CHAPTER_3_Forced_Vibration_of_Single_Degree_of_Freedom_System.pdf

Mechanical

1. CHAPTER 3: Forced Vibration of
Single Degree of Freedom System

2. Contents
3.1Introduction
3.2Equation of Motion
3.3Response of an Undamped System Under Harmonic Force
3.4Response of a Damped System Under Harmonic Force
3.5Response of a Damped System under
3.6Response of a Damped System Under the Harmonic
Motion of the Base
3.7Response of a Damped System Under Rotating Unbalance
3.8Application of SDOFSystemti
eFtF

0
)(=

3. 3.1Introduction
•Forced vibrationsoccurs when external energy
is supplied to the system during vibration
•The external force can be supplied through
either an applied force or an imposed
displacement excitation, which may be
harmonic, nonharmonicbut periodic,
nonperiodic, or random in nature.
•Harmonic responseresults when the system
responses to a harmonic excitation
•Transient responseis defined as the response
of a dynamic system to suddenly applied
nonperiodicexcitations

4. 3.2Equation of Motion
•From Figure, the equation of Motion Using
Newton’s Second Law of Motion:)1.3()(tFkxxcxm =++
The homogeneous
solution of the equation:)2.3(
0=++ kxxcxm 

5. The variations of homogeneous, particular, and general
solutions with time for a typical case are shown in
Figure below.

6. Consider an undamped system subjected to a
harmonic force. If a force acts on the
mass mof the system, )3.3(cos
0
tFkxxm =+ )4.3(sincos)(
21
tCtCtx
nnh
+=
The homogeneous solution is given by:
where is the natural frequency.
Because the exciting force and particular solution is
harmonic and has same frequency, we can assume a
solution in the form: 2/1
)/(mk
n
=
3.3 Response of an Undamped System Under Harmonic
ForcetFtF cos)(
0
=

7. where denotes the static deflection
where X is the max amplitude of x
p(t))5.3(cos)( tXtx
p
= )6.3(
1
22
0








−
=
−
=
n
st
mk
F
X



 kF
st
/
0
=
Thus,)7.3(cossincos)(
2
0
21
t
mk
F
tCtCtx
nn



−
++=

8. Using initial conditions ,
Hence,00
)0( and )0( xtxxtx  ==== )8.3(,
0
22
0
01
n
x
C
mk
F
xC


=
−
−= )9.3(cos
sincos)(
2
0
0
2
0
0
t
mk
F
t
x
t
mk
F
xtx
n
n
n












−
+








+





−
−=

The max amplitude can be expressed as:)10.3(
1
1
2








−
=
n
st
X




9. where the quantity is
called the magnification factor,
amplification factor, or
amplitude ratio. The variation of
the amplitude ratio with the
frequency ratio is shown in the
Figure. The response of the
system can be identified to be of
three types from the figure.st
X/

10. Case 1.When 0 < < 1, the denominator in
Eq.(3.10) is positive and the response is given by
Eq.(3.5) without change. The harmonic response of the
system is in phase with external force, shown in figure.n
/

11. where the amplitude,
Case 2.When > 1, the denominator in Eq.(3.10) is
negative and the steady-state solution can be
expressed asn
/ )11.3(cos)( tXtx
p
−= )12.3(
1
2
−








=
n
st
X



The variations are shown in
figure.

12. Case 3.When = 1, the amplitude X given by
Eq.(3.10) or (3.12) becomes infinite. This condition, for
which the forcing frequency is equal to the natural
frequency of the system, is called resonance. Hence,
the total response if the system at resonance becomes)15.3(sin
2
sincos)(
0
0
t
t
t
x
txtx
n
nst
n
n
n





+
+=
 n
/

13. Total Response
The total response of the system, Eq.(3.7) or Eq.(3.9),
can also be expressed as)16.3( 1for ;cos
1
)cos()(
2









−
+−=
n
n
st
n
ttAtx








14. )17.3( 1for ;cos
1
)cos()(
2









−
−−=
n
n
st
n
ttAtx






 and

15. If the forcing frequency is close to, but not exactly
equal to, the natural frequency of the system, beating
may occur. The phenomenon of beating can be
expressed as:)22.3(sinsin
2
/
)(
0
tt
mF
tx 







