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# Period and Frequency

## Horizontal distance traveled before y values repeat; number of complete waves in 2pi.

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Frequency and Period of Sinusoidal Functions

The transformation rules about horizontal stretching and shrinking directly apply to sine and cosine graphs.  If a sine graph is horizontally stretched by a factor of 12\begin{align*}\frac{1}{2}\end{align*} that is the same as a horizontal compression by a factor of 2.

How does the equation change when a sine or cosine graph is stretched by a factor of 3?

#### Watch This

http://www.youtube.com/watch?v=qJ-oUV7xL3w James Sousa: Amplitude and Period of Sine and Cosine

#### Guidance

The general equation for a sinusoidal function is:

f(x)=±asin(b(x+c))+d\begin{align*}f(x)={\pm} a \cdot \sin (b(x+c))+d\end{align*}

The ±\begin{align*}\pm\end{align*} controls the reflection across the x\begin{align*}x\end{align*}-axis.  The coefficient a\begin{align*}a\end{align*} controls the amplitude.  The constant d\begin{align*}d\end{align*} controls the vertical shift.  Here you will see that the coefficient b\begin{align*}b\end{align*} controls the horizontal stretch.

Horizontal stretch is measured for sinusoidal functions as their periods.  This is why this function family is also called the periodic function family.  The period of a sinusoid is the length of a complete cycle.  For basic sine and cosine functions, the period is 2π\begin{align*}2 \pi\end{align*}.  This length can be measured in multiple ways.  In word problems and in other tricky circumstances, it may be most useful to measure from peak to peak.

The ability to measure the period of a function in multiple ways allows different equations to model an identical graph.  In the image above, the top red line would represent a regular cosine wave.  The center red line would represent a regular sine wave with a horizontal shift.  The bottom red line would represent a negative cosine wave with a horizontal shift.  This flexibility in perspective means that many of the examples, guided practice and practice problems may have multiple solutions.  For now, try to always choose the function that has a period starting at x=0\begin{align*}x=0\end{align*}

Frequency is a different way of measuring horizontal stretch.  For sound, frequency is known as pitch.  With sinusoidal functions, frequency is the number of cycles that occur in 2π\begin{align*}2 \pi\end{align*}.  A shorter period means more cycles can fit in 2π\begin{align*}2 \pi\end{align*} and thus a higher frequency.  Period and frequency are inversely related by the equation:

period=2πfrequency\begin{align*}\text{period}={\frac{2 \pi}{\text{frequency}}}\end{align*}

The equation of a basic sine function is f(x)=sinx\begin{align*}f(x)={\sin} x\end{align*}.  In this case b\begin{align*}b\end{align*}, the frequency, is equal to 1 which means one cycle occurs in 2π\begin{align*}2 \pi\end{align*}.  This relationship is a common mistake in graphing sinusoidal functions.  Students find b=12\begin{align*}b={\frac{1}{2}}\end{align*} and then mistakenly conclude that the period is 12\begin{align*}\frac{1}{2}\end{align*} when it is in fact stretched to 4π\begin{align*}4 \pi\end{align*}.

Example A

Rank each wave by period from shortest to longest.

Solution:

The red wave has the shortest period.

The green and black waves have equal periods.  A common mistake is to see that the green wave has greater amplitude and confuse that with greater periods.

The blue wave has the longest period.

Example B

Identify the amplitude, vertical shift, period and frequency of the following function.  Then graph the function.

f(x)=2sin(x3)+1\begin{align*}f(x)=2 \sin \left(\frac{x}{3}\right)+1\end{align*}

Solution:  a=2,b=13,d=1\begin{align*}a=2, b=\frac{1}{3}, d=1\end{align*}. Since b=13\begin{align*}b=\frac{1}{3}\end{align*} (frequency), then the period must be 6π\begin{align*}6 \pi\end{align*}.

Often the most challenging part of graphing periodic functions is labeling the axes.  Since the period is 6π\begin{align*}6 \pi\end{align*}, start by drawing the sinusoidal axis shifted appropriately.  Then divide the 6π\begin{align*}6 \pi\end{align*} into four parts so that the 5 guiding points of the sine graph can be plotted with the amplitude and reflection in mind.  The very last thing to do is to draw and extend the curve.  Many students try to draw the curve too early and end up having to redo their work.

Example C

A measuring stick on a dock measures high tide to be 18 feet and low tide to be 6 feet.  It takes about 6 hours for the tide to switch between low and high tides.  Determine a graphical and algebraic model for the tides knowing that at t=0\begin{align*}t=0\end{align*} there is a high tide.

Solution:  Usually the best course of action for word problems is to identify information, plot points, sketch and then finally come up with an equation.

From the given information you can deduce the following points.  Notice how the sinusoidal axis can be assumed to be the average of the high and low tides.

 Time (hours) Water level (feet) 0 18 6 6 12 18 3 18+62=12\begin{align*}\frac{18+6}{2}=12\end{align*} 9 12

By plotting those points and filling in the sinusoidal axis you can observe a cosine graph.

