ENME 361 Fall 2008
Solution for Quiz #11
(December 2, 2008; Duration: 20 minutes)
1)
The displacement response of a single degree of freedom system subject to an initial
displacement
X
o
is given by
( )
2
2
()
s
in
1
1
n
t
o
n
Xe
xt
t
ζω
ω
ζϕ
ζ
−
=−
+
−
Given that the logarithmic decrement is defined as the natural logarithm of the ratio of any two
successive amplitudes of the response of a damped, linear system that occur one period apart,
determine the expression for the logarithmic decrement.
Solution
:
The logarithmic decrement is given by
ln
d
xt T
δ
⎛⎞
=
⎜⎟
+
⎝⎠
=
( )
( )
22
sin
1
/ 1
ln
sin
(
) 1
/ 1
n
nd
t
on
tT
d
t
t T
ωζ
ϕ
−
−+
−
+−
+
−
Since
2
2
1
d
n
T
π
=
−
, it follows that
( )
( )
( )
2
sin
1
sin
1
2
sin
1
n
n
t
t
ϕω
ϕπ
+
−+=
+
=
As an account of this, we have that
ln(
)
T
e
=
2
2
1
πζ
=
−
2)
A single degreeoffreedom system is subjected to a harmonic excitation
F
o
sin(
Ω
τ
), where
Ω
is the nondimensional excitation frequency.
Given that the steadystate response of the system
is of the form
( )
s
i
n
o
F
xH
k
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '11
 yoo
 Damping ratio, Logarithmic decrement, displacement response

Click to edit the document details