Math 2250-1
Tues Oct 23
5.4 Applications of 2nd order linear homogeneous DE's with constant coefficients, to unforced spring
(and related) configurations. We are focusing on this homogeneous differential equation for functions
x t , which can arise in a variety of physics situations and that we derived for a mass-spring configuration:
m x##C c x#C k x = 0 .
, Finish discussing the "simple harmonic motion" that occurs in case the damping coefficient c = 0, using
Monday's notes.
, Then discuss the possibilities that arise when the damping coefficient c O 0. There are three of them:
Case 2: damping
m x##C c x#C k x = 0
c
k
x##C
x#C
x=0
m
m
rewrite as
2
x##C 2 p x#C w0 x = 0.
(p =
c
2
k
, w0 =
2m
m
. The characteristic polynomial is
2
r2 C 2 p r C w0 = 0
which has roots
2 pG
r =K
2
4 p2 K 4w0
2
=Kp G
2
p2 K w0 .
2
2a) ( p2 O w0 , or c2 O 4 m k ). overdamped. In this case we have two negative real roots
r1 ! r2 ! 0
and
r t
r t
r t
r Kr
t
x t = c1 e 1 C c2 e 2 = e 2 c1 e 1 2 C c2 .
, solution converges to zero exponentially fast; solution passes through equilibrium location x = 0 at
most once.
c
.
2m
2
2b) ( p2 = w0 , or c2 = 4 m k ) critically damped. Double real root r1 = r2 =Kp =K
x t = eKp t c1 C c2 t .
, solution converges to zero exponentially fast, passing through x = 0 at most once, just like in the
overdamped case. The critically damped case is the transition between overdamped and underdamped:
2
2c) ( p2 ! w0 , or c2 ! 4 m k ) underdamped . Complex roots
r =Kp G
with w1 =
2
p2 K w0 =Kp G i w1
2
w0 K p2 ! w0 .
x t = eKp t A cos w1 t C B sin w1 t = eKp t C cos w1 t K a 1 .
, solution decays exponentially to zero, but oscillates infinitely often, with exponentially decaying
pseudo-amplitude eKp t C and pseudo-angular frequency w1 , and pseudo-phase angle a 1 .
Exercise 1) Classify by finding the roots of the characteristic polynomial. Then solve for x t :
1a)
x##C 6 x#C 9 x = 0
x 0 =1
3
x# 0 =
.
2
> with DEtools :
3
> dsolve x## t C 6$x# t C 9$x t = 0, x 0 = 1, x# 0 =
;
2
>
1b)
x##C 10 x#C 9 x = 0
x 0 =1
3
x# 0 =
.
2
> dsolve
3
2
x## t C 10$x# t C 9$x t = 0, x 0 = 1, x# 0 =
;
>
1c)
x##C 2 x#C 9 x = 0
x 0 =1
3
x# 0 =
.
2
> dsolve
>
x## t C 2$x# t C 9$x t = 0, x 0 = 1, x# 0 =
3
2
;
> with plots :
1
$sin 3$t , t = 0 ..4, color = red :
2
9
plot1a d plot exp K3$t $ 1 C $t , t = 0 ..4, color = green :
2
21
5
plot1b d plot
$exp Kt K
$exp K9$t , t = 0 ..4, color = blue :
16
16
5
plot1c d plot
$ 2 eKt $sin 2 2 $t C eKt $ cos 2 2 $ t , t = 0 ..4, color = black :
8
display plot0, plot1a, plot1b, plot1c , title = `IVP with all damping possibilities` ;
> plot0 d plot cos 3$t C
IVP with all damping possibilities
1
0.5
0
K0.5
K1
>
1
2
t
3
4
Small oscillation pendulum motion and vertical mass-spring motion are governed by exactly the "same"
differential equation that models the motion of the mass in our horizontal mass-spring configuration. The
nicest derivation for the pendulum depends on conservation of mass, as indicated below. Tomorrow we
will test both models with actual experiments (in the undamped cases), to see if the predicted periods
2p
T=
correspond to experimental reality.
w0