The amplitude is 6 so a=6\begin{align*}a=6\end{align*}.  There is no vertical reflection.  Since the period is 12 you can determine the frequency b\begin{align*}b\end{align*}:

12=2πbb=π6\begin{align*}12={\frac{2 \pi}{b}} \rightarrow b={\frac{\pi}{6}}\end{align*}

The vertical shift is 12 so d=12\begin{align*}d=12\end{align*}.   Thus you have all the pieces to make an algebraic model:

f(x)=+6cos(π6x)+12\begin{align*}f(x)=+6 \cdot \cos \left(\frac{\pi}{6}x\right)+12\end{align*}

Concept Problem Revisited

If a sine graph is horizontally stretched by a factor of 3 then the general equation has b=13\begin{align*}b={\frac{1}{3}}\end{align*}.  This is because b\begin{align*}b\end{align*} is the frequency and counts the number (or fraction) of a period that fits in a normal period of 2π\begin{align*}2 \pi\end{align*}.  Graphically, the sine wave will make a complete cycle in 6π\begin{align*}6 \pi\end{align*} units like Example B.

#### Vocabulary

Period is the distance it takes for a repeating function to make one complete cycle.

Frequency is the number of cycles a function makes in a set amount of time or distance on the x\begin{align*}x\end{align*} axis.  For sine and cosine graphs this distance is 2π\begin{align*}2 \pi\end{align*}

#### Guided Practice

1. A fish is caught in a water wheel by the side of a river.  Initially the fish is 2 feet below the surface of the water.  Twenty seconds later the fish is 14 feet in the air at the top of the water wheel.  Model the fish’s height in a graph and an equation.

2. Graph the following function: g(x)=cos(8x)+2\begin{align*}g(x)=-\cos (8x)+2\end{align*}.

3. Given the following graph, identify the amplitude, period, and frequency and create an algebraic model.

1. Use logic to identify five key points.  Use those key points to come up with a sketch.  Use the sketch to identify information for the equation.

 Time (seconds) Fish height (feet) 0 -2 20 14 40 -2 10 −2+142=6\begin{align*}\frac{-2+14}{2}=6\end{align*} 30 6

The amplitude is 8 so a=8\begin{align*}a=8\end{align*}.  The function looks like a negative cosine graph.  The vertical shift is d=6\begin{align*}d=6\end{align*} and the period is 40.

40f(x)=2πbb=π20=8cos(π20x)+6

Notice how the labeling on the graph is extremely deliberate.  On both the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} axes, only the most important intervals are labeled.  This keeps the sketch accurate, evenly spaced on your paper and easy to read.

2. The labeling is the most important and challenging part of this problem.  The amplitude is 1.  The shape is a negative cosine.  The vertical shift is 2.  The period is 2π8=π4\begin{align*}\frac{2 \pi}{8}={\frac{\pi}{4}}\end{align*}.  Working with this small period may be challenging at first, but remember that halving fractions is as simple as doubling the denominator.

3. The amplitude is 3.  The shape is a negative cosine.  The period is 5π2\begin{align*}\frac{5 \pi}{2}\end{align*} which implies that b=45\begin{align*}b={\frac{4}{5}}\end{align*}.  The vertical shift is 1. f(x)=3cos(45x)+1\begin{align*}f(x)=-3 \cdot \cos \left(\frac{4}{5}x\right)+1\end{align*}.

#### Practice

Find the frequency and period of each function below.

1. f(x)=sin(4x)+1\begin{align*}f(x)={\sin (4x)+1}\end{align*}

2. g(x)=3cos(2x)\begin{align*}g(x)=-3 \cos (2x)\end{align*}

3. \begin{align*}h(x)={\cos} \left(\frac{1}{2}x\right)+2\end{align*}

4. \begin{align*}k(x)=-2 \sin \left(\frac{3}{4}x\right)+1\end{align*}

5. \begin{align*}j(x)=4 \cos (3x)-1\end{align*}

Graph each of the following functions.

6. \begin{align*}f(x)=3 \sin (2x)+1\end{align*}

7. \begin{align*}g(x)=2.5 \cos (\pi x)-4\end{align*}

8. \begin{align*}h(x)=-\sin (4x)-3\end{align*}

9. \begin{align*}k(x)={\frac{1}{2}} \cos (2x)\end{align*}

10. \begin{align*}j(x)=-2 \sin \left(\frac{3}{4}x \right)-1\end{align*}

Create an algebraic model for each of the following graphs.

11.

12.

13.

14. At time 0 it is high tide and the water at a certain location is 10 feet high.  At low tide 6 hours later, the water is 2 feet high.  Given that tides can be modeled by sinusoidal functions, find a graph that models this scenario.

15. Find the equation that models the scenario in the previous problem.

### Vocabulary Language: English

Amplitude

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.
sinusoidal function

sinusoidal function

A sinusoidal function is a sine or cosine wave.
sinusoidal functions

sinusoidal functions

A sinusoidal function is a sine or cosine wave.
Vertical shift

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.