Amplitude, period, and phase shift
Sine curves describe many things that repeat: sound, light, tides, and seasons. These waves can be taller or shorter, faster or slower, and shifted in time. This page shows how each number in an equation like ${y=2\sin(3x-\pi)+1}$ changes the graph of ${y=\sin x}$.
Amplitude
Multiplying by a number ${a}$ multiplies every ${y}$-value by ${a}$. This stretches the graph up and down:
The amplitude of a sine curve is half the distance from its lowest point to its highest point. It tells how tall the wave is. For ${y=a\sin x}$ and ${y=a\cos x}$, the amplitude is ${|a|}$.
If ${a}$ is negative, the graph is also flipped over the ${x}$-axis. For example, ${y=-2\sin x}$ has amplitude ${|-2|=2}$, and it goes down first instead of up.
Example 1: Find the amplitude and the range of ${y=-\dfrac{3}{2}\cos x}$.
Solution:
The amplitude is
${\left|-\dfrac{3}{2}\right|=\dfrac{3}{2}}$
The values of ${\cos x}$ go from ${-1}$ to ${1}$, so the values of ${-\dfrac{3}{2}\cos x}$ go from ${-\dfrac{3}{2}}$ to ${\dfrac{3}{2}}$. The range is ${\left[-\dfrac{3}{2},\dfrac{3}{2}\right]}$.
Period
Now look at ${y=\sin 2x}$. As ${x}$ goes from ${0}$ to ${\pi}$, the inside, ${2x}$, goes from ${0}$ to ${2\pi}$. That is one full wave of sine. So ${y=\sin 2x}$ finishes a wave in only ${\pi}$ units. Its period is ${\pi}$, half the usual period.
In general, for a positive number ${k}$, ${kx}$ goes from ${0}$ to ${2\pi}$ as ${x}$ goes from ${0}$ to ${\dfrac{2\pi}{k}}$. So
${y=\sin kx}$ and ${y=\cos kx}$ have period ${\dfrac{2\pi}{k}}$
If ${k>1}$, the graph is squeezed together, and the waves come faster. If ${0<k<1}$, the graph is stretched out. For example, ${y=\cos\dfrac{x}{2}}$ has period ${\dfrac{2\pi}{1/2}=4\pi}$.
The frequency is the number of waves in one unit, ${\dfrac{1}{\text{period}}}$. A shorter period means a higher frequency.
Example 2: Find the amplitude and the period of ${y=4\sin 3x}$. Sketch one period.
Solution:
The amplitude is ${4}$. The period is ${\dfrac{2\pi}{3}}$.
Split the period ${\left[0,\dfrac{2\pi}{3}\right]}$ into four equal parts. Each part is
$\dfrac{1}{4}\cdot\dfrac{2\pi}{3}=\dfrac{\pi}{6}$
So the key points are at
${x=0}$, ${\dfrac{\pi}{6}}$, ${\dfrac{\pi}{3}}$, ${\dfrac{\pi}{2}}$, ${\dfrac{2\pi}{3}}$
Like ${\sin x}$, the curve goes: zero, high point, zero, low point, zero. The high point is ${4}$, and the low point is ${-4}$.
Phase shift
Replacing ${x}$ by ${x-b}$ moves a graph ${b}$ units to the right. (If ${b}$ is negative, it moves to the left.) For sine curves, this horizontal shift is called the phase shift.
The graph of ${y=\sin\left(x-\dfrac{\pi}{3}\right)}$ starts its wave at ${x=\dfrac{\pi}{3}}$, where the inside is ${0}$. One period ends ${2\pi}$ later, at ${x=\dfrac{7\pi}{3}}$.
Putting it together
For ${k>0}$, the curves
${y=a\sin k(x-b)}$ and ${y=a\cos k(x-b)}$
have these features:
| Feature | Value |
|---|---|
| Amplitude | ${|a|}$ |
| Period | ${\dfrac{2\pi}{k}}$ |
| Phase shift | ${b}$ |
One period starts at ${x=b}$ and ends at ${x=b+\dfrac{2\pi}{k}}$.
Watch out: To read the phase shift, first factor ${k}$ out of the inside. For example,
$\sin\left(2x-\dfrac{\pi}{2}\right)=\sin 2\left(x-\dfrac{\pi}{4}\right)$
So the phase shift is ${\dfrac{\pi}{4}}$, not ${\dfrac{\pi}{2}}$.
Vertical shift
Adding a number ${d}$ moves the whole graph up ${d}$ units (or down, if ${d}$ is negative). The wave is then centered on the horizontal line ${y=d}$, called the midline. It goes up to ${d+|a|}$ and down to ${d-|a|}$.
