3. Omega changes width: derive the period

Explain it like I am five

Picture the Ferris wheel again, but now it can spin faster or slower. A faster wheel finishes a lap sooner, so one full wave takes less width on the page. The number ω (omega, oh-MAY-guh) in y = sin(ωx) is that speed knob: it multiplies the input x inside the sine.

Example: y = sin(πx), so ω = π. One full wave, a cycle, needs the inside, πx, to grow by 2π. Solve π × T = 2π for the period T, the width of one cycle: divide both sides by π, and T = 2. Check: π × 2 = 2π. Plain sin x needs about 6.28 for a cycle, so this wave is squeezed.

Why T = $2 π over |ω|$ : a lap is 2π of turning, and the inside turns |ω| (the size of ω, sign dropped) for each 1 unit across, so a lap takes 2π ÷ |ω|. If ω = 0 nothing turns: y = sin 0 = 0 is a flat line, and 2π ÷ 0 has no value, since no number times 0 gives 2π.

In plain words

Imagine recording the same wheel rider while changing how quickly the rider moves around the circle. If the rider covers twice the turning distance during each step across your page, a full lap finishes in half as much page width. The wave becomes narrower. The number ω, pronounced oh-MAY-guh, multiplies the input inside sine and controls this width. It is another Constant. A large size of ω packs more laps into the same horizontal distance. A small nonzero size spreads a lap over more distance. A negative ω makes the rider move backward, so you must separate its direction from its speed.

Increasing the inside multiplier shortens the horizontal width of a cycle.

Reminder

One full turn. A full turn is 2π radians; a first top uses $π over 2$ radians inside sine.
Absolute value. |−4| = 4, so the period calculation uses 2π ÷ 4 rather than a negative divisor.
Dividing by a fraction. 2π ÷ $1 over 2$ = 2π × $2 over 1$ = 4π. The reciprocal reverses the original fraction multiplication.
Canceling a common factor. $2 π over 5 π$ = $2 over 5$ because both complete numerator and denominator contain the factor π.
Solving and substituting. 2T = 2π becomes T = π after division by 2; substitute 2 × π = 2π to verify the distance makes a complete lap.
Sine oddness. sin(−x) = −sin x; therefore sin(−4x) = −sin(4x). Opposite inside turns give opposite heights.
Nonzero division. 2π ÷ 0 has no answer because 0 multiplied by any real number is 0 rather than 2π.
Coefficient and input. In sin(5πx), the coefficient is 5π, not 5. At input $2 over 5$ , the entire inside becomes 5π × $2 over 5$ = 2π.

Why it works. One full sine lap requires the inside angle to change by 2π in size. When the outside input advances by T, the inside angle changes by ωT. Therefore |ω|T = 2π, and dividing by |ω| gives T = . If |ω| doubles, T halves; if |ω| halves, T doubles. Negative ω changes the travel direction, but one lap still requires a positive horizontal length. A only scales heights, so it cannot change that width when the graph remains nonconstant.

RuleRule: For y = A sin(ωx), with A ≠ 0 and ω ≠ 0: T = and |ω| = .
|ω| > 1 compresses horizontally; 0 < |ω| < 1 stretches horizontally; ω < 0 reverses the input direction. A = 0 or ω = 0 gives a constant graph without a smallest positive period.

The same idea, five ways

Say it

Say: period is two pi divided by the size of omega.

Write it

The inside multiplier changes how much horizontal distance is needed for one complete sine cycle.

In math

y = A sin(ωx)
|ω|T = 2π
T = $2 π over |ω|$
|ω| = $2 π over T$
A ≠ 0 and ω ≠ 0 for a nonconstant wave

A faster rider finishes a lap in less horizontal distance.

See it

With omega 2, the positive period distance is π.

The same idea, other ways

As a picture

In the width 2π, sin x makes one full wave while sin(2x) makes two. The latter's individual cycle width is π.

Two inside laps fit into the same outside width when ω = 2.

As travel speed

If you walk around the wheel twice as fast for each horizontal step, you finish a lap in half the distance. At half speed, the lap takes twice the distance.

With a complete-turn equation

The inside must advance 2π for a full lap. With ω = $1 over 2$ , the equation $1 over 2$ T = 2π requires T = 4π because $1 over 2$ × 4π = 2π.

As a rule of thumb

Inside lies: a bigger input multiplier makes a smaller width. Check with ω = 2 giving T = π, then ω = $1 over 2$ giving T = 4π.

The smaller inside multiplier stretches a single cycle over width 4π.

