Higher Engineering Mathematics

234 VECTOR GEOMETRY

Ordinates are added at 15◦intervals and the resultant

is shown by the broken line. The amplitude of the

resultant is 6.1 and itlagsy 1 by 25◦or 0.436 rad.

Hence the sinusoidal expression for the resultant

waveform is

yR= 6 .1 sin (ωt− 0 .436)

Problem 11. Determine a sinusoidal expres-

sion for y 1 −y 2 when y 1 =4 sinωt and

y 2 =3 sin (ωt−π/3).

y 1 andy 2 are shown plotted in Fig. 21.23. At 15◦

intervalsy 2 is subtracted fromy 1. For example:

at 0 ◦,y 1 −y 2 = 0 −(− 2 .6)=+ 2. 6

at 30◦,y 1 −y 2 = 2 −(− 1 .5)=+ 3. 5

at 150◦,y 1 −y 2 = 2 − 3 =−1, and so on.

Figure 21.23

The amplitude, or peak value of the resultant (shown
by the broken line), is 3.6 and it leadsy 1 by 45◦or
0.79 rad. Hence

y 1 −y 2 = 3 .6 sin(ωt+ 0. 79 )

Problem 12. Given y 1 =2 sinωt and

y 2 =3 sin (ωt+ω/4), obtain an expression

for the resultantyR=y 1 +y 2 , (a) by drawing

and (b) by calculation.

(a) When timet=0 the position of phasorsy 1 and

y 2 are as shown in Fig. 21.24(a). To obtain the

resultant,y 1 is drawn horizontally, 2 units long,

y 2 is drawn 3 units long at an angle ofπ/4 rads

or 45◦and joined to the end ofy 1 as shown in

Fig. 21.24(b).yRis measured as 4.6 units long

and angleφis measured as 27◦or 0.47 rad. Alter-

natively,yRis the diagonal of the parallelogram

formed as shown in Fig. 21.24(c).

Figure 21.24

Hence, by drawing,

yR= 4 .6 sin (ωt+ 0 .47)

(b) From Fig. 21.24(b), and using the cosine rule:

y^2 R= 22 + 32 −[2(2)(3) cos 135◦]

= 4 + 9 −[− 8 .485]= 21. 49

HenceyR=

√

(21.49)= 4. 64