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Mathematics LibreTexts

7.4 Modeling Changing Amplitude and Midline

While sinusoidal functions can model a variety of behaviors, it is often necessary to combine sinusoidal functions with linear and exponential curves to model real applications and behaviors.  We begin this section by looking at changes to the midline of a sinusoidal function.  Recall that the midline describes the middle, or average value, of the sinusoidal function.

Changing Midlines

 

Example 1

A population of elk currently averages 2000 elk, and that average has been growing by 4% each year.  Due to seasonal fluctuation, the population oscillates from 50 below average in the winter up to 50 above average in the summer.  Find a function that models the number of elk after t years.

SOLUTION

There are two components to the behavior of the elk population:  the changing average, and the oscillation.  The average is an exponential growth, starting at 2000 and growing by 4% each year.  Writing a formula for this:

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1063.gif

For the oscillation, since the population oscillates 50 above and below average, the amplitude will be 50.  Since it takes one year for the population to cycle, the period is 1.  We find the value of the horizontal stretch coefficientFile:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1065.gif.

Additionally, since we weren’t told when t was first measured we will have to decide if t = 0 corresponds to winter, or summer.  If we choose winter then the shape of the function would be a negative cosine, since it starts at the lowest value. 

Putting it all together, the equation would be:

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1067.gif

Since the midline represents the average population, we substitute in the exponential function into the population equation to find our final equation:

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1069.gif

This is an example of changing midline – in this case an exponentially changing midline.

Definition: Changing Midline

A function of the form File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1071.gif will oscillate above and below the average given by the function g(t).

Changing midlines can be exponential, linear, or any other type of function.   Here are some examples:

  Linear midline                       Exponential midline                Quadratic midline

Graphs   Graphs    GraphsFile:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1079.gif    File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1081.gif                         File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1083.gif

Example 2

Find a function with linear midline of the form File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1085.gif that will pass through the points given below. 

SOLUTION

Text Box: t	0	1	2	3
f(t)	5	10	9	8

Since we are given the value of the horizontal compression coefficient we can calculate the period of this function: File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1088.gif.

Since the sine function is at the midline at the beginning of a cycle and halfway through a cycle, we would expect this function to be at the midline at t = 0 and t = 2, since 2 is half the full period of 4.  Based on this, we expect the points (0, 5) and (2, 9) to be points on the midline. We can clearly see that this is not a constant function and so we use the two points to calculate a linear function: File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1090.gif.  From these two points we can calculate a slope:

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1092.gif

Combining this with the initial value of 5, we have the midline: File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1094.gif, giving a full function of the form File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1096.gif.  To find the amplitude, we can plug in a point we haven’t already used, such as (1, 10)

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1098.gif                          Evaluate the sine and combine like terms

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1100.gif

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1102.gif

A function of the form given fitting the data would be

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1104.gif

 

Alternative Approach

Notice we could have taken an alternate approach by plugging points (0, 5) and (2, 9) into the original equation.  Substituting (0, 5),

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1106.gif                         Evaluate the sine and simplify

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1108.gif

 

Substituting (2, 9)

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1110.gif                         Evaluate the sine and simplify

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1112.gif

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1114.gif

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1116.gif, as we found above.  Now we can proceed to find A the same way we did before.

Example 3

The number of tourists visiting a ski and hiking resort averages 4000 people annually and oscillates seasonally, 1000 above and below the average. Due to a marketing campaign, the average number of tourists has been increasing by 200 each year.  Write an equation for the number of tourists after t years, beginning at the peak season.

SOLUTION

Again there are two components to this problem:  the oscillation and the average. For the oscillation, the number of tourists oscillates 1000 above and below average, giving an amplitude of 1000. Since the oscillation is seasonal, it has a period of 1 year.  Since we are given a starting point of “peak season”, we will model this scenario with a cosine function. 

So far, this gives an equation in the form File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1118.gif

Description: Description: GraphsThe average is currently 4000, and is increasing by 200 each year.  This is a constant rate of change, so this is linear growth, File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1122.gif.

 

Combining these two pieces gives a function for the number of tourists:

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1124.gif

 

Try it Now: 1

Given the function File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1126.gif, describe the midline and amplitude using words.

