Ⅳ Functions · Stage 25 — Trigonometry · 25.6 Sinusoidal FunctionsAll lessons →
Stage 25 · Trigonometry

y = A·sin(ωx + φ) & Simple Harmonic Motion

Three dials — amplitude, frequency, phase — describe every vibration.

Ages 14–18 · Reasoning, one step at a time
The signature picture of this stage: y = A·sin(ωx + φ). Its amplitude A sets the height, its period 2π/ω sets how often it repeats, and its phase φ slides the whole curve sideways. Three numbers — one vibration.

Pluck a guitar string, push a child on a swing, watch the tide rise and fall — and the same shape keeps coming back. It is the sine curve from 25.4, but now we put it to work. A plain y = sin x has a fixed height of 1, a fixed repeat-length of 2π, and starts dead-on at the origin. Real vibrations are taller or shorter, faster or slower, and rarely start at the perfect moment. So we hang three dials on the wave — A for height, ω for how tightly it packs, and φ for how far it has slid — and write every one of them as y = A·sin(ωx + φ). Turn the three dials and you can match a heartbeat, a sound wave, or the swing of a pendulum. Let us learn what each dial does.

25.6.1 Amplitude A — the volume knob

Start with the cleanest change. Leave the inside of the sine alone and just multiply the whole thing by a number: y = A·sin x. Every output of sin x already lives between −1 and 1, so multiplying by A stretches those outputs to live between −A and A (when A > 0). The shape is untouched — same crests, same troughs, same crossings — only the height changes. We call |A| the amplitude: the distance from the calm middle line up to a crest.

Three amplitudes over the same midline y = 0. The blue wave y = 2 sin x reaches up to +2 and down to −2; halve A and it only reaches ±1; the range of y = A sin x is always [−|A|, |A|].
Key idea

In y = A·sin(ωx + φ), the number A out front is the amplitude. The wave swings between −|A| and +|A|; its amplitude is |A|, the distance from the midline to a peak. A is a vertical stretch — it never changes how often the wave repeats.

Watch out

Amplitude is |A|, not A. If A = −3, the amplitude is still 3 — the minus sign just flips the wave upside down (a crest becomes a trough), it does not give a "negative height."

Try it Turn the amplitude dial

Drag A and watch the wave grow and shrink against the ghost of y = sin x. The range updates live.

amplitude A

25.6.2 Frequency ω — packing the wave tighter

Now reach inside the sine and scale the input: y = sin(ωx). The plain wave finishes one full cycle when its angle has run through 2π. Here the angle is ωx, so one cycle finishes when ωx = 2π, that is when x = 2π/ω. That length is the period — how far along the x-axis the wave travels before it repeats. The bigger ω is, the sooner ωx reaches 2π, so a bigger ω makes a shorter period — the wave packs more cycles into the same stretch.

Read it as a one-line conclusion: period = 2π/ω. If ω = 2, the period is 2π/2 = π — the wave repeats twice as fast as y = sin x. If ω = ½, the period is 2π/(½) = 4π — it crawls, taking twice as long to come around.

Same height, different pace. y = sin 2x (blue) squeezes a full cycle into a period of π, while the ghost y = sin x needs the full 2π. Double ω, halve the period.
Key idea

The number ω inside the sine is the angular frequency. The period — the length of one full cycle — is T = 2π/ω. A larger ω squeezes more cycles into each unit of x, so the wave looks tighter.

Watch out

ω is not the period. They are reciprocals (up to the factor 2π): a bigger ω makes a shorter period, because period = 2π/ω. Confusing the two is the classic slip — if ω doubles, the wave gets faster, not slower.

25.6.3 Phase φ — sliding the wave sideways

The last dial lives inside too, but as an added constant: y = sin(ωx + φ). The wave normally begins its rising cycle where its angle is 0. Here the angle is ωx + φ, which equals 0 when x = −φ/ω. So the whole wave has slid: the cycle that used to start at x = 0 now starts at x = −φ/ω. The number φ is the phase; the amount −φ/ω is the phase shift.

Sign matters and it surprises people. A positive φ slides the wave left (because −φ/ω is negative), and a negative φ slides it right. For example y = sin(x + π/2) starts its cycle at x = −π/2 — shifted left by π/2 — which is exactly the cosine curve. That is the cleanest way to see cos x = sin(x + π/2), the shift we met in 25.4.

A phase of φ = π/2 slides y = sin(x + π/2) left by π/2 — landing exactly on the ghost cosine. The cycle now starts at x = −π/2 instead of 0.
Key idea

The added constant φ is the phase. The wave's cycle starts at x = −φ/ω. A positive φ shifts the graph left, a negative φ shifts it right — the opposite of what the plus sign first suggests.

Try it All three dials at once

Set A, ω, and φ and watch y = A·sin(ωx + φ) (blue) ride over the ghost y = sin x (slate). The readout reports the amplitude, the period 2π/ω, and the phase.

A
ω
φ

25.6.4 Finding the equation from a graph

Run it backward. If someone hands you a sinusoidal graph, you can recover all three dials by measuring three things:

Height ⇒ A. Measure from the midline up to a crest. That distance is the amplitude |A|. Period ⇒ ω. Measure how far the wave travels before it repeats — that is the period T — then turn it into ω by ω = 2π/T. Shift ⇒ φ. Find where a rising cycle crosses the midline going up; that x is the phase shift −φ/ω, so φ = −ω·(that x).

