SO
G
d
Th
u
LUTION
d
escribes a cir
c
u
s,
PR
A
6
vert
i
spri
n
(a)
t
the
s
c
le about the
a
OBLEM 1
9
6
0-lb disk is
a
i
cal shaft AB,
w
n
g constant
k
t
he angular v
e
s
haft when
f
ω
a
xis AB of rad
i
n
a
9
.125
a
ttached with
w
hich revolv
e
k
for horizont
a
e
locity
f
ω
at
1200 rpm.
f
=
i
us
.re+
2
()
f
re
ω
=+
an eccentric
i
e
s at a constan
t
a
l movement
which reson
a
i
ty
0.006
i
e=
t
angular velo
c
of the disk i
s
a
nce will occ
u
i
n.
to the mi
c
ity
.
f
ω
Kno
w
s
40,000 lb/f
t
u
r, (b) the de
f
dpoint of a
w
ing that the
t
, determine
f
lection r of
u
f
u
(a)
Resonanc
e occurs whe
n
PROB
n
f
ω
n
ω
LEM 19.12
5
, i.e.,
n
r
ω
=
kk
g
==
5
(Continu
e
∞
g
e
d)
PRO
B
A sma
l
suppor
t
a road
,
an am
p
b
etwe
e
trough
occur,
50 km
/
B
LEM 19.1
l
l trailer and
t
ed by two s
p
,
the surface
o
p
litude of 40
e
n successive
is 80 mm).
(b) the ampl
/
h.
26
its load have
p
rings, each o
f
o
f which can
b
mm and a
w
crests is 5 m
Determine (
a
itude of the
a total mass
f
constant 10
k
b
e approxim
a
w
avelength o
f
and the verti
c
a
) the speed
a
vibration of
t
of 250-kg. T
h
k
N/m, and is
a
ted by a sine
f
5 m (i.e., t
h
c
al distance f
r
a
t which res
o
t
he trailer at
h
e trailer is
pulled over
curve with
h
e distance
r
om crest to
o
nance will
a speed of
a
a
x
P
R
Sh
o
(a)
ar
b
S
O
Si
n
(a)
R
OBLEM 1
o
w that in th
e
if it is relea
s
b
itrary initial
v
O
LUTION
n
ce
,
c
cc>
we
u
0,tx==
From Eqs
.
9.127
e
case of hea
v
s
ed with no i
n
v
elocity.
u
se Equation
(
0
,
0
xv=
.
(1) and (2):
v
y damping
(
c
n
itial velocit
y
(
19.42), wher
e
1
λ
x
v
0
:
),
c
c
c>
a bod
y
from an arb
i
e
1
2
0, 0
λ
λ
<<
1
12
t
x
Ce C
λ
=+
11
dx
v
C
e
dt
λ
==
01
2
0
x
CC
C
C
λ
=
+
=
+
λ
y never pass
e
i
trary positio
n
2
t
e
λ
12
22
tt
e
Ce
λ
λ
λ
+
2
C
λ
e
s through its
n
or (b) if it
i
position of e
q
i
s started fro
m
q
uilibrium O
m
O with an
(1)
(2)
o
n
a
n
λ
λ
)
v
Copyrig
h
ht
© McGra
w
w
-Hill Educ
a
PROB
L
a
tion. Permis
s
L
EM 19.127
s
ion require
d
(Continue
d
for reprodu
d)
ction or disp
lay.
P
R
Sh
o
ini
t
S
O
Su
b
So
l
s
i
s
λ
l
w
s
s
m
t
t
m
m
R
OBLEM 1
o
w that in the
t
ial velocity c
a
O
LUTION
b
stitute the in
i
l
ving for
1
a
C
9.128
case of heav
y
a
nnot pass m
o
i
tial conditio
n
2
a
nd ,
C
y
damping
(c
o
re than once
t
n
s,
0
0,txx==
),
c
c>
a bod
y
t
hrough its eq
u
0
0
,vv=
in Eq
u
012
x
CC=+
0
1
λ
−
−
(
v
C
λ
λ
−
=−
−
1
y
released fro
m
u
ilibrium pos
i
u
ations (1) an
d
01
vC
λ
=
20
)
x
λ
λ
−
m
an arbitrary
i
tion.
d
(2) of Probl
e
122
C
λ
λ
+
position wit
h
e
m 19.127.
