1

VIBRATION ABSORBER

Revision E

By Tom Irvine Email: tomirvine@aol.com 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