Math 2280-001
Wed Feb 25
elements of 3.4, 3.5, 3.6
, Discuss the "variation of parameters" method of finding particular solutions, in Monday's notes.
, Then discuss the pendulum model below, and run the pendulum and mass-spring experiments.
, If time, give an overview of section 3.6: forced oscillations, to discuss what phenomena will arise.
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 a horizontal mass-spring configuration. The
nicest derivation for the pendulum depends on conservation of energy, as indicated below. Conservation
of energy is an important tool in deriving differential equations, in a number of different contexts. Today
we will test both the pendulum model and the mass-spring model with actual experiments (in the
undamped cases), to see if the predicted periods T = 2 p correspond to experimental reality.
w0
Pendulum: measurements and prediction (we'll check these numbers).
> restart :
Digits d 4 :
> L d 1.526;
g d 9.806;
g
wd
; # radians per second
L
f d evalf w / 2$ Pi ; # cycles per second
T d 1 / f; # seconds per cycle
Experiment:
Mass-spring:
compute Hooke's constant:
> 98.7 K 83.4; #displacement from extra 50g
.05$9.806
> kd
; # solve k$x=m$g for k.
.153
> m d .1; # mass for experiment is 100g
k
wd
; # predicted angular frequency
m
w
f d evalf
; # predicted frequency
2$Pi
1
T d ; # predicted period
f
Experiment:
We neglected the KEspring , which is small but could be adding intertia to the system and slowing down the
oscillations. We can account for this:
Improved mass-spring model
Normalize TE = KE C PE = 0 for mass hanging in equilibrium position, at rest. Then for system in
motion,
KE C PE = KEmass C KEspring C PEwork .
x
PEwork =
k s ds =
0
1 2
kx ,
2
KEmass =
1
m x# t
2
2
,
KEspring = ????
How to model KEspring ? Spring is at rest at top (where it's attached to bar), moving with velocity x# t at
bottom (where it's attached to mass). Assume it's moving with velocity µ x# t at location which is
fraction µ of the way from the top to the mass. Then we can compute KEspring as an integral with respect
to µ , as the fraction varies 0 % µ % 1 :
1
KEspring =
0
1
µ x# t
2
1
1
= mspring x# t
2
2
2
2
µ dµ=
0
mspring dµ
1
m
x# t
6 spring
2
.
Thus
TE =
1
2
mC
1
m
3 spring
x# t
2
C
1 2 1
k x = M x# t
2
2
2
C
1 2
kx ,
2
where
M= mC
1
m
3 spring
Dt TE = 0 0
M x# t x## t C k x t x# t = 0 .
x# t M x##C k x = 0 .
Since x# t = 0 only at isolated tKvalues, we deduce that the corrected equation of motion is
M x##C k x = 0
with
w0 =
k
=
M
k
1
m C mspring
3
Does this lead to a better comparison between model and experiment?
> ms d .0103; # spring has mass 10.3 g
1
M d m C $ms; # "effective mass"
3
k
; # predicted angular frequency
M
w
f d evalf
; # predicted frequency
2$Pi
1
T d ; # predicted period
f
> wd
>
.
Section 3.6: forced oscillations in mechanical systems 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 3.5 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 c1 cos w0 t C c2 sin w0 t
= 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 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.
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