How To Find Period Of A Sine Wave

Aamplitude and Period

Learning Objective(southward)

· Sympathize amplitude and flow.

· Graph the sine function with changes in amplitude and period.

· Graph the cosine function with changes in aamplitude and period.

· Match a sine or cosine part to its graph and vice versa.

Introduction

Y'all know how to graph the functions and . At present you'll larn how to graph a whole "family" of sine and cosine functions. These functions have the form or , where a and b are constants.

Periodic Functions

We used the variable previously to show an bending in standard position, and we also referred to the sine and cosine functions as and . Frequently the sine and cosine functions are used in applications that accept nothing to do with triangles or angles, and the letter x is used instead of for the input (also as to label the horizontal axis). Then from this betoken forward, we'll refer to these aforementioned functions as and . This change does not affect the graphs; they remain the same.

You know that the graphs of the sine and cosine functions have a pattern of hills and valleys that repeat. The length of this repeating pattern is . That is, the graph of (or ) on the interval looks like the graph on the interval or or . This pattern continues in both directions forever.

The graph below shows 4 repetitions of a pattern of length . Each 1 contains exactly one consummate copy of the "colina and valley" pattern.

If a function has a repeating design like sine or cosine, information technology is chosen a periodic function. The period is the length of the smallest interval that contains exactly one re-create of the repeating blueprint. Then the menstruation of or is . Any part of the graph that shows this pattern over one period is called a cycle. For example, the graph of on the interval is 1 cycle.

You know from graphing quadratic functions of the form that every bit you changed the value of a you changed the "width" of the graph. Now we'll expect at functions of the form and run across how changes to b will bear on the graph. For example, is periodic, and if so, what is the flow? Hither is a table with some inputs and outputs for this function.

*ten* (in radians)	210 (in radians)	sin2x
0	0	0

		one

		0



		0

Every bit the values of x go from 0 to , the values of become from 0 to . We tin run across from the graph that the office is a periodic function, and goes through one full bicycle on the interval [0, ], so its period is . If y'all substituted values of x from to , the values of would become from to , and would go through another complete cycle of the sine role.

Find that has two cycles on the interval [0, two ], which is the interval needs to complete one total cycle.

What is the smallest positive value for x where is at its minimum?

Show/Hide Answer

Incorrect. The function does attain its minimum value at this betoken, but is non a positive value. The correct respond is .

Wrong. Possibly you confused minimum and maximum. The correct answer is .

Correct. The minimum value for the sine function is . Look at the graph of . It attains this minimum at the bottom of every valley. The bottom of the first valley where x is positive is at .

Incorrect. You may have thought of 0 as the minimum value, but the sine role takes on negative values. The right reply is .

What is the menstruum of the function ? Here is a table with some inputs and outputs for this function.

x (in radians)	3x (in radians)	sin3x
0	0	0

		1

		0



		0

Equally the values of x go from 0 to, the values of become from 0 to . Nosotros can encounter from the graph that goes through 1 total cycle on the interval , so its period is .

Discover that has iii cycles on the interval [0, 2 ], which is the interval needs to complete 1 full cycle.

What is the period of the function ? Here is a table with some inputs and outputs for this function.

x (in radians)	x (in radians)	sin
0	0	0

		1

		0



		0

As the values of 10 go from 0 to , the values of go from 0 to .

We tin see from the graph that goes through 1 full cycle on the interval , and then its catamenia is .

Notice that has half of one full wheel on the interval [0, two ], which is the interval needs to complete i full bicycle.

Let's put these results into a table. For the first three functions nosotros have rewritten their periods with the numerator so that the pattern becomes clear. Can y'all see a relationship between the role and the denominator in the periods?

In each case, the period could be institute by dividing by the coefficient of x. In full general, the period of is , and the period of is . Since the period is the length of an interval, information technology must e'er be a positive number. Since information technology is possible for b to be a negative number, we must apply in the formula to be sure the period, , is ever a positive number.

You can think of the different values of b as having an "squeeze box" (or a spring) effect on the graphs of sine and cosine. A large value of b squeezes them in and a small value of b stretches them out.