${y=a\sin k(x-b)+d}$: amplitude ${|a|}$, period ${\dfrac{2\pi}{k}}$, phase shift ${b}$, midline ${y=d}$
How to graph one period
- Find the amplitude ${|a|}$, the period ${P=\dfrac{2\pi}{k}}$, the phase shift ${b}$, and the midline ${y=d}$.
- One period runs from ${x=b}$ to ${x=b+P}$. Split it into four equal parts of length ${\dfrac{P}{4}}$. This gives five ${x}$-values.
- Find the ${y}$-values. For sine, the pattern is: midline, high, midline, low, midline. For cosine, it is: high, midline, low, midline, high. If ${a<0}$, swap high and low.
- Plot the five points, and join them with a smooth wave.
Example 3: Graph one period of ${y=2\sin\left(2x-\dfrac{\pi}{2}\right)+1}$.
Solution:
Step 1: Factor ${2}$ out of the inside:
${y=2\sin 2\left(x-\dfrac{\pi}{4}\right)+1}$
So ${a=2}$, ${k=2}$, ${b=\dfrac{\pi}{4}}$, and ${d=1}$. The amplitude is ${2}$, the period is ${\dfrac{2\pi}{2}=\pi}$, the phase shift is ${\dfrac{\pi}{4}}$, and the midline is ${y=1}$.
Step 2: One period runs from ${\dfrac{\pi}{4}}$ to
${\dfrac{\pi}{4}+\pi=\dfrac{5\pi}{4}}$
Each quarter is ${\dfrac{\pi}{4}}$ long. The five ${x}$-values are
${\dfrac{\pi}{4}}$, ${\dfrac{\pi}{2}}$, ${\dfrac{3\pi}{4}}$, ${\pi}$, ${\dfrac{5\pi}{4}}$
Step 3: The high value is ${1+2=3}$, and the low value is ${1-2=-1}$. For sine, the ${y}$-values are
${1}$, ${3}$, ${1}$, ${-1}$, ${1}$
Step 4: Plot and join:
Example 4: Graph one period of ${y=-3\cos \pi x}$.
Solution:
The amplitude is ${3}$. Here ${k=\pi}$, so the period is ${\dfrac{2\pi}{\pi}=2}$. There is no phase shift, and the midline is ${y=0}$.
One period runs from ${0}$ to ${2}$, in quarters of ${0.5}$. The five ${x}$-values are ${0}$, ${0.5}$, ${1}$, ${1.5}$, ${2}$.
For cosine the pattern is high, midline, low, midline, high. Since ${a=-3}$ is negative, swap high and low. The ${y}$-values are
${-3}$, ${0}$, ${3}$, ${0}$, ${-3}$
This period is a plain number, ${2}$, not a multiple of ${\pi}$. That happens whenever ${k}$ contains ${\pi}$.
Finding an equation from a graph
You can also go backward: read the numbers ${a}$, ${k}$, ${b}$, and ${d}$ from a graph.
Example 5: Find an equation of the form ${y=a\cos k(x-b)+d}$ for this curve.
Solution:
Midline and amplitude. The highest value is ${5}$, and the lowest is ${-1}$. The midline is halfway between:
${d=\dfrac{5+(-1)}{2}=2}$
The amplitude is half the distance from top to bottom:
${a=\dfrac{5-(-1)}{2}=3}$
Period. One full wave goes from the high point at ${\dfrac{\pi}{3}}$ to the next high point at ${\dfrac{4\pi}{3}}$. So the period is
${\dfrac{4\pi}{3}-\dfrac{\pi}{3}=\pi}$
Then
${\dfrac{2\pi}{k}=\pi}$, so ${k=2}$
Phase shift. A cosine curve starts at a high point. This curve has a high point at ${x=\dfrac{\pi}{3}}$, so we can take ${b=\dfrac{\pi}{3}}$.
So an equation is
${y=3\cos 2\left(x-\dfrac{\pi}{3}\right)+2}$
Other answers are also correct. For example, ${b=\dfrac{4\pi}{3}}$ works too, because the curve repeats every ${\pi}$.
Summary
- For ${y=a\sin k(x-b)+d}$ or ${y=a\cos k(x-b)+d}$ with ${k>0}$: amplitude ${|a|}$, period ${\dfrac{2\pi}{k}}$, phase shift ${b}$, midline ${y=d}$.
- Factor ${k}$ out of the inside before reading the phase shift.
- To graph one period, split ${[b,\,b+\text{period}]}$ into four equal parts, and plot the five key points.
- To find an equation from a graph: midline and amplitude from the highest and lowest values, period from two high points, phase shift from a high point (cosine).
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