=

16. The time between the points of zero amplitude or the
points of maximum amplitude is called the period of
beatingand is given by





−==
−
==
nb
n
b
2
)23.3(
2
2
2
with the frequency of beating defined as

17. ( )( ) ( )  )26.3( cossincos
0
2
tFtctmkX  =−−−− ( ) 
( )  )27.3(0cossin
sincos
2
0
2
=−−
=+−


cmkX
FcmkX ( ) 
)28.3(
2/1
22
2
2
0
 cmk
F
X
+−
= The equation of motion can be derived as
Using trigonometric relations, we obtain
The solution gives)29.3(tan
2
1






−
=
−



mk
c
and
3.4Response of a Damped System Under Harmonic Force)24.3(tcosFkxxcxm
0
=++ )25.3()tcos(X)t(x
p
−=

18. The figure shows typical plots of the forcing function
and steady-state) response.
Substituting the following,;
m
k
n
= ;2
n
m
c
= ;
0
k
F
st
= n
r


=

19. and
We obtain)30.3(
)2()1(
1
21
1
222
2/1
2
2
2 rr
X
nn
st 





 +−
=
















+
















−
= )31.3(
1
2
tan
1
2
tan
2
1
2
1






−
=






















−
=
−−
r
r
n
n 







20. The following characteristics of the magnification
factor (M) can be noted from figure and as follows:
the quantity M = X/δstis called the magnification factor,
amplification factor, or amplitude ratio.

21. 1.For an undamped system , Eq.(3.30) reduces
to Eq.(3.10), and as .
2.Any amount of damping reduces the
magnification factor (M) for all values of the forcing
frequency.
3.For any specified value of r, a higher value of
damping reduces the value of M.
4.In the degenerate case of a constant force (when r=
0), the value of M= 1.)0(= →M 1→r )0( )30.3(
)2()1(
1
21
1
222
2/1
2
2
2 rr
X
nn
st 





 +−
=
















+
















−
=

22. 5.The reduction in M in the presence of damping is
very significant at or near resonance.
6.The amplitude of forced vibration becomes smaller
with increasing values of the forcing frequency (that
is, as ).
7.For , the maximum value of M occurs when
which can be seen to be lower than the undamped
natural frequency and the damped frequency
.2
1
0 →M 1→r )32.3(21or21
22
 −=−=
n
r 2
1 −=
nd )30.3(
)2()1(
1
21
1
222
2/1
2
2
2 rr
X
nn
st 





 +−
=
















+
















−
=

23. 8.The maximum value of X(when ) is given
by:
and the value of X at by
9.For when r= 0. For , the graph of
Mmonotonically decreases with increasing values
of r.2
21−=r )33.3(
12
1
2
max  −
=








st
X )34.3(
2
1


=








=
n
st
X 0,
2
1
==
dr
dM
 2
1
 )30.3(
)2()1(
1
21
1
222
2/1
2
2
2 rr
X
nn
st 





 +−
=
















+
















−
=

24. The following characteristics of the phase angle can be
observed from figure and Eq.(3.31) as follows:

25. 1.For an undamped system , Eq.(3.31) shows
that the phase angle is 0 for 0 < r < 1 and 180°for r
> 1. This implies that the excitation and response
are in phase for 0 < r < 1 and out of phase for r > 1
when .
2.For and 0 < r < 1, the phase angle is given by 0
< Φ< 90°, implying that the response lags the
excitation.
3.For and r > 1, the phase angle is given by 90°<
Φ< 180°, implying that the response leads the
excitation.0 )0(= 0= 0 )31.3(
1
2
tan
1
2
tan
2
1
2
1






−
=






















−
=
−−
r
r
n
n 







26. 4.For and r = 1, the phase angle is given by Φ=
90°, implying that the phase difference between the
excitation and the response is 90°.
5.For and large values of r, the phase angle
approaches 180°, implying that the response and the
excitation are out of phase.0 0 )35.3( )cos()cos()(
00


