B. Balachandran
ENME361 Vibrations
Fall 2008
Solution to Quiz #2
(September 11, 2008; Duration: 10 minutes)
1) Consider the differential equation
2
2
( )
( )
2
( )
( )
d w t
dw t
w t
u t
dt
dt
ζ
+
+
=
where
ζ
is a constant such that 0 <
ζ
< 1 and
u
(
t
) is the unit step function.
Given that the initial
conditions are zero; that is,
w
(0) =
dw
(0)/
dt
= 0, solve for the solution
w
(
t
) by using the Laplace
transform technique.
Some Laplace transform pairs are provided at the end.
Taking Laplace transforms of the different terms on both sides, the differential equation is
transformed into the algebraic equation
s
s
w
w
s
sW
w
sw
s
W
s
1
)
(
))
0
(
)
(
(
2
)
0
(
)
0
(
)
(
2
=
+
−
+
′
−
−
ζ
where
L
[
w
(
t
)] =
W
(
s
), the prime is used to indicate a time derivative, and we have made use of
the table for the Laplace transform of the unit step function with
t
o
=
0.
Applying the initial
conditions
0
)
0
(
)
0
(
=
′
=
w
w
and solving for
W
(
s
), we arrive at
)
1
2
(
1
)
(
2
+
+
=
s
s
s
s
W
ζ
Making use of the table, the inverse Laplace transform
L
1
[
W
(
s
)] is found to be
w
(
t
) =
2
1
1
sin(
);
1
;
cos
;
1
n
t
n
d
d
n
d
e
t
ζω
ω
ω
ϕ
ω
ω
ζ
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 Spring '11
 yoo
 Laplace, Complex number, ω − ω

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