FORCED VIBRATION OF A SINGLE DEGREE OF FREEDOM
SYSTEM - GENERAL EXCITATION.
3.1 The Impulse Response:
The impulse, sometimes called delta function, is a fictitious function, formally
defined in the following way
t≠η
 δ( t − η ) = 0,
∞
.
 δ( t − η ) dt = 1
∫
− ∞
(3.1.1)
It thus follows that this function has singularity at t=η,as shown below.
∞
↑
t=η
≈
−∞
≈
≈
t
∞
We may think that the delta function is the limit value, as ε→ 0, of the box
function of unit area, shown below.
ε
t=η
≈
−∞
≈
1/ε
t
∞
An unit impact at the time t=η may be expressed mathematically by δ
(t-η).
35
We now wish to determine the impulse response h(t-η) due to an impact δ
(tη), applied on a mass-spring-damper, shown below, which is initially at rest.
This event is characterised by the following equation
mh&&+ ch& + kh = δ( t − η)
(3.1.2)
together with the initial conditions
and
h = 0, t < η
(3.1.3)
h& = 0, t < η .
(3.1.4)
δ
(t-η)
x(t)
m
k
c
Obviously (3) and (4) imply that the system has no motion for t<η. Let us now
determine the dynamic of the system for t>η. Since no force is applied after
the impact, the response of the system after the impact is governed by
mh&&+ ch& + kh = 0, t > η ,
h(η + ) = A ,
h&(η + ) = B
(3.1.5)
(3.1.6)
(3.1.7)
+
where η represents the time immediately after the impact, and where A and
B are some constants representing the displacement and velocity of the mass
+
at the time t = η , respectively. We similarly denote the time immediately
−
before the impact by t = η .
We now show how to evaluate A and B. It follows from (1) that
∞
∫δ(t −
−∞
η+
η ) dt = ∫δ( t − η ) dt = 1 .
(3.1.8)
η−
Since the spring and damper forces are non-impulsive forces, we have by the
Newton’s second law that during the impact
36
δ( t − η) = mh&&.
(3.1.9)
Substituting (9) in (8) gives
η+
&& = 1 .
∫− mhdt
(3.1.10)
η
Integrating (10) yields
+
η
mh&t =η − = mh&(η + ) − mh&(η − ) = 1 .
(3.1.11)
−
But before the impact the system is at rest, ie. h&(η ) = 0 . Hence, (11) gives
mh&(η + ) = 1 ,
(3.1.12)
or
1
h&(η + ) = B =
m
.
(3.1.13)
This important result shows that the velocity gained by a unit impact equals
the inverse of the mass (independent of the spring constant k and the damper
constant c!). In fact, this is no more than the fundamental principle of impulse
and momentum: “The change in momentum equals the impulse of the acting
force”.
Integrating (12)
η+
η+
& = dt
∫− mhdt
∫−
η
(3.1.14)
η
we have
mh
η+
t =η −
=t
η+
t =η −
,
(3.1.15)
or
mh(η + ) − mh(η − ) = η + − η − = 0 .
−
But, by (3), h(η ) = 0 . Hence (16) implies that
mh(η + ) = 0 ,
(3.1.16)
(3.1.17)
or
h(η + ) = A = 0
.
(3.1.18)
This important result shows that the mass cannot gain finite displacement by
an impulsive force.
In summary, with (13) and (18) the impulse response is governed by the
equations
37
mh&&+ ch& + kh = 0, t > η
h(η + ) = 0
1
h&(η + ) =
m
(3.1.19)
.
The time shift t → t − η applied to (19) gives
mh&&( t − η ) + ch&( t − η ) + kh( t − η ) = 0, t > 0
h( 0) = 0
1
h&( 0) =
m
(3.1.20)
which, for ζ <1, has the solution (see (1.5.6) and (1.5.9a))
0,
t <η

