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# 2.15: Period and Frequency

Difficulty Level: At Grade Created by: CK-12
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While working on an assignment about sound in your science class, your Instructor informs you that what you know as the "pitch" of a sound is, in fact, the frequency of the sound waves. He then plays a note on a musical instrument, and the pattern of the sound wave on a graph looks like this:

He then tells you to find the frequency of the sound wave from the graph? Can you do it?

Don't worry. By the end of this Concept, you'll understand what frequency is and be able to find it from a plot like this one.

### Watch This

In the second part of this video you'll learn about the period of trigonometric functions.

### Guidance

The period of a trigonometric function is the horizontal distance traversed before the y\begin{align*}y-\end{align*}values begin to repeat. For both graphs, y=sinx\begin{align*}y = \sin x\end{align*} and y=cosx\begin{align*}y = \cos x\end{align*}, the period is 2π.\begin{align*}2\pi .\end{align*} As you may remember, after completing one rotation of the unit circle, these values are the same.

Frequency is a measurement that is closely related to period. In science, the frequency of a sound or light wave is the number of complete waves for a given time period (like seconds). In trigonometry, because all of these periodic functions are based on the unit circle, we usually measure frequency as the number of complete waves every 2π\begin{align*}2\pi\end{align*} units. Because y=sinx\begin{align*}y = \sin x\end{align*} and y=cosx\begin{align*}y = \cos x\end{align*} cover exactly one complete wave over this interval, their frequency is 1.

Period and frequency are inversely related. That is, the higher the frequency (more waves over 2π\begin{align*}2\pi\end{align*} units), the lower the period (shorter distance on the x\begin{align*}x-\end{align*}axis for each complete cycle).

After observing the transformations that result from multiplying a number in front of the sinusoid, it seems natural to look at what happens if we multiply a constant inside the argument of the function, or in other words, by the x\begin{align*}x\end{align*} value. In general, the equation would be y=sinBx\begin{align*}y=\sin Bx\end{align*} or y=cosBx\begin{align*}y=\cos Bx\end{align*}. For example, look at the graphs of y=cos2x\begin{align*}y = \cos 2x\end{align*} and y=cosx\begin{align*}y = \cos x\end{align*}.

Notice that the number of waves for y=cos2x\begin{align*}y = \cos 2x\end{align*} has increased, in the same interval as y=cosx\begin{align*}y = \cos x\end{align*}. There are now 2 waves over the interval from 0 to 2π\begin{align*}2\pi\end{align*}. Consider that you are doubling each of the x\begin{align*}x\end{align*} values because the function is 2x\begin{align*}2x\end{align*}. When π\begin{align*}\pi\end{align*} is plugged in, for example, the function becomes 2π\begin{align*}2\pi\end{align*}. So the portion of the graph that normally corresponds to 2π\begin{align*}2\pi\end{align*} units on the x\begin{align*}x-\end{align*}axis, now corresponds to half that distance—so the graph has been “scrunched” horizontally. The frequency of this graph is therefore 2, or the same as the constant we multiplied by in the argument. The period (the length for each complete wave) is π\begin{align*}\pi\end{align*}.

#### Example A

What is the frequency and period of y=sin3x\begin{align*}y = \sin 3x\end{align*}?

Solution: If we follow the pattern from the previous example, multiplying the angle by 3 should result in the sine wave completing a cycle three times as often as y=sinx\begin{align*}y = \sin x\end{align*}. So, there will be three complete waves if we graph it from 0 to 2π\begin{align*}2\pi\end{align*}. The frequency is therefore 3. Similarly, if there are 3 complete waves in 2π\begin{align*}2\pi\end{align*} units, one wave will be a third of that distance, or 2π3\begin{align*}\frac{2\pi}{3}\end{align*} radians. Here is the graph:

This number that is multiplied by x\begin{align*}x\end{align*}, called B\begin{align*}B\end{align*}, will create a horizontal dilation. The larger the value of B\begin{align*}B\end{align*}, the more compressed the waves will be horizontally. To stretch out the graph horizontally, we would need to decrease the frequency, or multiply by a number that is less than 1. Remember that this dilation factor is inversely related to the period of the graph.

Adding, one last time to our equations from before, we now have: y=D±Asin(B(x±C))\begin{align*}y=D \pm A \sin (B(x \pm C))\end{align*} or y=D±Acos(B(x±C))\begin{align*}y=D \pm A \cos (B(x \pm C))\end{align*}, where B\begin{align*}B\end{align*} is the frequency, the period is equal to 2πB\begin{align*}\frac{2\pi}{B}\end{align*}, and everything else is as defined before.

#### Example B

What is the frequency and period of y=cos14x\begin{align*}y=\cos \frac{1}{4}x\end{align*}?

