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Problems and Solutions Section 2.2 (2.16 through 2.31)
2.16
Calculate the constants
A
and
!
for arbitrary initial conditions,
x
0
and
v
0
,
in the case
of the forced response given by Equation (2.37). Compare this solution to the transient
response obtained in the case of no forcing function (i.e.
F
0
=
0).
Solution:
From equation (2.37)
x
(
t
)
=
Ae
!
"#
n
t
sin(
#
d
t
+
$
)
+
X
cos(
t
!
%
)
&
˙
x
(
t
)
=
!
n
Ae
!
n
t
sin(
d
t
+
)
+
A
d
e
!
n
t
cos(
d
t
+
)
!
X
sin(
t
!
)
Next apply the initial conditions to these general expressions for position and
velocity to get:
x
(0)
=
A
sin
+
X
cos
"
˙
x
(0)
=
#
$%
n
A
sin
+
A
d
cos
+
X
sin
Solving this system of two equations in two unknowns yields:
=
tan
"
1
(
x
0
"
X
cos
)
d
v
0
+
(
x
0
"
X
cos
)
%$
n
"
X
sin
&
’
(
)
*
+
A
=
x
0
"
X
cos
sin
Recall that
X
has the form
X
=
F
0
/
m
n
2
"
2
)
2
+
(2
#!
n
)
2
and
=
tan
"
1
2
n
n
2
"
2
%
&
’
(
)
*
Now if
F
0
= 0, then
X
= 0 and
A
and
φ
from above reduce to:
=
tan
"
1
x
0
d
v
0
+
x
0
$#
n
%
&
’
(
)
*
A
=
x
0
sin
=
(
v
0
+
n
x
0
)
2
+
(
x
0
d
)
2
d
2
These are identical to the values given in equation (1.38).
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View Full Document2.17
Show that Equations (2.28) and (2.29) are equivalent by verifying Equations
(2.29) and (2.30).
Solution:
From equation (2.28) and expanding the trig relation yields
x
p
=
X
cos(
!
t
"
#
)
=
X
cos
t
cos
+
sin
t
sin
[ ]
=
(
X
cos
)
A
s
!"
# $
#
cos
t
+
(
X
sin
)
B
s
# $
#
sin
t
Now with
A
s
and
B
s
defined as indicated, the magnitude is computed:
X
=
A
s
2
+
B
s
2
and
B
s
A
s
=
X
sin
X
cos
"
=
tan
#
1
B
s
A
s
$
%
&
’
(
)
2.18
Plot the solution of Equation (2.27) for the case that
m
= 1 kg,
= 0.01,
n
= 2
rad/s.
F
0
= 3 N, and
= 10 rad/s, with initial conditions
x
0
= 1 m and
v
0
= 1
m/s.
Solution:
The particular solution is given in equations (2.36) and (2.37).
Substitution of the values given yields:
x
p
=
0.03125cos(10
t
+
8.333
!
10
"
3
)
.
Then the total solution has the form:
x
(
t
)
=
Ae
!
0.02
t
sin(2
t
+
"
)
+
0.03125cos(10
t
+
0.008333)
=
e
!
0.02
t
A
sin2
t
+
B
cos2
t
( ) +
0.03125cos(10
t
+
0.008333)
Differentiating then yields
!
x
(
t
)
=
!
0.02
e
!
0.02
t
A
t
+
B
t
( ) +
sin(2
t
+
)
+
2
e
!
0.02
t
A
t
!
B
t
( )
!
0.3125sin(10
t
+
0.008333)
Apply the initial conditions to get:
x
(0)
=
1
=
B
+
0.03125cos(0.00833)
!
B
=
0.969
!
x
(0)
=
1
=
"
0.02
B
+
2
A
"
0.3125sin(0.00833)
!
A
=
0.489
So the solution and plot become (using Mathcad):
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View Full Document2.19
A 100 kg mass is suspended by a spring of stiffness 30
×
10
3
N/m with a viscous
damping constant of 1000 Ns/m. The mass is initially at rest and in equilibrium.
Calculate the steadystate displacement amplitude and phase if the mass is excited
by a harmonic force of 80 N at 3 Hz.
Solution:
Given
m
= 100kg,
k
=30,000 N/m,
c
= 1000 Ns/m,
F
0
= 80 N and
ω
=
6
π
rad/s:
f
0
=
F
0
m
=
80
100
=
0.8 m/s
2
,
!
n
=
k
m
=
17.32 rad/s
"
=
c
2
km
=
0.289
X
=
0.8
17.32
2
+
36
#
2
( )
2
+
2(0.289)(17.32)(6
)
( )
2
=
0.0041 m
Next compute the angle from
=
tan
"
1
188.702
"
55.323
#
$
%
&
’
(
Since the denominator is negative the angle must be found in the 4
th
quadrant. To
find this use Window 2.3 and then in Matlab type atan2(188.702,55.323) or use
the principle value and add
π
to it. Either way the phase is
θ
=1.856 rad.
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This note was uploaded on 03/31/2009 for the course MECHANICAL MAE351 taught by Professor J.g.lee during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 J.G.Lee

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