3.9: FORCED VIBRATIONS
KIAM HEONG KWA
1. Forced Vibrations with Damping
Continuing our investigation of the spring-mass system with damping, we now suppose that a periodic external force F (t) = F0 cos ωt is
applied to the mass. Here F0 and ω are numerically positive constants
representing the amplitude and the frequency, respectively, of the force.
The equation of motion is now
mu00 + γu0 + ku = F0 cos ωt,
(1.1)
where m, γ, and k are the mass, the damping coefficient, and the spring
constant of the spring-mass system. Thus the motion is described by
(1.2)
u(t) = uc (t) + U (t),
where uc (t) is the motion of the mass in the absence of the external
force and is the solution of the homogeneous equation
mu00 + γu0 + ku = 0,
(1.3)
while
(1.4)
U (t) =
F0
2
(k
−
mω
)
cos
ωt
+
γω
sin
ωt
(k − mω 2 )2 + γ 2 ω 2
is a particular solution of the full nonhomogeneous equation (1.1).
In other words, uc (t) is a damped free vibration which dies out after
sufficiently long time: limt→∞ uc (t) = 0. Hence it is called the transient
solution of the motion (1.2). Physically, the transient solution is the
response of the system to the initial conditions. With the presence of
damping, the energy put into the system by the initial conditions is
dissipated with increasing time and thus the transient solution dies
out eventually. In contrast, U (t) is the response of the system to the
presence of the external force F (t). It persists indefinitely as long as
the external force F (t) is present and oscillates with he same frequency
as F (t). It is thus called the steady-state solution or the forced response
of the motion (1.2).
Date: February 2, 2011.
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KIAM HEONG KWA
Writing
r cos(ωt − δ) = (k − mω 2 ) cos ωt + γω sin ωt,
where r ≥ 0, we have
r=
p
(k − mω 2 )2 + γ 2 ω 2
with the phase δ being determined by
γω
k − mω 2
p
p
sin δ =
and cos δ =
.
(k − mω 2 )2 + γ 2 ω 2
(k − mω 2 )2 + γ 2 ω 2
Thus
F0
cos(ωt − δ).
U (t) = p
(k − mω 2 )2 + γ 2 ω 2
By introducing the parameter
r
k
(1.5)
ω0 =
,
m
it is customary to write
(1.6)
F0
U (t) = p
cos(ωt − δ),
2
m2 (ω0 − ω 2 )2 + γ 2 ω 2
where
(1.7)
m(ω02 − ω 2 )
γω
p
and
cos
δ
=
.
sin δ = p
m2 (ω02 − ω 2 )2 + γ 2 ω 2
m2 (ω02 − ω 2 )2 + γ 2 ω 2
Exercise 1. Calculate a formula for the forced response U (t) similar to
(1.6) if the external
is replaced by F (t) = F0 sin ωt. (Hint: Note
force
π
π
that sin ωt = cos ωt −
. Use the translation of time τ = t −
.)
2
2ω
The Dependence of the Amplitude of the Forced Response on
the Excitation Frequency. The amplitude of the forced response
U (t) is given by
(1.8)
F0
R= p
.
m2 (ω02 − ω 2 )2 + γ 2 ω 2
It is interesting to consider the dependence of the amplitude R on the
frequence ω of the excitation.
F0
F0
• For low-frequency excitation, i.e., ω → 0, R →
=
.
2
mω0
k
3.9: FORCED VIBRATIONS
3
• For very high-frequency excitation, i.e., ω → ∞, R → 0.
Now treating R = R(ω) as a function of the excitation frequency ω
with ω taking values in the interval [0, ∞), we get
dR
2m2 (ω 2 − ω02 ) + γ 2
= −F0 ω × 2 2
.
dω
[m (ω0 − ω 2 )2 + γ 2 ω 2 ]3/2
Thus R has at most a critical point ωmax given by
r
r
2
γ
γ2
.