Solution: Using the generalization above, the frequency must be 14\begin{align*}\frac{1}{4}\end{align*} and therefore the period is 2π114\begin{align*}\frac{\frac{2\pi}{1}}{\frac{1}{4}}\end{align*}, which simplifies to: 2π14=2π1144141=8π1=8π\begin{align*}\frac{2\pi}{\frac{1}{4}}=\frac{\frac{2\pi}{1}}{\frac{1}{4}} \cdot \frac{\frac{4}{1}}{\frac{4}{1}} = \frac{8\pi}{1}=8\pi\end{align*}

Thinking of it as a transformation, the graph is stretched horizontally. We would only see 14\begin{align*}\frac{1}{4}\end{align*} of the curve if we graphed the function from 0 to 2π\begin{align*}2\pi\end{align*}. To see a complete wave, therefore, we would have to go four times as far, or all the way from 0 to 8π\begin{align*}8\pi\end{align*}.

#### Example C

What is the frequency and period of y=sin12x\begin{align*}y = \sin \frac{1}{2}x\end{align*}?

Solution:

Like the previous two examples, we can see that the frequency is 12\begin{align*}\frac{1}{2}\end{align*}, and so the period is 2π112\begin{align*}\frac{\frac{2\pi}{1}}{\frac{1}{2}}\end{align*}, which becomes 2π×21=4π\begin{align*}2\pi \times \frac{2}{1} = 4\pi\end{align*}

### Vocabulary

Period: The period of a wave is the horizontal distance traveled before the 'y' values begin to repeat.

Frequency: The frequency of a wave is number of complete waves every 2π\begin{align*}2 \pi\end{align*} units.

### Guided Practice

1. Draw a sketch of y=3sin2x\begin{align*}y = 3 \sin2x\end{align*} from 0 to 2π\begin{align*}2\pi\end{align*}.

2. Draw a sketch of y=2.5cosπx\begin{align*}y = 2.5 \cos \pi x\end{align*} from 0 to 2π\begin{align*}2\pi\end{align*}.

3. Draw a sketch of y=4sin12x\begin{align*}y=4 \sin \frac{1}{2} x\end{align*} from 0 to 2π\begin{align*}2\pi\end{align*}.

Solutions:

1. The "2" inside the sine function makes the function "squashed" by a factor of 2 in the horizontal direction.

2. The π\begin{align*}\pi\end{align*} inside the sine function makes the function "squashed" by a factor of π\begin{align*}\pi\end{align*} in the horizontal direction.

3. The 12\begin{align*}\frac{1}{2}\end{align*} inside the sine function makes the function "stretched" by a factor of 12\begin{align*}\frac{1}{2}\end{align*} in the horizontal direction.

### Concept Problem Solution

By inspecting the graph

You can see that the wave takes about 6.2 seconds to make one complete cycle. This means that the frequency of the wave is approximately 1 cycle per second (since 2π\begin{align*}2 \pi\end{align*} is approximately 6.28). (You should note that in a real wave of sound, you would need to use the speed of the wave and so the calculation would be different. But if you read the graph the same way you read trigonometric functions to find the frequency, this is the result you would find.)

### Practice

Find the period and frequency of each function below.

1. y=sin(4x)\begin{align*}y=\sin(4x)\end{align*}
2. y=cos(2x)\begin{align*}y=\cos(2x)\end{align*}
3. y=cos(12x)\begin{align*}y=\cos(\frac{1}{2}x)\end{align*}
4. y=sin(34x)\begin{align*}y=\sin(\frac{3}{4}x)\end{align*}
5. y=sin(3x)\begin{align*}y=\sin(3x)\end{align*}

Draw a sketch of each function from 0 to 2π\begin{align*}2\pi\end{align*}.

1. y=sin(3x)\begin{align*}y=\sin(3x)\end{align*}
2. y=cos(5x)\begin{align*}y=\cos(5x)\end{align*}
3. y=3cos(25x)\begin{align*}y=3\cos(\frac{2}{5}x)\end{align*}
4. y=12sin(34x)\begin{align*}y=\frac{1}{2}\sin(\frac{3}{4}x)\end{align*}
5. y=sin(2x)\begin{align*}y=-\sin(2x)\end{align*}
6. y=tan(3x)\begin{align*}y=\tan(3x)\end{align*}
7. y=sec(2x)\begin{align*}y=\sec(2x)\end{align*}
8. y=csc(4x)\begin{align*}y=\csc(4x)\end{align*}

Find the equation of each function.

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### Vocabulary Language: English

TermDefinition
Amplitude The amplitude of a wave is one-half of the difference between the minimum and maximum values of the wave, it can be related to the radius of a circle.
Frequency The frequency of a trigonometric function is the number of cycles it completes every $2 \pi$ units.
horizontal stretch Horizontal stretch describes the stretching of a graph from the $y$ axis. For a sinusoidal function with equation $f(x)=\pm a \cdot \sin (b(x+c))+d$, the coefficient $b$ controls horizontal stretch.
Period The period of a wave is the horizontal distance traveled before the $y$ values begin to repeat.
sinusoidal function A sinusoidal function is a sine or cosine wave.
sinusoidal functions A sinusoidal function is a sine or cosine wave.
Vertical shift A vertical shift is the result of adding a constant term to the value of a function. A positive term results in an upward shift, and a negative term in a downward shift.

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