Changing Amplitude

There are also situations in which the amplitude of a sinusoidal function does not stay constant. Back in Chapter 6, we modeled the motion of a spring using a sinusoidal function, but had to ignore friction in doing so.  If there were friction in the system, we would expect the amplitude of the oscillation to decrease over time.  In the equation File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1128.gif, A gives the amplitude of the oscillation, we can allow the amplitude to change by replacing this constant A with a function A(t).

Definition: Changing Amplitudes

A function of the form File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1130.gif will oscillate above and below the midline with an amplitude given by A(t).

When thinking about a spring with amplitude decreasing over time, it is tempting to use the simplest tool for the job – a linear function. But if we attempt to model the amplitude with a decreasing linear function, such as File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1132.gif, we quickly see the problem when we graph the equation File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1134.gif.

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1136.jpg

While the amplitude decreases at first as intended, the amplitude hits zero at t = 10, then continues past the intercept, increasing in absolute value, which is not the expected behavior.  This behavior and function may model the situation on a restricted domain and we might try to chalk the rest of it up to model breakdown, but in fact springs just don’t behave like this. 

A better model, as you will learn later in physics and calculus, would show the amplitude decreasing by a fixed percentage each second, leading to an exponential decay model for the amplitude.

Definition: Damped Harmonic Motion

Damped harmonic motion, exhibited by springs subject to friction, follows a model of the form

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1138.gif  or   File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1140.gif.

Example 4

A spring with natural length of 20 inches is pulled back 6 inches and released.  It oscillates once every 2 seconds.  Its amplitude decreases by 20% each second.  Find a function that models the position of the spring t seconds after being released.

SOLUTION

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1142.jpgSince the spring will oscillate on either side of the natural length, the midline will be at 20 inches.  The oscillation has a period of 2 seconds, and so the horizontal compression coefficient isFile:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1144.gif. Additionally, it begins at the furthest distance from the wall, indicating a cosine model.

Meanwhile, the amplitude begins at 6 inches, and decreases by 20% each second, giving an amplitude function ofFile:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1146.gif

 

Combining this with the sinusoidal information gives a function for the position of the spring:

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1148.gif

Example 5

A spring with natural length of 30 cm is pulled out 10 cm and released.  It oscillates 4 times per second.  After 2 seconds, the amplitude has decreased to 5 cm.  Find a function that models the position of the spring.

SOLUTION

The oscillation has a period of File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1150.gif second, so File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1152.gif.  Since the spring will oscillate on either side of the natural length, the midline will be at 30 cm.  It begins at the furthest distance from the wall, suggesting a cosine model.  Together, this gives

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1154.gif.

For the amplitude function, we notice that the amplitude starts at 10 cm, and decreases to 5 cm after 2 seconds.  This gives two points (0, 10) and (2, 5) that must be satisfied by an exponential function:  File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1156.gif and File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1158.gif.  Since the function is exponential, we can use the form File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1160.gif.  Substituting the first point, File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1162.gif, so a = 10.  Substituting in the second point,

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1164.gif                   Divide by 10

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1166.gif                      Take the square root

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1168.gif

This gives an amplitude function of File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1170.gif.  Combining this with the oscillation,

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1172.gif

Try it Now: 2

A certain stock started at a high value of $7 per share and has been oscillating above and below the average value, with the oscillation decreasing by 2% per year. However, the average value started at $4 per share and has grown linearly by 50 cents per year.

Find a function \(S(t)\) that models the value of the stock after \(t\) years.

  1. Find a formula for the midline
  2. Find a formula for the amplitude.
Example 6

In AM (Amplitude Modulated) radio, a carrier wave with a high frequency is used to transmit music or other signals by applying the to-be-transmitted signal as the amplitude of the carrier signal.  A musical note with frequency 110 Hz (Hertz = cycles per second) is to be carried on a wave with frequency of 2 KHz (KiloHertz = thousands of cycles per second).  If the musical wave has an amplitude of 3, write a function describing the broadcast wave.

SOLUTION

The carrier wave, with a frequency of 2000 cycles per second, would have period File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1174.gif of a second, giving an equation of the form File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1176.gif.  Our choice of a sine function here was arbitrary – it would have worked just was well to use a cosine.

The musical tone, with a frequency of 110 cycles per second, would have a period of File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1178.gif of a second.  With an amplitude of 3, this would correspond to a function of the form File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1180.gif.  Again our choice of using a sine function is arbitrary.