(1.10)
ωmax = ω02 −
=
ω
1
−
0
2m2
2mk
(1.9)
More explicitly, R has precisely one critical point when
has no critical point when
γ2
> 2.
mk
γ2
≤ 2 and it
mk
γ2
≤ 2, R is increasing on [0, ωmax ] and is decreasing on
mk
(ωmax , ∞). Hence
F
q 0
Rmax = R(ωmax ) =
2
γω0 1 − 4mγ2 ω2
• When
0
F
q 0
=
γ2
γω0 1 − 4mk
γ2
F0
(1.11)
1+
≈
γω0
8mk
is the maximum value of R, where the last expression is an
γ2
approximation for small
.
mk
γ2
• When
> 2, R is decreasing on [0, ∞). Hence
mk
F0
(1.12)
R(0) =
k
is the maximum value of R.
One thing to note is that
(1.13)
as γ → 0, then ωmax → ω0 and Rmax → ∞
by (1.10) and (1.11). Since R is a continuous function of ω, this implies
that for a lightly damped system, the forced response U (t) tends to
oscillate at a very large amplitude when ω is in the vicinity of ω0
even for relatively small external forces. This phenomenon is known as
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KIAM HEONG KWA
resonance, i.e., the tendency of a system to oscillate at larger amplitude
at some frequencies than at others. The frequency ωmax is called the
resonant frequency of the system.
2. Forced Vibrations without Damping
Removing the damping from the system, the equation of motion is
mu00 + ku = F0 cos ωt,
(2.1)
the equation for an undamped forced vibration. The behavior of the
motion depends on whether the forcing r
frequency ω is equal to or difk
ferent from the natural frequency ω0 =
of the system.
m
Case ω 6= ω0 . The motion is described by
F0
u00
F0 cos ωt
(2.2) u(t) = u0 −
cos
ω
t
+
sin ω0 t +
,
0
2
2
m(ω0 − ω )
ω0
m(ω02 − ω 2 )
where u0 = u(0) and u00 = u0 (0). The motion is thus periodic if and
ω
ω
p
only if the ratio
is rational, i.e.,
= for some positive integers
ω0
ω0
q
2πq
2πp
p and q, and the period is
=
. However, the motion is always
ω0
ω
bounded regardless of whether it is periodic or not.
An interesting instance occurs when u0 = u00 = 0. In this case, the
motion is given by
F0 (cos ωt − cos ω0 t)
(2.3)
u(t) =
.
m(ω02 − ω 2 )
Making use of the identities
cos(A ± B) = cos A cos B ∓ sin A sin B
for any A, B ∈ R, the substitutions A =
turn (2.3) into
(ω0 + ω)t
(ω0 − ω)t
and B =
2
2
2F0
(ω0 − ω)t
(ω0 + ω)t
(2.4)
u(t) =
sin
sin
.
2
2
m(ω0 − ω )
2
2
This shows that the motion is a rapidly oscillation with frequency
ω0 + ω
with a slowly varying sinusoidal amplitude
2
2F0
(ω0 − ω)t (2.5)
m(ω 2 − ω 2 ) sin
.
2
0
3.9: FORCED VIBRATIONS
5
This phenomenon is more pronounced when |ω0 − ω| is small, so that
ω0 + ω is much greater than |ω0 − ω|. This exhibits a beat. The variation of the amplitude with time is called amplitude modulation.
Case ω = ω0 . In this case, the motion is given by
u0
F0 t sin ω0 t
(2.6)
u(t) = u0 cos ω0 t + 0 sin ω0 t +
,
ω0
2mω0
where u0 = u(0) and u00 = u0 (0). Regardless of the initial conditions,
the motion is unbounded eventually in view of the unboundedness of
the term t sin ω0 t. In other words, the motion is a growing oscillation.
Remark 1. Eq.(2.6) can be derived from (2.2) as one passes to the
limit ω → ω0 . Rearrranging the right-hand side of (2.2) and treating
it as a function of t and ω yields
u0
F0 (cos ωt − cos ω0 t)
(2.7)
u(t, ω) = u0 cos ω0 t + 0 sin ω0 t +
.
ω0
m(ω02 − ω 2 )
By l’Hôpital’s rule, one has
−F0 t sin ωt
F0 (cos ωt − cos ω0 t)
F0 t sin ω0 t
lim
= lim
=
.
2
2
ω→ω0
ω→ω0
m(ω0 − ω )
−2mω
2mω0
Thus
F0 t sin ω0 t
u00
lim u(t, ω) = u0 cos ω0 t +
sin ω0 t +
.
ω→ω0
ω0
2mω0