Quarry School

4. Read A and omega before doing arithmetic

Explain it like I am five

Picture a music player with two separate knobs: one sets how loud, the other sets how fast the beat repeats. In y = A sin(ωx), A is the loudness knob. It sits outside the sine and multiplies every output, so the amplitude, the height from the middle to a peak, is |A|. ω (omega) is the tempo knob. It sits inside, multiplying x, so the period, the width of one wave, is T = 2π|ω|.

Example: y = −6 sin(56x). Point before you compute. Outside the sine: A = −6. Inside, multiplying x: ω = 56. Amplitude = |−6| = 6; the minus only flips the wave. Period: T = 2π ÷ 56. Dividing by a fraction means multiplying by its reciprocal, the fraction turned upside down: T = 2π × 65 = 12π5. Check: 56 × 12π5 = 60π30 = 2π, one full lap.

Why name them first: A changes outputs and ω changes inputs, so each has its own formula. Mixing them up gives answers like period 6, which uses the loudness knob to set the tempo.

In plain words

Picture a music player with two separate knobs. One changes how far the speaker moves. The other changes how quickly its movement repeats. A sine equation has two numbers with those separate jobs. The number outside sine, called A, multiplies each output and controls the height. The number multiplying x inside sine, called ω (omega), controls how much input you need for a complete wave. Before calculating anything, point to each number and name its job. A minus sign with no written number means −1. A fraction inside the parentheses belongs to the width knob, even when a different fraction sits outside.

π/2π−4−3−2−11234midline y = 0amplitude 4one period
Read height from the amplitude mark and one complete cycle from the period mark.
Reminder
  • Absolute value. Absolute value is distance from zero: |−34| = 34.
  • Fraction multiplication. Multiply tops and bottoms: 23 × 35 = 615 = 25.
  • Fraction division. Keep the first number and flip the divisor: 2π ÷ 13 = 2π × 3 = 6π.
  • Canceling factors. π is nonzero, so 2π3π = 23. Cancel factors multiplying the entire top and bottom.
  • Solving and substituting. 2T = 2π becomes T = π after division by 2. Put it back: 2 × π = 2π.
  • Sine values and odd symmetry. sin(π2) = 1 and sin(−π2) = −1 because sin(−u) = −sin u.
Why it works. Sine itself produces outputs from −1 to 1. Multiplying those outputs by A makes their largest distance from zero |A|. Inside the function, the angle ωx must travel a full 2π for the pattern to repeat. A positive input length T therefore satisfies |ω|T = 2π, so T = 2π|ω|. These are different calculations because one number multiplies the output and the other multiplies the input. Naming A and ω first prevents you from using the right formula on the wrong number.
RuleRule: For y = A sin(ωx), with A ≠ 0 and ω ≠ 0, amplitude = |A| and T = 2π|ω|. A is outside sine; ω multiplies x inside sine.
The same idea, five ways
Say it

Say: amplitude is the size of the outside number; period is two pi divided by the size of the inside number.

Write it

The outside coefficient controls height, and the inside coefficient controls the input length of a cycle.

In math
  • y = A sin(ωx)
  • amplitude = |A|
  • T = 2π|ω|
  • |ω|T = 2π
  • range: [−|A|, |A|]
Like

Find the height knob and the speed knob before turning either one.

See it
y = A sin(ωx)
Outside A → amplitude |A|
Inside ω → period 2π ÷ |ω|
Check: |ω|T = 2π
The coefficients have different jobs.
The same idea, other ways
As two machines

First multiply input x by ω. Next sine converts that angle to a height. Last multiply the height by A. In 6 sin(5x), 5 belongs to the first machine and 6 belongs to the last.

xmultiply by 55x6 sin6 sin(5x)firstsecond
Inside multiplication happens before outside multiplication.
With two small cases

In 6 sin x, the inside coefficient is the unwritten 1. Height becomes 6 but period stays 2π. In sin(5x), height stays 1 but the angle reaches 2π at x = 2π5.

π/2π3π/22π−6−4−2246amplitude 6one period
Read height from the amplitude mark and one complete cycle from the period mark.
.1Outside coefficient A

Treat sine as a machine that has already delivered its height. A multiplies that delivered number. If sine delivers 1 and A = −6, the final height is −6. If sine delivers −1, the final height is 6. This flips the wave and makes its largest distance from the middle 6. The sign tells direction; absolute value tells size.

  • Rule: A multiplies every sine output.
  • Rule: Amplitude = |A| and range = [−|A|, |A|].
  • Rule: A < 0 reflects heights across the x-axis, whose equation is y = 0.
−6−4−2246amplitude 6one period
Read height from the amplitude mark and one complete cycle from the period mark.
Reminder
  • Signed multiplication. A negative times a negative is positive: −6 × (−1) = 6.
The same idea, five ways
Say it

Say: A multiplies the height after sine is evaluated.

