitude-Phase: A sin(ωt + φ)
Ease of applying initial conditions	Very easy — plug in t = 0 directly	Requires converting back to C₁, C₂ first
Physical interpretation	Harder — two terms obscure the amplitude	Excellent — A, ω, φ are directly readable
Graphing	Need to add two curves	Single curve — sketch immediately
Works for damped systems?	Yes, with exponential factor	Yes, as Ae^−αt sin(ωt + φ)
Best for...	Algebraic manipulation, finding constants	Physical insight, communication of results

Comparing the two forms of the solution

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KEY TAKEAWAY

Think of the sum form as an ingredient list and the amplitude-phase form as a finished dish photo. The ingredient list (C₁ and C₂) is what you need to build the solution, but the photo (A, ω, φ) tells you what the result actually looks and behaves like. Good problem-solving means knowing when to use each form.

SECTION 8

Connection to Advanced Theory

The simple harmonic oscillator (no friction, no external force) is just the starting point. In more realistic models, damping and forcing terms appear, and the amplitude-phase interpretation extends gracefully to those cases.

Concept	This Lesson (Undamped, Free)	Advanced Extension
Amplitude	Constant A — oscillation goes forever	Decays as Ae^−αt in damped systems
Frequency	Natural frequency ω₀ = √(k/m)	Damped frequency ω_d = √(ω₀² − α²) is slightly lower
Phase	Determined by initial conditions only	In forced systems, phase depends on driving frequency (resonance!)
ODE form	y″ + ω²y = 0	y″ + 2αy′ + ω²y = F(t)

How amplitude, frequency, and phase concepts extend to more complex ODEs

When you encounter damped oscillations in a future course, you will see the amplitude shrink over time while the frequency and phase still follow the same interpretive rules. In forced (driven) oscillations, the phase difference between the driving force and the response becomes critical — it determines whether the system absorbs or reflects energy. The phenomenon of resonance occurs when the driving frequency matches the natural frequency, causing the amplitude to grow dramatically. All of these advanced ideas build directly on the amplitude-frequency-phase framework you learned today.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

In the solution y(t) = 7 sin(4t + π/6), identify the amplitude, angular frequency, and phase. Then explain in your own words what would happen to the graph if the amplitude were doubled while everything else stayed the same.

Show answer

The amplitude is A = 7, the angular frequency is ω = 4 rad/s, and the phase is φ = π/6 radians (30°). If the amplitude were doubled to 14, the wave would reach twice as high above and twice as far below the center line, but it would still complete its cycles at the same rate (same frequency) and still start at the same point in its cycle (same phase). Graphically, the curve is stretched vertically while remaining unchanged horizontally.

Question 2: BASIC CALCULATION

Find the period T and the ordinary frequency f for the solution y(t) = 3 sin(10t − π/4).

Show answer

From the solution, ω = 10 rad/s. The period is T = 2π/ω = 2π/10 = π/5 ≈ 0.628 seconds. The ordinary frequency is f = 1/T = 5/π ≈ 1.592 Hz. This means the system completes about 1.59 full oscillations every second.

Question 3: INTERMEDIATE

Convert y(t) = 3 cos(2t) + 4 sin(2t) to amplitude-phase form y(t) = A sin(2t + φ). Find A and φ.

Show answer

Matching A sin(2t + φ) = A sin(φ) cos(2t) + A cos(φ) sin(2t) with 3 cos(2t) + 4 sin(2t), we get A sin(φ) = 3 and A cos(φ) = 4. The amplitude is A = √(3² + 4²) = √(9 + 16) = √25 = 5. For the phase, tan(φ) = 3/4 and since both sin(φ) and cos(φ) are positive, φ is in the first quadrant: φ = arctan(3/4) ≈ 0.6435 rad (about 36.87°). The final answer is y(t) = 5 sin(2t + 0.6435).

Question 4: APPLIED

A car's suspension is modeled by y″ + 16y = 0 with y(0) = 0.05 m (initial compression) and y′(0) = 0 (released from rest). Find the complete amplitude-phase solution and determine the maximum speed |y′(t)| the suspension mass reaches.

Show answer

From ω² = 16 we get ω = 4 rad/s. The general solution is y(t) = C₁ cos(4t) + C₂ sin(4t). Applying y(0) = 0.05 gives C₁ = 0.05. The derivative y′(t) = −4C₁ sin(4t) + 4C₂ cos(4t), so y′(0) = 4C₂ = 0 gives C₂ = 0. Thus y(t) = 0.05 cos(4t), which in amplitude-phase (sine) form is y(t) = 0.05 sin(4t + π/2). The amplitude is A = 0.05 m, ω = 4 rad/s, and φ = π/2. The velocity is y′(t) = −0.2 sin(4t), so the maximum speed is |y′|_max = Aω = 0.05 × 4 = 0.2 m/s.

Question 5: CRITICAL THINKING

Two solutions to the same ODE y″ + 25y = 0 are given: y₁(t) = 2 sin(5t + π/3) and y₂(t) = 2 sin(5t + 4π/3). Show that y₁(t) + y₂(t) is also a sinusoid, find its amplitude and phase, and explain geometrically why the resulting amplitude is smaller than 4 (the sum of the individual amplitudes).

Show answer

Expanding using the sine addition formula: y₁ = 2[sin(5t)cos(π/3) + cos(5t)sin(π/3)] = sin(5t) + √3 cos(5t), and y₂ = 2[sin(5t)cos(4π/3) + cos(5t)sin(4π/3)] = −sin(5t) − √3 cos(5t). Adding: y₁ + y₂ = 0 · sin(5t) + 0 · cos(5t) = 0. Remarkably, the sum is zero — the two waves cancel completely! This happens because their phase difference is exactly π radians (180°), making them perfectly out of phase. The resulting amplitude is 0, which is less than 4 because the waves destructively interfere. In general, two sinusoids of the same frequency add to give an amplitude that depends on their phase difference: when phases differ by π, cancellation is total. This is the mathematical basis of noise-canceling headphones.

SUMMARY

Summary & Review

Summary

The general solution to the undamped harmonic oscillator y″ + ω²y = 0 is y(t) = C₁ cos(ωt) + C₂ sin(ωt), which can always be rewritten as y(t) = A sin(ωt + φ). The amplitude A = √(C₁² + C₂²) measures the maximum displacement from equilibrium. The angular frequency ω determines how fast the oscillation repeats, with period T = 2π/ω and ordinary frequency f = ω/(2π). The phase φ = arctan(C₁/C₂) encodes where in the cycle the motion begins at t = 0.

The sum form is best for applying initial conditions, while the amplitude-phase form is best for physical interpretation and graphing. These three parameters — amplitude, frequency, and phase — completely characterize any simple oscillation and form the foundation for understanding damped, forced, and coupled oscillating systems in more advanced courses.