2 
1
Problems and Solutions Section 2.1 (2.1 through 2.15)
2.1
To familiarize yourself with the nature of the forced response, plot the solution of a
forced response of equation (2.2) with
ω
= 2 rad/s, given by equation (2.11) for a variety
of values of the initial conditions and
ω
n
as given in the following chart:
Case
x
0
v
0
f
0
ω
n
1
0.1
0.1
0.1
1
2
0.1
0.1
0.1
1
3
0.1
0.1
1.0
1
4
0.1
0.1
0.1
2.1
5
1
0.1
0.1
1
Solution:
Given:
!
= 2 rad/sec.
From equation (2.11):
x
(
t
) =
n
v
!
0
sin
n
!
t
+ (
x
0

2
2
0
!
!
"
n
f
) cos
n
!
t
+
2
2
0
!
!
"
n
f
cos
!
t
Insert the values of
x
0
,
v
0
,
f
0
, and
!
n
for each of the five cases.
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2 
2
2.2
Repeat the calculation made in Example 2.1.1 for the mass of a simple springmass
system where the mass of the spring is considered and known to be 1 kg.
Solution:
Given:
m
sp
= 1 kg, Example 1.4.4 yields that the effective mass is
m
e
=
m
+
3
sp
m
= 10 +
3
1
= 10.333 kg.
Thus the natural frequency,
X
and the coefficients in equation (2.11) for the system now
become
!
n
=
1000
10
+
1
3
=
9.837 rad/s,
!
=
2
!
n
=
19.675 rad/s
X
=
f
0
!
n
2
"
!
2
=
2.338
9.837
2
"
19.675
2
=
"
8.053
#
10
"
3
m,
v
0
!
n
=
0.02033 m
Thus the response as given by equation (2.11) is
x
(
t
)
=
0.02033sin9.837
t
+
8.053
!
10
"
3
(cos9.837
t
"
cos19.675
t
) m
2.3
A springmass system is driven from rest harmonically such that the displacement
response exhibits a beat of period of 0.2
!
s. The period of oscillation is measured to be
0.02
!
s. Calculate the natural frequency and the driving frequency of the system.
Solution:
Given: Beat period:
T
b
= 0.2
!
s, Oscillation period:
T
0
= 0.02
!
s
Equation (2.13):
x
(
t
) =
2
2
0
2
!
!
"
n
f
sin
!
n
"
!
2
t
#
$
%
&
’
(
sin
!
n
+
!
2
t
"
#
$
%
&
’
So,
T
b
= 0.2
!
=
!
!
"
#
n
4
!
!
"
n
=
!
!
2
.
0
4
= 20 rad/s
T
0
= 0.02
!
=
!
!
"
+
n
4
!
!
+
n
=
!
!
02
.
0
4
= 200 rad/s
Solving for
n
!
and
!
gives:
Natural frequency
:
n
!
= 110 rad/s
Driving frequency
:
!
= 90 rad/s