Chapter 5. Applications of Higher-Order Differential Equations
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322
Chapter 5 Applications of Higher-Order Differential Equations
acting on the mass, then we determine the differential equation that models this
situation in the following way:
m
At equilibrium ks
d2x
dt 2
forces acting on the system
ks
x
mg
ks
kx
mg.
mg, so after simpliﬁcation, we obtain the differential equation
m
d2x
dt 2
kx
or
m
d2x
dt 2
kx
0.
The two initial conditions that are used with this problem are the initial displacement x 0
Α and initial velocity dx/ dt 0
Β. Hence, the function x t that
describes the displacement of the mass with respect to the equilibrium position
is found by solving the initial-value problem
d2x
kx 0
dt 2
dx
x0
Α,
0
dt
m
Β.
(5.1)
The differential equation in initial-value problem (5.1) disregards all retarding
forces acting on the motion of the mass.
The solution x t to this problem represents the displacement of the mass at
time t. Based on the assumptions made in deriving the differential equation (the
positive direction is down), positive values of x t indicate that the mass is beneath
the equilibrium position while negative values of x t indicate that the mass is
above the equilibrium position.
EXAMPLE 5.1.1: A mass weighing 60 lb stretches a spring 6 inches.
Determine the function x t that describes the displacement of the mass
if the mass is released from rest 12 inches below the equilibrium
position.
SOLUTION: First, the spring constant k must be determined from the
given information. By Hooke’s law, F
ks, so we have 60
k 0.5.
Therefore, k 120 lb/ft. Next, the mass, m, must be determined using
F mg. In this case, 60 m 32, so m 15/ 8 slugs. Because k/ m 64
5.1 Harmonic Motion
323
1
0.5
0.25
0.5
0.75
1
1.25
1.5
-0.5
-1
Figure 5-1 Simple harmonic motion
and 12 inches equals 1 foot, the initial-value problem that needs to be
solved is
64x 0
x
x0
1, x 0
0.
This problem is now solved with DSolve, and the resulting output is
named de1.
In[833]:= Clear x, t, de1
de1
64 x t
DSolve x t
x 0
0 ,x t ,t
Out[833]=
x t
Cos 8 t
0, x 0
1,
We graph the solution with Plot in Figure 5-1.
In[834]:= Plot x t /.de1, t, 0,
Π
2
In order to better understand the relationship between the formula
obtained in this example and the motion of the mass on the spring,
an alternate approach is taken here. We begin by deﬁning sol to be the
solution to the initial-value problem: given t, sol[t] returns the value
of cos 8t.
In[835]:= Clear sol
In[836]:= sol t
Out[836]= Cos 8 t
de1
1, 1, 2
Then, the function zigzag is deﬁned to produce a list of points joined
by line segments to represent the graphics of a spring. Given ordered
324
Chapter 5 Applications of Higher-Order Differential Equations
pairs a, b and c, d , a positive integer n, and a “small” number Ε,
zigzag[{a,b},{c,d},n,eps] connects the set of points
a, b , a
Ε, b
d
b
n
, a
Ε, b
2
d
b
n
a
Note that we will always have
a c.
,..., a
1
n 1
Ε, b
1 i Ε, b
n
1
i
d
d
b
n
b
n
,...
, c, d
with line segments.
In[837]:= Clear spring, zigzag, length, points, pairs
zigzag
a ,b , c ,d ,n ,Ε
Module length, points, pairs ,
length d b
i length
, i, 1, n 1
points Table b
n
i
pairs Table a
1 Ε, points i ,
i, 1, n 1
PrependTo pairs, a, b
AppendTo pairs, c, d
Line pairs
The function spring produces the graphics of a point (the mass
attached to the end of the spring) as well as that of the spring obtained
with zigzag. The result of entering spring[t] when displayed with
Show looks like a spring with a mass attached.
In[838]:= spring t
Show
Graphics zigzag 0, sol t ,
0, 1 , 20, 0.05 , PointSize 0.075 ,
Point 0, sol t
, Axes Automatic,
AxesStyle GrayLevel 0.5 , Ticks None,
3 3
PlotRange
1, 1 ,
,
,
2 2
AspectRatio 1, DisplayFunction Identity
A list of graphics is produced in somegraphs for values of t from t
to t Π/ 2 using increments of Π/ 16.
Π Π
,
2 16
-Graphics-, -Graphics-, -Graphics-,
-Graphics-, -Graphics-, -Graphics-,
-Graphics-, -Graphics-, -Graphics-
In[839]:= somegraphs
Out[839]=
Table spring t , t, 0,
0
5.1 Harmonic Motion
325
This list of nine graphics objects is then partitioned into groups of three
with Partition in toshow for use with GraphicsArray.
