1

THE TIME-DOMAIN RESPONSE OF A SINGLE-DEGREE-OF-FREEDOM

SYSTEM SUBJECTED TO AN IMPULSE FORCE

Revision C

By Tom Irvine

Email: tom@vibrationdata.com

December 19, 2013

Consider a single-degree-of-freedom system.

Where

m = mass

c = viscous damping coefficient

k = stiffness

y = displacement of the mass

f(t) = applied force

Note that the double-dot denotes acceleration.

The free-body diagram is

m

k y yc

y

m

k c

y

f(t)

f(t)

2

Summation of forces in the vertical direction

ymF (1)

)t(fkyycym (2)

)t(fkyycym (3)

Divide through by m,

)t(fm

1y

m

ky

m

cy

(4)

By convention,

n2)m/c( (5)

2

n)m/k( (6)

where

n is the natural frequency in (radians/sec)

is the damping ratio

By substitution,

)t(fm

1y

2nyn2y (7)

Now apply an impulse force at t=0 via a pair of step functions.

)t(u)t(uF)t(f (8)

where is a very small time step.

3

The governing equation becomes

)t(u)t(uFm

1yy2y

2nn (9)

Now consider that the system undergoes oscillation with < 1.

Take the Laplace transform of each side.

)t(u)t(uFm

1Lyy2yL

2nn (10)

)sexp(s

1

s

1

m

F)s(Y

)0(y2)s(sY2

)0(y)0(sy)s(Ys

2n

nn

2

(11)

)sexp(

s

1

s

1

m

F)0(y)0(y2s)s(Ys2s n

2nn

2 (12)

)0(y)0(y2s)sexp(s

1

s

1

m

F)s(Ys2s n

2nn

2

(13)

2n

2n

2n

2nn

2 ss2s (14)

22n

2n

2nn

2 1ss2s (15)

4

Let

2nd 1 (16)

Substitute equation (16) into (15).

2d

2n

2nn

2 ss2s (17)

)0(y)0(y2s)sexp(s

1

s

1

m

F)s(Ys n

2d

2n

(18)

)0(y

s

1)0(y

s

2s

s

)sexp(

sm

F

s

1

sm

F

)s(Y

2d

2n

2d

2n

n

2d

2n

2d

2n

(19)

Divide the right-hand-side of equation (19) into three parts.

)s(Y)s(Y)s(Y)s(Y 210 (20)

where

2d

2n

0s

1

sm

F)s(Y (21)

5

2d

2n

1s

)sexp(

sm

F)s(Y (22)

)0(y

2d

2ns

1)0(y

2d

2ns

n2s)s(2Y

(23)

Consider Y0(s) from equation (21).

2d

2n

0s

1

sm

F)s(Y (24)

Expand into partial fractions using Reference 1.

2d

2n

d

d

n2n

2d

2n

n2n

2n

2nn

2

s m

F

s

s

m

F

s

1

m

F

s2s

1

sm

F

(25)

6

The inverse Laplace transform per Reference 1 is

0t,tsintcostexpm

F)t(u

m

F)t(y d

d

ndn2

n2n

0

(26)

0t,tsintcostexp)t(um

F)t(y d

d

ndn2

n

0

(27)

Now consider the term

2d

2n

1s

)sexp(

sm

F)s(Y (28)

Expand into partial fractions using Reference 1.

)sexp(

s m

F

)sexp(s

s

m

F

)sexp(s

1

m

F

s2s

)sexp(

sm

F

2d

2n

d

d

n2n

2d

2n

n2n

2n

2nn

2

(29)

7

Take the inverse Laplace transform from Appendices A, B & C.

tfor

tsintexpm

F

tcostexpm

Ftu

m

F)t(y

dnd

n2n

dn2n

2n

1

(30)

tfor

tsintcostexptum

F)t(y d

d

ndn2

n

1

(31)

The inverse Laplace transform from Reference 2 for the natural response is

tsintexp)0(y)0(y

tcostexp)0(y)t(y dnd

ndn2

(32)

8

tdsintnexpd

1)0(y

tdsind

ntdcostnexp)0(y)t(2y

(33)

The final displacement solution is

)t(y)t(y)t(y)t(y 210 (34)

t0

,tsintcostexp)t(um

F

tsintexp1

)0(y

tsintcostexp)0(yty

dd

ndn2

n

dnd

dd

ndn

(35)

9

t

,tsintcostexptum

F

tsintcostexp)t(um

F

tsintexp1

)0(y

tsintcostexp)0(yty

dd

ndn2

n

dd

ndn2

n

dnd

dd

ndn

(36)

The response for a Dirac Delta impulse is derived in Appendix D.

References

1. T. Irvine, Partial Fractions in Shock and Vibration Analysis, Revision G,

Vibrationdata, 2011.

2. T. Irvine, Table of Laplace Transforms, Revision I, Vibrationdata, 2011.

3. T. Irvine, Free Vibration of a Single-Degree-of-Freedom System, Revision B,

Vibrationdata, 2005.

10

APPENDIX A

tum

F)sexp(

s

1

m

FL

2n

2n

1 (A-1)

APPENDIX B

Consider the generic term

)sexp(

s

s)s(F

22

(B-1)

Shift in the s-plane.

ss (B-2)

sexp

s

s)s(F

22 (B-3)

expsexp

s

s)s(F

22 (B-4)

expsexp

s expsexp

s

s)s(F

2222 (B-5)

11

t

,tsintexpexptcostexpexp)t(f

(B-6)

t

tsintcostexpexp)t(f

(B-7)

f(t) = 0 t < ρ (B-8)

Note that the texp term accounts for the phase shift.

