84
3 Forced Vibration of Single Degree of Freedom System
The solution of Eq. (3.54) is
x = C 1 e
λ 1 t
+ C 2 e
λ 2 t
(3.58)
The system is unstable, as the solution indicates a diverging non-oscillating
motion.
Case II
c − F 0
2m
2
=
k
m
(3.59)
In this case, both λ 1 and λ 2 are real, equal and positive. The system is unstable,
as it indicates a diverging non-oscillating motion.
Case III
c − F 0
2m
2
<
k
m
(3.60)
Here, λ 1 and λ 2 are complex conjugate.
The solution is expressed as
x = A e
F 0 −c
2m
t sin
⎧
⎨
⎩
k
m
−
c − F 0
2m
2
t+ ∈
⎫
⎬
⎭
(3.61)
Equation (3.61) indicates a diverging oscillating motion, as the exponent is
positive. Thus, the system is unstable in this case as well.
Therefore, the condition for the dynamic stability is given by
F 0 ≤ c
(3.62)
Example 3.10 A cantilever pipe of length L, cross-sectional area A 0 , flexural rigidity
E I 0 and mass is shown in Fig. 3.15. If the fluid of mass density ρ flows through the
pipe, determine the velocity of the fluid v at which the system will be unstable.
The cantilever pipe is treated as a single degree of freedom system. The freebody
diagram is shown in Fig. 3.15c. The spring constant k is
k =
8E I 0
L 3
(a)
The force in the pipe due to fluid velocity is ρ A 0 v
2 . As the pipe deflects, the
component of the force at an angle α is
F = 2ρ A 0 v
2 sin α
(b)