Solution 8.103
2
:
x
FmxkxcxFmx
=−−+=
==
Solution 8.104
4
:4 0
x
ck
Fmx kxcxmxorx x x
+↓ = − − = + + =
2kx 2kx
+
x
Solution 8.105
For seismic instruments,
()
2
12
n
x
ωω
=
()
180 3
60
3
n
ω
Solution 8.106
ny
Rotation about G:
Force changes for position of static equilibrium.
k
δ
st = 600(10
θ
) = 6000
θ
lb
(
θ
small)
2
10 48.0 4.60
n
+
kδ
st
10″10″
Solution 8.107
p
12
2
12
ωω
δ
−+
Gy
k(rθ – roϕ)
Solution 8.109
–
wt
Solution 8.110
O
O
y
–50
0123456
50
0
Time, t, sec
38.8°
28.0°
Solution 8.111
()( )
ω
δ
ω
== =
100 7.071 rad s
2227.071
n
n
k
m
1
2
2.192
λ
=−
–0.09
Solution 8.112
11 2 2B
or
22
222
sin
mm
θθθωω
++=
Steady-state amplitude:
ω
2
1
ω
3
2n
Set up computer program to determine range of k for which A ≤ 1.5b. Note that
2
11
2
,2
nn
c
k
mm
ωδω
==
or
2
1
2
c
km
ft
Solution 8.114
18 0.208
c
δω
== =
1.408 rad
ψ
Solution 8.115
==
6.25 8.33 N s
b
b
Solution 8.116
Equation 8.9: 2
2sin 0
nn
yyy
δω ω
++=
tt
λλ
n
n
(a)
()( )
124 1.096
24 14.14
δ
==
; (b)
()( )
80 0.707
24 14.14
δ
==
() ( )
()
()
δ
λ
λ
−
−
>
=−+ −=−
=−+ −=−
21
1
21
2
a 1 overdamped
14.14 1.096 1.096 1 9.16 s
14.14 1.096 1.096 1 21.8 s
Initial condition considerations:
012 1
2
01122
0.1 0.1722 m
0.0722 m
0
yAAA
A
yAA
λλ
==+ =
=−
== +
Solution: y = 0.1722 e–9.16t – 0.0722 e–21.8t m
(b)
δ
< 1 (underdamped)
Equation 8.12:
[
]
sin
nt
d
yCe t
δω
ω
ψ
−
=+
ω
ωδ
=−= − =
22
1 14.14 1 0.707 10 rad s
dn
Initial condition considerations:
0
0
0.1 sin 0.1414 m
0 sin cos 0.785 rad
nd
yC C
yCC
ψ
δω ψ ω ψ ψ
== =
==− + =
Solution:
()
[
]
[
]
0.707 14.14 10
0.1414 sin 10 0.785 0.1414 sin 10 0.785 m
tt
ye t e t
−−
=+=+
0.1
y (m)
0.01
(a)
(b)
0
0 1.0
t (s)