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◦.
- y=cos3A [1, 120◦]
- y=2sin
5 x
2
[2, 144◦]
- y=3sin4t [3, 90◦]
- y=3cos
θ
2
[3, 720◦]
- y=
7
2
sin
3 x
8
[
7
2
, 960 ◦
]
- y=6sin(t− 45 ◦) [6, 360◦]
- y=4cos( 2 θ+ 30 ◦) [4, 180◦]
- y=2sin^22 t [2, 90◦]
- 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