Write it

The outside coefficient scales the outputs.

In math
  • y = A × sin(ωx)
  • amplitude = |A|
Like

Multiply every reading on a height gauge by the same number.

See it
sin(ωx)multiply by −6−6 sin(ωx)inputoutput
A acts on the output.
Worked exampleRead a negative outside coefficient

For y = −6 sin(5x), find A, amplitude, and range. This asks how the outputs are multiplied and which heights are achieved.

−6−4−2246amplitude 6one period
Read height from the amplitude mark and one complete cycle from the period mark.
  1. A = −6.−6 is outside sine and multiplies its output.
  2. Amplitude = |−6| = 6.Amplitude is a distance, so the sign does not make it negative.
  3. At x = π10, 5x = π2 and y = −6 × 1 = −6.Substitution exhibits the lowest output.
  4. At x = 3π10, 5x = 3π2 and y = −6 × (−1) = 6. The range is [−6, 6].These inputs reach both endpoints, and sine never exceeds its own range.
Answer
  • A = −6.
  • Amplitude = 6.
  • Range = [−6, 6].
Check Half the full height is 6−(−6)2 = 122 = 6, agreeing with |A|.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: Amplitude of −6 sin(5x) is −6.
−6 is a signed coefficient, but amplitude measures distance.
✓ Instead: A = −6 and amplitude = 6.
Tips and tricks
  • Tip: copy A with its sign, then use absolute value for amplitude.
.2Inside coefficient ω

The inside coefficient changes the angle before sine sees it. If x moves forward 1 unit and ω = 25, the angle moves forward only 25 of a radian. More input distance is needed for a full turn. In sin(4πx), the entire product 4π is ω. Keep π because it also multiplies x.

  • Rule: ω is the entire factor multiplying x inside sine.
  • Rule: For a nonconstant wave, T = 2π ÷ |ω| and |ω|T = 2π.
  • Rule: Negative ω reverses the angle's direction; period length stays positive.
π2π3π4π5π−11amplitude 1one period
Read height from the amplitude mark and one complete cycle from the period mark.
Reminder
  • Fraction division. 2π ÷ 25 = 2π × 52 = 5π.
The same idea, five ways
Say it

Say: omega gives the angle change per input unit.

Write it

The inside coefficient controls how quickly the angle completes a turn.

In math
  • angle = ωx
  • T = 2π|ω|
Like

A slower wheel needs more time to complete a revolution.

See it
ω = 25
(25)T = 2π
T = 2π × 52 = 5π
A smaller inside coefficient produces a longer cycle.
Worked exampleSolve for a full cycle length

Find ω and the period of y = sin(25x). This asks how quickly the angle changes and how much x must change for one cycle.

π2π3π4π5π−11amplitude 1one period
Read height from the amplitude mark and one complete cycle from the period mark.
  1. ω = 25.The entire fraction multiplies x.
  2. 25T = 2π.This equation finds the positive input length needed for one angular turn.
  3. Multiply both sides by 52: T = 2π × 52 = 5π.The product 52 × 25 is 1, removing the coefficient of T.
  4. Put T back: 25 × 5π = 2π.Substitution verifies that this length supplies exactly a full turn.
Answer
  • ω = 25.
  • Period = 5π.
Check At x = 0 the output is 0. At x = 5π the angle is 2π and the whole graph has returned to its starting height and direction after a top and a bottom.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: In sin(4πx), ω = 4.
π also multiplies x, so omitting it changes the angle.
✓ Instead: ω = 4π, so T = 2π4π = 12.
Tips and tricks
  • Tip: circle every factor multiplying x inside the parentheses. Together they form ω.
Strategy: step by step
  1. 1. Write any missing coefficient: −sin(...) means −1 × sin(...), and sin x means 1 × sin(1x).
  2. 2. Copy A from outside sine and the entire ω from the factor multiplying x inside. Keep their signs here.
  3. 3. Use |A| for amplitude and |ω| for the period calculation.
  4. 4. Calculate T = 2π ÷ |ω|. Dividing by a fraction multiplies by its reciprocal, meaning its upside down version.
  5. 5. Put T back into |ω|T = 2π to check the complete turn. The range is [−|A|, |A|].
Strategy
Strategy: find amplitude and period from the coefficients
1
Is a coefficient unwritten?
YesWrite 1, or −1 if there is a minus sign.
NoCopy the written coefficient with its sign.
↓
2
Is A = 0 or ω = 0?
YesThe sine expression is constant. Do not assign a smallest positive period with the wave formula.
NoUse the nonconstant wave formulas.
↓
3
Is |ω| a fraction?
YesMultiply 2π by the reciprocal of the whole fraction.
NoDivide and reduce common factors.
  1. Name A and ω before arithmetic.
  2. Use |A| for amplitude and 2π ÷ |ω| for period.
  3. Multiply |ω| by the proposed T to recover 2π.
Worked exampleA buoy in long swells: read A and ω before the arithmetic

A marker buoy rides long ocean swells. Its height above its calm-water level, in meters, x seconds after a lifeguard starts a stopwatch, is modeled by y = −sin(2x5). A negative y means the buoy is below calm-water level. Match the model to y = A sin(ωx) and state A and ω before doing any arithmetic. Then find the amplitude, the period T (the time for one complete cycle) and the range of the buoy's heights.

