Harmonic excitation (damped)
x
EOM:
k
m
c
F(t)=F0cosωt
m&x& + cx& + kx = F0 cos ωt
&x& + 2ζω n x& + ωn2 x = f 0 cos ωt
The response x(t) (solution) can be separated into 2 part;
2
&
&
&
x
+
2
ζω
x
+
ω
1. Homogeneous solution xh(t)
n
nx = 0
2. Particular solution xp(t)
&x& + 2ζω n x& + ω2n x = f 0 cos ωt
x(t ) = xh (t ) + x p (t )
Homogeneous solution
Same as free vibration
Homogeneous solution xh(t)
Under damped
0 < ζ <1
xh (t ) = Ae − ζωnt sin(ωd t + φ)
Over damped 1 < ζ
xh (t ) = A1e
Critically damped ζ = 1
where
ωd = ωn 1 − ζ 2
⎛⎜ − ζ + ζ 2 −1 ⎞⎟ ω t
⎝
⎠ n
+ A2 e
⎛⎜ − ζ − ζ 2 −1 ⎞⎟ ω t
⎝
⎠ n
xh (t ) = ( A1 + A2t )e − ωnt
Particular solution
From EOM
&x& + 2ζω n x& + ωn2 x = f 0 cos ωt
The particular solution xp(t) can be written in the form:
x p (t ) = X cos(ωt − θ)
or
x p (t ) = As cos(ωt ) + Bs sin(ωt )
Substitution xp(t) into EOM, the coefficients X and θ (As and
Bs) can be determined. Then the particular solution is
2ζω n ω
x p (t ) =
cos(ωt − tan
)
2
2
2
2 2
2
ωn − ω
(ωn − ω ) + (2ζω n ω)
f0
X
−1
θ
Response of harmonic excitation (1)
Response of harmonic excitation
x(t ) = xh (t ) + x p (t )
Depend on ζ
Ex Underdamped
x(t ) = Ae
− ζω n t
sin(ωd t + φ) + X cos(ωt − θ)
xh(t)
Decrease with time
Transient response
xp(t)
Amplitude X is constant
Steady-state response
Response of harmonic excitation (2)
Ex Underdamped
x(t ) = Ae
− ζω n t
sin(ωd t + φ) + X cos(ωt − θ)
xh(t)
xp(t)
As t → ∞, transient
response dies out and
total response
x(t) → xp(t)
Steady-state response (1)
From steady-state response
x p (t ) =
Xk Xω2n
1
=
=
F0
f0
(1 − r 2 ) 2 + (2ζr ) 2
Xω2n
f0
,
f0
(ω2n − ω2 ) 2 + (2ζω n ω) 2
θ = tan −1
Amplitude response
2ζ r
1− r 2
θ(r )
cos(ωt − tan −1
where
2ζω n ω
)
2
2
ωn − ω
r=
ω
ωn
Phase response
180º
4
90º
2
0º
0
1
r
2
0
1
r
2
Steady-state response (2)
From steady-state response
x p (t ) =
1
Xk Xω2n
=
=
F0
f0
(1 − r 2 ) 2 + (2ζr ) 2
Xω2n
f0
Amplitude response
f0
(ω2n − ω2 ) 2 + (2ζω n ω) 2
where
r=
cos(ωt − tan −1
2ζω n ω
)
2
2
ωn − ω
ω
ωn
Maximum amplitude occurs when
1
d ⎛⎜
dr ⎜ (1 − r 2 ) 2 + (2ζr ) 2
⎝
⎞
⎟=0
⎟
⎠
r = 1 − 2ζ 2
ω = ω n 1 − 2ζ 2
4
2
0
1
r
2
or
⎛ Xk ⎞
1
⎜⎜
⎟⎟ =
2
⎝ F0 ⎠ max 2ζ 1 − ζ
Frequency response method (1)
Euler’s formula
e jωt = cos ωt + j sin ωt
EOM
m&x&(t ) + cx& (t ) + kx(t ) = F0 cos ωt
EOM (complex form)
m&z&(t ) + cz& (t ) + kz (t ) = F0 e jωt
Real part of the complex solution corresponds to the solution
of EOM.
x p (t ) = Re( z )
where
z = Ze jωt
Z is a complex-valued constant
Substituting z into EOM (complex form).
(−ω2 m + jcω + k ) Ze jωt = F0 e jωt
Frequency response method (2)
(−ω2 m + jcω + k ) Ze jωt = F0 e jωt
1
Z=
⋅ F0 = H ( jω) F0
2
(k − mω ) + (cω) j
H(jω) = (complex) frequency
response function
F0
− jθ
, where
Z=
⋅
e
2 2
2 12
[(k − mω ) + (cω) ]
cω
θ = tan
(k − mω2 )
−1
F0
j ( ωt − θ )
z=
⋅
e
[(k − mω2 ) 2 + (cω) 2 ] 1 2
x p (t ) = Re( z )
F0
x p (t ) =
⋅ cos(ωt − θ)
2 2
2 12
[(k − mω ) + (cω) ]
Frequency response method (3)
F0
cω
−1
x p (t ) =
⋅ cos(ωt − θ) ; θ = tan
2 2
2 12
[(k − mω ) + (cω) ]
(k − mω2 )
or
2ζω n ω
x p (t ) =
cos(ωt − θ) ; θ = tan
2
2
2
2 2
2
ω
−
ω
(ωn − ω ) + (2ζω n ω)
n
f0
xp(t) in the form of frequency response function
Z = H ( jω) F0 = H ( jω) F0 ⋅ e − jθ
z = H ( jω) F0 ⋅ e j ( ωt −θ)
x p (t ) = Re( z )
x p (t ) = H ( jω) F0 ⋅ cos(ωt − θ)
−1
Example 1
The block of mass m = 45 kg is suspended by two springs each of
stiffness k = 3 kN/m and is acted upon by the force F = 350cos(15t) N
where t is the time in seconds. Determine the amplitude X of the steadystate motion if the viscous damping coefficient c is (a) 0 and (b) 900
Ns/m. [J. L. Meriam & L. G. Kraige 8/52]
Example 2
For a vibrating system, m = 10 kg, k = 2500 N/m, and c = 45 Ns/m. A
harmonic force of amplitude 180 N and frequency 3.5 Hz acts on the
mass. If the initial displacement and velocity of the mass are 15 mm and
5m/s, find the complete solution representing the motion of the mass.
[Singiresu S. Rao, Mechanical Vibrations 4th edition in SI units 3/33]
Example 3
A machine part of mass 1.95 kg vibrates in a viscous medium.
Determine the damping coefficient when a harmonic exciting force of
24.46 N results in a resonant amplitude of 1.27 cm with a period of 0.2 s.
[William T. Thomson & Marie Dillon Dahleh, Theory of Vibration with Applications 5th
edition 3/1]
Example 4
A weight attached to a spring of stiffness 525 N/m has a viscous
damping device. When the weight is displaced and released, the period
of vibration is 1.80 s, and the ratio of consecutive amplitudes is 4.2 to
1.0. Determine the amplitude and phase when a force F = 2cos(3t) acts
on the system. [William T. Thomson & Marie Dillon Dahleh, Theory of Vibration with
Applications 5th edition 3/3]
Example 5
A spring-mass is excited by a force F0cos(ωt). At resonance, the
amplitude is measured to be 0.58 cm. At 0.8 resonant frequency, the
amplitude is measured to be 0.46 cm. Determine the damping factor ζ of
the system. [William T. Thomson & Marie Dillon Dahleh, Theory of Vibration with
Applications 5th edition 3/5]