h
an arbitrary
r
n
g
r
g
)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.129
In the case of light damping, the displacements 1,
x
2,
x
3,
x
shown in Figure 19.11 may be assumed equal to
the maximum displacements. Show that the ratio of any two successive maximum displacements n
x
and 1n
x
+
is a constant and that the natural logarithm of this ratio, called the logarithmic decrement, is
2
1
2(/ )
ln
1(/)
nc
nc
xcc
xcc
π
+
=−
SOLUTION
For light damping,
Equation (19.46):
(
)
2
00
sin( )
c
mt
xxe t
ωφ
−
=
+
At given maximum displacement,
()
2
0
0
,
sin( ) 1
cn
m
nn
n
t
n
tt xx
t
xxe
ωφ
−
=
=
+=
=
At next maximum displacement,
()
1
2
11
01
10
,
sin( ) 1
cn
m
nn
n
t
n
tt xx
t
xxe
ωφ
+
+
+
+
−
+
==
+=
=
But 1
1
2
2
Dn Dn
nn
D
tt
tt
ω
ωπ
π
ω
+
+
−=
−=
Ratio of successive displacements:
()
2
1
2
2
2
1
2
0
10
cn
m
cn
m
c
cm
nn
m
D
t
n
t
n
tt
xxe
xxe
ee
π
ω
+
+
−
−
+
+
−−
=
==
Thus,
1
ln n
nD
xc
xm
π
ω
+
= (1)
From Equations (19.45) and (19.41):
2
2
1
1
2
Dn
c
c
D
c
c
cc
mc
ωω
ω
⎛⎞
=−
⎜⎟
⎝⎠
⎛⎞
=−
⎜⎟
⎝⎠
Thus, 2
1
21
ln
1
n
nc
c
xcm
xmc c
c
π
+
=
⎛⎞
−⎜⎟
⎝⎠
(
)
()
2
1
2
ln Q.E.D.
1
c
c
c
c
n
nc
c
x
x
π
+
=
−
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.130
In practice, it is often difficult to determine the logarithmic decrement of a system with light damping defined
in Problem 19.129 by measuring two successive maximum displacements. Show that the logarithmic
decrement can also be expressed as (1/ ) ln ( / ),
nnk
kxx
+ where k is the number of cycles between readings of
the maximum displacement.
SOLUTION
As in Problem 19.129, for maximum displacements n
x
and nk
x
+
at 0
and t , sin( ) 1
nnk n
tt
ωφ
+
+
=
and sin( ) 1.
nn k
t
ωφ
++=
(
)
()
2
2
0
()
0
cn
m
cnk
m
t
n
t
nk
xxe
xxe +
−
−
+
=
=
Ratio of maximum displacements:
(
)
()
()
2
2
2
0
0
cn
mcnnk
m
cnk
m
t
tt
n
t
nk
xxe e
xxe
−
−+
−+
−
+
==
But (2 )
2
Dn k Dn
nnk
D
ttk
tt k
ω
ωπ
π
ω
+
+
−=
−=
Thus, 2
2
n
nk D
xck
xm
π
ω
+
⎛⎞
=+ ⎜⎟
⎝⎠
ln n
nk D
xc
k
xm
π
ω
+
= (2)
But from Problem 19.129, Equation (1):
1
log decrement ln n
nD
xc
xm
π
ω
+
==
Comparing with Equation (2), 1
log decrement ln Q.E.D.
n
nk
x
kx
+
=
P
R
In
a
d
τ
=
lim
i
dis
p
dis
p
gre
a
SO
Eq
u
(a)
R
OBLEM 1
9
a
n underdam
p
2/
d
πω
=
corr
i
ting curves
s
p
lacement an
d
p
lacements is
a
ter than
1
4
.
d
τ
LUTION
u
ation (19.46)
:
Maxima (
p
9
.131
p
ed system
(
e
sponding to
s
hown in Fig
u
d
the followi
n
1
2
,
d
τ
(c)
b
et
w
:
p
ositive or ne
g
(
),
c
cc<
the
p
two successi
v
u
re 19.11. S
h
n
g maximum
w
een a maxi
m
xx
=
g
ative) when
x
p
eriod of vi
b
v
e points wh
e
h
ow that the
negative dis
p
m
um positive
d
()
2
0
sin(
c
m
t
d
x
e
ω
−
0:
x
=
c
b
ration is co
m
e
re the displa
c
interval of ti
m
p
lacement is
1
2
d
isplacement
a
)
d
t
φ
+
m
monly defi
n
c
ement-time
c
m
e (a)
b
etw
e
2
,
d
τ
(b)
b
et
w
a
nd the follo
w
c
n
ed as the ti
m
c
urve touche
s
e
en a maxim
u
w
een two suc
c
w
ing zero dis
p
m
e interval
s
one of the
u
m positive
c
essive zero
p
lacement is
c
o
o
o
g
c
h
o
o
e
h
e
c
c
a
a
PROB
LEM 19.13
1
1
(Continu
e
e
d)