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THE VIBRATION RESPONSE OF SOME SPRING-DAMPER SYSTEMS

WITH AND WITHOUT MASSES

By Tom IrvineEmail: tomirvine@aol.com

April 26, 2012

Introduction

Consider the single-degree-of-freedom system subjected to an applied force in Figure 1.

Figure 1.

where

Summation of forces in the vertical direction yields the equation of motion.

)t(Fkxxc (1)

Derive the transfer function by taking the Laplace transform.

)t(FLkxxcL (2)

F(t) = Applied force

c = viscous damping coefficient

k = stiffness

x = displacement

k c

)t(F

x

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)s(F)s(Xk)0(xc)s(Xsc (3)

)0(xc)s(F)s(Xksc (4)

ksc

)0(xc)s(F)s(X

(5)

Now assume that the initial displacement is zero.

ksc

)s(F

)s(X (6)

ksc

1

)s(F

)s(X

(7)

The dynamic stiffness in the Laplace domain is

ksc)s(X

)s(F

(7)

The dynamic stiffness in the frequency domain is

cjk

)(X

)(F (8)

The initial value problem for the free response is

ksc

)0(xc)s(X

(9)

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c/ks)0(x

)s(X

(10)

The inverse Laplace transform yields the time domain response.

tc/kexp)0(x)t(x (11)

APPENDIX A

Consider the single-degree-of-freedom system subjected to an applied force.

Figure A-1.

The equation of motion is

)t(Fkxxcxm (A-1)

Derive the transfer function by taking the Laplace transform.

)t(FLkxxcxmL (A-2)

)s(F)s(Xk)0(xc)s(Xsc)0(msx)0(xm)s(Xms2 (A-3)

k c

)t(F

xm

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)0(msx)0(xm)0(xc)s(F)s(Xkscms2 (A-4)

kscms

)0(msx)0(xm)0(xc)s(F)s(X

2 (A-5)

Now assume that the initial displacement is zero.

kscms

)s(F)s(X

2 (A-6)

kscms

1

)s(F

)s(X

2 (A-7)

The dynamic stiffness in the Laplace domain is

kscms)s(X

)s(F 2 (A-8)

The dynamic stiffness in the frequency domain is

cjmk

)(X

)(F 2 (A-9)

The mechanical impedance in the frequency domain is

j

cjmk

)(V

)(F 2 (A-10)

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m

kjc

)(V

)(F (A-11)

The apparent mass in the frequency domain is

j

mk

jc

)(A

)(F (A-12)

c

j

k

m)(A

)(F

2

(A-13)

m

cj

m

k1m

)(A

)(F

2 (A-14)

Note that

m

kn (A-15)

m2c n (A-16)

2 (A-17)

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By substitution,

n2

2n 2j1m

)(A

)(F (A-18)

n2

2n j1m

)(A

)(F (A-19)

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APPENDIX B

Consider the following two-degree-of-freedom system subjected to an applied force.

Figure B-1.

The free-body diagrams are

Figure B-2.

k

c

)t(F

x1

x2

k(x2-x1)

k(x1-x2)

)t(F

x2

1xc

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The equations of motion are

0)t(F)x-k(x 21 (B-1)

)t(F)x-k(x 12 (B-2)

0)x-k(xxc 211 (B-3)

Solve for 2x using Equation (B-2).

12 xk

)t(Fx (B-4)

By substitution,

0)x-k(xxc 211 (B-5)

0xk

)t(F

-xkxc 111

(B-6)

0k

)t(F-kxc 1

(B-7)

)t(Fxc 1 (B-8)

Take the Laplace transform.

)t(FLxcL 1 (B-9)

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)s(F)0(xc)s(Xsc 11 (B-10)

)0(xc)s(F)s(Xsc 11 (B-11)

sc

)0(xc)s(F)s(X 11

(B-12)

Recall

)t(F)x-k(x 12 (B-13)

Take the Laplace transform.

)t(FL)x-k(xL 12 (B-14)

)s(F)s(Xk)s(Xk 12 (B-15)

)s(Xk)s(F)s(Xk 12 (B-16)

sc

)0(xc)s(Fk)s(F)s(Xk 12 (B-17)

)0(xs

k

sc

k1)s(F)s(Xk 12

(B-18)

)0(xs

1

sc

1

k

1)s(F)s(X 12

(B-19)

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Now assume the initial displacement is zero.

sc

1

k

1)s(F)s(X2 (B-20)

skc

kcs)s(F)s(X2 (B-21)

The receptance in the Laplace domain is

skc

kcs

)s(F

)s(X2

(B-22)

The dynamic stiffness in the Laplace domain is

kcs

skc

)s(X

)s(F

2 (B-22)

The dynamic stiffness in the frequency domain is

cjk

kcj

)(X

)(F

2

(B-22)

The initial value problem for the free response is

)0(xs

1)s(X 12 (B-23)

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The inverse Laplace transform yields the time domain response in terms of the unit step

function u(t).

