2.7: FORCED OSCILLATIONS
KIAM HEONG KWA
1. Forced Oscillations
As an application, we use Fourier series method to solve the nonhomogeneous differential equation
dy
d2 y
+ c + ky = F (t),
2
dt
dt
where µ, c, k ∈ R are constants such that µ > 0, c ≥ 0, and k > 0,
while F is a given nonzero 2p-periodic function1. Following the text,
we will only consider the case c > 0 and leave the case c = 0 as an
exercise.
(1.1)
µ
Let y be the general solution of (1.1). If y = yh + yp and yh is a
solution of the homogeneous equation
d2 y
dy
+ c + ky = 0,
2
dt
dt
then we call yh a homogeneous term of y and yp a particular solution of (1.1)2. More efficient is the decomposition
(1.2)
(1.3)
µ
y = ytr + ys ,
where ytr is a homogeneous term of y and ys is a particular solution of
(1.2) that contains no homogeneous term. Since limt→∞ ytr (t) = 0, so
that
(1.4)
y(t) ≈ ys (t) for all sufficiently large t > 0,
we call ytr the transient solution and ys the steady-state solution
of (1.1).
Date: April 12, 2011.
1The middle term c dy is customarily called the damping term, so that the
dt
coefficient c is called the damping coefficient, while the non-homogeneous term
F is usually called the forcing term.
2It should be noted that such a decomposition of the general solution y of (1.1)
into the sum of a homogeneous term and a particular solution is not unique because
if yeh is also a solution of (1.2), then yh + yeh and yp − yeh are also a homogeneous
term of y and a particular solution of (1.1) respectively.
1
2
KIAM HEONG KWA
Suppose F has a convergent Fourier series and
∞ X
nπt
nπt
(1.5)
F (t) = F[F ](t) = a0 +
an cos
+ bn sin
.
p
p
n=1
The nth term of the Fourier series,
nπt
nπt
+ bn sin
,
(1.6)
fn (t) = an cos
p
p
is a simple sinusoidal component of the input function F . Its fundamental frequency is
nπ
(1.7)
ωn =
,
p
measured in radians per unit time. By elementary theory of second
order linear homogeneous equations, if F = fn , then ys has the form
nπt
nπt
(1.8)
yn (t) = αn cos
+ βn sin
,
p
p
where αn and βn can be determined from
"
2 #
nπ
cnπ
+ βn
= an ,
αn k − µ
p
p
"
2 #
cnπ
nπ
−αn
+ βn k − µ
= bn ,
p
p
so that
αn =
an An − bn Bn
an Bn + bn An
and βn =
,
2
2
An + Bn
A2n + Bn2
where
An = k − µ
nπ
p
2
and Bn =
cnπ
.
p
More is true: if F is the sum of fn for n in a finite set {n1 , n2 , · · · , nN } ⊂
P
PN
N, say F = N
i=1 fni , then ys has the form
i=1 yni , where
ni πt
ni πt
yni (t) = αni cos
+ βni sin
p
p
with
an An − bni Bni
an Bn + bni Ani
αni = i 2 i
and βni = i 2 i
,
2
Ani + Bni
Ani + Bn2i
where
2
ni π
cni π
Ani = k − µ
and Bni =
.
p
p
for each i ∈ {1, 2, · · · , N }.
2.7: FORCED OSCILLATIONS
3
In view of the uniqueness as well as term-by-term differentiation of
the Fourier series of a smooth function, this method can be generalized
to yield:
Theorem 1.1. The steady-state solution of (1.1) (with µ > 0, c > 0,
k > 0, and the forcing (1.5)) has the Fourier series representation
∞ X
nπt
nπt
(1.9)
ys (t) = F[ys ](t) = α0 +
αn cos
+ βn sin
p
p
n=1
with
(1.10)
α0 =
an An − bn Bn
a0
an Bn + bn An
, αn =
and βn =
,
2
2
k
An + Bn
A2n + Bn2
where
An = k − µ
(1.11)
nπ
p
2
and Bn =
cnπ
p
for each n ∈ N.
Example 1 (Exercise 2.7.5 in the text). Consider the equation
y 00 + 4y 0 + 5y = sin t −
1
sin 2t.
2
The forcing function
1
sin 2t
2
equals its Fourier series and is 2π-periodic. So the steady-state solution
has a Fourier series expansion
F (t) = sin t −
ys (t) = α0 +
∞
X
(αn cos nt + βn sin nt)
n=1
by theorem 1.1, so that
ys0 (t) =
∞
X
(−nαn sin nt + nβn cos nt) ,
n=1
ys00 (t) =
∞
X
n=1
−n2 αn cos nt − n2 βn sin nt .
4
KIAM HEONG KWA
Plugging the right-hand sides of these equations into the differential
equation yields
ys00 (t) + 4ys0 (t) + 5ys (t)
∞
∞
X
X
2
2
=
−n αn cos nt − n βn sin nt + 4
(−nαn sin nt + nβn cos nt)
n=1
n=1
+ 5α0 + 5
∞
X
(αn cos nt + βn sin nt)
n=1
= 5α0 +
∞
X
(−n2 αn + 4nβn + 5αn ) cos nt
n=1
+(−n2 βn − 4nαn + 5βn ) sin nt
= sin t −
1
sin 2t,
2
so that
5α0 = 0, −n2 αn + 4nβn + 5αn = 0,
and


