HARMONIC EXCITATION OF A SINGLE DEGREE OF FREEDOM
SYSTEM.
2.1 Harmonic Excitation:
The differential equation of motion, for the system shown below, is given by
mx&&+ cx& + kx = f ( t )
(2.1.1)
where f(t) is the applied force.
c
k
x
.
cx
kx
m
Static equilibrium
position
m
f(t)
f(t)
If
f ( t ) = Fo sin ω t ,
(2.1.2)
where F0 is a constant, then the system is said to be a harmonically excited
system. The differential equation of motion for a harmonically excited system
is thus
mx&&+ cx& + kx = Fo sin ω t .
(2.1.3)
We now wish to find the general solution of (2.1.3). Then the response of the
system can be determined by the initial conditions x( 0) and x&( 0) .
21
Let xH be the general solution of the homogeneous problem
mx&&H + cx&H + kx H = 0 ,
(2.1.4)
and let xP be a particular solution of
mx&&P + cx&P + kx P = F0 sin ω t .
(2.1.5)
x ( t ) = x H (t ) + x P ( t )
(2.1.6)
Then
Theorem 2.1.1
is the general solution of (3).
Proof.
Since xH(t) is the general solution of (4), it is expressed in terms of a linear
combination of two linearly independent functions. Hence, by (6), x(t) is also
expressed in terms of a linear combination of two linearly independent
functions. Thus, all we need to show is that x(t) in (6) satisfies the differential
equation (3).
By (6)
x& = x&H + x&P
(2.1.7)
&x& = &x&H + x&&P .
(2.1.8)
and
Using (6)-(8) we have
mx&&+ cx& + kx
= m(&x&H + &x&P ) + c( x&H + x&P ) + k ( x H + x P ) ,
(2.1.9)
= ( mx&&H + cx&H + kx H ) + ( mx&&P + cx&P + kx P )
and by (4) and (5)
( mx&&H + cx&H + kxH ) + ( mx&&P + cx&P + kx P )
.
= 0 + F0 sin ω t
(2.1.10)
It thus follows from (9) and (10) that x(t), given by (6), satisfies (3).
22
The general solution xH(t), of the homogeneous problem (4) has been derived
in section 1.5. It is given by (see (1.5.6))
 A1e − ζ ω n t sin ω d t + B1e − ζ ω n t cosω d t ; ζ < 1

x H (t ) = 
A2 e − ω n t + B2 te − ω n t ;
ζ = 1.
 ( − ζ + ζ 2 − 1)ω n t
( − ζ − ζ 2 − 1)ω n t
A
e
B
e
; ζ >1
+
 3
3
(2.1.11)
We now show how to determine a particular solution xP of (5). We try a
solution of the form
x P ( t ) = X sin(ω t − φ) ,
(2.1.12)
where X and φare constants and, as in (3), ω is the frequency of the exciting
force. Substituting (12) into (5) gives
− mω 2 X sin(ω t − φ) + cω X cos(ω t − φ) + kX sin(ω t − φ)
,
= F0 sin ω t
or, using the identity cos(α ) = sin(α +
(2.1.13)
π
) , we have
2
π
)
2
+ kX sin(ω t − φ) = F0 sin ω t .
− mω 2 X sin(ω t − φ) + cω X sin(ω t − φ+
(2.1.14)
This is a vector equation with the graphical interpretation shown below.
mω 2X
cω X
F0
φ
ωt
kX
23
This polygon gives
F02 = ( kX − mω 2 X ) 2 + ( cω X ) 2 ,
(2.1.15)
F0
(2.1.16)
or
X=
( k − mω ) + ( cω )
2 2
2
.
We also have
tanφ =
cω X
cω
=
,
kX − mω 2 X k − mω 2
(2.1.17)
or
φ = tan − 1
cω
.
k − mω 2
(2.1.18)
In summary, a particular solution xP of (5) is given by (12), with X and φas in
(16) and (18), respectively.
Ex. 2.1.1: (a) Show that, say, X sin(ω t − φ) + 1.23456e
another particular solution which solves (5).
− ζ ω nt
sin(ω d t ) , is
(b) Can you find another particular solution of (5) ?
By Theorem 2.1.1 the general solution of (3) is
 A1e − ζ ω n t sin ω d t + B1e − ζ ω n t cosω d t

