m

k c

y

f(t)

THE TIME-DOMAIN RESPONSE OF A SINGLE-DEGREE-OF-FREEDOM SYSTEM SUBJECTED TO AN IMPULSE FORCE

Revision C

By Tom IrvineEmail: tom@vibrationdata.com

December 19, 2013

Consider a single-degree-of-freedom system.

Where

m = massc = viscous damping coefficientk = stiffnessy = displacement of the mass

f(t) = applied force

Note that the double-dot denotes acceleration.

The free-body diagram is

m

k y c y

y

f(t)

1

Summation of forces in the vertical direction

∑ F = m y (1)

m y=−c y−ky+ f ( t ) (2)

m y+c y+ky= f ( t ) (3)

Divide through by m,

y +( cm ) y+( km ) y=( 1m ) f ( t )

(4)

By convention,

( c/m)=2ξωn (5)

(k /m )=ωn2 (6)

where

n is the natural frequency in (radians/sec) is the damping ratio

By substitution,

y +2 ξ ωn y+ωn 2 y= 1

mf ( t )

(7)

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

f ( t )=F [u ( t )−u( t−ε )] (8)

where ε is a very small time step.

2

The governing equation becomes

y +2 ξ ωn y+ωn 2 y= 1

mF [u( t )−u ( t−ε )]

(9)

Now consider that the system undergoes oscillation with ξ < 1.

Take the Laplace transform of each side.

L { y +2 ξ ωn y+ωn 2 y }=L { 1

mF [u ( t )−u( t−ε )]}

(10)

s2Y ( s )−sy (0 )− y ' (0)+2 ξ ωnsY (s )−2 ξ ωn y (0 )

+ωn 2Y ( s )= F

m [1s −1s

exp(−ε s )]

(11)

{s2+2 ξ ωn s+ωn 2}Y ( s )−{s+2 ξ ωn} y (0 )− y ' (0 )= Fm [1s−1

sexp (−ε s )]

(12)

{s2+2 ξ ωn s+ωn 2}Y ( s )= Fm [1s −1

sexp (−ε s )]+ {s+2 ξ ωn } y (0)+ y ' (0 )

(13)

s2+2 ξ ωn s+ωn 2=(s+ξ ωn)

2−(ξ ωn )2+ωn 2 (14)

s2+2 ξ ωn s+ωn 2=(s+ξ ωn)2+ωn 2 (1−ξ2 ) (15)

3

Let

ωd=ωn√1−ξ2 (16)

Substitute equation (16) into (15).

s2+2 ξ ωn s+ωn 2=(s+ξ ωn)

2+ωd2 (17)

{ (s+ξ ωn)2+ωd2}Y (s )= Fm [1s −1

sexp(−ε s )]+{s+2 ξ ωn} y (0 )+ y '(0 )

(18)

Y ( s )=

Fm s {1(s+ξ ωn)2+ωd2 }−Fm s {exp (−ε s )

(s+ξ ωn )2+ωd2 }+{s+2 ξ ωn

(s+ξ ωn )2+ωd2 }y (0 )+{1(s+ξ ωn )2+ωd2 }y ' (0 )

(19)

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

Y ( s )= Y 0 (s )+Y 1( s )+Y 2( s ) (20)

where

Y 0 (s )=Fm s {1 (s+ξ ωn )2+ωd2 }

(21)

4

Y 1( s )=− Fm s {exp(−ε s )

(s+ξ ωn )2+ωd2 } (22)

Y 2 (s )= {s+2 ξ ωn(s+ξ ωn)2+ωd2 } y (0)+{1(s+ξ ωn)2+ωd2 } y ' (0)

(23)

Consider Y0(s) from equation (21).

Y 0 (s )=Fm s {1 (s+ξ ωn )2+ωd2 }

(24)

Expand into partial fractions using Reference 1.

{Fm s }{1 s2+2 ξ ωns+ωn 2 } =

Fm ωn

2 {1s }−(Fm ωn 2 ){s+ξωn (s+ξ ωn)2+ωd 2 }−(Fm ωn 2 )(ξωn

ωd ){ωd (s+ξ ωn )2+ωd 2 } (25)

5

The inverse Laplace transform per Reference 1 is

y0 ( t )=F

m ωn 2 u( t )−

Fm ωn

2 exp (−ξ ωn t ) [cos (ωd t )+ξωn

ωdsin (ωd t )] , t≥0

(26)

y0( t )=F

m ωn 2 {u ( t )−exp (−ξ ωn t ) [cos (ωd t )+

ξωn

ωdsin (ωd t )]} , t≥0

(27)

Now consider the term

Y 1( s )=− Fm s {exp(−ε s )

(s+ξ ωn )2+ωd2 } (28)

Expand into partial fractions using Reference 1.