 − ζ ω n (t − η)
h( t − η) = e
ζ < 1 . (3.1.21)
sin(ω d ( t − η ) ), t > η

 mω d
Similar expressions can be simply obtained from (1.5.6) and (1.5.9) for the
cases where ζ ≥ 1 .
Equation (21) determines the response of an underdamped system (ζ <1) due
to an impulse at the time t=η.
38
3.2 System’s Response to General Excitation by Convolution.
We now focus our attention on the general excitation by the force f(t) of the
mass-spring-damper shown below.
f(t)
x(t)
m
k
c
For a small time interval ∆η, an arbitrary function f(t) can be regarded as an
impulse of magnitude f(η)∆η, as shown below. Hence
∆f ( t , η ) = f (η ) ∆ηδ( t − η ) .
(3.2.1)
It thus follows that an arbitrary function f(t) can be represented by the
convoluted sum
f ( t ) = ∑ f (η ) ∆ηδ( t − η ) .
(3.2.2)
f(t)
f( η)
t
η
∆η
39
By (1), the response of the system to the excitation at t=η is thus
∆x ( t , η ) = f (η ) ∆ηh( t − η ) ,
(3.2.3)
and, by (2), the response to the total function f(η) is
x ( t ) = ∑ f (η ) ∆ηh( t − η ) .
(3.2.4)
At the limit, when ∆η→ 0, the summation (4) is replaced by the convolution
integral
t
x ( t ) = ∫f (η ) h( t − η ) dη
.
(3.2.5)
0
Since η is a dummy variable of integration, we may write (5) in the following
equivalent form
t
x ( t ) = ∫f ( t − η ) h(η ) dη
.
(3.2.6)
0
Example 3.2.1.
The system shown below it initially at rest. At the time t=0 a constant force F0
is applied on the system. Determine the response of the system.
F0
x(t)
m
k
c
40
Solution.
The differential equation of motion is
mx&&+ kx = F0
(3.2.7)
and the initial condition are
x ( 0) = 0 = x&( 0) .
(3.2.8)
Obviously, for this undamped system ζ =0<1 and hence ω d=ω n. We thus have
by (3.1.21)
h ( t − η) =
1
sin(ω n (t − η )) .
mω n
(3.2.9)
The convolution integral (5) gives
t
F
x ( t ) = ∫ 0 sin(ω n (t − η )) dη
mω n
0
t
F 1
= 0
cos(ω n (t − η ))
mω n ω n
η =0
[
(3.2.10)
]
F0
cos(ω n (t − t ))− cos(ω n t )
2
mω n
F
= 0 2 [1 − cos(ω n t )] .
mω n
=
With ω n =
2
k
we have
m
x (t ) =
F0
[1 − cos(ω nt )] .
k
41
(3.2.11)
Example 3.2.2.
The system shown below is initially at rest. At the time t=0 a harmonic force
f(t)=F0sin(ωt) is applied on the system. Determine the response of the system.
f(t)=F0 sin(ω t)
x(t)
m
k
c
Solution.
The differential equation of motion is
mx&&+ kx = F0 sin(ω t ),
t>0
(3.2.12)
and the initial conditions are
x ( 0) = 0 = x&( 0) .
(3.2.13)
By (3.1.21) the impulse response is
h ( t − η) =
1
sin(ω n (t − η )) ,
mω n
(3.2.14)
and the convolution integral (5) is given by
F sin(ωη )
x (t ) = ∫ 0
sin(ω n (t − η )) dη .
ω
m
n
0
t
(3.2.15)
Using the trigonometric relation
1
sin α sin β = [cos(α − β ) − cos(α + β )]
2
we have
42
(3.2.16)
F0
x (t ) =
2mω n
t
∫{cos[(ω +
0
ω n )η − ω n t ]
.
(3.2.17)
− cos[(ω − ω n )η + ω n t ]}dη
Hence
x (t ) =
F0 sin[(ω + ω n )η − ω n t ]

2mω n 
ω + ωn
t
sin[(ω − ω n )η + ω n t ] 
−

ω − ωn
η =0
=
F0
2mω n
sin(ω t ) sin(ω t )
−

ω + ω n ω − ω n
+
which, with ω n =
x (t ) =
k
(3.2.18)
sin(ω n t ) sin(ω n t ) 
+

ω + ωn
ω − ωn 
m , simplifies to


ω 
ω
ω
−
sin(
t
)
sin(
t
)



n  .
2
ω n 
  ω  



1
k −  
 ω  
n


F0
(3.2.19)
An alternative approach. This problem may be solved alternatively as
follows. Substituting ζ =0 in equations (2.1.19)-(2.1.21) we find that the
general solution of (12) is given by
x (t ) =
F0 sin(ω t )
+ A sin(ω n t ) + B cos(ω n t) , (3.2.20)
  ω 2 
k 1 −   
 ω  
n