+ X sin(ω t − φ); ζ < 1


x ( t ) =  A2 e − ω n t + B2 te − ω n t + X sin(ω t − φ); ζ = 1
 ( − ζ + ζ 2 − 1)ω n t
( − ζ − ζ 2 − 1)ω n t
A
e
+
B
e
3
 3

+ X sin(ω t − φ); ζ > 1

24
(2.1.19)
where X and φare given by (16) and (18), respectively. Alternatively, using
(1.5.2) and (1.5.3), we may write X and φin terms of ω n and ζ ,as
X=
F0
k
2 2
 ω 
1 −  
 ω 
n

  ω 
 + 2ζ
  ω 
n

(2.1.20)
2
and
φ = tan − 1
2ζ
ω
ωn
2
ω 
1−  
ω n 
.
(2.1.21)
Ex. 2.1.2: Show that (16) and (18) imply (20) and (21), respectively .
We see that if c=0 and the exciting frequency ω = ω n =
k
then the
m
amplitude X → ∞ . This is the resonance phenomenon. To avoid resonance
we should introduce damping and avoid harmonic excitation with frequency
equal to the natural frequency of the system.
The general solution (19) shows that for ζ >0 the homogeneous part of the
solution vanishes as t → ∞ . We therefore say that the particular solution xP,
given by (12), is the steady state response of the system.
25
Example 2.1.1
We have mentioned (see sections 1.4 and 1.6) that
x ( t ) = A sin(ω t ) + B cos(ω t )
(2.1.22)
can be expressed equivalently in the form
x ( t ) = C sin(ω t − α ) .
(2.1.23)
Determine the explicit relations between the constants A, B and the constants
C, α by using the graphical method described in this section.
Solution.
Equating (22) and (23) we have
C sin(ω t − α ) = A sin(ω t ) + B cos(ω t )
or
C sin(ω t − α ) = A sin(ω t ) + B sin(ω t +
(2.1.24)
π
),
2
(2.1.25)
which has the graphical interpretation shown below.
B
C
-α
A
ωt
We therefore have
C=
A2 + B 2
(2.1.26)
and
α = − tan − 1
B
A
.
26
(2.1.27)
Alternatively, we may arrive at these expression in the following way. Using
the trigonometric identity
sin( β + γ) = sin β cosγ+ cos β sin γ
(2.1.28)
we have
C sin(ω t − α ) = C cos( − α ) sin(ω t ) + C sin( − α ) cos(ω t )
= C cosα sin(ω t ) − C sin α cos(ω t ).
(2.1.29)
Thus, comparing (29) and (22) yields
A = C cosα
(2.1.30)
B = − C sinα .
(2.1.31)
and
Adding the squares of (30) and (31) gives
A2 + B 2 = C 2 cos2 α + C 2 sin 2 α = C 2 ,
(2.1.32)
which implies (26).
Dividing (31) by (30) gives
− tanα =
B
,
A
(2.1.33)
which implies (27) .
Q. Which of the above methods appears to be more direct and simpler to
apply?
27
Example 2.1.2: Rotating unbalanced.
Consider the system of total mass M, with the rotating mass W shown below.
The mass W rotates about o with constant angular velocity ω and radius of
rotation e. The system is supported by a spring of constant k and a damper of
constant c. Determine the amplitude of vibration of M at steady state.
W
e
W
e
ωt
o
Static equilibrium
position for o
ωt
o
x
Total mass: M
Total mass: M
k
kx
c
.
cx
Solution:
Without lost of generality we may assume that the mass of the non-rotating
structure is lumped at o (Why this is so?). Then we have
Position of M-W → x .
Position of W → x + e sin(ωt).
The differential equation of motion is, therefore,
d2
( M − W )&x&+ W 2 [ x + e sin(ω t )] + cx& + kx = 0
dt
(2.1.34)
or
Mx&&− Wx&&+ Wx&&− Wω 2 e sin(ω t ) + cx& + kx
Mx&&− Wω e sin(ω t ) + cx& + kx = 0
2
.
(2.1.35)