{Fm s }{exp(−ε s ) s2+2 ξ ωns+ωn

2 } =−Fm ωn

2 {1s }exp(−ε s )

+(Fm ωn 2 ){s+ξωn (s+ξ ωn )2+ωd 2 }exp (−ε s )

+(Fm ωn 2 )(ξωn

ωd ){ωd (s+ξ ωn)2+ωd 2 }exp(−ε s )

(29)

6

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

y1( t )= −Fm ωn

2 u (t−ε )+ (Fm ωn 2 ) exp [−ξ ωn (t−ε ) ] cos [ ωd (t−ε ) ]

+(Fm ωn 2 )(ξωn

ωd ) exp [ −ξ ωn (t−ε ) ] sin [ ωd ( t−ε ) ]

for t≥ε(30)

y1( t )=Fm ωn

2 {−u (t−ε )+exp [−ξ ωn ( t−ε ) ] {cos [ ωd ( t−ε ) ]+(ξωn

ωd ) sin [ ωd (t−ε ) ]}}for t≥ε

(31)

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

y2 (t )=[ y (0 )]exp (−ξωn t )cos (ωd t )+[ ξωn y (0 )+ y ' (0)ωd ]exp (−ξωn t )sin (ωd t )

(32)

7

y2 (t )= y (0 )exp (−ξωn t ){cos (ωd t )+[ξωnωd ]sin (ωd t )}+ y ' (0 )[1ωd ]exp (−ξωn t )sin (ωd t )

(33)

The final displacement solution is

y ( t )= y0( t )+ y1( t )+ y2 ( t ) (34)

y (t )= y (0 ) exp (−ξ ωn t ){cos (ωd t )+[ξωnωd ]sin (ωd t )}+ y ' (0 )[1ωd ] exp (−ξ ωn t )sin (ωd t )

+Fm ωn

2 {u ( t )−exp (−ξ ωn t ) [cos (ωd t )+ξωn

ωdsin (ωd t )]} ,

0≤t<ε

(35)

8

y (t )= y (0 ) exp (−ξ ωn t ){cos (ωd t )+[ξωnωd ]sin (ωd t )}+ y ' (0 )[1ωd ] exp (−ξ ωn t )sin (ωd t )

+Fm ωn

2 {u ( t )−exp (−ξ ωn t ) [cos (ωd t )+ξωn

ωdsin (ωd t )]}

+Fm ωn

2 {−u ( t−ε )+exp [−ξ ωn ( t−ε ) ] {cos [ ωd ( t−ε ) ]+ξωn

ωdsin [ ωd (t−ε ) ]}} ,

t≥ε

(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.

9

APPENDIX A

L−1{ Fm ωn

21s exp (−ε s )}= F

m ωn 2 u (t−ε )

(A-1)

APPENDIX B

Consider the generic term

F (s )={ s+ λ (s+α )2+β2 }exp(− ρ s )

(B-1)

Shift in the s-plane.

s=s+α (B-2)

F (s )={ s−α+λ { s2

+β2¿}exp [−ρ ( s−α ) ]

(B-3)

F (s )={ s−α+λ { s2

+β2¿}exp [−ρ s ]exp [ ρ α ]

(B-4)

F (s )={ s { s2

+β2¿}exp [−ρ s ] exp [ ρ α ]+{−α+λ

{ s2 + β2¿}exp [−ρ s ] exp [ ρ α ]

(B-5)

10

f ( t )=exp (ρ α ) exp (−α t ) cos [ β (t− ρ ) ] + [−α+λβ ] exp ( ρ α ) exp (−α t ) sin [ β ( t−ρ ) ] ,

t>ρ

(B-6)

f ( t )=exp ( ρ α ) exp (−α t ) {cos [ β ( t−ρ ) ] + [−α+λβ ]sin [ β ( t−ρ ) ]}t>ρ (B-7)

f(t) = 0 t < ρ (B-8)

Note that the exp (−α t ) term accounts for the phase shift.