.1Horizontal compression

A Horizontal compression packs the same height record into less width. Imagine folding more repeated wallpaper designs into the same wall length. The inside factor 2 makes the circle rider travel 2 radians whenever x advances 1 radian. The full lap therefore needs x to advance only π instead of 2π. Each height feature arrives sooner, but the amplitude stays 1 when no outside height multiplier is added.

Rule: |ω| > 1 produces Horizontal compression.
Rule: Horizontal distances are multiplied by $1 over |ω|$ .
Rule: For sin(2x), T = π and amplitude remains 1.

The peaks have the same height but occur closer together.

Reminder

Canceling a common factor. $2 π over 6$ = $π over 3$ because numerator and denominator share the factor 2.

The same idea, five ways

Say it

Say: a larger inside factor squeezes the wave's width.

Write it

When |ω| is greater than 1, one cycle takes less than 2π of horizontal distance.

In math

|ω| > 1
T = $2 π over |ω|$ < 2π
For ω = 2: T = π

Pack more identical wallpaper designs into the same wall width.

See it

The compressed wave finishes its first lap at π.

Worked exampleCompress width with an original inside factor

This asks for the cycle width of y = sin(6x), then checks whether the height changed.

The compressed cycle has width and keeps peak height 1.

ω = 6, so T = $2 π over 6$ = $π over 3$ .The inside angle moves six times as far as the outside input.
Substitute T: 6 × $π over 3$ = 2π. The amplitude is |1| = 1.The product checks one full inside turn, while the unchanged outside coefficient leaves heights unchanged.

Answer

Period = $π over 3$ .
Amplitude = 1.
Horizontal distances are one-sixth their original size.

Check At x = , the inside angle is 6 × = , so the graph still reaches height 1.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: sin(6x) has amplitude 6.

The 6 is inside and changes which input reaches each height, rather than multiplying the height.

✓ Instead: Its amplitude is 1 and period is .

Tips and tricks

Tip: Memory device: Inside lies. Factor 6 inside produces width one-sixth as large.

.2Horizontal stretch

A Horizontal stretch spreads a full height record over more width. Imagine pulling a wallpaper strip longer while keeping its height unchanged. In sin( $1 over 2$ x), each step in x advances the inside angle only half as far. The rider needs twice the horizontal distance to complete a lap. The first peak moves from $π over 2$ to π, and one full cycle moves from width 2π to width 4π. The peak height is still 1.

Rule: 0 < |ω| < 1 produces Horizontal stretch.
Rule: For sin( $1 over 2$ x), T = 4π.
Rule: Width changes do not change amplitude when A stays fixed.

The source half-speed sine wave spreads one cycle across width 4π.

Reminder

Dividing by a fraction. 2π ÷ $1 over 2$ = 2π × $2 over 1$ = 4π. Multiplication by the reciprocal undoes multiplication by the original fraction.

The same idea, five ways

Say it

Say: a smaller nonzero inside factor spreads the wave wider.

Write it

An inside multiplier between zero and one in size makes the period larger than 2π.

In math

0 < |ω| < 1
T > 2π
For ω = $1 over 2$ : T = 4π

Pull a wallpaper design longer without increasing its height.

See it

A half-size inside coefficient produces double-size horizontal distances.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: sin(x) has period π because 2π × = π.

That multiplies the old width by the input factor, but the needed input distance changes by its reciprocal.

✓ Instead: T = 2π ÷ = 4π.

Tips and tricks

Tip: Substitute the period into the inside. If it does not give one full turn in size, recompute.

.3Negative omega and nonzero assumptions

A negative ω sends the wheel rider backward, like reversing a walking direction around a track. The traveled direction changes, but the distance for one lap stays positive. Sine's oddness makes sin(−2x) equal −sin(2x), so its graph starts downward. If ω = 0, the rider never moves: the inside angle stays at 0 and every sine output is 0. If A = 0, every height is erased. Either case is a constant line with no smallest positive repeating distance.

Rule: For sine, sin(−|ω|x) = −sin(|ω|x). Negative omega reflects the input across the y-axis; sine's oddness makes the result also look like an output reflection across the x-axis.
Rule: T uses |ω|, so ω = −2 gives T = π, never −π.
Rule: The wave period formula requires A ≠ 0 and ω ≠ 0.
Rule: A = 0 or ω = 0 makes y = A sin(ωx) identically 0. Every positive shift repeats it, with no least positive one.

Reversing omega reverses the first height direction while retaining period π.

