7.11 Frequency Domain Analysis …
299
The solution of eq. (7.87) is assumed to be of the following form
ξ r (t) = ¯
ξ r e
iωt
=
¯
ξ
(r )
R + i ¯
ξ
(r )
I
e
iωt
(7.94)
Therefore,
˙
ξ r (t) = iω ¯
ξ r e
iωt
(7.95)
¨
ξ r (t) = −ω
2 ¯
ξ r e
iωt
(7.96)
Combining Eqs. (7.94), (7.95), (7.96) and (7.87), yields
−ω
2 ¯
ξ r + 2iωp r ζ r ¯
ξ r + p
2
r
¯
ξ r = ¯
f
∗
r
or,
−ω
2
+ 2iωp r ζ r + p
2
r
¯
ξ r = ¯
f
∗
r
(7.97)
Equation (7.97) combining with Eq. (7.94), yields
−ω
2
+ 2iωp r ζ r + p
2
r
¯
ξ
(r )
R + i ¯
ξ
(r )
I
= ¯
f
(r )
R + i ¯
f
(r )
I
(7.98)
Equating the real parts and imaginary parts separately in Eq. (7.98) yields the
following two equations
−ω
2
+ p
2
r
¯
ξ
(r )
R − 2ωp r ζ r ¯
ξ
(r )
I = ¯
f
(r )
R
(7.99)
2ωp r ζ r ¯
ξ
(r )
R +
−ω
2
+ p
2
r
¯
ξ
(r )
I = ¯
f
(r )
I
(7.100)
From the solution of Eqs. (7.99) and (7.100), real and imaginary parts of the modal
displacement amplitude are obtained. They are
¯
ξ
(r )
R =
−ω
2
+ p
2
r
¯
f
(r )
R + 2ωp r ζ r ¯
f
(r )
I
−ω 2 + p 2
r
2 + (2ωp r ζ r )
2
(7.101a)
¯
ξ
(r )
I
=
−ω
2
+ p
2
r
2 ¯
f
(r )
R − 2ωp r ζ r ¯
f
(r )
I
−ω 2 + p 2
r
2 + (2ωp r ζ r )
2
(7.101b)
Substituting Eq. (7.94) into Eq. (7.86), we get a complex vector of displacement
{x}. It is to be noted that the solution of the displacements is the real part of the
time-dependent solution. From Eq. (7.9), we find