The musical wave is acting as the amplitude of the carrier wave, so we will multiply the musical tone’s function by the carrier wave function, resulting in the function

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1182.gif

 

File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1184.jpg

Important Topics of This Section

  • Changing midline
  • Changing amplitude
    • Linear Changes
    • Exponential Changes
    • Damped Harmonic Motion

 

Try it Now Answers

  1. The midline follows the path of the quadraticFile:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1186.gifand the amplitude is a constant value of 8.

 

  1. File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1188.gif

      S(t)=File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1190.gif

Section 7.4 Exercises

Find a possible formula for the trigonometric function whose values are given in the following tables.

1.

x

0

3

6

9

12

15

18

y

-4

-1

2

-1

-4

-1

2

2.

x

0

2

4

6

8

10

12

y

5

1

-3

1

5

1

-3

  1. The displacement File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1192.gif, in centimeters, of a mass suspended by a spring is modeled by the function File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1194.gif, where t is measured in seconds.  Find the amplitude, period, and frequency of this displacement.

 

  1. The displacement File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1192.gif, in centimeters, of a mass suspended by a spring is modeled by the function File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1197.gif, where t is measured in seconds.  Find the amplitude, period, and frequency of this displacement.
  2. A population of rabbits oscillates 19 above and below average during the year, reaching the lowest value in January. The average population starts at 650 rabbits and increases by 160 each year. Find a function that models the population, P, in terms of the months since January, t.
  3. A population of deer oscillates 15 above and below average during the year, reaching the lowest value in January. The average population starts at 800 deer and increases by 110 each year. Find a function that models the population, P, in terms of the months since January, t.
  4. A population of muskrats oscillates 33 above and below average during the year, reaching the lowest value in January. The average population starts at 900 muskrats and increases by 7% each month. Find a function that models the population, P, in terms of the months since January, t.
  5. A population of fish oscillates 40 above and below average during the year, reaching the lowest value in January. The average population starts at 800 fish and increases by 4% each month. Find a function that models the population, P, in terms of the months since January, t.
  6. A spring is attached to the ceiling and pulled 10 cm down from equilibrium and released. The amplitude decreases by 15% each second. The spring oscillates 18 times each second. Find a function that models the distance, D, the end of the spring is below equilibrium in terms of seconds, t, since the spring was released.
  7. A spring is attached to the ceiling and pulled 7 cm down from equilibrium and released. The amplitude decreases by 11% each second. The spring oscillates 20 times each second. Find a function that models the distance, D, the end of the spring is below equilibrium in terms of seconds, t, since the spring was released.
  8. A spring is attached to the ceiling and pulled 17 cm down from equilibrium and released. After 3 seconds the amplitude has decreased to 13 cm. The spring oscillates 14 times each second.  Find a function that models the distance, D the end of the spring is below equilibrium in terms of seconds, t, since the spring was released.
  9. A spring is attached to the ceiling and pulled 19 cm down from equilibrium and released. After 4 seconds the amplitude has decreased to 14 cm. The spring oscillates 13 times each second.  Find a function that models the distance, D the end of the spring is below equilibrium in terms of seconds, t, since the spring was released.
     

Match each equation form with one of the graphs.           

13. a. File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1199.gif               b. File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1201.gif                

14. a. File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1203.gif                   b. File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1205.gif

IGraphs  IIGraphs  IIIGraphs  IVGraphs

 

Find a function of the form File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1215.gif that fits the data given.

15.

x

0

1

2

3

y

6

29

96

379

16.

x

0

1

2

3

y

6

34

150

746

Find a function of the form File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1217.gif that fits the data given.

17.

x

0

1

2

3

y

7

6

11

16

18.

x

0

1

2

3

y

-2

6

4

2

Find a function of the form File:/C:\Users\JIMHOR~1\AppData\Local\Temp\msohtmlclip1\01\clip_image1219.gif that fits the data given.

19.

x

0

1

2

3

y

11

3

1

3

20.

x

0

1

2

3

y

4

1

-11

1

References

[1] You technically can divide by sin(x) as long as you separately consider the case where sin(x) = 0.  Since it is easy to forget this step, the factoring approach used in the example is recommended.

Contributors

  • David Lippman (Pierce College)
  • Melonie Rasmussen (Pierce College)