Civil Engineering Reference
In-Depth Information
or
k
m
1
=
(10.41)
6
k
m
2
=
In mathematical rigor, there are four roots, since the negative values correspond-
ing to Equation 10.41 also satisfy the frequency equation. The negative values
are rejected because a negative frequency has no physical meaning and use of the
negative values in the assumed solution (Equation 10.35) introduces only a phase
shift and represents the same motion as that corresponding to the positive root.
The 2 degrees-of-freedom system of Figure 10.4 is found to have two natural
circular frequencies of oscillation. As is customary, the numerically smaller of
the two is designated as
1
and known as the
fundamental
frequency. The task
remains to determine the amplitudes
A
2
and
A
3
in the assumed solution. For this
purpose, Equation 10.37 is
5
k
A
2
A
3
0
0
2
−
m
−
2
k
=
(10.42)
2
−
2
k
2
k
−
m
As Equation 10.42 is a set of homogeneous equations, we can find no absolute
values of the amplitudes. We can, however, obtain information regarding the
numerical relations among the amplitudes as follows. If we substitute
2
2
=
1
=
k
m
into
either
algebraic equation, we obtain
A
3
=
2
A
2
, which defines the
amplitude ratio
A
3
/
A
2
=
2
for the first, or fundamental, mode of vibration. That
is, if the system oscillates at its fundamental frequency
1
, the amplitude of
oscillation of
m
2
is twice that of
m
1
. (Note that we are unable to calculate the
absolute value of either amplitude; only the ratio can be determined. The absolute
values depend on the initial conditions of motion, as is subsequently illustrated.)
The displacement equations for the fundamental mode are then
U
2
(
t
)
/
A
(1
2
sin(
=
1
t
+
1
)
(10.43)
A
(1
3
2
A
(1
2
U
3
(
t
)
=
sin(
1
+
1
)
=
sin(
1
t
+
1
)
where the superscript on the amplitudes is used to indicate that the displacements
correspond to vibration at the fundamental frequency.
Next we substitute the second natural circular frequency
2
2
m
into either equation and obtain the relation
A
3
=−
0
.
5
A
2
, which defines the sec-
ond amplitude ratio as
A
3
/
A
2
=−
0
.
5
.
So, in the second natural mode of vibra-
tion, the masses move in opposite directions. The displacements corresponding
to the second frequency are then
=
2
=
6
k
/
A
(2
2
sin(
=
2
t
+
2
)
U
2
(
t
)
(10.44)
A
(2
3
5
A
(2
2
U
3
(
t
)
=
sin(
2
+
2
)
=−
0
.
sin(
2
t
+
2
)
where again the superscript refers to the frequency.