)t(u)0(x)t(x 12 (B-24)

This can be simplified as

)0(x)t(x 12 (B-25)

Likewise

)0(x)t(x 11 (B-26)

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APPENDIX C

Consider the following two-degree-of-freedom system subjected to an applied force.

Figure C-1.

Figure C-2.

k1(x2-x1)

k1(x1-x2)

)t(F

x2

)t(F

k1

c

x1

x2

k2

-k2x2

1xc

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The equations of motion are

0)t(Fxk)x-(xk 22211 (C-1)

)t(Fxkxkk 11221 (C-2)

0)x-(xkxc 2111 (C-3)

Solve for 2x using Equation (C-2).

121

1

212 x

kkk

kk)t(Fx

(C-4)

By substitution,

0)x-(xkxc 2111 (C-5)

0xkk

kkk)t(F-xkxc 1

21

1

21111

(C-6)

0)t(Fkk

k-x

kk

k1kxc

21

11

21

111

(C-7)

0)t(Fkk

k-xkk

kkkkxc21

11

21

12111

(C-8)

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21

21

1

21

1

21

2121

1

212

kk

kksc

)0(xc

kk

k)s(F

kk

kksc

1

kk

k

kk

1)s(X

(C-21)

21

21

1

21

1

21

21

1

212

kk

kk

sc

)0(xc

kk

k)s(F

kk

kk

sc

k1

kk

1)s(X

(C-22)

Now assume the initial displacement is zero.

)s(F

kk

kksc

k1

kk

1)s(X

21

21

1

212

(C-23)

The receptance in the Laplace domain is

21

21

1

21

2

kk

kksc

k1

kk

1

)s(F

)s(X (C-24)

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2121

1

21

2

kkskkc

k

kk

1

)s(F

)s(X (C-25)

21212

21

21121212

kkkkskkc

kkkkkskkc

)s(F

)s(X

(C-26)

21212

21

211212

kkkkskkc

k2kkskkc

)s(F

)s(X

(C-27)

The dynamic stiffness in the Laplace domain is

21121

21212

21

2 k2kkskkc

kkkkskkc

)s(X

)s(F

(C-28)

The initial value problem for the free response is

21

21

1

21

12

kk

kksc

)0(xc

kk

k)s(X (C-29)

21

21

1

21

12

kk

kk

c

1s

)0(x

kk

k)s(X (C-30)

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The inverse Laplace transform yields the time domain response.

t

kk

kk

c

1exp

kk

k)0(x)t(x

21

21

21

112 (C-31)

0xkxkk 11221 (C-32)

22111 xkkxk (C-33)

21

211 x

k

kkx

(C-34)

t

kk

kk

c

1exp)0(x)t(x

21

2111 (C-35)

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APPENDIX D

Consider the following two-degree-of-freedom system subjected to an applied force.

Figure D-1.

Figure D-2.

k1(x2-x1)

1xc

k1

c

x1

x2

k2

F t

m

k1(x1-x2)

x2

-k2x2

m

F(t)

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The equations of motion are

222112 xk)x-(xk)t(Fxm (D-1)

)t(Fxkxkkxm 112212 (D-2)

0)x-(xkxc 2111 (D-3)

Take the Laplace transform of equation (D-2).

)t(FLxkxkkxmL 112212 (D-4)

)s(F)s(Xk)s(Xkk)0(xsm)0(xm)s(Xsm 112212222 (D-5)

)0(xsm)0(xm)s(F)s(Xk)s(Xkksm 22112212 (D-6)

Take the Laplace transform of equation (D-3).

0)x-(xkxcL 2111 (D-7)

0)s(Xk)s(Xk)0(xc)s(Xsc 221111 (D-8)

)0(xc)s(Xk)s(Xksc 12211 (D-9)

Consider equations (D-6) and (D-9) for the case of zero initial conditions.

)s(F)s(Xk)s(Xkksm 112212 (D-10)

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0)s(Xk)s(Xksc 2211 (D-11)

)s(Xk)s(Xksc 2211 (D-12)

)s(Xksc

k)s(X 2

1

21

(D-13)

)s(F)s(Xksc

kk)s(Xkksm 2

1

21221

2

(D-14)

)s(F)s(Xksc kkkksm 2121

212

(D-15)

1

2121

22

ksc