There is some other way to depict this effect. In the interval , goes through one cycle while goes through ii cycles. If you go dorsum and cheque all of the examples in a higher place, you will encounter that has cycles in the interval . Likewise, has cycles in the interval .

Example
Problem	What are the periods of and ?
		For , . Substitute this value into the formula.
		For , .
Answer	The period of is , and the period of is .

Aamplitude

Equally you lot have seen, the graphs of all of these sine and cosine functions alternate between hills and valleys. The height of i hill (which equals the depth of one valley) is called the amplitude. You can come across that for all the graphs we have looked at so far, the amplitude equals 1.

The formal fashion to say this for whatsoever periodic function is:

You know that the maximum value of or is i and the minimum value of either is . So if yous applied the above definition, yous would become:

This result agrees with what was observed from the graph.

You accept seen that changing the value of b in or either stretches or squeezes the graph like an accordion or a jump, but it does non change the maximum or minimum values. For all of these functions, the maximum is 1 and the minimum is . And then if yous applied the definition of aamplitude, you would be doing the exact same calculation as we just did above. The amplitude of whatever of these functions is 1.

Let'south look at a unlike kind of change to a function past graphing the function . Here's a tabular array with some values of this function.

x (in radians)	sinx	2sinx
0	0	0
		one


	one	ii


		ane
	0	0

We'll have the beginning and third columns to make part of the graph and then extend that pattern to the left and to the right.

Now you tin can utilise this graph in the following example.

Example
Problem	What is the amplitude of ?
		You can find the maximum and minimum values of the function from the graph. For example, at the value is 2, and at the value is .
		Use the definition of amplitude.
	Notice that the height of each colina is 2, and the depth of each valley is 2. This is equal to the amplitude, equally nosotros mentioned at the start.
	Notice also that the aamplitude is equal to the coefficient of the function: This is not a coincidence.
Answer	The amplitude is two.

Let'due south compare the graph of this office to the graph of the sine function.

The effect of multiplying by 2 is to stretch the graph vertically by a cistron of 2. Because it has been stretched vertically by this cistron, the amplitude is twice as much, or 2. If nosotros had looked at , the graph would take been stretched vertically by a factor of iii, and the aamplitude of this function is 3. Once again, this is equal to the coefficient of the function. In general, we take the following rule.

The aamplitude of or is .

As the terminal instance, , shows, multiplying past a constant on the exterior affects the amplitude. If you multiply past a constant on the outside and on the inside, as in , y'all will affect both the amplitude and the period. Here is the graph of :

Case
Problem	Determine the amplitude and flow of .
		Apply the formula for amplitude, with .
		Use the formula for catamenia, with .
Respond	The amplitude is three and the flow is .

In this example, you lot could accept found the period past looking at the graph above. I consummate wheel is shown, for case, on the interval , so the period is .

In the functions and , multiplying by the constant a merely affects the amplitude, non the catamenia. Every bit nosotros said earlier, changing the value of b only affects the period, non the amplitude. The general event is every bit follows.

The aamplitude of or is .

The period of or is .

To help you empathize changes in aamplitude and period for both the sine function and cosine function, endeavour the following interactive practice:

This is a Java Applet created using GeoGebra from world wide web.geogebra.org - it looks similar yous don't accept Java installed, please go to world wide web.coffee.com

What are the aamplitude and period of ?

A) The amplitude is , and the menstruation is .

B) The aamplitude is , and the menstruum is .

C) The aamplitude is 1, and the period is .

D) The aamplitude is one, and the period is .

Show/Hide Answer

A) The amplitude is , and the period is .

Right. In this function, , and then this is the amplitude. The menstruation equals .

B) The amplitude is , and the menstruum is .

Wrong. The amplitude is correct, but the period is not. Y'all probably multiplied by four instead of dividing. The correct reply is A.

C) The aamplitude is ane, and the flow is .

Incorrect. The period is correct, just the amplitude is not. You may have idea the aamplitude is the maximum minus the minimum, but it is half of this. The right answer is A.

D) The amplitude is 1, and the menstruation is .