Recall the term

)sexp(

s

s

m

F)s(Y

2d

2n

n2n

12

(B-9)

Compare the term in equation (B-9) to the generic term from (B-1).

)sexp(

s

s)s(F

22

(B-10)

12

The inverse Laplace transform of equation (B-9) is

tcostexpexp

m

F)t(y

dnn2n

12 (B-11)

tcostexp

m

F)t(y

dn2n

12 (B-12)

13

APPENDIX C

Consider the generic term

)sexp(

s )s(F

22

(C-1)

Shift in the s-plane.

ss (C-2)

sexp

s )s(F

22 (C-3)

expsexp

s )s(F

22 (C-4)

t

tsintexpexp)t(f

(C-5)

t

tsintexp)t(f

(C-6)

14

f(t) = 0 t < ρ (C-7)

Note that the texp term accounts for the phase shift.

)sexp(

s m

F)s(Y

2d

2n

d

d

n2n

13

(C-8)

Compare with the generic term

)sexp(

s )s(F

22

(C-9)

expsexp

s )s(F

22 (C-10)

t

tsintexpexp)t(f

(C-11)

t

tsintexpm

F)t(y dn

d

n2n

13

(C-12)

15

APPENDIX D

Dirac Delta Impulse

)t(fm

1y

2nyn2y (D-1)

Let I = total impulse

)t(I)t(f (D-2)

.

)t(m

Iyy2y

2nn (D-3)

m

I)s(Y

)0(y2)s(sY2

)0(y)0(sy)s(Ys

2n

nn

2

(D-4)

m

I)0(y)0(y2s)s(Ys2s n

2nn

2 (D-5)

)0(y)0(y2sm

I)s(Ys n

2d

2n (D-6)

16

)0(y

s

1)0(y

s

2s

s

1

m

I

)s(Y

2d

2n

2d

2n

n

2d

2n

(D-7)

The displacement is

tsintexp1

m

I

tsintexp1

)0(y

tsintcostexp)0(yty

dnd

dnd

dd

ndn

(D-8)

The displacement for zero initial conditions is

tsintexp1

m

Ity dn

d

(D-9)

The impulse response function is

tsintnexpm

1=)t(h d

dd

(D-10)

17

The corresponding Laplace transform is

2d

22n

2d

ns

1

sn2s

1=)s(H

m

1

m

1

(D-11)

tcostexpm

I

tsintexpm

I

tcostexp)0(y

tsintexp)0(y

tcostsintexp)0(y

tsintcostexp)0(yty

dn

dnd

n

dn

dnd

n

dnddn

dd

ndnn

(D-12)

tsintcostexpm

I

tsintcostexp)0(y

tsintexp)0(yty

dd

ndn

dd

ndn

ddd

2n

2

n

(D-13)

18

tsintcostexpm

I

tsintcostexp)0(y

tsintexp)0(yty

dd

ndn

dd

ndn

d2d

2d

2n

2

n

(D-14)

The velocity for zero initial conditions is

tsintcostexp

m

Ity d

d

ndn

(D-15)

The impulse response function is

tsintcos)tnexp(-=)t(h d

d

ndv

m

1

(D-16)

The corresponding Laplace transform is

2d

22n

2v

ns

s

sn2s

s=)s(H

m

1

m

1

(D-17)

19

The acceleration for zero initial conditions is

)t(y)t(y2)t(m

Ity 2

nn

(D-18)

tsintexp1

tsintcostexp2)t(m

I

ty

dnd

2nd

d

ndnn

(D-19)

tsin

2tcos2texp)t(

m

Ity d

d

2n

2n

dnn

(D-20)

tsin12tcos2texp)t(

m

Ity d

2

d

2n

dnn

(D-21)

The Laplace transform is

2n

2

22n

2n

2n

asn2s

122s21

m

IsY

(D-22)

2n

2

2nn

asn2s

s21

m

IsY

(D-23)

20

2n

2

2nn

2n

2

2n

2

asn2s

s2

sn2s

sn2s

m

IsY

(D-24)

2n

2

2

asn2s

s

m

IsY

(D-25)

The corresponding transfer function in terms of the Laplace transform is

2d

2

2

2n

2

2

a

ns

s

sn2s

s=)s(H

m

1

m

1 (D-26)

21

APPENDIX E

Dirac Delta Impulse, Transmitted Force

Again, the impulse response function for displacement is

tsintnexpm

1=)t(h d

dd

(E-1)

The impulse response function for velocity is

tsintcos)tnexp(-=)t(h d

d

ndv

m

1

(E-2)

The transmitted force f t (t) is

kyyc (t) f t (E-3)

y

m

ky

m

cm (t) f t (E-4)

yy2m (t) f

2nnt (E-5)

22

The impulse response function for transmitted force is

tsintnexp

tsintcos)tnexp(-2=)t(h

dd

2n

dd

ndnft

(E-6)

tsintnexp

tsin2

tcos2)tnexp(-=)t(h

dd

2n

dd

2n

2

dnft

(E-7)

tsin

2tcos2)tnexp(-=)t(h d

d

2n

2

d

2n

dnft

(E-8)

The final transmitted force impulse response function is

tsin

2tcos2)tnexp(-=)t(h d

d

22

ndnft1

(E-9)

Note that the transmitted force impulse response function is essentially the same as the

absolute acceleration impulse response function for the case of base excitation.