43
where A and B are arbitrary constant which may be determined uniquely by
the initial conditions. The initial condition x(0)=0 gives B=0. Hence
x (t ) =
F0 sin(ω t )
+ A sin(ω n t ) .
  ω 2 
k 1 −   
 ω  
n


(3.2.21)
Differentiating (21) w.r.t. the time t, we have
x&( t ) =
F0ω cos(ω t )
+ Aω n cos(ω n t ) .
  ω 2 
k 1 −   
 ω  
n


(3.2.22)
Therefore, the initial condition
x&( 0) =
F0ω
  ω 2 
k 1 −   
 ω  
n


+ Aω n = 0
(3.2.23)
gives
A=−
ω
.
  ω 2  ω n
k 1 −   
 ω  
n


F0
(3.2.24)
Substituting (24) in (22), the response (19) is obtained.
44
3.3 System’s Response to General Excitation by the Laplace Transform.
The Laplace transform X(s) of a function x(t) is defined by
∞
ℑ[ x ( t )] = ∫x ( t ) e − st dt ≡ X ( s) .
(3.3.1)
0
Recalling the general differential relation (from the theory of ordinary
differential equations)
d n x ( t ) 
ℑ n 
 dt 
= sn X ( s) − sn − 1 x (0) − sn − 2
n− 1
(3.3.2)
dx ( 0)
d x ( 0)
− ...−
dt
dt n − 1
we find that (setting n=1)
ℑ[]
x& = sX ( s) − x ( 0) ,
(3.3.3)
and (setting n=2)
ℑ[&x&( t ) ]= s2 X ( s) − sx( 0) − x&( 0) .
(3.3.4)
So, using (3) and (4) the Laplace transform of our mass-spring-damper system
mx&&+ cx& + kx = f ( t )

x&( 0) = v0
x ( 0) = x0 ,
(3.3.5)
is given by
m[ s2 X ( s) − sx0 − v0 ] + c[ sX ( s) − x0 ] + kX ( s) = F ( s) , (3.3.6)
where F(s) is the Laplace transform of the excitation f(t). We may write (6) in
the form
45
X ( s) =
F ( s)
( ms + c) x0 + mv0
+
.
ms2 + cs + k
ms2 + cs + k
(3.3.7)
In practice F(s) is usually determined from a table of Laplace transform. Then
X(s) is obtained by (7). Finally the response x(t), called the inverse Laplace
transform of X(s), is determined from the table of the transformation. The
following example demonstrates the use of this method.
Example 3.3.1
The system, shown below, is dropped from hight H. At t=0 it touches the
ground. If x(t) is the displacement of the mass, measured from its position at
t=0 (ie. x(0)=0), determine the ensuing motion.
Initially
m
F.B.D.
At t=0
k
m
H
mg
x(t)
m
k
kx
Solution.
The free-body diagram gives
mx&&= − kx + mg , t>0,
(3.3.8)
and conservation of energy (initial position - just before the spring touches
the ground) yields x&( 0) = 2 gH . Hence the equations of motion are
mx&&+ kx = mg
.

x
(
0
)
=
0
,
x
&(
0
)
=
2
gH

46
(3.3.9)
Here c=0, x(0)=0 and x&( 0) =
X ( s) =
2 gH . We thus have by (7)
m 2 gH
F ( s)
+
.
2
2
ms + k ms + k
The Laplace transform of 1 is
(3.3.10)
1
(see relation 2 in the Table of Laplace
s
Transform Pairs). Therefore
F ( s) = ℑ[ mg ] =
mg
,
s
(3.3.11)
and (10) can thus be written in the form
X ( s) =
m 2 gH
mg
+
.
2
2
s( ms + k ) ms + k
(3.3.12)
Relation 21 of the Table of Laplace Transform Pairs reads
f (t ) =
1
1
(
1
−
cos
ω
t
)
⇔
ℑ
[
f
(
t
)]
=
.
ω2
s( s2 + ω 2 )
(3.3.13)
We therefore have

 mg

g
− 1
=
ℑ
ℑ− 1 
 2

2
2 
s( ms + k ) 
s( s + ω n )  .
g
= 2 (1 − cosω n t )
ωn
(3.3.14)
Relation 17 of the Table of Laplace Transform Pairs reads
f ( t ) = sin ω t ⇔ ℑ[ f ( t )] =
ω
.
s2 + ω 2
Hence
47
(3.3.15)
m 2 gH 
ωn 
− 1  2 gH
ℑ− 1  2
=
ℑ