Incorrect. You may have idea the amplitude is the maximum minus the minimum, but information technology is half of this. You lot probably multiplied by iv, instead of dividing, to notice the period. The right answer is A.

Graphs of Sine Functions

You know the function has amplitude and catamenia . You can apply these facts to draw the graph of any office in the form by starting with the graph of and modifying it.

For example, suppose yous wanted the graph of . Since , this role has the same period equally . Since , the amplitude is four. Therefore, you would have the graph of and simply stretch information technology vertically by a factor of 4. Here is ane bicycle for these two functions.

Note that the points that were on the ten-axis "remain" on the x-axis. At these points (where ), the value of is 0. If you lot multiply 0 past iv (or anything else), y'all will still have a value of 0. So the points volition still exist on the ten-axis. On the other mitt, the highest and lowest points take moved away from the 10-axis. They had y-values of 1 and for , and they take y-values of four and for .

Example
Trouble	Graph on the interval .
	The value of b is 1, so the graph has a period of , as does .
	The value of a is , and so the graph has an amplitude of 1, every bit does .
	Though the amplitude and the catamenia are the same as the role , the graph is not exactly the same. The consequence of multiplying by is to replace y-values past their opposites. So the graph of gets reflected over the x-axis.
Answer

If you lot want to check these graphs with a graphing calculator, make sure that the graphing window has the correct settings. For this last example, you would use and . In full general, yous would probably desire to adjust the 10-values to show i total cycle and the y-values using the amplitude.

If the values of a and b are both different from 1, then you demand to combine the effects of the two changes.

Instance
Problem	Graph on the interval .
	The value of a is three, so the graph has an amplitude of 3. This has the effect of taking the graph of and stretching it vertically by a factor of 3.
	The value of b is , then the graph has a period of . This is twice the period of . This has the outcome of taking the graph of and stretching it horizontally by a gene of 2. (The alternative way to say this is that has of a bicycle on the interval .)
	To make the graph of , you must combine the two effects described above.

Answer

Sometimes you demand to stretch the graph of the sine function, and sometimes you need to shrink it.

Instance
Problem	Graph 2 cycles of a sine function whose amplitude is and whose period is.
	In that location are dissimilar functions of the form that fit this description considering a and b could exist positive or negative. We will draw the graph assuming these are positive.
	Because the amplitude is , this has the outcome of taking the graph of and shrinking information technology vertically by a factor of 2.
	The period is , which is the menses of . This has the effect of taking the graph of and shrinking information technology horizontally by a factor of 3.
	To make the graph, you must combine the two effects described in a higher place.
Answer

Even without knowing the specific value of a abiding, y'all can sometimes yet narrow down the possibilities for the shape of a graph.

The graph of a function , where a is a abiding, is drawn on the interval .

Which of the following options could be this graph?

A) B)

C) D)

Show/Hide Respond

Incorrect. This has the correct shape and period, but information technology is in the wrong position. Regardless of the value of a, the graph must pass through the ten-centrality at , which it does not. The right reply is D.

Incorrect. This is the graph of a cosine function. Regardless of the value of a, the graph must pass through the x-axis at . The correct answer is D.

Incorrect. Because the coefficient of ten is 1, the graph should have a period of , only this graph has a catamenia of . The correct answer is D.

Right. Because the coefficient of x is i, the graph has a period of , which this option has. The factor a could stretch or compress the graph, but it must still pass through the 10-axis at the points , which it does.

Graphs of Cosine Functions

You know the function has aamplitude and period . But as you did with sine functions, y'all can use these facts to draw the graph of any office in the class by starting with the graph of and modifying it.

For case, suppose you lot wanted the graph of . Since , information technology has the aforementioned amplitude as . And, considering , the menses is given by:

Since this is twice the period of , you would take the graph of and stretch it horizontally past a cistron of 2. Hither is a side-by-side comparing of these two graphs.

Annotation that in the interval , the graph of has one total cycle. Since , the graph of has of a cycle in that interval.

