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# Period of a Function

Period means the time interval between two waves. Periodic function is a function that repeats its values at regular intervals or periods. In other words, a function which repeats its values after every particular interval, is known as a periodic function. This particular interval is termed as the period of that function.

A function f is said to be periodic with period m, if we have
f (x + m) = f (x), For every m
> 0.

It means that the function f(x) possess same values after an interval of "m". One can say that the function f repeats all its values after every interval of "m".

For example - The sine function i.e. sin x has a period 2
$\pi$ because 2 $\pi$ is the smallest number for which sin (x + 2$\pi$) = sin x,  for all x.

We may also calculate the period using the formula derived from the basic sine and cosine equations. The period for function y = A sin( B x - c ) and y = A cos( B x - c ) is $\frac{2 \pi}{B}$ radians. The reciprocal of period of a function is equal to its frequency. Frequency is defined as the number of cycles completed in one second. If the period of a function is denoted by P and f be its frequency, then -

f = $\frac{1}{P}$.

 Related Calculators Frequency Period Calculator Periodic Table Calculator Calculator Functions Calculate Exponential Function

## Fundamental Period of a Function

The fundamental period of a function is the period of the function which are of the form,${f}(x+k)={f}(x)$, then k is called the period of the function and the function f is called a periodic function.

Let us define the function h(t) on the interval [0,2] as follows:
$h(t) = \begin{cases} 0 & \text{ if } x= 0\leq t\leq 1 \text{or if} \frac{5}{3}\leq t\leq 2\\ 3t-1 & \text{ if } x=\frac{1}{3}\leq t\leq \frac{2}{3} \\ 1& \text{ if } x= \frac{2}{3}\leq t\leq \frac{4}{3}\\ -3t+5& \text{ if } x=\frac{4}{3}\leq t\leq\frac{5}{3} \end{cases}$
If we extend the function h to all of R by the equation, ${h}(t+2)={h}(t)$
=> h is periodic with the period 2.
The graph of the function is shown below.

## Finding the Period of a Function

If a function repeats over at a constant period we say that is a periodic function. Basically it is represented like f(x) = f(x + p), p is the real number and this is the period of the function. Period means the time interval between the two occurrences of the wave. To find the period of the periodic function we have to use the following formula, Where Period = 2pb, here b is the co - efficient of x

Sine and cosine functions have the forms of a periodic wave:

Period: It is represented as "T", Period is the distance among two repeating points on the graph function.

Amplitude: It is represented as "A" It is the distance between the middle point to highest or lowest point on the graph function.

$\sin (a \theta)$ = $\frac{2 \pi}{a}$

and $\cos (a \theta)$ = $\frac{2 \pi}{a}$

### Solved Examples

Question 1: Find the period of the given periodic function. Where f(x) = 9sin(6px7 + 5)
Solution:

Given periodic function is f(x) = 9sin(6px7+ 5)

To find the period we have the formulas

period = 2pb

Where period of the periodic function = 2p(6p7) = 146 = 73

Question 2: Find the period of the periodic function f(x) . Where f(x) = 9 Cos x
Solution:

The given periodic function is f(x) = 9 Cos x

To find the period, we have the formula.

period = $\frac{2 \pi}{b}$

Where period of the periodic function = 2p $\times$ 1 = 2p

As we are aware that sin (2$\pi$ + x) = sin x and cos (2$\pi$ + x) = cos x.., we see that the periods of sine and cosine functions are 2$\pi$..
Also, tan ($\pi$ + x) = tan x, hence the period of tan x is $\pi$
Let us graph the primary trigonometric functions.
The following graph shows the function y = sin x
Let us find the co-ordinates of the points to graph.

 x -2 π -3 π/2 - π - π/2 0 π/2 π 3 π/2 2π 5 π/2 y 1 1 0 -1 0 1 0 -1 0 1

Period = 2 $\pi$
Axis: y = 0 [x-axis ]
Amplitude : 1
Maximum value = 1
Minimum value = -1
domain = { x : x $\in$ R }
Range = [-1, 1]

The following graph shows the trigonometric function y = cos x

Let us prepare the table to values

 x -2 π -3 π/2 - π - π/2 0 π/2 π 3 π/2 2π 5 π/2 y 1 0 -1 0 1 0 -1 0 1 0

Period: 2 π
Axis: y = 0
Amplitude = 1
Maximum value = 1
Minimum value = -1
Domain: { x : x $\in$ R }
Range = [ -1, 1]

## Period of a Sine Function

If we have a function f(x) = sin (xs), where s > 0, then the graph of the function makes s complete cycles between 0 and $2\pi$ and each of the function have the period, p = $\frac{2\pi}{s}$.

