Problems and Solutions for Section 4.6 (4.67 through 4.76)
4.67
Calculate the response of the system of Figure 4.16 discussed in Example 4.6.1 if
F
1
(
t
) =
δ
(
t
) and the initial conditions are set to zero. This might correspond to a
twodegreeoffreedom model of a car hitting a bump.
Solution:
From example 4.6.1, with
F
1
(
t
) =
δ
(
t
), the modal equations are
!!
r
1
+
0.2
!
r
1
+
2
r
1
=
0.7071
!
(
t
)
!!
r
2
+
0.4
!
r
2
+
4
r
2
=
0.7071
!
(
t
)
Also from the example,
!
n
1
=
2
rad/s
"
1
=
0.07071
!
d1
=
1.4106
rad/s
!
n
2
=
2
rad/s
"
2
=
0.1
!
d2
=
1.9899
rad/s
The solution to an impulse is given by equations (3.7) and (3.8):
r
i
(
t
)
=
ˆ
F
m
i
!
di
e
"
#
i
!
ni
t
sin
!
di
t
This yields
r
(
t
)
=
0.5012
e
!
0.1
t
sin1.4106
t
0.3553
e
!
0.2
t
sin1.9899
t
"
#
$
%
&
’
The solution in physical coordinates is
x
(
t
)
=
M
!
1/2
P
r
(
t
)
=
.2357
!
.2357
.7071
.7071
"
#
$
%
&
’
0.167
e
!
0.1
t
sin1.4106
t
!
0.118
e
!
.02
t
sin1.9899
t
"
#
$
%
&
’
x
(
t
)
=
0.0394
e
!
0.1
t
sin1.4106
t
+
0.0279
e
!
0.2
t
sin1.9899
t
0.118
e
!
0.1
t
sin1.4106
t
!
0.0834
e
!
0.2
t
sin1.9899
t
"
#
$
%
&
’
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4.68
For an undamped twodegreeoffreedom system, show that resonance occurs at
one or both of the system’s natural frequencies.
Solution:
Undamped twodegreeoffreedom system:
M
!!
x
+
K
x
=
F
(
t
)
Let
F
(
t
)
=
F
1
(
t
)
0
!
"
#
$
%
&
Note: placing
F
1
on mass 1 is one way to do this. A second force could be placed
on mass 2 with or without
F
1
.
Proceeding through modal analysis,
I
!!
r
+
!
r
=
P
T
M
"
1/2
F
(
t
)
Or,
!!
r
1
+
!
1
2
r
1
=
b
1
F
1
(
t
)
!!
r
2
+
!
2
2
r
2
=
b
2
F
1
(
t
)
where
b
1
and
b
2
are constants from the matrix
P
T
M
1/2
.
If
F
1
(
t
) =
a
cos
ω
t
and
ω
=
ω
1
then the solution for
r
1
is (from Section 2.1),
r
1
(
t
)
=
!
r
10
!
1
sin
!
1
t
+
r
10
cos
!
1
t
+
b
1
a
2
!
1
t
sin
!
1
t
The solution for
r
2
is
r
2
(
t
)
=
!
r
20
!
2
sin
!
2
t
+
r
20
"
b
2
a
!
2
2
"
!
1
2
#
$
%
&
’
(
cos
!
2
t
+
b
2
a
!
2
2
"
!
1
2
t
sin
!
1
t
If the initial conditions are zero,
r
1
(
t
)
=
b
1
a
2
!
1
t
sin
!
1
t
r
2
(
t
)
=
b
2
a
!
2
2
"
!
1
2
cos
!
1
t
"
cos
!
2
t
(
)
Converting to physical coordinates
X
(
t
) =
M
1/2
Pr
(
t
) yields
x
1
(
t
)
=
c
1
r
1
(
t
)
+
c
2
r
2
(
t
)
x
2
(
t
)
=
c
3
r
1
(
t
)
+
c
4
r
2
(
t
)
where
c
i
is a constant from
M
1/2
P
.
So, if the driving force contains just one natural frequency, both masses will be
excited at resonance. The driving force could contain the other natural frequency
(
ω
=
ω
n2
), which would cause
r
1
and
r
2
to be
r
1
(
t
)
=
b
1
a
!
1
2
"
!
2
2
cos
!
2
t
"
cos
!
1
t
(
)
r
2
(
t
)
=
b
2
a
2
!
2
t
sin
!
2
t
and
x
1
(
t
)
=
c
1
r
1
(
t
)
+
c
2
r
2
(
t
)
x
2
(
t
)
=
c
3
r
1
(
t
)
+
c
4
r
2
(
t
)
so both masses still oscillate at resonance.
Also, if
F
1
(
t
) =
a
1
cos
ω
1
t
+
a
2
cos
ω
2
t
where
ω
1
=
ω
n1
and
ω
2
=
ω
n2
, then both
r
1
and
r
2
would be at resonance, so
x
1
(
t
) and
x
2
(
t
) would also be at resonance.
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4.69
Use modal analysis to calculate the response of the drive train system of Problem
4.44 to a unit impulse on the car body (i.e., and location
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 Spring '09
 J.G.Lee
 Force, SEPTA Regional Rail, 0 $, $ 3, $ 2, $0 0

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