  1. Write the hidden coefficients: y = −sin(2x5) is y = −1 × sin(25 × x).A minus sign in front of sin with no number shown means −1 × sin(...). Inside, 2x5 means 2x ÷ 5, which equals 25 × x. So the number multiplying x is the fraction 25.
  2. Copy the coefficients with their signs: A = −1 (outside sine) and ω = 25 (the entire factor multiplying x inside sine).The rule needs the whole factor on x, not just the 2 on top or the 5 underneath. Reading ω as 2 or as 5 would give the wrong period. Writing A = −1 with its sign records exactly what the model says before deciding what the sign does.
  3. Amplitude = |A| = |−1| = 1 m. For the period, use |ω| = 25.Amplitude is a distance from the center line (calm-water level), so it is never negative. The minus sign only flips the wave: the buoy dips first instead of rising first, but it still moves 1 m each way. ω is already positive, so |ω| is the same number.
  4. T = 2π ÷ |ω| = 2π ÷ 25 = 2π × 52 = 10π2 = 5π s ≈ 15.7 s.Dividing by the fraction 25 is the same as multiplying by its reciprocal 52, which is the fraction turned upside down. Because |ω| is less than 1, the cycle is stretched to longer than 2π.
  5. Check the complete turn: |ω|T = 25 × 5π = 10π5 = 2π ✓. Range = [−|A|, |A|] = [−1, 1].Over one period the input to sine, 25x, must grow by exactly 2π, which is one complete turn. It does, so T = 5π is right. The outputs of sine stay between −1 and 1, and multiplying by A = −1 only swaps their signs. So y also stays between −1 and 1, which means the buoy is never more than 1 m below or above calm-water level.
Answer
A = −1 and ω = 25. Amplitude = 1 m, period T = 5π s ≈ 15.7 s, range [−1, 1]. The buoy's height stays between 1 m below and 1 m above calm-water level.
Check A quarter of a period in, at x = 5π4 ≈ 3.9 s, the input is 25 × 5π4 = π2, so y = −sin(π2) = −1, the lowest point. At x = 15π4 ≈ 11.8 s the input is 3π2, so y = −sin(3π2) = 1, the highest point. At x = 5π the input is 2π, so y = −sin(2π) = 0 and the pattern starts over. The extremes −1 and 1 match the range, and the cycle closes at 5π. About 15.7 s per swell suits long ocean swells. The misreadings ω = 2 (T = π ≈ 3.1 s) and ω = 5 (T = 2π5 ≈ 1.3 s) would be far too quick.

Work to write

  1. y = −1 × sin(25x), so A = −1 and ω = 25
  2. amplitude = |A| = |−1| = 1 m
  3. T = 2π ÷ 25 = 2π × 52 = 5π s ≈ 15.7 s
  4. check: 25 × 5π = 2π
  5. range = [−1, 1] (from 1 m below to 1 m above calm-water level)

A = −1 and ω = 25. Amplitude = 1 m, period T = 5π s ≈ 15.7 s, range [−1, 1]. The buoy's height stays between 1 m below and 1 m above calm-water level.

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 unwritten inside coefficient

Find amplitude and period of y = 6 sin x. Find the height above the middle and the length of a cycle.

π/2π3π/22π−6−4−2246amplitude 6one period
Read height from the amplitude mark and one complete cycle from the period mark.
  1. Write 6 sin(1x), so A = 6 and ω = 1.An unwritten coefficient is 1.
  2. Amplitude = |6| = 6.The outside size controls height.
  3. T = 2π1 = 2π.The ordinary sine angle needs a full 2π.
  4. Put T back: 1 × 2π = 2π.This verifies the full turn.
Answer
  • Amplitude = 6.
  • Period = 2π.
Check At π2 the output is 6; at 3π2 it is −6. Both lie 6 units from the middle.
Rung 2Rung 2: two whole number coefficients

Find amplitude and period of y = 7 sin(5x). The two requested lengths have different coefficients.