Reminder

Oddness and absolute value. sin(− $π over 2$ ) = −1 by sin(−x) = −sin x; |−2| = 2 is a positive size.
Division by zero. 2π ÷ 0 is undefined because no number multiplied by 0 gives the nonzero number 2π.

The same idea, five ways

Say it

Say: use omega's size for width and its sign for travel direction.

Write it

A negative inside coefficient reverses input direction but gives a positive period for a nonconstant wave.

In math

T = $2 π over |ω|$ > 0
sin(−|ω|x) = −sin(|ω|x)
A ≠ 0 and ω ≠ 0
ω = 0 implies y = A sin 0 = 0

A backward lap has a reversed direction but the same track length.

See it

The two directions finish a full lap after the same positive input distance.

Worked exampleReverse the source compression's direction

This asks for the positive period and first valley of y = sin(−2x). Explain the minus sign separately from the width.

The first valley replaces the first peak, with no change in cycle width.

ω = −2, so |ω| = 2 and T = $2 π over 2$ = π.Period is a distance and therefore uses the coefficient's size.
Use oddness: sin(−2x) = −sin(2x). At x = $π over 4$ , the inside angle is −2 × $π over 4$ = − $π over 2$ , giving output −1.The backward quarter turn reaches the bottom instead of the top.
Check the full-lap size: |−2| × π = 2π.The proposed positive width must advance the inside by one turn in size.

Answer

Period = π.
First valley: ( $π over 4$ , −1).
Amplitude = 1.
The graph equals −sin(2x).

Check At input π, the inside is −2π, a full backward lap with height 0; adding π to any input subtracts 2π from the inside and preserves its sine height.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: sin(−2x) has period −π.

A period is a positive input distance, even when the inside travels backward.

✓ Instead: T = 2π ÷ |−2| = π.

✗ Not this: Counterexample: ω = 0 gives period 2π ÷ 0 = 0.

Division by zero has no value, and this graph is constant rather than a nonconstant wave.

✓ Instead: sin(0x) = 0 at every input and has no smallest positive period.

Tips and tricks

Tip: Check for zero first, then remove omega's sign for the period calculation.

Strategy: step by step

1. Read ω, the entire coefficient multiplying x inside sine. In sin(2x), ω = 2; in sin( $1 over 2$ x), ω = $1 over 2$ ; in sin(5πx), ω = 5π.
2. Check A ≠ 0 and ω ≠ 0 before calling the graph a nonconstant wave.
3. Take |ω| to remove direction. Period (T) measures a positive horizontal length.
4. Compute T = 2π ÷ |ω|. Divide by a fraction by multiplying by its reciprocal. Cancel π only when it is a common factor of the full numerator and denominator.
5. Compare |ω| with 1. Larger than 1 means Horizontal compression; between 0 and 1 means Horizontal stretch; equal to 1 keeps the original width.
6. Check by multiplying the proposed period by |ω|. The product must be 2π, exactly one inside lap.
7. If a graph gives T and you need ω, use |ω| = $2 π over T$ . Choosing ω > 0 gives the convenient representative ω = $2 π over T$ ; period alone does not determine its sign.

Strategy

Strategy: turn an inside coefficient into a period

Is A = 0 or ω = 0?

YesThe sine expression is constant zero. It has no smallest positive period, so stop the period formula.

NoCompute T = .

↓

Is |ω| a fraction?

YesMultiply 2π by its reciprocal, then simplify.

NoDivide directly, canceling only common nonzero factors.

↓

Are you given T instead of ω?

YesCompute |ω| = ; choose the positive representative if a signed ω is not specified.

NoKeep the period you computed and verify |ω|T = 2π.

Identify the entire inside coefficient ω.
Check that A and ω are nonzero.
Take |ω| and divide 2π by it.
Reduce fractions and common factors.
Multiply |ω| by the proposed T to check for one full turn 2π.

Worked exampleWind turbine blade tip: period from ω, then ω from a period

A wind turbine's blades turn at a steady rate. One blade is 42 m long from the center of the hub to its tip. At time x = 0 seconds that tip is level with the hub's center and moving upward. Its height above the hub's center, in meters, is modeled by y = 42 sin( $4 π over 9$ x). The sine input u = $4 π over 9$ x is the angle, in radians, that the blades have turned. The figure graphs u (vertical axis) against x as a straight line through (0, 0) and (9, 4π). (a) Find the period T of the tip's height, say whether the graph is a horizontal compression or a horizontal stretch of y = 42 sin x, and check T. (b) In lighter wind the rotor slows until one full turn takes 7 seconds, and the tip again starts level with the hub's center and moving upward. Find ω for the new model y = 42 sin(ωx), and say whether its graph is a horizontal compression or stretch of y = 42 sin x.

