Basic Engineering Mathematics, Fifth Edition

152 Basic Engineering Mathematics

From para (a), page 150, if T=kLn

then lgT=nlgL+lgk

and comparing with Y=mX+c

shows that lgTis plotted vertically against lgLhor-

izontally, with gradientnand vertical-axis intercept

lgk.

A table of values for lgTand lgLis drawn up as shown

below.

T 1.0 1.3 1.5 1.8 2.0 2.3

lgT 0 0. 114 0. 176 0. 2550. 301 0. 362

L 0.25 0.42 0.56 0.81 1.0 1.32

lgL− 0. 602 − 0. 377 − 0. 252 − 0. 0920 0. 121

A graph of lgTagainst lgLis shown in Figure 18.5 and

thelawT=kLnis true since a straight line results.

A

C B

2 0.60 2 0.50 2 0.40 2 0.30 2 0.20 0 0.20

0.10

0.20

0.30

0.25

0.05

lg L

lg T

0.40

2 0.10 0.10

Figure 18.5

From the graph, gradient of straight line,

n=

AB

BC

=

0. 25 − 0. 05

− 0. 10 −(− 0. 50 )

=

0. 20

0. 40

=

1

2

Vertical axis intercept, lgk= 0 .30. Hence,

k=antilog 0. 30 = 100.^30 =2.0

Hence,the law of the graph isT= 2. 0 L^1 /^2 or

T= 2. 0

√

L.

When lengthL= 0 .75m,T= 2. 0

√

0. 75 = 1 .73s

Problem 6. Quantitiesxandyare believed to be

related by a law of the formy=abx,whereaandb

are constants. The values ofxand corresponding

values ofyare

x 0 0.6 1.2 1.8 2.4 3.0

y 5.0 9.67 18.736.1 69.8135.0

Verify the law and determine the approximate

values ofaandb. Hence determine (a) the value of

ywhenxis 2.1 and (b) the value ofxwhenyis 100

From para (b), page 150, if y=abx

then lgy=(lgb)x+lga

and comparing with Y=mX+c

shows that lgyis plotted vertically andxhorizontally,

with gradient lgband vertical-axis intercept lga.

Another table is drawn up as shown below.

x 0 0.6 1.2 1.8 2.4 3.0

y 5. 0 9. 67 18. 7 36. 1 69. 8 135. 0

lgy 0. 70 0. 99 1. 27 1. 56 1. 84 2. 13

A graph of lgy againstx is shown in Figure 18.6

and,since a straight line results, the lawy=abxis

verified.

C

B

A

lg

y

x

2.50

2.13

2.00

1.50

1.00

1.17

0.50

0 1.0 2.0 3.0

0.70

Figure 18.6