1


1
−n2 βn − 4nαn + 5βn = −


0 2
if n = 1,
if n = 2,
if n 6= 1, 2.
It is easy to see that α0 = 0 and αn = βn = 0 for n 6= 1, 2. For n = 1,
we have
4α1 + 4β1 = 0 and 4β1 − 4α1 = 1,
1
from which it follows that α1 = −β1 = − . For n = 2, we have
8
1
α2 + 8β2 = 0 and β2 − 8α2 = − ,
2
4
1
from which it follows that α2 =
and β2 = −
. Hence
65
130
1
1
4
1
ys (t) = − cos t + sin t +
cos 2t −
sin 2t.
8
8
65
120
The function yn in (1.8) is called the nth normal mode of the
vibration. It is the response of the system to the nth component fn of
the Fourier series of F . Using the trigonometric identity
cos(a − b) = cos a cos b + sin a sin b for all a, b ∈ R,
2.7: FORCED OSCILLATIONS
5
it can be expressed as
(1.12)
yn (t) = Cn cos
nπt
− φn ,
p
where the amplitude is
(1.13)
Cn =
p
αn2 + βn2
and the phase angle or phase lag is determined from
(1.14)
cos φn =
αn
βn
and sin φn =
Cn
Cn
whenever Cn > 0. In consequence, (1.9) can also be expressed as
∞
X
nπt
Cn cos
− φn .
(1.15)
ys (t) = F[ys ](t) = α0 +
p
n=1
Example 2 (Adapted from exercise 2.7.9 in the text). Consider the
spring-mass system
y 00 + 0.05y 0 + 10.01y = F (t),
where the 2π-periodic forcing function F is such that
(
1
if 0 < t < π,
F (t) =
−1 if π < t < 2π.
It has the Fourier series representation
∞
4 X sin(2n + 1)t
4 X 1
sin nt =
.
F[F ](t) =
π n=1 n
π n=0 2n + 1
n odd
The natural frequency3 of the spring is given by
r
10.01
w0 =
≈ 3.1638584039112749.
1
d2 y
dy
natural frequency w0 of the system µ 2 + c
+ ky = F (t), where
dt
dt
µ, c, k ∈ R are constants such that µ > 0, c ≥ 0, and k > 0, is the frequency of the
d2 y
nontrivial solutions of the corresponding free undamped system µ 2 + ky = 0.
dt
r
k
That is, w0 =
.
µ
3The
6
KIAM HEONG KWA
By theorem 1.1, the steady-state solution has a Fourier series expansion
∞
X
ys (t) = α0 +
(αn cos nt + βn sin nt) ,
n=1
so that
ys0 (t)
=
ys00 (t) =
∞
X
(−nαn sin nt + nβn cos nt) ,
n=1
∞
X
−n2 αn cos nt − n2 βn sin nt .
n=1
Plugging the right-hand sides of these equations into the differential
equation yields
ys00 (t) + 0.05ys0 (t) + 10.01ys (t)
∞
∞
X
X
2
2
(−nαn sin nt + nβn cos nt)
=
−n αn cos nt − n βn sin nt + 0.05
n=1
n=1
+ 10.01α0 + 10.01
∞
X
(αn cos nt + βn sin nt)
n=1
= 10.01α0 +
∞
X
(−n2 αn + 0.05nβn + 10.01αn ) cos nt
n=1
+(−n2 βn − 0.05nαn + 10.01βn ) sin nt
=
4 X 1
sin nx,
π n=1 n
n odd
from which it follows that
α0 = 0, −n2 αn + 0.05nβn + 10.01αn = 0,
and

 4
2
−n βn − 0.05nαn + 10.01βn = nπ
0
if n odd,
if n is even.
It is then easy to see that
(
(
6= 0 if n is odd,
6= 0 if n is odd,
αn
and βn
= 0 if n is even.
= 0 if n is even.
Hence the first six nonzero normal modes must be
yn (t) = αn cos nt + βn sin nt for n = 1, 3, 5, 7, 9, 11
2.7: FORCED OSCILLATIONS
7
and their frequencies are respectively
ωn = n for n = 1, 3, 5, 7, 9, 11.
The one closest to the natural frequency w0 is ω3 = 3. Hence we expect
that the dominant normal mode is y3 .
More explicitly, for all odd n,
0.05n
10.01 − n2
4
4
αn = −
and
β
=
·
· ,
n
2
2
2
2
2
2
(10.01 − n ) + (0.05n) nπ
(10.01 − n ) + (0.05n) nπ
so that the amplitude of each normal mode is given by

1
2
p
p
·√
if n is odd,
2
2
2
2
2
nπ
(10.01 − n ) + (0.05n)
Cn = αn + βn =

0
if n is even.
We will not do this rigorously, but point out that one can calculate the
n-value that maximizes Cn using elementary calculus techniques. In
fact, as we can see from the formula for Cn , Cn decays to zero as n
becomes large. Hence it suffices to compare the nonzero values of Cn
for the first few n-values to find out the dominant normal mode:
C1 = 0.125234383226,
C3 = 0.638021878839, ← maximum amplitude that corresponds
to the dominant normal mode
C5 = 0.0336595289182,
C7 = 0.0109379342144,
C9 = 0.00529819456993,
C11 = 0.00306527563502,
C13 = 0.00196838453564,
C15 = 0.00135515361343.