The blades' turning angle u = x (radians, vertical axis) against time x (seconds). It is a straight line through (0, 0) and (9, 4π), so the blades make two full turns, 4π radians, in 9 seconds.

Read ω as the whole coefficient of x inside sine. In y = 42 sin( $4 π over 9$ x), ω = $4 π over 9$ , not $4 over 9$ and not 4π.The width depends on the entire number multiplying x; the π and the 9 both belong to ω.
Check that A = 42 ≠ 0 and ω = $4 π over 9$ ≠ 0.Only then is the height a nonconstant wave with a smallest positive period. A = 42 sets how high the tip goes, not how wide one cycle is, so it plays no part in T.
Take |ω| = $4 π over 9$ ; ω is already positive.A period is a positive horizontal length, so only the size of ω matters, not its sign.
T = 2π ÷ |ω| = 2π ÷ $4 π over 9$ = 2π · $9 over 4 π$ = $18 π over 4 π$ = $18 over 4$ = $9 over 2$ = 4.5 seconds.Sine repeats when its input grows by 2π, and the input grows by |ω| radians per second, so one cycle lasts 2π ÷ |ω| seconds. Dividing by a fraction means multiplying by its reciprocal. π is a factor of both 18π and 4π, so it cancels, and $18 over 4$ then reduces to $9 over 2$ .
Compare |ω| with 1: $4 π over 9$ ≈ 1.40 > 1, so the graph is a horizontal compression of y = 42 sin x. Its period shrinks from 2π ≈ 6.28 s to 4.5 s.|ω| > 1 squeezes one cycle into less than 2π. Use the full value including π: $4 over 9$ by itself is below 1 and would wrongly suggest a stretch.
Check: T · |ω| = $9 over 2$ · $4 π over 9$ = $36 π over 18$ = 2π.Across one period the input must advance exactly one lap, 2π radians; any other product means T is wrong.
Read the line u = $4 π over 9$ x: it rises from 0 at x = 0 to 4π at x = 9. 4π is two laps, so one lap takes 9 ÷ 2 = 4.5 seconds, which matches T.The input u is the angle turned, a straight line with slope ω. Each rise of 2π is one full turn of the blade and one cycle of its height.
For (b), T = 7, so |ω| = $2 π over T$ = $2 π over 7$ . The tip again starts level and rising, so ω > 0. That gives ω = $2 π over 7$ and the model y = 42 sin( $2 π over 7$ x).|ω| = $2 π over T$ undoes T = 2π ÷ |ω|. The period alone fixes only |ω|. With A = 42 > 0, a negative ω would make the tip drop first, so the rising start picks ω > 0.
Compare: $2 π over 7$ ≈ 0.90, between 0 and 1, so this graph is a horizontal stretch of y = 42 sin x. Check: 7 · $2 π over 7$ = 2π.A turn that takes longer than 2π ≈ 6.28 s needs |ω| < 1. The product 2π confirms exactly one lap per period.

Answer

(a) T = $9 over 2$ = 4.5 seconds
|ω| = $4 π over 9$ ≈ 1.40 > 1, so the graph is a horizontal compression of y = 42 sin x. (b) ω = $2 π over 7$ ≈ 0.90, giving y = 42 sin( $2 π over 7$ x), a horizontal stretch of y = 42 sin x.

Check · = = 2π, so 4.5 s is exactly one lap of the input: 42 sin((x + )) = 42 sin(x + 2π) = 42 sin(x). The line u = x reaches 4π, two full turns, at x = 9, which is 4.5 s per turn. For (b), 7 · = 2π. The order 4.5 < 2π ≈ 6.28 < 7 agrees with a compression in (a) and a stretch in (b).

Work to write

ω = $4 π over 9$ ; A = 42 ≠ 0 and ω ≠ 0, so the graph is a nonconstant wave
|ω| = $4 π over 9$
T = 2π ÷ $4 π over 9$ = 2π · $9 over 4 π$ = $18 π over 4 π$ = $9 over 2$ = 4.5 s
|ω| = $4 π over 9$ ≈ 1.40 > 1 → horizontal compression of y = 42 sin x
Check: $9 over 2$ · $4 π over 9$ = 2π
(b) |ω| = $2 π over 7$ ; the tip rises first, so ω = $2 π over 7$ and y = 42 sin( $2 π over 7$ x)
$2 π over 7$ ≈ 0.90 < 1 → horizontal stretch; check: 7 · $2 π over 7$ = 2π

(a) T = $9 over 2$ = 4.5 seconds; |ω| = $4 π over 9$ ≈ 1.40 > 1, so the graph is a horizontal compression of y = 42 sin x. (b) ω = $2 π over 7$ ≈ 0.90, giving y = 42 sin( $2 π over 7$ x), a horizontal stretch of y = 42 sin x.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: an inside coefficient containing π

This asks for the period of y = sin(5πx). Keep the whole factor 5π as omega, then simplify.