Now, lets discuss about some examples based on sin function:

Let us discuss the graph of y = sin 2x

Period = $\pi$
Axis: y = 0 [x-axis ]
Amplitude: 1
Maximum value = 1
Minimum value = -1
Domain: { x : x $\in$ R }
Range = [ -1, 1]

## Period of a Trig Function

The distance between the repetition of any function is called the period of  function.  For a trigonometric function, the length of one complete cycle is called period. For the graph of any trigonometric function, we can take x = 0 as the starting point.
In general we have three basic trigonometric functions like sin, cos and tan function, having 2$\pi$, 2$\pi$ and $\pi$ period respectively.

### Solved Examples

Question 1: Graph of y = 4 sin(x/2)
Solution:

Period = 4$\pi$
Axis: y = 0 [x-axis ]
Amplitude: 4
Maximum value = 4
Minimum value = -4
Domain: { x : x $\in$ R }
Range = [ -4, 4]

Question 2: Graph of y = 4 sin2x
Solution:

 x - $\pi$ - 3$\pi$/4 - $\pi$/2 - $\pi$/4 0 $\pi$/4 $\pi$/2 3$\pi$/4 $\pi$ y 0 4 0 -4 0 4 0 -4 0

Period = $\pi$
Axis: y = 0 [x-axis ]
Amplitude: 4
Maximum value = 4
Minimum value = -4
Domain: { x : x $\in$ R }
Range = [ -4, 4]

From the above examples, we observe that the period and amplitude of the function changes as there is change in the numerical co-efficient and the angle [argument].
Let us tabulate the argument and the amplitude and compare.

 Function Amplitude Period y = sin x 1 2  $\pi$ y = 4sin 2x 4 $\pi$ y = 4 sin(x/2) 4 4  $\pi$

Hence, as the argument increases the period decreases and as the argument decreases the period increases.
Let us discuss the following general formula.
y = f(x) = a sin( k (x - b) + c and
y = f(x) = a cos ( k( x - b) ) + c

Since the primary functions sin x and cos x are of period 2  $\pi$, from the above general equation, the period = $\frac{2\pi}{\left | k \right |}$
Here, 'a' represents the amplitude
x = b, represents the horizontal shift
and y = c represents the axis.

## Period of a Tangent Function

If we have a function f(x) = tan (xs), where s > 0, then the graph of the function makes s complete cycles between $-\frac{\pi}{2}$, 0 and $\frac{\pi}{2}$ and each of the function have the period of p = $\frac{\pi}{s}$.

### Solved Example

Question: Determine the period of the function $f(\theta ) = \tan \left ( \frac{3\theta }{2} \right )$
Solution:

For the trigonometric tan function the period p is given as follows:
P = $\frac{\pi}{s}$ = $\frac{\pi}{\frac{3}{2}}$ = $\frac{2\pi}{3}$

Below is the graph of the trigonometric function y = tan x, where x = 3$\theta$/2

Period = $\frac{2\pi}{3}$
Axis y = 0
Amplitude: undefined
No maximum or minimum values.

## Period of a Function Examples

Given below are some of the examples on periodic functions.

### Solved Examples

Question 1: What is the period of the following periodic function?
Where f(x) = 6 sin 5x
Solution:
The given periodic function is f(x) = 6 sin 5x

We have the formula for the period of a function.

period = $\frac{2 \pi}{b}$

Where period of the periodic function = $\frac{2 \pi}{5}$

Question 2: The frequency of the periodic function is 2phi. Find the period of the periodic function.
Solution:
The frequency of the periodic function will be f = $\frac{1}{p}$

So, period p = $\frac{1}{f}$

= $\frac{1}{2 \pi}$

Question 3: Find the period, amplitude, horizontal shift, vertical shift and the equation of the axis for the following functions.
y = 3 cos(5x - 5$\pi$) + 2
Solution:

We have y = 3 cos(5x - 5 $\pi$) + 2 = 3 cos(5(x - $\pi$))+ 2
Comparing this with the general equation,
y = a cos (k(x - b)) + c
we get a = 3, k = 5, b = $\pi$, c = 2
Hence, the Period of the function is $\frac{2\pi}{\left | k \right |} = \frac{2\pi}{5}$

Amplitude of the function is a = 3
As b = $\pi$, the graph is shifted $\pi$ units to the right.
As c = 2, the equation of the axis is y = 2.

We can draw the graph as follows.

Question 4: The following trigonometric function have the basic function y = sin x. Determine the equation of each of the function a. The graph of the trigonometric function has period $\pi$ and amplitude 12. The equation of the axis is y = -6.
Solution:

From the given data, a = 12, b = $\pi$, c = -6
Period = 2$\pi$/|k|
k = 2 $\pi$/period = 2 $\pi$/$\pi$ = 2

The general equation is, y = a sin( k (x-b) + c
Hence, our required equation is, y = 12 sin(2x) - 6

The graph of the function will be as below.

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