Recall the term

Y 12(s )=( Fm ωn

2 ){ s+ξωn

(s+ξ ωn)2+ωd 2 }exp(−ε s ) (B-9)

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

F (s )={ s+ λ (s+α )2+β2 }exp(− ρ s )

(B-10)

11

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

y12( t )= ( Fm ωn

2 )exp (ε ξ ωn ) exp (−ξ ωn t ) cos [ ωd (t−ε ) ] (B-11)

y12( t )= ( Fm ωn

2 ) exp [−ξ ωn (t−ε ) ] cos [ ωd (t−ε ) ] (B-12)

12

APPENDIX C

Consider the generic term

F (s )={ λ (s+α )2+β2 }exp(− ρ s )

(C-1)

Shift in the s-plane.

s=s+α (C-2)

F (s )={ λ { s2

+β2¿}exp [−ρ ( s−α ) ]

(C-3)

F (s )={ λs2+β2 }exp [−ρ s ] exp [ ρ α ]

(C-4)

f ( t )= [λβ ] exp (ρ α )exp (−α t ) sin [ β (t−ρ ) ]

t>ρ(C-5)

f ( t )= [λβ ] exp [ −α ( t− ρ ) ] sin [ β (t−ρ ) ]

t>ρ

13

(C-6)

f(t) = 0 t < ρ (C-7)

Note that the exp (−α t ) term accounts for the phase shift.

Y 13(s )=( Fm ωn

2)( ξωn

ωd ){ ωd (s+ξ ωn )2+ωd 2 }exp(−ε s)

(C-8)

Compare with the generic term

F (s )={ λ (s+α )2+β2 }exp(−ρ s )

(C-9)

F (s )={ λs2+β2 }exp [−ρ s ] exp [ ρ α ]

(C-10)

f ( t )= [λβ ] exp (ρ α )exp (−α t ) sin [ β (t−ρ ) ]

t>ρ (C-11)

y13( t )= −(Fm ωn 2 )(ξωn

ωd ) exp [ −ξ ωn (t−ε ) ] sin [ ωd (t−ε ) ]

t>ε (C-12)

14

APPENDIX D

Dirac Delta Impulse

y +2 ξ ωn y+ωn 2 y= 1

mf ( t )

(D-1)

Let I = total impulse

f ( t )= I δ ( t ) (D-2)

.

y +2 ξ ωn y+ωn 2 y= I

mδ( t )

(D-3)

s2Y ( s )−sy (0 )− y ' (0)+2 ξ ωnsY (s )−2 ξ ωn y (0 )

+ωn 2Y ( s )= I

m

(D-4)

{s2+2 ξ ωns+ωn 2}Y ( s )− {s+2 ξ ωn } y ( 0)− y '(0 )= I

m (D-5)

{ (s+ξ ωn)2+ωd2}Y (s )= Im

+ {s+2 ξ ωn } y (0 )+ y' (0 ) (D-6)

15

Y ( s )=

Im {1(s+ξ ωn)2+ωd2 }+{s+2 ξ ωn

(s+ξ ωn )2+ωd2 }y (0 )+{1(s+ξ ωn )2+ωd2 }y ' (0 )

(D-7)

The displacement is

y ( t )= y (0 ) exp (−ξ ωn t ){cos (ωd t )+[ξωnωd ]sin (ωd t )}+ y ' (0 )[1ωd ] exp (−ξ ωn t )sin (ωd t )

+Im [1ωd ] exp (−ξ ωn t )sin (ωd t )

(D-8)

The displacement for zero initial conditions is

y (t )= Im [ 1ωd ] exp (−ξ ωn t ) sin (ωd t )

(D-9)

The impulse response function is

16

hd ( t )= 1mωd

[exp (−ξωn t )] [sin ωd t ]

(D-10)

The corresponding Laplace transform is

Hd ( s )= 1m [ 1

s2+2ξωn s+ωn2 ]= 1m [ 1

( s+ξωn )2+ωd2 ]

(D-11)

y (t )=−ξ ωn y (0) exp (−ξ ωn t ){cos (ωd t )+[ξωnωd ]sin (ωd t )}+ y (0) exp (−ξ ωn t ) {−ωd sin (ωd t )+ξωncos (ωd t )}

− y ' (0 )[ξ ωnωd ] exp (−ξ ωn t )sin (ωd t )

+ y '( 0) exp (−ξ ωn t )cos (ωd t )

−Im [ξ ωnωd ] exp (−ξ ωn t ) sin (ωd t )

+Im exp (−ξ ωn t )cos (ωd t )

(D-12)

