VIBRATIONS OF 1 DOF SYSTEMS
VIBRATION: a periodic motion about an equilibrium position, such as the regular displacement of guitar string playing.
Vibrations are in general dangerous: they are associated with noise, malfunctioning and possibly fatigue failure.
Vibrations analysis of continuous systems is usually performed by means of lumped parameter model, i.e by substituting a continuous system with a discrete one with a finite number of DOF. We will thus study only the simple 1 DOF case as representative analysis for the most general case.
VIBRATIONS OF 1 DOF SYSTEMS
FREE VIBRATIONS
0i d el+ + =F F F 0mx cx kx+ + =
D’Alembert equation
i mx= −F Inertia Force
d cx= −F Damper Force
el kx= −F Spring Force
x
k
c
m
lumped parameters
m
continuous
VIBRATIONS OF 1 DOF SYSTEMS
FREE VIBRATIONS
x
m
k0mx kx+ =
The solution of this second order differential equation will have the following form
1 21 2( ) t tx t X e X eλ λ= +
where λ1,2 are the roots of the characteristic equation
1,24
2 nmk ki i
m mλ ω± −
= = ± = ±2 0m kλ + =
Let us start considering a system with no dumping. The dynamic equilibrium equation will be:
VIBRATIONS OF 1 DOF SYSTEMS
FREE VIBRATIONS
x
m
k
1 2( ) n ni t i tx t X e X eω ω−= +
Applying the Euler rule: cos sinize z i z= +
1 2 1 2( ) ( ) cos( ) ( ) sin( )n nx t X X t i X X tω ω= + + −
cos( ) sin( )n nA t iB tω ω= +
Exploiting the rotating vector representation, we have
( ) cos( )nx t C tω ϕ= +
Im
Re
ωnt
AB
C
φ
VIBRATIONS OF 1 DOF SYSTEMS
FREE VIBRATIONS
x
m
k The value of C and φ can be determined by the initial conditions of the system
( ) cos( )nx t C tω ϕ= +
0
0
(0)(0)
x xx x
=⎧⎨ =⎩
x
t
x0
C
2 / n Tπ ω =
0 0
00
(0) cossin(0) n
x x x Cx Cx x
ϕω ϕ
= =⎧ ⎧⇒⎨ ⎨ = −= ⎩⎩
0 0 0 0
0 00
(0) cos /sin / cos(0)
n
n
x x x C tg x xx C C xx x
ϕ ϕ ωω ϕ ϕ
= = = −⎧ ⎧ ⎧⇒ ⇒⎨ ⎨ ⎨= − == ⎩ ⎩⎩
VIBRATIONS OF 1 DOF SYSTEMS
x
m
k
c
0mx cx kx+ + =
The solution of this second order differential equation will have the same form
1 21 2( ) t tx t X e X eλ λ= +
where λ1,2 are the roots of the characteristic equation
2
1,24
2c c mk
mλ − ± −
=2 0m c kλ λ+ + =
FREE VIBRATIONS WITH DAMPING
In this case, the dynamic equilibrium equation will be:
VIBRATIONS OF 1 DOF SYSTEMS
x
m
k
c
21) 4c mk≥
23) 4c mk<
According to the sign of the term , it is possible to distinguish three cases:
2 4c mk−
FREE VIBRATIONS WITH DAMPING
The root λ1,2 are real number, producing an aperiodic damped motion of the mass m.
22) 4c mk=The root λ1,2 are coincident and real, producing an aperiodic damped motion of the mass m. Rare limit case
The root λ1,2 are complex and conjugate, producing a periodic damped motion of the mass m. Vibrations.