The differential equation of motion can be written in the form
Mx&&+ cx& + kx = Wω 2 e sin(ω t ) .
28
(2.1.36)
Note that Wω e is constant. Hence substituting Wω e for F0 and M for m
in (16), we find that the steady state amplitude of vibration is given by
2
X=
2
Wω 2 e
( k − Mω ) + ( c ω )
2 2
29
2
.
(2.1.37)
2.2 Harmonic Base Excitation.
The base of the system shown below moves according to the function
y ( t ) = Y sin(ω t ) ,
(2.2.1)
where Y is the constant amplitude of the base motion.
m
x(t)
k
m
x(t)
. .
k(x-y)
c
c(x-y)
y(t)=Ysin(ω t)
Let x(t) be the displacement of the mass m from its static position. Then, the
spring force is k(x-y), the damper force is c( x& − y&) , and the differential
equation of motion is
mx&&+ c( x& − y&) + k ( x − y ) = 0 .
(2.2.2)
Denote
z=x-y.
(2.2.3)
Then (2) gives
mz&&+ cz& + kz = − my&&.
(2.2.4)
From (1) we have
&y& = − Yω 2 sin(ω t ) .
(2.2.5)
Hence, substituting (5) in (4) yields
mz&&+ cz&+ kz = mYω 2 sin(ω t ) ,
(2.2.6)
which is a differential equation of motion similar to (2.1.3), with
F0 = mYω 2 . By (2.1.16) the steady state amplitude for this case is
X=
mYω 2
( k − mω ) + ( c ω )
2 2
30
2
,
(2.2.7)
and the phase lag between the motion of the base and the mass is given by
(2.1.18).
31
2.3 Transmissibility of Vibration.
One important aspect of vibration analysis in application, is determination of
the steady state amplitude of vibration under harmonic excitation. The other
important factor is determination of the force fT(t) transmitted at steady state
from the vibrating system to its supporting structure. We now address the later
issue.
The force transmitted from the mass-spring-system, shown below, to the
ground consists of two components, the damper and the spring forces, ie.
f T ( t ) = cx& + kx .
k
m
x(t)
m
x(t)
(2.3.1)
c
kx
.
cx
It has been shown in section 2.1 that at steady state the system vibrates with
harmonic motion
x ( t ) = X sin(ω t − φ) ,
(2.3.2)
and that the following equation is satisfied (see (2.1.14))
π
)
2
+ kX sin(ω t − φ) = F0 sin ω t .
(2.3.3)
π
) + kX sin(ω t − φ) .
2
(2.3.4)
− mω 2 X sin(ω t − φ) + cω X sin(ω t − φ+
Hence, by (1) we have
f T = cω X sin(ω t − φ+
The force transmitted to the ground is, therefore, harmonic. Let
f T ( t ) = FT sin(ω t − α )
32
(2.3.5)
where FT is constant. Then equations (3)-(5) have the graphical interpretation
shown below.
mω 2X
α
F0
β
φ
cω X
FT
kX
ωt
It follows from this polygon that
FT = ( kX ) 2 + ( cω X ) 2 = X k 2 + ( cω ) 2
(2.3.6)
and
α = φ− β = tan − 1
cω
− 1 cω
−
tan
k
k − mω 2
.
(2.3.7)
Define the transmissibility TR as the ratio
TR =
FT
.
F0
(2.3.8)
Then by (2.1.15) and (6) we have
k 2 + ( cω ) 2
TR =
( k − mω 2 ) 2 + ( cω ) 2
.
(2.3.9)
We may write this equation in terms of ω and ζ as
1 + ( 2ζ
TR =
ω 2
)
ωn
ω
ω 2
[1 − ( ) 2 ]2 + ( 2ζ
)
ωn
ωn
33
.
(2.3.10)
It follows from (9) that increasing m, such that ( k − mω ) becomes
dominant, reduces the transmissibility TR, since m appears only in the
denominator. The problem of reducing TR while keeping ω n constant, is an
interesting optimisation problem (see (10)).
2 2
34