17

y (t )= − y (0) exp (−ξ ωn t )[ξ2ωn2

ωd+ωd ]sin (ωd t )

+ y '( 0) exp (−ξ ωn t )[cos (ωd t )−ξ ωnωd

sin (ωd t )]+ Im

exp (−ξ ωn t )[cos (ωd t )−ξ ωnωd

sin (ωd t )](D-13)

y (t )= − y (0) exp (−ξ ωn t )[ξ2ωn2+ω

d2

ωd2 ]sin (ωd t )

+ y '( 0) exp (−ξ ωn t )[cos (ωd t )−ξ ωnωd

sin (ωd t )]+ Im

exp (−ξ ωn t )[cos (ωd t )−ξ ωnωd

sin (ωd t )](D-14)

The velocity for zero initial conditions is

y (t )= Im

exp (−ξ ωn t ) [cos (ωd t )−ξ ωnωd

sin (ωd t )]

(D-15)

The impulse response function is

hv( t )= 1m

exp (−ξωn t ) [cos (ωd t )−ξ ωnωd

sin (ωd t )]

(D-16)

18

The corresponding Laplace transform is

H v(s )= 1m [ s

s2+2ξωn s+ωn2 ]=1m [ s

(s+ξωn)2+ωd2 ]

(D-17)

19

The acceleration for zero initial conditions is

y (t )= Imδ ( t ) −2ξωn y ( t )−ωn2 y ( t )

(D-18)

y ( t )=

Im { δ ( t )−2ξωnexp (−ξ ωn t )[cos (ωd t )−

ξ ωnωd

sin (ωd t ) ]−ωn2[1ωd ] exp (−ξ ωn t )sin (ωd t ) }

(D-19)

y (t )= Im { δ ( t )+exp (−ξ ωn t )[−2 ξωn cos (ωd t )+

(2 ξ ωn )2−ωn2

ωdsin (ωd t )] }

(D-20)

y (t )= Im { δ( t )+exp (−ξ ωn t )[−2ξωn cos (ωd t )+

ωn2

ωd[ (2ξ )2−1 ]sin (ωd t ) ] }

(D-21)

The Laplace transform is

20

Y a (s )= Im {1+

−2ξωn s−2ξ2ωn2+ωn2 [ (2ξ )2−1 ]

s2+2ξωn s+ωn2 }

(D-22)

Y a (s )= Im {1+

−2ξωn s−ωn2

s2+2ξωn s+ωn2

}

(D-23)

Y a (s )= Im {s

2+2 ξωn s+ωn2

s2+2 ξωn s+ωn2

+−2ξωns−ωn2

s2+2ξωn s+ωn2}

(D-24)

Y a (s )= Im { s2

s2+2ξωn s+ωn2}

(D-25)

The corresponding transfer function in terms of the Laplace transform is

Ha (s )= 1m [ s2

s2+2ξωn s+ωn2 ]= 1

m [ s2

(s+ξωn )2+ωd2 ]

(D-26)

21

APPENDIX E

Dirac Delta Impulse, Transmitted Force

Again, the impulse response function for displacement is

hd ( t )= 1mωd

[exp (−ξωn t )] [sin ωd t ]

(E-1)

The impulse response function for velocity is

hv( t )= 1m

exp (−ξωn t ) [cos (ωd t )−ξ ωnωd

sin (ωd t )]

(E-2)

The transmitted force f t (t) is

f t ( t )= c y+ky (E-3)

f t ( t )=m [ cm y+ km y ] (E-4)

f t ( t )=m [2ξωn y+ωn2 y ] (E-5)

22

The impulse response function for transmitted force is

hft( t )= 2ξωnexp (−ξωnt ) [cos (ωd t )−ξ ωnωd

sin (ωd t )]+ωn2

ωd[exp (−ξωnt ) ] [sin ωd t ]

(E-6)

hft ( t )= exp(−ξωnt ) [ 2ξωn cos (ωd t )− 2ξ2 ω

n2

ωdsin (ωd t )]

+exp (−ξωnt ) [ωn2

ωdsin ωd t ]

(E-7)

h ft( t )= exp(−ξωnt ) [ 2ξωn cos (ωd t )+(ωn2

ωd− 2ξ2 ω

n2

ωd )sin (ωd t )] (E-8)

The final transmitted force impulse response function is

h ft( t )= exp(−ξωnt ) [ 2ξωn cos (ωd t )+ωn2( 1−2 ξ2 ωd )sin (ωd t )]

(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.

23