VIBRATIONS OF 1 DOF SYSTEMS
2crc mk=
cr
cc
ξ =
21,2 1
s
n niω
λ ξω ω ξ= − ± −
nkm
ω =
FREE VIBRATIONS WITH DAMPING
2
1,24
2c c mk
mλ − ± −
=
It can be convenient to rearrange the characteristic equation, introducing some significant term:
the damping value distinguish between periodic and aperiodic response of the system
the natural pulse of the undamped system
damping factor: the ration between the actual value of damping and its critical value
VIBRATIONS OF 1 DOF SYSTEMS
FREE VIBRATIONS WITH DAMPING
x
m
k
( ) cos( )nsx t Ce tξω ω ϕ−= +
x
t
x0
ne ξω−
ne ξω−−
c
VIBRATIONS OF 1 DOF SYSTEMS
FORCED VIBRATIONS
x
m
k
0( ) cos( )F t F tω=
c
( )( ) cos( ) sin( )i i i iF t A t B tω ω= +∑
Let us consider the effect of a periodic force acting on the system. Every periodic force can be decomposed in Fourier series, i.e
Thus, we may exploit the superposition principle and consider the system as loaded by a single sinusoidal force, without loss of generality
VIBRATIONS OF 1 DOF SYSTEMS
( ) g px t x x= +
FORCED VIBRATIONS
x
m
k0 cos( )F tω
c
( )i d el t+ + =F F F F
D’Alembert equation
0 cos( )mx cx kx F tω+ + =
gx = solution of the free motion
sx = specific integral solution
We assume the specific integral to be sinusoidal, with the same pulse of the applied force with a delay of φ and we limit our study to steady state vibrations
( ) cos( )g px t x x X tω ϕ= + = −
VIBRATIONS OF 1 DOF SYSTEMS
FORCED VIBRATIONS
x
m
k0 cos( )F tω
c
20cos( ) sin( ) cos( ) cos( )m X t c X t kX t F tω ω ϕ ω ω ϕ ω ϕ ω− − − − + − =
( ) cos( )x t X tω ϕ= −
( ) sin( )x t X tω ω ϕ= − −
2( ) cos( )x t X tω ω ϕ= − −
0 cos( )mx cx kx F tω+ + =
VIBRATIONS OF 1 DOF SYSTEMS
FORCED VIBRATIONS
x
m
k0 cos( )F tω
c
20sin( cos( ) cos( ))cos( ) F tkm t c t X tX Xωω ω ωω ϕϕ ω ϕ− − =+ −− −
Im
Re
ωt
F0
φ
-F0
kX
c Xω2m Xω
VIBRATIONS OF 1 DOF SYSTEMS
FORCED VIBRATIONS( ) ( )2 22 2 2
0X k m c Fω ω⎡ ⎤− − =⎢ ⎥⎣ ⎦( ) ( )2 22 2 20
2
X k m c F
ctgk m
ω ω
ωϕω
⎧ ⎡ ⎤− − =⎢ ⎥⎪ ⎣ ⎦⎪⎨⎪ =⎪ −⎩
2crc mk=cr
cc
ξ = nkm
ω =
Im
Re
ωt
F0
φ
-F0
kX
c Xω2m Xω
( ) ( )0
2 22
FX
k m cω ω=
− +
2
ctgk m
ωϕω
=−
0
222
/
1 cr
cr
F k
cm ck c kω ω
=⎛ ⎞⎛ ⎞
− + ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
2
1
cr
cr
ccc k
mk
ω
ω=
−
VIBRATIONS OF 1 DOF SYSTEMS
FORCED VIBRATIONS
0
2 22
2
/
1 2nn
F kX
ω ωξωω
=⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
2
2
2
1
n
n
tg
ωξω
ϕωω
=−
2crc mk=cr
cc
ξ = nkm
ω =
Im
Re
ωt
F0
φ
-F0
kX
c Xω2m Xω
VIBRATIONS OF 1 DOF SYSTEMS
FORCED VIBRATIONS
0
2 22
2
/
1 2nn
F kX
ω ωξωω
=⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
2
2
2
1
n
n
tg
ωξω
ϕωω
=−
VIBRATIONS OF 1 DOF SYSTEMS
VIBRATIONS: ISOLATION OF THE BASEx
m
k0 cos( )F tω
c
sin( ) cos( )c X t kX tT ω ωω= − +
( ) cos( )x t X tω=
( ) sin( )x t X tω ω= −
T cx kx= +
Let us see how to design the suspension in order to reduce the force transmitted to the base.
