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VIBRATION ABSORBER
Revision E
By Tom Irvine Email: [email protected] May 21, 2012 _______________________________________________________________________ Two-degree-of-freedom System Consider the system in Figure 1.
Figure 1.
A free-body diagram of mass 1 is given in Figure 2. A free-body diagram of mass 2 is
given in Figure 3.
k1
k2
x1
m2
m1
x2
f1(t)
2
Figure 2.
Determine the equation of motion for mass 1.
11 xmF (1)
11122111 xk)xx(k)t(fxm (2)
)t(fxk)xx(kxm 11112211 (3)
)t(fxk)xx(kxm 11112211 (4)
)t(fxkx)kk(xm 12212111 (5)
x1
k1x1
m1
f1(t)
k2 (x2-x1)
3
Figure 3.
Derive the equation of motion for mass 2.
22 xmF (6)
)xx(kxm 12222 (7)
0)xx(kxm 12222 (8)
0xkxkxm 122222 (9)
Assemble the equations in matrix form.
0
)t(f
x
x
kk
kkk
x
x
m0
0m 1
2
1
22
221
2
1
2
1
(10)
The homogenous form is
0
0
x
x
kk
kkk
x
x
m0
0m
2
1
22
221
2
1
2
1
(11)
x2
m2
f2(t) k3 x2
k2 (x2-x1)
4
Seek a solution of the form
)tjexp(qx (12)
The q vector is the generalized coordinate vector.
Note that
)tjexp(qjx (13)
)tjexp(qx 2 (14)
The generalized eigenvalue problem is
0)tjexp(q
q
m0
0m
kk
kkk
2
1
2
12
22
221
(15)
0m0
0m
kk
kkkdet
2
12
22
221
(16)
0mkk
kmkkdet
22
22
212
21
(17)
0kmkmkk 222
221
221 (18)
0mmkmkmkmkkk 214
222
122
2122
221 (19)
0kkkmkmkmmm 212212212
214 (20)
5
0kkkkmkmmm 21212212
214 (21)
The eigenvalues are the roots of the polynomial.
a2
ac4bb 22
1
(22)
a2
ac4bb 222
(23)
where
21mma (24)
21221 kkmkmb (25)
21kkc (26)
Frequency Response Function
Let
tjexpF)t(f 11 (27)
tjexpX)t(x 11 (28)
tjexpX)t(x 22 (29)
6
tjexpFtjexpXktjexpX)kk(tjexpXm 122121112 (30)
122121112 FXkX)kk(Xm (31)
122112
21 FXkXm)kk( (32)
Recall the equation for mass 2.
0xkxkxm 122222 (33)
0XkXkXm 1222222 (34)
0XkXmk 12222
2 (35)
1
22
2
22 X
mk
kX
(36)
By substitution,
11
22
2
22
112
21 FXmk
kXm)kk(
(37)
22
2112
2122
212
21 mkFXkXmkm)kk( (38)
22
2112
222
212
21 mkFXkmkm)kk( (39)
7
2
222
212
21
22
211
kmkm)kk(
mkFX
(40)
222
221
221
22
2
1
1
kmkm)kk(
mk
F
X
(41)
1
22
22
21
12
1
21
22
2
1
1
k
kmk
k
m
k
k1k
mk
F
X (42)
1
22
22
21
12
1
2
22
2
1
11
k
kmk
k
m
k
k1
mk
F
Xk (43)
1
2
2
22
1
12
1
22
22
2
1
11
k
k
k
m1
k
m
k
k1k
mk
F
Xk (44)
1
2
2
22
1
12
1
2
2
22
1
11
k
k
k
m1
k
m
k
k1
k
m1
F
Xk (45)
8
Let
1
1211 m
k (46)
2
2222 m
k (47)
By substitution,
1
22
22
2
111
2
2
22
1
11
k
k1
k
k1
1
F
Xk
(48)
1
22
2
22 X
mk
kX
(49)
1
22
22
2
111
2
2
22
1
11
k
k1
k
k1
1
k
FX (50)
9
1
22
22
2
111
2
2
22
1
2
22
2
12
k
k1
k
k1
1
k
k
mk
FX
(51)
1
22
22
2
111
2
2
22
12
2
2
2
2
12
k
k1
k
k1
1
km
k
m
k
FX
(52)
1
22
22
2
111
2
2
22
2222
112
222
k
k1
k
k1
1k/F
X (53)
1
22
22
2
111
2
2
22
2
22
112
k
k1
k
k1
1
1
k/FX
(54)
10
1
22
22
2
111
2
2
22
2
22
112
k
k1
k
k1
1
1
k/FX
(55)
1
22
22
2
111
2
112
k
k1
k
k1
k/FX
(56)
Note that for 2222 m/k ,
0X1 (57)
212 k/FX (58)
The force acting on m2 is: 1F
11
0
1
2
3
4
5
6
7
8
0 0.5 1.0 1.5 2.0 2.50.8 1.25
/ 22
X1 k
1 /
Fo
TRANSFER MAGNITUDE
Figure 4.
The curve in Figure 4 is taken from equation (50).
Let
1
2
m
m (59)
2
11
22
1
2
k
k
(60)
μ = 0.20
ω22/ω11 = 1.0
12
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
MASS RATIO
(
/
22 )
Figure 5.
The natural frequencies are shown as a function of mass ratio in Figure 5.
System with Damping
The derivation for a system with damping is given in Appendix A.
References
1. W. Thomson, Theory of Vibrations with Applications, Second Edition, Prentice-
Hall, New Jersey, 1981.