−8−6−4−22468amplitude 7one period
Read height from the amplitude mark and one complete cycle from the period mark.
  1. A = 7 and ω = 5.7 is outside sine, and 5 is inside.
  2. Amplitude = 7.The outside coefficient is positive.
  3. T = 2π5.Five times the input needs one fifth of the ordinary length for a turn.
  4. Put T back: 5 × 2π5 = 2π.This verifies the complete turn.
Answer
  • Amplitude = 7.
  • Period = 2π5.
Check At x = π10, the angle is π2 and y = 7; at x = 2π5, the angle is 2π and the cycle finishes.
Rung 3Rung 3: negative height and fractional width

Find amplitude and period of y = −34 sin(27x). Both answers are positive lengths.

π2π3π4π5π6π7πamplitude 0.75one period
Read height from the amplitude mark and one complete cycle from the period mark.
  1. A = −34 and ω = 27.Outside and inside fractions have separate jobs.
  2. Amplitude = 34.The negative sign reflects heights, but distance is positive.
  3. T = 2π ÷ 27 = 2π × 72 = 7π.Flip the divisor and cancel the factor 2.
  4. Put T back: 27 × 7π = 2π.This confirms the proposed period.
Answer
  • Amplitude = 34.
  • Period = 7π.
Check At x = 7π4, the inside angle is π2 and y = −34. Three times that input gives y = 34, confirming both extremes.
Rung 4Rung 4: an inside coefficient containing π

Find amplitude and period of y = 56 sin(4πx). Keep the entire 4π as ω.

−11amplitude 0.833333one period
Read height from the amplitude mark and one complete cycle from the period mark.
  1. A = 56 and ω = 4π.π is a factor multiplying x.
  2. Amplitude = 56.The outside number is positive.
  3. T = 2π4π = 24 = 12.Cancel π, then reduce by 2.
  4. Put T back: 4π × 12 = 2π.This verifies that one half input unit gives a full angular turn.
Answer
  • Amplitude = 56.
  • Period = 12.
Check At x = 18, the angle is π2, so y = 56. A quarter of the period reaches the top.
Rung 5Rung 5: two negative coefficients

Find amplitude and period of y = −75 sin(−5π3x). Separate both signs from the sizes before calculating.

−11amplitude 1.4one period
Read height from the amplitude mark and one complete cycle from the period mark.
  1. A = −75 and ω = −5π3.The first sign belongs to the output and the second to the input.
  2. Amplitude = 75 and |ω| = 5π3.Lengths use absolute values.
  3. T = 2π ÷ 5π3 = 2π × 35π = 65.Flip the whole divisor, then cancel π.
  4. Put T back: 5π3 × 65 = 6π3 = 2π.This confirms the positive full turn distance.
  5. Rewrite as y = 75 sin(5π3x).Sine is odd, so the negative inside gives another minus, canceling the outside one.
Answer
  • Amplitude = 75.
  • Period = 65.
Check At x = 310, the original inside angle is −π2, so y = −75 × (−1) = 75. The equivalent positive equation gives the same top.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: y = 12 sin(3πx) has period 2π3.
This leaves out the π in the inside coefficient.
✓ Instead: ω = 3π and T = 2π3π = 23.
✗ Not this: Counterexample: 0 sin(5x) has fundamental period 2π5.
The function is constantly zero. Every positive shift leaves it unchanged, so no smallest positive period exists.
✓ Instead: Its amplitude is 0. The nonconstant wave period formula does not assign its fundamental period.
Tips and tricks
  • Tip: outside controls height; inside controls how quickly the angle turns.
  • Tip: write A and ω before arithmetic, and check |ω|T = 2π afterward.
  • Tip: a missing coefficient is 1, and a lone minus sign means −1.
Trap. Using the outside number as period, or treating π as decoration inside sine. Copy A and the entire ω before applying their separate formulas.
Keep in mind
  • A minus sign with no number in front means A = −1: y = −sin(5x) has amplitude 1.
  • π inside belongs to ω: in y = 9 sin(4πx), ω = 4π and T = 2π4π = 12.
  • A fraction outside sets height and a fraction inside sets width: y = 49 sin(16x) has amplitude 49 and period 2π × 6 = 12π.
  • Signs never make these answers negative: amplitude uses |A| and the period uses |ω|.
Memory hookOutside sets height, inside sets width: A stands alone outside the sine (amplitude |A|); ω sits with x inside (period 2π|ω|).
Flash cards: say the answer out loud, then flip
In y = A sin(ωx), which number sets the height, and where does it sit?
  • A, outside the sine
  • the amplitude is |A|.
Find the amplitude and period of y = −sin(7x).
  • Amplitude: 1
  • Period: 2π7
What is the reciprocal of 56?
65, the fraction turned upside down
Find the amplitude and period of y = 15 sin(2πx).
  • Amplitude: 15
  • Period: 2π2π = 1
In y = 10 sin(πx), can you ignore the π?
No. ω = π, so T = 2ππ = 2.