2
2
ms + k 
 ωn s + ωn
=
.
(3.3.16)
2 gH
sin ω n t
ωn
Using (14) and (16) we find that the inverse Laplace transform of X(s), given
by (12), is
x (t ) =
g
(1 − cosω n t ) +
ω 2n
2 gH
sin ω n t .
ωn
48
(3.3.17)
3.4 The Principle of Superposition.
A system is called a linear system if its response to the excitation
f(t)=f1(t)+f2(t) can be obtained by superimposing the response due to f1(t) and
that corresponding to f2(t).
Consider now the mass-spring-damper system with two excitations, f1(t) and
f2(t), shown below.
f1(t)
f2(t)
x(t)
m
k
c
The differential equation of motion is
mx&&+ cx& + kx = f1 ( t ) + f 2 (t ) .
(3.4.1)
Let x1(t) be the solution of
mx&&1 + cx&1 + kx1 = f1 ( t )
(3.4.2)
and let x2(t) be the solution of
mx&&2 + cx&2 + kx2 = f 2 (t ) .
(3.4.3)
x ( t ) = x1 ( t ) + x2 ( t )
(3.4.4)
Then
is the solution of (1).
Proof. We will show that (2)-(4) implies (1). Adding (2) to (3) gives
49
( mx&&1 + cx&1 + kx1 ) + ( mx&&2 + cx&2 + kx2 )
.
= f1 (t ) + f 2 (t )
(3.4.5)
Rearranging (5) we have
m(&x&1 + &x&2 ) + c( x&1 + x&2 ) + k ( x1 + x2 ) = f 1 ( t ) + f 2 ( t ) .
(3.4.6)
Hence by (4)
mx&&+ cx& + kx = f1 ( t ) + f 2 (t )
(3.4.7)
which is the same as (1).
We thus see that our basic mass-spring-damper, modelled by (1.5.1), is a
linear system. It should be noted that most real life systems are non-linear. To
simplify the analysis we usually make “linearisation”.
Example 3.4.1. The Simple Pendulum
Consider the simple pendulum shown below.
θ
θ
T
A
mg sinθ
L
A
mg
m
θ
Applying the Newton’s second law along the line A-A (perpendicular to the
tension T) we have
&& = − mg sin θ ,
mLθ
(3.4.8)
The differential equation of motion for the simple pendulum is therefore
50
&& + g sin θ = 0 .
Lθ
(3.4.9)
This is a non-linear system in the sense that the superposition principle does
not hold with respect to equation (9). If θ1 and θ2 are solutions of
&& + g sin θ = 0
Lθ
1
1
(3.4.10)
&& + g sin θ = 0 ,
Lθ
2
2
(3.4.11)
and
respectively, then in general θ=θ1+θ2 is not a solution of (9), since
&& + θ
&& ) + g sin(θ + θ )
L(θ
1
2
1
2
.
&&1 + g sin θ1 + Lθ
&&2 + g sin θ 2 = 0
≠ Lθ
(3.4.12)
With the assumption of small vibrations the system (9) becomes linear. For
small angle θ
sinθ ≅θ ,
(3.4.13)
and we obtain the linear system
&& + gθ = 0 .
Lθ
(3.4.14)
Clearly, if we are interested in the analysis of the pendulum for large
amplitude of vibrations then we must use equation (9), rather than (14). It
requires a non-linear treatment which is beyond the scope of this course.
Note that the natural period of vibration associated with the linear system (14)
is
1 L
cycles per second, independent of the initial conditions. On the
2π g
other hand the period of vibrations for the non-linear pendulum (9) depends on
the initial conditions of the problem.
The principle of superposition may be applied with respect to the initial
conditions, as shown in the following example.
51
Example 3.4.1. (Example 3.3.1 revisited)
In Example 3.3.1 it was required to determine the solution x(t) of
mx&&+ kx = mg

x&( 0) = 2 gH
x ( 0) = 0,
.
(3.4.15)
An alternative solution may be obtained by breaking (15) down into the
following two simpler problems
mx&&1 + kx1 = mg