If you are using a graphing computer, you need to suit the settings for each graph to get a graphing window that shows all the features of the graph. For the last example, yous would utilize and . In full general, you would probably want to adjust the ten-values to show i full wheel and the y-values to prove the highest and lowest points. In the adjacent example, you would use and for the graphing window because you lot are specifically asked to graph it over the domain and the graph will have an aamplitude of 2, going as depression equally -2 and equally loftier equally +2.

Example
Trouble	Graph on the interval .
	The amplitude equals . This has the issue of taking the graph of and stretching information technology vertically by a factor of two. The negative sign on the "exterior" has an additional effect: the y-values are replaced past their opposites, so the graph is too flipped over the 10-axis.
	The value of b is ane, so the graph has a period of , as does .
	When the merely change is a vertical stretch, compression, or flip, the x-intercepts remain the same. So the graph will pass through the x-axis at and .
Answer

Again, if the values of a and b are both different from ane, you need to combine the effects of the 2 changes. In the next instance, you will run into a variation that you lot take not seen before. Up to this point, all of the values of b take been rational numbers, merely here we are using the irrational number . This situation does not really modify the process, but you lot will see that information technology changes the calibration on the x-axis in a new way.

Example
Problem	Graph on the interval .
	The value of a is 4, so the graph has an amplitude of 4. This has the effect of taking the graph of and stretching information technology vertically by a factor of 4.
	The value of b is , and then the graph has a period of . This has the effect of shrinking the graph of horizontally past a factor of , causing it to complete i complete wheel on the interval [0, 2].
	Notation the result on the ten-values of the intercepts, the high points, and the depression points. Because the period is 2, the first cycle of the graph will accept loftier points at and 2. The low point volition even so be midway between these, and so it is at . The x-intercepts are still midway betwixt the high and the low points, so they volition be at and . The 2nd cycle of the graph has all of these points shifted to the right 2 units.
	To make the graph of , yous must combine the furnishings described above.
Answer

In two previous examples ( and ) you saw that a negative sign on the outside (a negative value of a) has the effect of flipping the graph around the x-axis. In the next instance, you will run into the effect of a negative sign on the "inside" (a negative value of b).

I concluding hint: also trying to figure out the overall outcome of the value of a or b on the graph, you might desire to check specific points. For instance, just substitute into the function and see where that signal will end up.

Which of the following options is the graph of on the interval ?

A) B)

C) D)

Bear witness/Hibernate Answer

Incorrect. This is the graph of . Remember to cheque specific points like . At that point, . So the point should be on the graph. The correct answer is C.

Incorrect. You confused the effects of a and b. This is the graph of . The correct answer is C.

Correct. The value of a is , which volition stretch the graph vertically by a factor of . The period of the graph is , every bit is the period of . The effect of the negative sign on the inside is to supervene upon x-values past their opposites. This will flip the graph around the y-axis. However, because the graph of cosine is symmetric about the y-axis, this has no effect at all. Then the only modify to the graph of is the vertical stretch.

Incorrect. This graph has the right period and aamplitude. However, you have confused the effect of a minus sign on the within with a minus sign on the outside. The correct respond is C.

Matching Graphs and Functions

Given any part of the form or , you know how to find the amplitude and period and how to employ this information to graph the functions.

Given a graph of a sine or cosine part, you lot besides tin can decide the aamplitude and menstruation of the part. From this information, you can find values of a and b, and then a function that matches the graph.

Remember that along with finding the amplitude and catamenia, information technology's a adept idea to look at what is happening at . If a and b are any nonzero constants, the functions and will have the post-obit values at :

This tells you that the graph of passes through regardless of the values of a and b, and the graph of never passes through regardless of the values of a and b. So, for case, if you lot are given a graph passing through the origin and are asked to determine which office information technology represents, you know correct abroad that it is not in the course .

You lot need to be careful about the sign of a. You might decide that a function has an aamplitude of 4, for case. Though it is possible that , it is also possible that . Here is an example of each of these two possibilities.

You will demand to compare the graph to that of or to see if, in add-on to any stretching or shrinking, there has been a reflection over the x-axis. The graph higher up on the right can be thought of every bit the upshot of stretching and reflecting the graph of across the ten-centrality. If in that location has been a reflection, then the value of a will be negative. Once you take determined if a is positive or negative, you can always cull a positive value of b.