2. R. Vierck, Vibration Analysis, 2nd Edition, Harper Collins, New York, 1979.
0.111
22
13
APPENDIX A
Two-degree-of-freedom System with Damping Consider the system in Figure A-1.
Figure A-1.
A free-body diagram of mass 1 is given in Figure A-2. A free-body diagram of mass
2 is given in Figure A-3.
f1(t) c
k1
k2
x1
m2
m1
x2
14
Figure A-2.
Determine the equation of motion for mass 1.
11 xmF (A-1)
1112212111 xk)xx(kxxc)t(fxm (A-2)
)t(fxk)xx(kxxcxm 1111221211 (A-3)
)t(fxk)xx(kxxcxm 1111221211 (A-4)
)t(fxkx)kk(xcxcxm 1221212111 (A-5)
x1
k1x1
m1
f1(t)
k2 (x2-x1)
12 xxc
15
Figure A-3.
Derive the equation of motion for mass 2.
22 xmF (A-6)
)xx(kxxcxm 1221222 (A-7)
0)xx(kxxcxm 1221222 (A-8)
0xkxkxcxcxm 12221222 (A-9)
Assemble the equations in matrix form.
0
)t(f
x
x
kk
kkk
x
x
cc
cc
x
x
m0
0m 1
2
1
22
221
2
1
2
1
2
1
(A-10)
x2
m2
f2(t) k3 x2
k2 (x2-x1) 12 xxc
16
The homogenous, undamped form is
0
0
x
x
kk
kkk
x
x
m0
0m
2
1
22
221
2
1
2
1
(A-11)
Seek a solution of the form
)tjexp(qx (A-12)
The q vector is the generalized coordinate vector.
Note that
)tjexp(qjx (A-13)
)tjexp(qx 2 (A-14)
The generalized eigenvalue problem is
0)tjexp(q
q
m0
0m
kk
kkk
2
1
2
12
22
221
(A-15)
0m0
0m
kk
kkkdet
2
12
22
221
(A-16)
0mkk
kmkkdet
22
22
212
21
(A-17)
0kmkmkk 222
221
221 (A-18)
17
0mmkmkmkmkkk 214
222
122
2122
221 (A-19)
0kkkmkmkmmm 212212212
214 (A-20)
0kkkkmkmmm 21212212
214 (A-21)
The eigenvalues are the roots of the polynomial.
a2
ad4bb 22
1
(A-22)
a2
ad4bb 222
(A-23)
where
21mma (A-24)
21221 kkmkmb (A-25)
21kkd (A-26)
Frequency Response Function
Let
tjexpF)t(f 11 (A-27)
tjexpX)t(x 11 (A-28)
18
tjexpX)t(x 22 (A-29)
tjexpFtjexpXktjexpcXj
tjexpX)kk(tjexpcXjtjexpXm
1222
1211112
(A-30)
12221211112 FXkcXjX)kk(cXjXm (A-31)
122112
21 FXcjkXcjm)kk( (A-32)
Recall the equation for mass 2.
0xkxkxcxcxm 12221222 (A-33)
0XkXjcXkXjcXm 122222222 (A-34)
0XjckXjcmk 12222
2 (A-35)
1
22
2
22 X
jcmk
jckX
(A-36)
19
By substitution,
11
22
2
2211
221 FX
jcmk
jckkXjcm)kk(
(A-37)
jcmkF
XjckkXjcmkjcm)kk(
22
21
122122
212
21
(A-38)
jcmkF
Xjckkjcmkjcm)kk(
22
21
12222
212
21
(A-39)
jckkjcmkjcm)kk(
jcmkFX
2222
212
21
22
211
(A-40)
jckkjcmkjcm)kk(
jcmk
F
X
2222
212
21
22
2
1
1 (A-41)
11
222
22
11
12
1
21
22
2
1
1
k
cj
k
kkjcmk
k
cj
k
m
k
k1k
jcmk
F
X
(A-42)
20
11
222
22
11
12
1
2
22
2
1
11
k
cj
k
kkjcmk
k
cj
k
m
k
k1
jcmk
F
Xk
(A-43)
11
22
22
22
11
12
1
22
22
2
1
11
k
cj
k
kk
k
cj
k
m1
k
cj
k
m
k
k1k
jcmk
F
Xk
(A-44)
11
2
22
22
11
12
1
2
22
22
1
11
k
cj
k
k
k
cj
k
m1
k
cj
k
m
k
k1
k
cj
k
m1
F
Xk
(A-45)
Let
1
1211 m
k (A-46)
2
2222 m
k (A-47)
11
2
22
22
2
12
11
2
1
2
22
22
2
1
11
k
cj
k
k
k
cj1
k
cj
k
k1
k
cj1
F
Xk
(A-48)
21
Let
1
2
m
m (A-49)
2
11
22
1
2
k
k
(A-50)
1
2
11
22
22
22
2
12
11
22
11
22
22
22
2
1
11
k
cj
k
cj1
k
cj1
k
cj1
F
Xk
(A-51)
22
11
2
11
22
22
222
22
2
22
112
11
22
11
22
22
222
22
2
1
11
m
cj
m
cj1
m
cj1
m
cj1
F
Xk
(A-52)
Let
222
2m
c (A-53)
22
(A-54)
1111
222
11
22
222
22
2
1111
222
11
22
11
22
222
22
2
1
11
2j2j12j1
2j1
F
Xk
23
0
1
2
3
4
5
6
7
8
9
10
0 0.5 1.0 1.5 2.0 2.50.8 1.25
/ 22
X1 k
1 /
Fo
TRANSFER MAGNITUDE
Figure A-4.
Equation (A-54) is plotted in Figure A-4 for a particular case.
μ = 0.20
ω22/ω11 = 1.0
= 0.05