Higher Engineering Mathematics, Sixth Edition

Trigonometric waveforms 143

0

7

y

(^908180827083608) A 8
y 5 7cos^2 2 A
Figure 14.28
Now try the following exercise
Exercise 62 Further problemson sine and
cosine curves
In Problems 1 to 9 state the amplitude and period
of the waveform and sketch the curve between
0 ◦and 360◦.

1. y=cos3A [1, 120◦]


2. y=2sin

5 x

2

[2, 144◦]

3. y=3sin4t [3, 90◦]


4. y=3cos

θ

2

[3, 720◦]

5. y=

7

2

sin

3 x

8

[

7

2

, 960 ◦

]

6. y=6sin(t− 45 ◦) [6, 360◦]


7. y=4cos( 2 θ+ 30 ◦) [4, 180◦]


8. y=2sin^22 t [2, 90◦]


9. y=5cos^2

3

2

θ [5, 120◦]

14.5 Sinusoidal formAsin(ωt±α)

In Fig. 14.29, letORrepresent a vector that is free to

rotate anticlockwise aboutOat a velocity ofωrad/s.

A rotating vector is called a phasor. After a time

tseconds OR will have turned through an angle

ωtradians (shown as angleTORin Fig. 14.29). IfSTis

constructed perpendicular toOR,thensinωt=ST/TO,

i.e.ST=TOsinωt.

If all such vertical components are projected on to a

graph ofyagainstωt, a sine wave results of amplitude

OR(as shown in Section 14.3).

If phasorORmakes one revolution (i.e. 2πradians)

inTseconds, then the angular velocity,

ω= 2 π/Trad/s, from which,T= 2 π/ωseconds.

Tis known as theperiodic time.

The number of complete cycles occurring per second

is called thefrequency,f

Frequency=

number of cycles

second

=

1

T

=

ω

2 π

i.e. f=

ω

2 π

Hz

Henceangular velocity,ω= 2 πf rad/s

Amplitudeis the name given to the maximum or peak

value of a sine wave, as explained in Section 14.3. The

amplitude of the sine wave shown in Fig. 14.29 has an

amplitude of 1.

A sine or cosine wave may not always start at 0◦.

To show this a periodic function is represented by

y=sin(ωt±α)ory=cos(ωt±α),whereαis a phase

displacement compared with y=sinAor y=cosA.

A graph of y=sin(ωt−α)lagsy=sinωtby angle

0

1.0

y

908

/2  3 /2

1808 2708 3608

1.0

y sin t

t

0 S R t t

T

 rads/s

2 

Figure 14.29