Im
Reωt
kX
c Xω
T
φ
0( ) cos( )T t T tω ϕ= +
VIBRATIONS OF 1 DOF SYSTEMS
VIBRATIONS: ISOLATION OF THE BASEx
m
k0 cos( )F tω
c
2 2 20T X k c ω= +
2crc mk=cr
cc
ξ =nkm
ω =
20 2
41 mkT kXk
ξ ω= +
22
2 21 cr
cr
cckXc k
ω= +
2
1 2n
kX ωξω
⎛ ⎞= + ⎜ ⎟
⎝ ⎠
2n
ctgkω ωϕ ξ
ω= =
TIm
Reωt
kX
c Xω
φ
VIBRATIONS OF 1 DOF SYSTEMS
VIBRATIONS: ISOLATION OF THE BASEx
m
k0 cos( )F tω
c
2
0 1 2n
T kX ωξω
⎛ ⎞= + ⎜ ⎟
⎝ ⎠
0
2 22
2
/
1 2nn
F kX
ω ωξωω
=⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
2
0
2 220
2
1 2
1 2
n
nn
TF
ωξω
ω ωξωω
⎛ ⎞+ ⎜ ⎟⎝ ⎠=
⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
TIm
Reωt
kX
c Xω
φ
VIBRATIONS OF 1 DOF SYSTEMS
2
2 2n
ωω
>>
VIBRATIONS: ISOLATION OF THE BASE
0
0
1TF
<< 2 2nω ω>> k
2
0
2 220
2
1 2
1 2
n
nn
TF
ωξω
ω ωξωω
⎛ ⎞+ ⎜ ⎟⎝ ⎠=
⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
0
0
1TF
<
VIBRATIONS OF 1 DOF SYSTEMS
VIBRATIONS: ISOLATION OF THE MASS
x
m
k
c
( ) cos( )y t Y tω=
Let us see how to design the suspension in order to isolate the suspended mass from the vibration of the frame.
y
( )el k x y= − −F
( ) ( ) 0mx c x y k x y+ − + − =
( )d c x y= − −F
i mx= −F
VIBRATIONS OF 1 DOF SYSTEMS
VIBRATIONS: ISOLATION OF THE MASS
x
m
k
c
y( ) ( ) 0mx c x y k x y+ − + − =
( )s x y= −
ms cs ks my+ + = −
( ) cos( )y t Y tω=
( ) sin( )y t Y tω ω= −2( ) cos( )y t Y tω ω= −
2 cos( )ms cs ks m Y tω ω+ + =
VIBRATIONS OF 1 DOF SYSTEMS
Im
Re
ωtφ kS
c Sω2m Sω
VIBRATIONS: ISOLATION OF THE MASS
2 cos( )ms cs ks m Y tω ω+ + =
x
m
k
c
y
cos( )s S tω ϕ= −
2 2sin cococos( s( ) ()) s )(m S kS tS m Yc tt tω ω ϕ ϕω ω ω ω ωϕ −− −− + =−
2m Yω
2m Yω−
VIBRATIONS OF 1 DOF SYSTEMS
VIBRATIONS: ISOLATION OF THE MASS
x
m
k
c
yIm
Re
ωtφ kS
c Sω2m Sω
2m Yω
2m Yω−
2
2
2 22
21 2
n
nn
Yωω
ω ωξωω
=⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
2
2 22
2
/
1 2nn
m Y kS ω
ω ωξωω
=⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
VIBRATIONS OF 1 DOF SYSTEMS
2
2
2
1
n
n
tg
ωξω
ϕωω
=−
VIBRATIONS: ISOLATION OF THE MASS
2
2
2 22
21 2
n
nn
YS
ωω
ω ωξωω
=⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
VIBRATIONS OF 1 DOF SYSTEMS
VIBRATIONS: ISOLATION OF THE MASS
s y= −
0x =
S Y=
ϕ π=
S Y=2
2 1n
ωω
>> k
ϕ π= 1ξ << c
s x y= −
EQUIVALENT STIFFNESS PARAMETERS
EQUIVALENT STIFFNESS OF SPRINGS IN PARALLEL
1 1
2 2
1 2 1 2
F k xF k xF F F k x k x
= Δ= Δ= + = Δ + Δ
1 2eqk k k= +
1k
2kF
xΔ
1 1 2
1
( )
eq
F x k kF k x= Δ += Δ 1
n
eq ii
k k=
= ∑
EQUIVALENT STIFFNESS OF SPRINGS IN SERIES
1 1
2 2
1 2( )eq
F k xF k xF k x x
= Δ= Δ= Δ + Δ
1k 2k
1xΔ 2xΔ
1 2
( )eq
F F Fk k k
= + 1 2
1 2eq
k kk
k k=
+
F
1
1 1n
ieq ik k=
= ∑
EQUIVALENT STIFFNESS PARAMETERS