Solution 8.1
()
39.81 736 N m
736 4.20 lb in.
m 175.13 N m
4.20 50.4 lb ft
in. ft
W
k
k
k
== =
==
==
Solution 8.2
()
rad 1 cycle 6
12 Hz 1.910 Hz
sec 2 rad
n
f
ππ
==
Solution 8.3
()
12 12
22
sin
n
xC t
ωψ
=+
Solution 8.4
()
δ
== =
29.81 0.200 m
st
W
Solution 8.5
()
ω
=
7 rad s from Prob. 8.4
n
y
()
=− =+
=
2
2
max
4.9 cos 7.3 2.68 m s
4.9 m s
a
a
Solution 8.6
km
2, 2
n
nn
m
kf
m
ω
ωππ
==
Solution 8.7
()
ω
== =
2 180,000
260 rad s
n
k
my
SE
(Dynamic forces
only)
m
Solution 8.8
2230
n
ππ
Solution 8.9
()
ωωπ
+= = = =
20 where 2 2.23 14.01 rad s
nn
xx km
()
ωωωωωω
ω
=+ =− +
===+ =
== =×=
=
=
cos sin , sin cos
w
hen 0, 0,0 0 , 0
w
hen 0, 0.025 m, 0.025 1, 0.025 m
0.025 cos14.01 m
or 25 cos14.01 mm in seconds
nn nnnn
n
xA tB tx A tB t
tx BB
tx AA
xt
xtt
Solution 8.10
93
n
kk
mm
ω
==
Solution 8.11
(Dynamic forces only)
3 kx 3 kx 3 kx
m
x
Solution 8.12
n
k
ω
Solution 8.13
12
FFF
=+
12
12 12 12
111
or Thus or
kk
FFF k
kkk kkk kk
=+⋅ =+ =+
Solution 8.14
ω
==
Tot
1km
n
n
f
Solution 8.15
Equivalent system:
3
33
24 6
2
n
mmL
EI EI
mL mL
ω
==
x
Solution 8.18
Dynamic force only, with s2 and s2 being inertial displacements and x is the displacement of
(1) from equilibrium.
12
2 constants
ss L
++ =
:Fma=
21
Eliminate T from Eqs. 2 and 3:
93m
n
m
kx
m
Solution 8.19
3
2
0
ky
my L
+=
Solution 8.20
()( )
ω
δω
== =
== =
392 2 14 rad s
42 0.75
22214
n
n
km
C
m
Solution 8.21
2
22
1
d
dn
ππ
τωωδ
== −
Solution 8.22
nn
Solution 8.23
2
1.25
22
1.25 , 0.6
1
d
n
n
ττ
ππ
δ
ω
ωδ
=
==
−
Solution 8.24
=
154.4 lb-sec ft
n
Solution 8.25
δω
== =
42 0.75 underdamped
n
c
Solution 8.26
0
n
t
δω δω τ
−
()
22
2N
N
πδ
Solution 8.27
4.65 0.0783
x
()( )( )
ω
π
⋅
=⋅ =
2
s
2 1.1 10 2 0.01245 1.721
n
n
m
N
cm
Solution 8.28
ω
××
== =
3
3 474 10 18.85 rad s
4000
n
k
m
Solution 8.29
Ignore weight.
21 21
aa
Kinetics:
()
1
11 11 11 11 11
1
xA A
F mx F kx mx mx kx F
==−−=→+=−
F
B
x
F
B
b
x
A
θ
k
2
x
2
y
1
m
2
C
2
x
B
B
Solution 8.30
()()
()
()
()
ω
δω
ωω ω
δω ω
δω ω
== =
== =
+− +
==− + =
== =
22
12
12
21 0
108 3 6 rad s
18 0.5
2236
sin cos
00
0.5 6 5.196 0.577
n
n
dd d
nd
nd
km
c
m
AtA te
xt A A
AA x x
[]
()
τ
−
=+
=== =
0
3
0
0
so cos5.196 0.577 sin5.196
a
nd 0.605 0.1630
2
t
xx t te
d
xt xt x
Solution 8.31
n
t
ω
−
0000 00 0
nn n
c
Solution 8.32
12 2rad sec
nkm
ω
== =
12
12
tt
xAe Ae
λλ
=+
Determine A1 and A2 in usual fashion:
() ( )
12
20 10
21 12
21
1,2
0.764 5.24
,where
0.7639, 5.236
0.585 0.085
0.5 0.393 ft 4.72 in.
tt
n
tt
xx
xe e
xe e
xt
λλ
λλ
λλ λλ
λωδδ
−
−−
=+
−−
=−±=−−
=−
==
Solution 8.33
()
1
3
where ln ln 1.792
21.792
x
δ
δδ
====
+
2232.2
3
Solution 8.34
ω
δω
===
== =
98 2 7 rad s
42 1.5
n
km
c
Then 2.67 18.33
01.171 0.1708
xx e e
=−
Solution 8.35
2
,
B
B
x
xb
Tcx cx
===
aaa a
1
2
:
x
FmxTkxmx
b
Cxkxmx
=−−=
−−=
A
m
Oa
T
1
x
O
y
Solution 8.36
µ
ωµ
=−± =
+=
2
:mg
g
xk
nk
Fmxkx mx
xx
Solution:
µ
ωω
ω
=+ 2
g
cos sin k
nn
n
xA tB t
Initial conditions for first half-cycle 0@0
0
xx t
x
==
=
Results:
µ
ω
=− =
02
g,0
k
n
A
xB
µµ
gg
kk
Solution 8.37
x
0
x4μ
k
g
ω
n2
2π
ω
n
m
k
t
Linear envelope
τ
τ == 2π
μ
k
mg (motion )
(motion )
k
2
xmg
N = mg
μ
k
mg
xk
2x
Solution 8.38
()( ) ()
ω
δω
== =
== =
==
12 22
22
100,000 100 rad s
10
500 0.25 for a
2210100
1000 100,000
n
n
o
k
m
c
m
Fk
X
2
−
Solution 8.39
2
64.4
Solution 8.40
6rad sec
n
ω
= (from Prob. 8.39)
()
2
22
2.4 0.1
2226
o
n
Fk c
Xm
δω
====
Solution 8.41
6rad sec
n
ω
= (from Prob. 8.39)
2
O
Fk
X
=
Solution 8.42
100 32.2
nkm
()
eff
2
a0:
111.35
O
Fk
X
δω
==
−
Solution 8.43
12
2
22
112
212
2
1
12.5
10.3304
M
M
M
R
=
′==
==
=
()
2
100 2.52%
M=
Solution 8.44
The condition for the maxima is
12
nn
δ
ωω ωω
−+
n
ω