Here are some examples of this procedure.

Example
Problem	Determine a function of the form or whose graph is shown beneath.
	Starting time, observe that the graph passes through the origin, and so you are looking for a function of the form .
	Next, observe that the maximum value of the role is 2 and the minimum is , and then the amplitude is 2. The graph has the aforementioned "orientation" as . (It starts with a hill to the correct of the y-axis.) This implies that a is positive, and in item, .
	Finally, detect that the graph shows two cycles and that one entire bike is contained in the interval . Therefore, the menses is . Since , . According to our process, once you have determined if a is positive or negative, you can always choose a positive value of b . And so .
	Combine these iii pieces of information.
Answer

Instance
Problem	Determine a role of the course or whose graph is shown below.
	First, detect that the graph does not pass through the origin, but rather crests, reaching a maximum when 10 = 0, then you are looking for a part of the form .
	Side by side, observe that the maximum value of the function is and the minimum is , and then the amplitude is . The graph has the aforementioned "orientation" as . (Information technology has a hill with the y-axis running through the centre.) This implies that a is positive, and in particular, .
	Finally, observe that the graph shows 2 cycles and that ane entire wheel is contained in the interval . Therefore, the period is . Considering , . According to our process, once you take determined if a is positive or negative, yous can always choose a positive value of b . So .
	Combine these iii pieces of information.
Answer

Make certain that you recognize where a bike starts and ends. The period is the length of the interval over which the one cycle runs.

Which of the following functions is represented past the graph below?

Bear witness/Hide Answer

Incorrect. You lot correctly recognized the graph as a reflected sine function, but the period is incorrect. Mayhap you saw the on the right and used that every bit the length of i bicycle. Yet, the entire graph is one cycle, and the period equals . The right answer is D.

Incorrect. You correctly establish the amplitude and period of this sine part. However, you too demand to check the orientation of the graph. Observe that to the right of the y-centrality yous take a valley instead of a colina. The correct respond is D.

Incorrect. Yous correctly establish the amplitude and the orientation of this sine office. All the same, the period is incorrect. Possibly you recognized that the menstruation of the graph is twice the period of , and thought that the value of b would exist 2. But to detect the value of b, you must set the period equal to . The correct reply is D.

Correct. The graph passes through the origin, so the office could accept the form , but not . The amplitude is 1. The graph has a valley on the right, which could be the result of a reflection of over the x-axis. Therefore, . The graph shows ane cycle, so the flow is . Since , the value of b could be . So this could exist the graph of .

Remember that when writing a role y'all tin can use the notation in place of the variable y.

Which of the following graphs represents ?

A) B)

C) D)

Testify/Hide Answer

Incorrect. This is the graph of a office of the form . Think to cheque the value of the office at . Since , the office passes through , non the origin every bit shown in this graph. The correct answer is B.

Correct. Get-go, this graph has the shape of a cosine function. 2nd, because in the equation, the aamplitude is 3. Finally, because , the menses of this function is . The graph in this answer completes ane full cycle between and so its period is every bit needed.

Incorrect. This graph does have the shape of a cosine function, and the amplitude is three, which is right. Still, the menstruation is incorrect. Yous probably multiplied past instead of dividing. The correct answer is B.

Incorrect. This graph does take the shape of a cosine function. However, in determining the graph, it appears that y'all switched the values of a and b. The correct answer is B.

Summary

The functions and are periodic functions: their graphs have a repeating design of hills and valleys that continues in both directions forever. The height of the hill or the depth of the valley is called the aamplitude, and is equal to . Any ane full design in the graph is called a cycle, and the length of an interval over which a bicycle occurs is chosen the catamenia. The period is equal to the value .

You tin can use this information to graph any of these functions by starting with the bones graph of or and so doing a combination of stretching or shrinking the graph vertically based on the value of a, stretching or shrinking the graph horizontally based on the value of b, or reflecting it based on the signs of a and b.

You tin can as well kickoff with a graph, determine the values of a and b, and and then determine a part that it represents.

Source: http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L2_T3_text_final.html

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