Civil Engineering Reference
In-Depth Information
indicating that the mass oscillates 1.5 in. above and below the static equilibrium position
continuously in time and completes one cycle every
2
/
21
.
98
sec. Therefore, the cyclic
frequency is
2
=
21
.
98
2
f
=
=
3
.
5 cycles/sec (Hz)
The cyclic frequency is often simply referred to as the
natural frequency.
The time required
to complete one cycle of motion is known as the
period
of oscillation, given by
1
f
=
1
3
.
5
=
=
0
.
286 sec
10.2.1 Forced Vibration
Figure 10.3 shows a simple harmonic oscillator in which the mass is acted on by
a time-varying external force
F
(
t
). The resulting motion is known as
forced
vibration,
owing to the presence of the external forcing function. As the only dif-
ference in the applicable free-body diagrams is the external force acting on the
mass, the finite element form of the system equations can be written directly
from Equation 10.13 as
00
0
k
m
k
1
u
1
u
2
F
(
t
)
x
u
1
¨
¨
−
1
f
1
+
=
(10.21)
m
u
2
−
11
mg
+
F
(
t
)
Figure 10.3
Simple
harmonic oscillator
subjected to external
force
F
(
t
).
While the constraint equation for the reaction force at node 1 is unchanged, the
differential equation for the motion of node 2 is now
F
(
t
)
(10.22)
The complete solution for Equation 10.22 is the sum of the homogeneous solu-
tion and two particular solutions, since two nonzero terms are on the right-hand
side. As we already obtained the homogeneous solution and the particular solu-
tion for the
mg
term, we focus on the particular solution for the external force.
The particular solution of interest must satisfy
m
u
2
+
¨
ku
2
=
mg
+
(10.23)
m
u
2
+
¨
ku
2
=
F
(
t
)
exactly for all values of time. Dividing by the mass, we obtain
F
(
t
)
m
2
u
2
=
(10.24)
u
2
+
¨
2
where
m
is the square of the natural circular frequency. Of particular
importance in structural dynamic analysis is the case when external forcing func-
tions exhibit sinusoidal variation in time, since such forces are quite common.
Therefore, we consider the case in which
F
(
t
)
=
k
/
=
F
0
sin
f
t
(10.25)