−+−=
−
tXteXtx
d
t
n )36.3(1
2
nd
 −=
Total response
For an underdampedsystem,
where
For the initial conditions, Eq.(3.35) yields)37.3(sinsincos
coscos
000000
000


XXXx
XXx
dn
++−=
+=


27. •Quality factor and bandwidth:)38.3(
2
1
max
Q
XX
n
stst
==

















=

 )39.3(
2
XcW=
From figure, R
1and R
2is the bandwidth of the system.
Set , half power point hence,2//QX
st
=  22
1
2)2()1(
1
222
==
+−
Q
rr

28. Solving the equation for small values of ,
or)40.3(0)81()42(
2224
=−+−− rr )44.3(2
12 n
 =+
Thus, we obtained)46.3(
2
1
12


 −

n
Q
Using the relation,)42.3(21,21
2
22
2
2
2
2
12
1
2
1






+








==−








==
nn
RrRr )43.3( 4)())((
222
1
2
21212
2
1
2
2 nn
RR  −=−+=−

29. 3.5Response of a Damped System Under
The equation of motion becomes)47.3(
0
ti
eFkxxcxm

=++
Assuming the particular solution
Substituting,ti
eFtF

0
)(= )48.3()(
ti
p
Xetx

= )49.3(
)(
2
0
icmk
F
X
+−
= )50.3(
)()(
22222222
2
0 





+−
−
+−
−
=




cmk
c
i
cmk
mk
FX

30. The absolute value becomes,
•Frequency Response)56.3( )()(


i
eiHiH
−
= )54.3(
21
1
)(
2
0
rirF
kX
iH


+−
=
where )57.3(
1
2
tan
2
1






−
=
−
r
r

The complex frequency response is given by:
Thus, the steady-state solution becomes, )58.3()()(
)(0 

−
=
ti
p
eiH
k
F
tx

31. If , the corresponding steady-state solution
is given by the imaginary part of Eq.(3.53)
If , the corresponding steady-state
solution is given by the real part of Eq.(3.53) 
)59.3()(Re)(Re
)cos(
)()(
)(
)(00
2/1
222
0






=






=
−
+−
=
−



titi
p
eiH
k
F
eiH
k
F
t
cmk
F
tx tFtF cos)(
0
= tFtF sin)(
0
=  
)60.3()(Im
)sin(
)()(
)(
)(0
2/1
222
0






=
−
+−
=
−



ti
p
eiH
k
F
t
cmk
F
tx

32. •Complex Vector Representation of Harmonic
Motion.
Differentiating Eq.(3.58) with respect to time,
The various terms of the
equation of motion can be
represented in the figure, )61.3()()()()(onAccelerati
)()()(Velocity
2)(02
)(0
txeiH
k
F
itx
txieiH
k
F
itx
p
ti
p
p
ti
p




−===
===
−
−



33. If ,
3.6Response of a Damped System Under the Harmonic
Motion of the Base)64.3(0)()( =−+−+ yxkyxcxm  tYty sin)(= )65.3()sin(
cossin


−=
+=+=++
tA
tYctkYyckykxxcxm 
From the figure, the equation of motion is
where 





−=+=
−
k
c
ckYA


122
tan and )(

34. The steady-state response of the mass can be
expressed as 
)66.3()sin(
)()(
)(
)(
12/1
222
22



−−
+−
+
= t
cmk
ckY
tx
p 





−
=
−
2
1
1
tan



mk
c )67.3()sin()( −= tXtx
p
where
or
where )68.3(
)2()1(
)2(1
)()(
)(
2/1
222
2
2/1
22
22






+−
+
=






+−
+
=
rr
r
cmk
ck
Y
X




and )69.3(
)14(1
2
tan
)()(
tan
22
3
1
22
3
1






−+
=






+−
=
−−
r
r
cmkk
mc






35. The variations of displacement transmissibilityis
shown in the figure below.)70.3(
2r-1
r21
Re)(
2














+
+
=
ti
p
Ye
ri
i
tx


   )71.3()()2(1
2/1
2
 iHrT
Y
X
d
+==

36. 1.The value of T
dis unity at r = 0and close to unity for
small values of r.
2.For an undamped system (ζ= 0), T
d→∞at
resonance (r = 1).
3.The value of T
dis less than unity (T
d< 1) for values of
r >√2 (for any amount of damping ζ).
4.The value of T
d= 1 for all values of ζat r =√2.
The following aspects of displacement transmissibility
can be noted from the figure:

37. 5.For r <√2, smaller damping ratios lead to larger
values of T
d. On the other hand, for r >√2, smaller
values of damping ratio lead to smaller values of T
d.
6.The displacement transmissibility, T
d, attains a
maximum for 0 < ζ< 1 at the frequency ratio r = r
m<
1 given by:  
2/1
2
181
2
1
−+= 

m
r

38. •Force transmitted:)72.3()()( xmyxcyxkF  −=−+−= )73.3()sin()sin(
2
 −=−= tFtXmF
T
The force transmissibility is given
by:)74.3(
)2()1(
)2(1
2/1
222
2
2






+−
+
=
rr
r
r
kY
F
T



39. The equation of motion can be written as 
)76.3()sin(
)()(
)sin(
)(
12/1
222
1
2



−=
+−
−
= tZ
cmk
tYm
tz )75.3(sin
2
tYmymkzzczm =−=++  )77.3(
)2()1()()(
222
2
222
2
rr
r
Y
cmk
Ym
Z


+−
=
+−
=
The steady-state solution is given by:
where, the amplitude,
•Relative Motion:
and 





−
=





−
=
−−
2
1
2
1
1
1
2
tantan
r
r
mk
c 




41. 41
Example 3.1
Vehicle Moving on a Rough Road
The figure below shows a simple model of a motor
vehicle that can vibrate in the vertical direction while
traveling over a rough road. The vehicle has a mass of
1200kg. The suspension system has a spring constant of
400 kN/m and a damping ratio of ζ= 0.5. If the vehicle
speed is 20 km/hr, determine the displacement
amplitude of the vehicle. The road surface varies
sinusoidallywith an amplitude of Y = 0.05mand a
wavelength of 6m.

42. 42rad/s 290889.0
6
1
3600
1000
22 v
v
f =





==  rad/s 2574.18
1200
10400
2/1
3
=








==
m
k
n

The frequency can be found by
For v = 20 km/hr, ω= 5.81778 rad/s. The natural
frequency is given by,
Hence, the frequency ratio is318653.0
2574.18
81778.5
===
n
r



43. 43
Thus, the displacement amplitude of the vehicle is
given by
The amplitude ratio can be found from Eq.(3.68):469237.1
)318653.05.02()318653.01(
)318653.05.02(1
)2()1(
)2(1
2/1
22
2
2/1
222
2
=






+−
+
=






+−
+
=
rr
r
Y
X

 m 073462.0)05.0(469237.1469237.1 === YX
This indicates that a 5cm bump in the road is
transmitted as a 7.3cm bump to the chassis and the
passengers of the car.

44. ❑Unbalanceinrotatingmachinesisacommonsourceofvibration
excitation.
❑Weconsiderhereaspring-masssystemconstrainedtomoveinthe
verticaldirectionandexcitedbyarotatingmachinethatisunbalanced,
asshowninFig.below.
❑Theunbalanceisrepresentedbyaneccentricmassmwith
eccentricityethatisrotatingwithangularvelocityw.
❑Bylettingxbethedisplacementofthenonrotatingmass(M-m)from
thestaticequilibriumposition,thedisplacementofmis
3.7 Response of a Damped System Under Rotating
Unbalance

45. 45)78.3(sin
2
tmekxxcxM =++ )79.3( )(Im)sin()(
)(
2
















=−=
−




ti
n
p
eiH
M
me
tXtx  
)80.3()(
)()(
2
2/1
222
2





iH
M
me
cMk
me
X
n








=
+−
=
The equation of motion can be derived by the usual
procedure:
The solution can be expressed as
The amplitude and phase angle is given by)81.3(tan
2
1






