Math 2280-001
Fri Feb 27
Section 3.6: forced oscillations in mechanical systems (and as we shall see in section 3.7, also in electrical
circuits) overview:
We study solutions x t to
m x##C c x#C k x = F0 cos w t
using section 3.5 undetermined coefficients algorithms.
,
undamped c = 0 :
In this case the complementary homogeneous differential equation for x t is
m x##C k x = 0
k
x##C
x=0
m
2
x##C w0 x = 0
which has simple harmonic motion solutions
xH t = c1 cos w0 t C c2 sin w0 t = C0 cos w0 t K a .
So for the non-homongeneous DE the section 5.5 method of undetermined coefficients implies we can find
particular and general solutions as follows:
k
, w s w0 d
0 xP = A cos w t because only even derivatives, we don't need
m
sin w t terms !!
0 x = xP C xH = A cos w t C C0 cos w0 t K a 0 .
,
w s w0 but w z w0 , A z C0 Beating!
,
w = w0 case 2 section 3.5 undetermined coefficients; since
2
p r = r2 C w0 = r C iw0 1 r K iw0 1
our undetermined coefficients guess is
xP = t1 A cos w0 t C B sin w0 t
0 x = xP C xH = C t cos w t K a C C0 cos w0 t K a 0 .
("pure" resonance!)
,
damped c O 0 :
in all cases xP = A cos w t C B sin w t = C cos w t K a (because the
roots of the characteristic polynomial are never purely imagninary G i w when c O 0 ).
, underdamped:
x = xP C xH = C cos w t K a C eKp t C1 cos w1 t K a 1 .
,
critically-damped: x = xP C xH = C cos w t K a C eKp t c1 t C c2 .
,
over-damped:
Kr t
x = xP C xH = C cos w t K a C c1 e
1
Kr t
C c2 e
2
.
, in all three cases on the previous page, xH t /0 exponentially and is called the transient solution
xtr t (because it disappears as t/N).
xP t as above is called the steady periodic solution xsp t (because it is
what persists as t/N, and because it's periodic).
F0
, if c is small enough and w z w0 then the amplitude C of xsp t can be large relative to
, and
m
the system can exhibit practical resonance. This can be an important phenomenon in electrical circuits,
where amplifying signals is important. We don't generally want pure resonance or practical resonance in
mechanical configurations.
Forced undamped oscillations:
Exercise 1a) Solve the initial value problem for x t :
x##C 9 x = 80 cos 5 t
x 0 =0
x# 0 = 0 .
1b) This superposition of two sinusoidal functions is periodic because there is a common multiple of their
(shortest) periods. What is this (common) period?
1c) Compare your solution and reasoning with the display at the bottom of this page.
> with plots :
> plot1 d plot K5$cos 5$t , t = 0 ..10, color = green, style = point :
plot2 d plot 5$cos 3$t , t = 0 ..10, color = blue, style = point :
plot3 d plot K5$cos 5$t C 5$cos 3$t , t = 0 ..10, color = black :
display plot1, plot2, plot3 , title ='superposition' ;
superposition
8
4
0
K4
K8
2
4
t
6
8
10
In general:
There is an interesting beating phenomenon for w z w0 (but still with w s w0 ). This is explained
analytically via trig identities, and is familiar to musicians in the context of superposed sound waves
(which satisfy the homogeneous linear "wave equation" partial differential equation):
cos a K b K cos a C b = cos a cos b C sin a sin b
K cos a cos b K sin a sin b
= 2 sin a sin b .
1
1
Set a =
w C w0 t, b =
w K w0 t in the identity above, to rewrite the first term in x t as a
2
2
product rather than a difference:
F
xt =
0
2
2
m w K w0
2 sin
1
2 w C w0 t sin
v
1
0
w
K
w
t
C
x
cos
w
t
C
sin w0t
0
0
2
0
w0
.
In this product of sinusoidal functions, the first one has angular frequency and period close to the original
angular frequencies and periods of the original sum. But the second sinusoidal function has small angular
frequency and long period, given by
1
4p
angular frequency:
w K w0 ,
period:
.
2
w K w0
We will call half that period the beating period, as explained by the next exercise:
2 F0
2p
beating period:
,
beating amplitude:
2
2
w K w0
m w Kw
.
0
Exercise 2a) Use one of the formulas on the previous page to write down the IVP solution x t to
x##C 9 x = 80 cos 3.1 t
x 0 =0
x# 0 = 0 .
2b) Compute the beating period and amplitude. Compare to the graph shown below.
> plot 262.3$sin 3.05$t sin .05$t , t = 0 ..100, color = black, title = `beating` ;
beating
200
0
K200
20
40
60
t
Resonance:
(You can also get this solution by letting w/w0 in the beating formula.)