x&1 ( 0) = 0
x1 ( 0) = 0,
(3.4.16)
mx&&2 + kx2 = 0

x&2 ( 0) = 2 gH
x2 ( 0) = 0,
(3.4.17)
and then superimposing the solutions:
x ( t ) = x1 (t ) + x2 ( t ).
(3.4.18)
It is easily shown that (18) is indeed the solution of (15). Adding (16) to (17)
gives
( mx&&1 + kx1 ) + ( mx&&2 + kx2 ) = mg + 0 = mg

x1 ( 0) + x2 ( 0) = 0, x&1 ( 0) + x&2 ( 0) = 2 gH
(3.4.19)
Rearranging (19) and using (18), we find that x(t) given by (18) satisfies (15).
The solution to (16) is given by (3.2.11) (with F0 = mg )
x1 ( t ) =
mg
[1 − cos(ω n t )] ,
k
(3.4.20)
and the solution to (17) is given by (1.2.18) (with x0=0, v0 =
x2 ( t ) =
2 gH
sin(ω n t ) .
ωn
2 gH )
(3.4.21)
Hence, by (18) the solution x(t) to (15) is
x1 ( t ) =
mg
[1 − cos(ω n t )] +
k
2 gH
sin(ω n t )
ωn
(3.4.22)
which is the same as obtained in (3.3.17) by the method of Laplace transform.
3.5 The Transfer Function.
52
A system may be characterised by its differential equation of motion. In our
case
mx&&+ cx& + kx = f ( t ) .
(3.5.1)
If we know m, c and k then the system response x(t) to the forcing function f(t)
can be determined be solving (1).
The system can alternatively be characterised by its impulse response. If we
know the impulse response h(t-η) then the response to the excitation f(t) may
be obtained by the convolution
t
x ( t ) = ∫f ( t ) h( t − η ) dη .
(3.5.2)
0
This relation can be expressed in terms of the following block diagram:
f(t)
Time domain
h(t-η)
Input
System
x(t)
Output
As noted above the input-output relation is determined in this case by the
convolution (2), or symbolically
x = f ⊗ h.
(3.5.3)
A third way to characterise the system is by its transfer function H(s),
formally defined as
H ( s) =
X ( s)
,
F ( s)
(3.5.4)
where X(s) is the Laplace transform of the system response with zero initial
conditions, and F(s) is the Laplace transform of the exciting force. The inputoutput relation in the Laplace domain is a simple multiplication
X ( s) = H ( s) F ( s)
(3.5.5)
which can be expressed by the following block diagram form:
53
F(s)
H(ω )
Laplace domain
Input
System
X(s)
Output
The use of Laplace transform is so popular because the multiplication (5) is
much easer to perform than the convolution (3) in the time domain. This is
based on the well known convolution theorem that states: The convolution of
two time domain functions is equal to the inverse Laplace transformation of
the product of their two transforms.
Note that H(s) is the Laplace transform of the impulse response h(t). You can
confirm this statement by using relation 24a of the Table of Laplace
Transform Pairs.
54
3.6 Composite Function Excitation
A function that can be expressed by several, different, analytical expressions
in some of its intervals of dependence is called a composite function. For
example, the function below,
 t , 0 < t ≤t 0
f (t ) = 
t > t0
t0 ,
(3.6.1)
is by the above definition a composite function.
f(t)
t0
t
t0
Consider the mass-spring-damper system
 mx&&+ cx& + kx = f ( t )
,

x
(
0
)
=
x
;
x
&(
0
)
=
v

0
0
(3.6.2)
excited by the composite function
0 < t < t1
 f1 ( t ),
 f ( t ), t < t < t
 2
1
2
f (t ) = 
.
M
M


 f n ( t ), tn − 1 < t < ∞
(3.6.3)
We first may find the response x(t) for 0<t<t1. It is simply the solution of
55
0 < t < t1
mx&&+ cx& + kx = f1 ( t ),
.

x
(
0
)
=
x
;
x
&(
0
)
=
v

0
0
(3.6.4)
Once (4) is solved, we know x ( t1 ) and x&( t1 ) . These values can be used as
the initial conditions to the dynamic of the system in the second interval
t1<t≤t2, ie.
t1 < t < t2
mx&&+ cx& + kx = f 2 ( t ),
.