−
=
−



Mk
c

46. 46
By defining
,
and  
)82.3()(
)2()1(
2
2/1
222
2


iHr
rr
r
me
MX
=
+−
= )83.3(
1
2
tan
2
1






−
=
−
r
r
 ncc
Mccc  2 and / ==

48. 48
The following observations can be made from Eq.(3.82)
and the figure above:
1.All the curves begin at zero amplitude. The
amplitude near resonance is markedly affected by
damping. Thus if the machine is to be run near
resonance, damping should be introduced
purposefully to avoid dangerous amplitudes.
2.At very high speeds (ωlarge), MX/meis almost
unity, and the effect of damping is negligible.
3.For 0 < ζ< 1/√2 , the maximum of MX/meoccurs
when)84.3(0=





me
MX
dr
d

49. 49
The solution gives:
With corresponding maximum value 1
21
1
2

−
=

r 2
max 12
1
−
=





me
MX
Thus the peaks occur to the right of the resonance
value of r = 1.
4.For , does not attain a maximum. Its
value grows from 0 at r = 0 to 1 at r → ∞ . 
me
MX
,21

50. 50
Example 3.2
Francis Water Turbine
The schematic diagram of a Francis water turbine
is shown in the figure in which water flows from A
into the blades B and down into the tail race C. The
rotor has a mass of 250 kg and an unbalance (me)
of 5kg-mm. The radial clearance between the rotor
and the stator is 5mm. The turbine operates in the
speed range 600 to 6000rpm. The steel shaft
carrying the rotor can be assumed to be clamped
at the bearings. Determine the diameter of the
shaft so that the rotor is always clear of the stator
at all the operating speeds of the turbine. Assume
damping to be negligible.

51. 51

52. 52(E.1)
)1()(
2
2
2
2
rk
me
Mk
me
X
−
=
−
=


 rad/s 200
60
2
60006000rpm
rad/s 20
60
2
600rpm600




==
==
The max amplitude can be obtained from Eq.(3.80) by
setting c = 0 as
The value of ωranges from:
While the natural frequency is given by
to(E.2) rad/s 625.0
250
k
k
M
k
n
===

53. 53(E.3)N/m 1004.10
10
2
004.0
)20(
1
)20()100.5(
005.0
24
25
2
2
23





=
−
=






−

=
−
k
k
k
k (E.4)N/m 1004.10
10
200
004.0
)200(
1
)200()100.5(
005.0
26
27
2
2
23





=
−
=






−

=
−
k
k
k
k
For ω= 20π, Eq.(E.1) gives
The stiffness of the cantilever beam is given by
For ω= 200π, Eq.(E.1) gives(E.5)
64
33
4
33 







==
d
l
E
l
EI
k


54. 54
and the diameter of the beam can be found:
or(E.6)mm 127m 1270.0
m 106005.2
)1007.2(3
)2)(1004.10)(64(
3
64
44
11
3243
4
==
=


==
−
d
E
kl
d




55. The End!

56. Reasons for the measurement of vibrations
❑Somemachinesarerunningathighspeedswhichmaycause
resonantconditionandtheymaygetfail.
❑Insomesituationstheexcessivevibrationsmaytransfertothe
nearbymachinesorstructures.
➢Tocheckthehealthofthemachines.
➢Tounderstandthedynamicbehavioritisnecessarytomeasurethe
vibrations.
❑Ithelpstoidentifyimportantparametersofsystemsuchasmass,
stiffness,damping.
3.8 Application of SDOFSystem
(Theory of Vibration measuring instruments)
❑Displacement measuring instrument (Vibrometer)
❑Velocity measuring instrument (Velometer)
❑Acceleration measuring instrument (Accelerometer)
Vibration Measuring Instruments

57. 3.8.1 Introduction

59. 3.8.2 Theory of Vibration measuring instruments

66. 3.8.3 Displacement Measuring Instruments (Vibrometer)

67. 3.8.4 Velocity measuring instrument (Velometer)

69. 3.8.5 Acceleration Measuring Instrument (Accelerometer)
Accelerometer: instrument with high natural frequency.
❑When the natural frequency of the instrument is high compared to
that of the vibration to be m measured, the instrument indicates
acceleration

72. The End!