A coefficient containing π can produce a period with no π remaining.

Read ω = 5π; it is positive, so |ω| = 5π.Both 5 and π multiply x inside sine.
T = $2 π over 5 π$ = $2 over 5$ .The numerator and denominator have the common nonzero factor π, which cancels.
Substitute: 5π × $2 over 5$ = 2π.The computed outside distance must generate one full inside turn.

Answer

ω = 5π.
Period = $2 over 5$ .
Amplitude = 1.

Check The first peak is at input , because 5π × = . Four such quarter-period distances give = .

Rung 2Rung 2: negative direction with positive width

This asks for the period and direction of y = sin(−4x). Keep width and sign separate.

The inside minus sign reverses the height direction while preserving positive cycle width.

ω = −4, so |ω| = 4 and T = $2 π over 4$ = $π over 2$ .The absolute value supplies travel size rather than direction.
Use oddness: sin(−4x) = −sin(4x). The curve first goes downward.Backward turns reverse sine's height sign.
Substitute the period: |−4| × $π over 2$ = 2π. At x = $π over 8$ , the inside is −4 × $π over 8$ = − $π over 2$ , giving −1.The first calculation confirms one full turn in size; the second confirms the first valley's direction.

Answer

Period = $π over 2$ .
Amplitude = 1.
First valley: ( $π over 8$ , −1).

Check At input , the inside is −2π and the height is 0 after one backward lap. The positive width is four times the first-valley input .

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: a bigger ω makes a longer period.

A bigger inside multiplier finishes the same full turn in less outside distance.

✓ Instead: The period varies inversely with |ω|: ω = 2 gives T = π; ω = gives T = 4π.

✗ Not this: Counterexample: period T determines the signed coefficient ω uniquely.

Both positive and negative coefficients with the same size have the same positive period.

✓ Instead: T determines |ω| = ; choose ω > 0 when naming a representative equation.

✗ Not this: Counterexample: T = is the smallest positive period even when A = 0.

Zero A erases all output changes, so every positive shift repeats the constant line.

✓ Instead: Use the least-period formula for A ≠ 0 and ω ≠ 0. Constant graphs have no least positive period.

Tips and tricks

Tip: Know cold: divide 2π by the size of omega. Memory device: Inside lies. Larger inside factor means smaller width; check 2 and $1 over 2$ against periods π and 4π.
Tip: Understand, then rebuild: write |ω|T = 2π, meaning one complete inside lap, then divide by |ω| to recover the period formula.
Tip: Put on the cheat sheet: T = $2 π over |ω|$ , |ω| = $2 π over T$ , and the nonzero conditions A ≠ 0, ω ≠ 0.
Tip: A period gives omega's size, not its sign. Choosing the positive representative makes later graph-reading decisions consistent.
Tip: A affects height and ω affects width. Confirm which side of the function name each multiplier occupies before calculating.

Trap. The inside multiplier is not the period. For sin(2x), ω = 2 but T = π. Divide 2π by |ω|. A negative ω also gives a positive period because width is a distance, not a direction.

Keep in mind

The period is not ω itself: y = sin(5x) has ω = 5 but T = $2 π over 5$ .
Bigger |ω| means a shorter period: |ω| > 1 squeezes the wave and 0 < |ω| < 1 stretches it.
A negative ω still gives a positive period, because width is a distance: y = sin(−5x) has T = $2 π over 5$ .
Dividing by a fraction means multiplying by its flip: for ω = $2 over 7$ , T = 2π × $7 over 2$ = 7π.

Memory hookω is the speed knob: faster spin, shorter wave. One lap is 2π, so the period is T = .

Flash cards: say the answer out loud, then flip

What is the period of a wave?

The width of one full cycle, after which the pattern repeats.

State the period formula for y = A sin(ωx).

T =

Find the period of y = sin(8x).

A cycle is 6 units wide. Find the positive ω.

ω = =

Does y = sin(−8x) have period −?

No. Use |ω| = 8, so T = $π over 4$
a period is always positive.