80
100
Exercise 3a) Solve the IVP
x##C 9 x = 80 cos 3 t
x 0 =0
x# 0 = 0 .
Either use the general solution formula above this exercise and substitute in the appropriate values for the
various terms, or use this as a chance to practice variation of parameters for the particular solution.
3b) Compare the solution graph below with the beating graph in exercise 2.
> plot
40
3 t$sin 3$ t , t = 0 ..40, color = black, title = `resonance` ;
resonance
400
0
K400
>
10
20
t
30
40
Damped forced oscillations c O 0 for x t :
m x##C c x#C k x = F0 cos w t
Undetermined coefficients for xP t :
k xP = A cos w t C B sin w t
C c xP #=KA w sin w t C B w cos w t
2
2
C m xP ##=KA w cos w t K B w sin w t
.
_________________________________________
2
L xP = cos w t
k ACc B wKm A w
2
C sin w t
k BKc A wKm B w
.
Collecting and equating coefficients yields the matrix system
kKm w
2
Kc w
cw
A
kKm w
B
2
F0
=
,
0
which has solution
A
B
=
kKm w
1
kKm w
2 2
2
C c2w
2
cw
Kc w
F0
kKm w
2
F0
=
0
kKm w
kKm w
2 2
C c2w
2
2
cw
In amplitude-phase form this reads
xP = A cos w t C B sin w t = C cos w t K a
with
F0
C=
. (Check!)
kKm w
cos a =
2 2
C c2w
2
kKm w
kKm w
sin a =
2
2 2
C c2w
2
cw
kKm w
2 2
.
2
C c2w
And the general solution x t = xP t C xH t is given by
x = xsp C xtr = C cos w t K a C eKp t C1 cos w1 t K a 1 .
,
underdamped:
,
critically-damped: x = xsp C xtr = C cos w t K a C eKp t c1 t C c2 .
, over-damped:
Important to note:
Kr t
x = xsp C xtr = C cos w t K a C c1 e
F0
1
Kr t
C c2 e
2
.
if w z w0 and c z 0 , because the
m
denominator will then be close to zero. This phenomenon is practical resonance.
, The phase angle a is always in the first or second quadrant.
,
The amplitude C in xsp can be quite large relative to
Exercise 4) ( a cool M.I.T. video.) Here is practical resonance in a mechanical mass-spring demo. Notice
that our math on the previous page exactly predicts when the steady periodic solution is in-phase and when
it is out of phase with the driving force, for small damping coefficient c! Namely, for c small, when
2
2
2
2
w !! w0 we have a near zero (in phase) for xsp ,because sin a z 0, cos a z 1; when w OO w0
we have a near p (out of phase), because sin a z 0, cos a zK1; for w z w0 , a is near
p
,
2
because sin a z 1, cos a z 0 .
http://www.youtube.com/watch?v=aZNnwQ8HJHU
Exercise 5) Solve the IVP for x t :
x##C 2 x#C 26 x = 82 cos 4 t
x 0 =6
x# 0 = 0 .
Solution:
x t =
41 cos 4 t K a C 10 eKt cos 5 t K b
a = arctan 0.8 , b = arctan K3 .
> with DEtools :
> dsolve x## t C 2$x# t C 26$x t = 82$cos 4$t , x 0 = 6, x# 0 = 0 ;
x t = K3 eKt sin 5 t C eKt cos 5 t C 5 cos 4 t C 4 sin 4 t
(1)
Practical resonance: The steady periodic amplitude C for damped forced oscillations (two pages back) is
F0
C w =
.
kKm w
Notice that as w/0, C w /
F0
k
2 2
C c2w
2
and that as w/N, C w /0. The precise definition of practical
resonance occuring is that C w have a global maximum greater than
F0
k
, on the interval 0 ! w !N .
2
(Because the expression inside the square-root, in the denominator of C w is quadratic in w it will have
2
at most one minimum in the variable w , so C w will have at most one maximum for non-negative w. It
will either be at w = 0 or for w O 0, and the latter case is practical resonance.)
Exercise 6a) Compute C w for the damped forced oscillator equation related to the previous exercise,
except with varying damping coefficient c:
x##C c x#C 26 x = 82 cos w t .
6b) Investigate practical resonance graphically, for c = 2 and for some other values as well. Then use
Calculus to test verify practical resonance when c = 2.
> with plots :
82
> C d w, c /
2 2
:
2
26 K w
C c2 $w
> plot C w, 2 , w = 0 ..20, color = black, title = `practical resonance when c=2` ;
practical resonance when c=2
8
5
3
1
0
>
5
10
w
15
20