Math 2280-001
Fri Apr 24
, Finish Wednesday's notes first. Then ...
9.4 Forced oscillation problems via Fourier Series. Today we will revisit the forced oscillation problems
of last Friday, where we predicted whether or not resonance would occur, and then tested our predictions
with the convolution solutions. Using Fourier series expansions for the forcing function one can say
precisely whether or not there will be resonance. We will be studying the differential equations
2
x## t C c x# t C w0 x t = f t
for various forcing functions f t . (We have divided the original mass-spring DE by the mass m and
relabeled the damping coefficient and forcing functions.) For most of the lecture we consider undamped
configurations, c = 0.
Warm-up Exercise 1) (This was the final exercise last Friday.)
Use superposition to find particular solutions, and discuss whether or not resonance will occur in the
following two forced oscillation problems. Notice that the period of the forcing function is 6 p in a (not
the natural period). In b the period of the forcing function is 2 p. And yet, the resonance occurs in a, and
not in b.
a)
x## t C x t = cos t C sin
t
3
.
b)
x## t C x t = cos 2 t K 3 sin 3 t C cos 6 t .
Hint: There's a table of particular solutions at the end of today's notes.
Exercise 1 is indicative of how we can understand resonance phenomena for forced oscillation problems
with general periodic forcing functions f : Consider the undamped forced DE
2
x## t C w0 x t = f t ,
where f t has period P = 2 L. Compute the Fourier series for f:
N
N
a0
p
p
fw
C
an cos n
t C
bn sin n t
2
L
L
n=1
n=1
with
L
L
a0
1
1
a0 =
f t dt
(so
=
f t dt is the average value of f)
L
2
2L
>
>
KL
an d
bn d
KL
f, cos n
f, sin n
p
t
L
=
p
t
L
=
1
L
1
L
L
f t cos n
p
t dt, n 2 ;
L
f t sin n
p
t du, n 2 ;
L
KL
L
KL
As long as no (non-zero) term in the Fourier series has an angular frequency of w0 , there will be no
resonance. In fact, in this case the infinite sum of (undetermined coefficients) particular solutions will
converge to a bounded particular solution. For sure there will NOT be resonance if it's true for all n 2 ;
that
p
wn d n
s w0
L
but even if some wn does equal w0 there won't be resonance unless either an or bn is nonzero.
Conversely, if the Fourier series of f does contain cos w0 t or sin w0 t terms, those terms will cause
resonance.
Recall the first "resonance game" example from last Friday:
x## t C x t = square t
with
square t =
K1 Kp ! t ! 0
0!t!p
1
and 2 pKperiodic. This forcing function appeared to cause resonance:
Here's a formula for square t valid for 0 % t % 11 p , and last Friday's results:
> with plots :
10
> square d t/1 C 2$
>
K1 n $Heaviside t K n$Pi
:
n=1
plot1a d plot square t , t = 0 ..30, color = green :
display plot1a, title = `square wave forcing at natural period` ;
square wave forcing at natural period
1
K1
10
20
30
t
Convolution solution formula and graph:
t
> x1 d t/ sin t $square t K t dt :
0
plot1b d plot x1 t , t = 0 ..30, color = black :
display plot1a, plot1b , title = `resonance response ?` ;
resonance response ?
10
0
K10
10
20
30
t
Exercise 2 Use the Fourier series for square t that we've found before
4
1
sin n t
p n odd n
and infinite superposition to find a particular solution to
x## t C x t = square t
that explains why resonance occurs. Make use of the undetermined coefficients particular solution
formulas at the end of today's notes.
square t =
>
If we remove the sin t term from the square wave forcing function, and re-use the convolution formula,
we see that we've eliminated the resonance:
t
4
> x2 d t/ sin t $ square t K t K $sin tKt
p
dt :
0
plot x2 t , t = 0 ..30 ;
0.3
0
K0.3
10
20
t
30
Exercise 2) Understand Example 3 from last Friday, using Fourier series:
x## t C x t = f3 t
Example 3) Forcing not at the natural period, e.g. with a square wave having period T = 2 .
20
> f3 d t/1 C 2$
>
K1 n $Heaviside t K n :
n=1
plot3a d plot f3 t , t = 0 ..20, color = green :
display plot3a, title = `out of phase square wave forcing` ;
out of phase square wave forcing
1
0
K1
5
10
t
15
20
15
20
This forcing function did not cause resonance:
t
> x3 d t/ sin t $f3 t K t dt :
0
plot3b d plot x3 t , t = 0 ..20, color = black :
display plot3a, plot3b , title = `resonance response ?` ;
1
K1
resonance response ?
5
10
t
Hint: By rescaling we can express f3 t = square p t =
4
p
>
n odd
1
sin n p t .
n
Brute force tech check of Fourier coefficients in previous example:
> f d t/K1 C 2$Heaviside t ;
plot f t , t =K1 ..1 ;
L d 1;
f := t/K1 C 2 Heaviside t
1
K1
K0.5
K1
0.5
t
1
L := 1
1
> a0 d
$
L
(1)
L
f t dt;
KL
assume n, integer ; # this will let Maple attempt to evaluate the integrals
1
a d n/ $
L
b d n/
1
$
L
L
f t $cos
n$p
t dt :
L
f t $sin
n$p
t dt :
L
KL
L
KL
a n ;
b n ;
a0 := 0
0
K K1 n~ C 2
K
n~ p
>
>
K1 n~
n~ p
(2)
Practical resonance example:
Exercise 3 The steady periodic solution to the differential equation
x## t C .2$x# t C 1 x t = square t
exhibits practical resonance. Explain this with Fourier series. Hint: Use
4
1
sin n t
p n odd n
the table of particular solutions at the end of today's notes.
square t =
>
Particular solutions from Chapter 3 or Laplace transform table:
2
x## t C w0 x t = A sin w t
A
xP t = 2
sin w t
when w s w0
2
w0 K w
t
xP t =K
A cos w0 t
when w = w0
2 w0
................................................................................................................
2
x## t C w0 x t = A cos w t
A
xP t = 2
cos w t
when w s w0
2
w0 K w
t
xP t =
A sin w0 t
when w = w0
2 w0
............................................................................................................
2
x##C c x#C w0 x = A cos w t
cO0
xP t = xsp t = C cos w t K a
with
A
C=
2
2
w0 K w
2
.
2
C c2w
2
2
w0 K w
cos a =
2
w0 K w
2
2
C c2w
2
cw
sin a =
2
2
w0 K w
2
.
C c2w
2
.......................................................................................................
2
x##C c x#C w0 x = A sin w t
cO0
xP t = xsp t = C sin w t K a
with
C=
A
2
2
w0 K w
2
.
2
C c2w
2
cos a =
2
w0 K w
sin a =
2
w K w0
2
2
C c2w
2
cw
2
2
w0 K w
2
.
C c2w
2