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3.4: Mechanical Vibrations
±
Mass  Spring  Dashpot
x
Equilibrium Point
m
c
Setup: Attach a mass,
m
, to a spring and a dashpot.
x
H
t
L
is the distance from equilibrium. If
x
H
t
L
>
0, then the spring is stretched. If
x
H
t
L
<
0, then the spring
is compressed.
A d
a
s
h
p
o
t
is a piston that always opposes motion. Assume that the force of opposition is proportional
to velocity with constant of proportionality
c
>
0. Hence, we have
F
R
= 
c v
= 
c
dx
dt
.
If
c
=
0, then the motion is u
n
d
a
m
p
e
d
. If
c
„
0, then the motion is d
a
m
p
e
d
.
Hooks
Law
: The restorative force,
F
S
, is proportional to the distance stretched and opposes motion.
So, for
k
>
0,
F
S
= 
k x
.
External
Force
F
E
=
F
H
t
L
. If
F
E
=
F
H
t
L
=
0, then the motion is f
r
e
e
. If
F
E
=
F
H
t
L
„
0, then the motion if f
o
r
c
e
d
.
Newton's
Law
:
F
T
=
m a
=
m
d
2
x
dt
2
.
Total
Forces
on
Mass
:
F
T
=
F
S
+
F
R
+
F
E
.
Hence, we need to solve
m
d
2
x
dt
2
+
c
dx
dt
+
k x
=
F
E
=
F
H
t
L
.
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Free Undamped Motion:
F
H
t
L
=
0,
c
=
0.
Then, we need to solve the basic second order, linear, homogeneous differential equation with constant
coefficients:
m
d
2
x
dt
2
+
k x
=
0
±
d
2
x
dt
2
+
k
m
x
=
0
±
d
2
x
dt
2
+ Ω
0
2
x
=
0,
where we define
Ω
0
=
k
m
. Thus, we have a characteristic equation of
r
2
+ Ω
0
2
=
0
±
r
= –Ω
0
i
.
Thus, our solution can be written as a linear combination of sines and cosines,
x
H
t
L
=
A
cos
H
Ω
0
t
L
+
B
sin
H
Ω
0
t
L
=
C
A
A
C
cos
H
Ω
0
t
L
+
B
C
sin
H
Ω
0
t
LE
, where
C
=
A
2
+
B
2
=
C
@
cos
H
Α
L
cos
H
Ω
0
t
L
+
sin
H
Α
L
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This note was uploaded on 01/13/2012 for the course MATH 306 taught by Professor Keithemmert during the Spring '11 term at Tarleton.
 Spring '11
 KeithEmmert

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