Page 480 - Dynamics of Mechanical Systems

0593_C13_fm Page 461 Monday, May 6, 2002 3:21 PM

Introduction to Vibrations 461

We can express cosθ and cosθ in terms of k and z by using the identity:
0
sin θ ( 2) ≡ ( 1 2)−( 1 2) cosθ (13.9.12)
2

That is,

2
2
cosθ =−12 sin θ ( ) =− 2k 2 sin z (13.9.13)
2
1
Then, in view of Eqs. (13.9.10) and (13.9.11), we have:
cosθ =− 2 (13.9.14)
12k
0
Using these results, we can develop the integrand of Eq. (13.9.9) as:

[
− )]
( cosθ − cosθ ) −12 = 12k 2 sin z −(12k 2 / 12
/
−
2
0
[ 2 sin )] 12
/
2
= 2k − (1 z
(13.9.15)
−12
/
2
2
= (2k cos ) z
= ( 2 k cos ) z
1
Consider next the differential dθ. From Eq. (13.9.11), we have:
d sin( θ 2) = ( 1 2) cos( θ 2)
dθ
= d ( ksin z) (13.9.16)
dθ

= kcos z dz
dθ

Observe that in view of Eq. (13.9.11), (1/2)cos(θ/2) may be expressed as:

[
( 12)cos θ 2) = ( ) − sin θ 2)] 12
(
(
/
2
1 2 1
(13.9.17)
= ( 12 1 ) − ( k 2 sin 2 ) z 12
/
Then, by combining the terms of Eqs. (13.9.16) and (13.9.17) we have:
dz = ( ) 2 2 z) / 12
1 2
2
kcos z 12 cos θ ( ) = ( ) − (1 k sin (13.9.18)
dθ