and
as
obtained
from
(4)
x
(
t
)
x
&(
t
)
 1
1
(3.6.5)
Then the solution of (5) determines the initial conditions x ( t2 ) and x&( t2 )
required for the formulation of the problem in the third time interval t2<t≤t3.
Continuing in this manner determines the response of (2) for all time t. The
following example demonstrate the application of the method.
Example 3.6.1
Determine the response x(t) of
 mx&&+ cx& + kx = f ( t )

x ( 0) = 0 ; x&( 0) = 0
(3.6.6)
where f(t) is the step function
F0 , 0 < t ≤t1
f (t ) = 
.
0
,
t
>
t

1
(3.6.7)
f(t)
F0
t
t1
56
Solution.
The response of the system during the time interval 0<t≤t1 is determined by
0 < t < t1
mx&&+ cx& + kx = F0 ,
.

0
0
;
0
0
x
(
)
=
x
&(
)
=

(3.6.7)
It was found in example 3.2.1 (see equation (3.2.11)) that the solution to (7) is
x (t ) =
F0
[1 − cos(ω nt )] ,
k
0 < t < t1 .
(3.6.8)
In particular we have
x ( t1 ) =
F0
[1 − cos(ω nt1 )],
k
(3.6.9)
x&( t1 ) =
F0ω n
sin(ω n t1 ) .
k
(3.6.10)
and
Hence, the response of the system for t≥t1 is governed by
mx&&+ cx& + kx = F0 ,
t ≥ t1


x ( t ) = F0 1 − cos ω t , x&( t ) = F0ω n sin ω t .
( n 1)
[ ( n 1 )]
1
1


k
k
(3.6.11)
It is always recommended to have the initial conditions expressed in terms of
the time origin. We therefore introduce the time shift
τ = t − t1 .
(3.6.12)
The system (11) is thus
mx&&+ cx& + kx = F0 ,
τ≥0


x ( 0) = F0 1 − cos ω t , x&( 0) = F0ω n sin ω t .
( n 1)
[ ( n 1 )]


k
k
57
(3.6.13)
The solution of (13) is given by (1.2.18)
F0
sin(ω n t1 )sin(ω nτ )
k
.
F0
+
[1 − cos(ω nt1 )]cos(ω nτ )
k
x=
(3.6.14)
We may write (14) in the form
x=
F0
[cos(ω nτ )
k
. (3.6.15)
+ sin(ω n t1 )sin(ω nτ ) − cos(ω n t1 )cos(ω nτ ) ]
Hence by the trigonometric relation
cos(α + β ) = cosα cos β − sin α sin β ,
(3.6.16)
equation (15) is simplified to
x=
[
]
F0
cos(ω nτ ) − cos (ω n (τ + t1 )) .
k
(3.6.17)
We now shift the time back, from τ to t. Using (12) we have
x=
[
]
F0
cos(ω n ( t − t1 )) − cos(ω n t ) , t ≥ t1 .
k
(3.6.18)
In summary, by (8) and (18), the response of (6) to the excitation (7) is given
by
 F0
 k [1 − cos(ω n t )] ,
x=
F
 0 cos(ω n ( t − t1 )) − cos(ω n t ) ,
k
[
]
0 < t < t1
.
(3.6.19)
t ≥ t1
It is worthwhile to understand the physical meaning of the solution (19). The
dynamic of the system for this case can be regarded as the sum of two uniform
excitations: one with constant amplitude F0 starting at t=0, the other of
58
amplitude -F0 starting at t=t1. This is demonstrated graphically in the figure
below.
f(t)
F0
t
t1
=
f(t)
First excitation
F0
+
t
0
t1
-F0
Second excitation
The response x1(t) to the first excitation is (see (3.2.11) again)
x1 ( t ) =
F0
[1 − cos(ω nt )] , 0 < t < ∞ ,
k
(3.6.20)
and the response x2(t) to the second excitation (by the same argument) is
0,
0 < t ≤t1


x2 ( t ) =  F0
.
−
1
−
cos
ω
t
−
t
,
t
<
t
<
∞
(
)
(
)
n
1
1

 k
[
]
(3.6.21)
The total response x(t)=x1(t)+x2(t) is precisely the response given in equation
(19).
59