Worked example

A wave rises to +3 and falls to −3, repeats every π units, and its upward midline crossing sits at x = −π/4. Then A = 3; ω = 2π/π = 2; and the start is −φ/ω = −π/4, so φ = −2·(−π/4) = π/2. The equation is y = 3 sin(2x + π/2). Check: at x = −π/4 the angle is 2(−π/4) + π/2 = 0, so y = 0 and rising — exactly as drawn.

Try it Read the equation — match the target

A target wave is drawn in green. Dial your A, ω, φ until your blue wave lands on top of it. When all three match, the readout turns green.

A
ω
φ

25.6.5 Simple harmonic motion

Here is the payoff. A mass bouncing on a spring, a pendulum swinging through a small angle, the air pressure of a pure musical note, the height of the tide — all of them, when nothing drains their energy, trace a sine curve in time. We call this simple harmonic motion (SHM), and its template is exactly our wave with x replaced by time t:

A spring's displacement over time, y = A·sin(ωt + φ). The amplitude A is how far it swings from rest; the period T = 2π/ω is the seconds per bounce; the frequency f = 1/T is bounces per second.
Key idea — the SHM template

Displacement y = A·sin(ωt + φ), with t in seconds. Then amplitude = |A| (the maximum swing, in metres), period T = 2π/ω (seconds per cycle), and frequency f = 1/T = ω/2π (cycles per second, in hertz). Same three dials — now wearing real units.

Worked example

A spring obeys y = 0.05 sin(4πt) metres. Read off the dials: amplitude A = 0.05 m (it swings 5 cm each way); ω = 4π, so the period is T = 2π/(4π) = 0.5 s; and the frequency is f = 1/0.5 = 2 Hz — two full bounces every second. The phase φ = 0 means it starts at rest, moving up.

The same template describes a tuning fork at 440 Hz (ω = 2π·440), a tide cresting every 12.4 hours, or a pendulum clock ticking once a second. Learn the three dials once and you can model them all. In Stage 26 · Sequences, we will see periodic patterns return in a different costume — as terms that repeat — but that is a story for the next stage.

What to carry forward

Three dials reshape the sine curve, each in its own way. Keep this table and the SHM template at your fingertips.

DialIn y = A·sin(ωx + φ)What it does to the graph
Amplitude Amultiplies the whole sinevertical stretch to height |A|; range [−|A|, |A|]; A < 0 flips it
Frequency ωscales the input xsets the period = 2π/ω; bigger ω ⇒ shorter period (tighter)
Phase φadds inside the sineslides sideways; cycle starts at x = −φ/ω; φ > 0 shifts left

SHM template:  y = A·sin(ωt + φ)  ·  amplitude |A|  ·  period T = 2π/ω  ·  frequency f = ω/2π.

Exercises

  1. State the amplitude and the range of y = 4 sin x.

    Answer

    The amplitude is |4| = 4. The wave swings between −4 and 4, so the range is [−4, 4].

  2. Find the period of y = sin(3x) and of y = sin(½x).

    Answer

    Period = 2π/ω. For ω = 3: 2π/3. For ω = ½: 2π/(½) = 4π. The bigger ω gives the shorter period.

  3. By how much, and in which direction, does the graph of y = sin(x − π/3) shift compared with y = sin x?

    Answer

    Here φ = −π/3 and ω = 1, so the cycle starts at x = −φ/ω = π/3. A negative φ shifts the wave right by π/3.

  4. A wave reaches a maximum of 2 and a minimum of −2, repeats every 4π units, and crosses the midline going up at x = 0. Write its equation y = A sin(ωx + φ).

    Answer

    Amplitude A = 2. Period 4π ⇒ ω = 2π/4π = ½. The upward crossing at x = 0 means −φ/ω = 0, so φ = 0. Equation: y = 2 sin(½x).

  5. A spring's displacement is y = 0.1 sin(πt) metres, t in seconds. Give its amplitude, period, and frequency in real units.

    Answer

    Amplitude 0.1 m (10 cm each way). ω = π, so period T = 2π/π = 2 s. Frequency f = 1/T = 0.5 Hz — one full bounce every two seconds.

  6. True or false: in y = A sin(ωx + φ), doubling ω doubles the period. Explain.

    Answer

    False. Period = 2π/ω, so doubling ω halves the period — the wave gets faster, not slower. ω and the period are inversely related.

🎯 Quick check

Six questions to lock it in. Tap the answer you think is right.

§ For teachers and parents

This lesson covers the sinusoidal model y = A·sin(ωx + φ) and simple harmonic motion, aligned with CCSS HSF-TF.B.5 (model periodic phenomena with specified amplitude, frequency, and midline) and HSF-IF.C.7e (graph trigonometric functions, showing period, midline, and amplitude). It builds directly on graphing sine and cosine (Stage 25.4) and on transformations of functions (Stage 20). Encourage learners to drive the interactive dials and read the amplitude, period = 2π/ω, and phase aloud — the most common error is treating ω as the period rather than its reciprocal partner.

eastmath.com · Stage 25 · 25.6 Sinusoidal Functions · Reasoning, one step at a time