Amplitude and Period: y = A sin(Bx)

April 12, 2026

Problem

Compare y = sin(x), y = 2 sin(x), and y = sin(2x). How do amplitude and frequency change?

Explanation

The general sine wave has the form y=Asin(Bx)y = A\sin(Bx). The two parameters control completely separate things:

  • A|A| is the amplitude — the maximum height. The wave swings between A-A and +A+A.
  • BB controls the frequency (how many oscillations per 2π2\pi). The period is T=2π/BT = 2\pi/B.

Pulling these apart with sliders is the fastest way to internalize what each one does.

The Formulas

For y=Asin(Bx)y = A\sin(Bx):

Amplitude=APeriod=T=2πBFrequency=B2π\text{Amplitude} = |A| \qquad \text{Period} = T = \dfrac{2\pi}{|B|} \qquad \text{Frequency} = \dfrac{|B|}{2\pi}

Step-by-Step Solution

Compare three waves:

  1. y1=sinxy_1 = \sin x (the baseline)
  2. y2=2sinxy_2 = 2\sin x (doubled amplitude)
  3. y3=sin(2x)y_3 = \sin(2x) (doubled frequency)

Step 1 — Analyze y1=sinxy_1 = \sin x.

  • Amplitude: A=1|A| = 1
  • Period: T=2π6.28T = 2\pi \approx 6.28
  • Maximum value: 11, minimum: 1-1
  • Zeros at x=0,π,2π,x = 0, \pi, 2\pi, \ldots

Step 2 — Analyze y2=2sinxy_2 = 2\sin x.

  • A=2A = 2, B=1B = 1
  • Amplitude: A=2|A| = 2 (doubled)
  • Period: T=2π/1=2πT = 2\pi/1 = 2\pi (unchanged)
  • Maximum value: 22, minimum: 2-2
  • Zeros: same as y1y_1 (at x=0,π,2π,x = 0, \pi, 2\pi, \ldots)

The wave is stretched vertically — it reaches further up and down, but takes the same amount of xx to complete one cycle.

Step 3 — Analyze y3=sin(2x)y_3 = \sin(2x).

  • A=1A = 1, B=2B = 2
  • Amplitude: A=1|A| = 1 (unchanged)
  • Period: T=2π/2=π3.14T = 2\pi/2 = \pi \approx 3.14 (halved)
  • Maximum value: 11 (unchanged)
  • Zeros: 2x=0,π,2π,3π,2x = 0, \pi, 2\pi, 3\pi, \ldots, so x=0,π/2,π,3π/2,x = 0, \pi/2, \pi, 3\pi/2, \ldots (twice as many!)

The wave is compressed horizontally — it completes two full cycles in the same horizontal distance that y1y_1 takes for one.

Step 4 — Tabulate values at x=π/4x = \pi/4 for comparison.

  • y1(π/4)=sin(π/4)=220.707y_1(\pi/4) = \sin(\pi/4) = \dfrac{\sqrt{2}}{2} \approx 0.707
  • y2(π/4)=2sin(π/4)=21.414y_2(\pi/4) = 2\sin(\pi/4) = \sqrt{2} \approx 1.414 (twice as tall)
  • y3(π/4)=sin(π/2)=1y_3(\pi/4) = \sin(\pi/2) = 1 (already at the peak — twice as fast)

Step 5 — Combined transformation: y=2sin(2x)y = 2\sin(2x).

If you use both transformations together, the wave has amplitude 2 and period π\pi simultaneously: tall and fast.


Answer:

  • y1=sinxy_1 = \sin x: amplitude 1, period 2π2\pi
  • y2=2sinxy_2 = 2\sin x: amplitude 2, period 2π2\pi (vertical stretch)
  • y3=sin(2x)y_3 = \sin(2x): amplitude 1, period π\pi (horizontal compression)

The two parameters AA and BB act on completely independent aspects of the wave. Adjusting one never affects the other.

Try It

  • Slide A to see the wave grow taller without changing how often it oscillates.
  • Slide B to see it oscillate faster without changing height.
  • The HUD shows the current amplitude, period, and frequency.

Interactive Visualization

Parameters

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Amplitude and Period: y = A sin(Bx) | MathSpin