In[840]:= toshow
Out[840]=
Partition somegraphs, 3
-Graphics-, -Graphics-, -Graphics- ,
-Graphics-, -Graphics-, -Graphics- ,
-Graphics-, -Graphics-, -Graphics-
We then display the array of graphics objects toshow with Show and
GraphicsArray in Figure 5-2. We see that the plots displayed show
Figure 5-2 Simple harmonic motion: a spring
326
Chapter 5 Applications of Higher-Order Differential Equations
the displacement of the mass at the values of time from t
using increments of Π/ 16.
0 to t
Π/ 2
In[841]:= Show GraphicsArray toshow
In order to achieve an animation so that we can see the motion of the
spring, we use a Do loop. For example, entering
In[842]:= Do Show spring t ,
DisplayFunction
Π
Π
t, 0, ,
2 118
$DisplayFunction ,
displays spring[t] for t-values from t
0 to t
Π/ 2 using increments of Π/ 118. To animate these graphs, select the cell bracket of the
graphs to be animated, go to the menu under Cell and select Animate
Selected Graphics. Alternatively, after selecting the graphs to be animated, you can use the keyboard shortcut Command-Y to animate the
selected graphics.
When these graphs are animated, as indicated in the following screen
shot, we can see the motion of the spring.
Remember that positive values of x t indicate that the mass is beneath
the equilibrium position while negative values of x t indicate that the
5.1 Harmonic Motion
327
1
1
1
1
-1
-1
1
1
1
1
-1
-1
1
1
1
1
-1
-1
Figure 5-3 Simple harmonic motion illustrated with a spring and a plot
mass is above the equilibrium position. To see this, we graph x t in
Figure 5-3.
In[843]:= graph
Π
,
2
GrayLevel 0.3 ,
GrayLevel 0.6 ,
Plot sol t , t, 0,
PlotStyle
AxesStyle
Ticks
1 ,
PlotRange
AspectRatio
1, 1 ,
3 3
,
,
2 2
1,
DisplayFunction
Identity
Then, we deﬁne p. Given t, p[t] generates a graphics object consisting
of the graph of x t on the interval 0, Π/ 2 , which is named graph, and
a “small” point placed at t, x t .
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Chapter 5 Applications of Higher-Order Differential Equations
In[844]:= p t
Module dp ,
dp
Graphics PointSize 0.07 ,
Point t, sol t
Show graph,
dp, DisplayFunction Identity
We then use Table and GraphicsArray to generate a set of graphics
objects consisting of the graphs of spring[t] and p[t], shown sideby-side, for t-values from t 0 to t Π/ 2 using increments of Π/ 10.
In[845]:= moregraphs
Table GraphicsArray
Π Π
p t
, t, 0, ,
2 10
spring t ,
The list moregraphs is then partitioned into two element subsets and
displayed using Show and GraphicsArray in Figure 5-3.
In[846]:= toshow
Partition moregraphs, 2
In[847]:= Show GraphicsArray toshow
As before, we can use a Do loop to generate several graphs and animate
the result to see the motion of the spring, as indicated in the following
screen shot.
In[848]:= graphs
Do Show GraphicsArray spring t , p t
DisplayFunction $DisplayFunction ,
Π
Π
t, 0, ,
2 118
,
5.1 Harmonic Motion
329
Notice that the displacement function x t
cos 8t indicates that the
spring–mass system never comes to rest once it is set into motion. The
solution is periodic, so the mass moves vertically, retracing its motion.
Hence, motion of this type is called simple harmonic motion.
EXAMPLE 5.1.2: An object with mass m 1 slug is attached to a spring
with spring constant k 4. (a) Determine the displacement function of
0. Plot the solution for Α
1, 4, 2.
the object if x 0
Α and x 0
How does varying the value of Α affect the solution? Does it change the
values of t at which the mass passes through the equilibrium position?
(b) Determine the displacement function of the object if x 0
0 and
Β. Plot the solution for Β 1, 4, 2. How does varying the value
x 0
of Β affect the solution? Does it change the values of t at which the mass
passes through the equilibrium position?
SOLUTION: For (a), the initial-value problem we need to solve is
x
x0
4x
0
Α, x 0
0
for Α
1, 4, 2. We now determine the solution to each of the three
problems with DSolve. For example, entering
In[849]:= Clear x
de2
Out[849]=
DSolve x t
4x t
x 0
0 ,x t ,t
x t
Cos 2 t
0, x 0
1,
solves the initial-value problem if Α 1 and names the result de2. Note
that the formula for the solution is the second part of the ﬁrst part of
the ﬁrst part of de2 and is extracted from de2 with Part ([[...]])
by entering de2[[1,1,2]]. Alternatively, if you are using Version 5
you can select and copy the formula in the output and paste it to any
location. Similarly, entering
In[850]:= de3
Out[850]=
DSolve x t
4x t
x 0
0 ,x t ,t
x t
4 Cos 2 t
0, x 0
4,
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Chapter 5 Applications of Higher-Order Differential Equations
4
2
0.5
1
1.5
2
2.5
3
-2
-4
Figure 5-4 Simple harmonic motion: varying the initial displacement
In[851]:= de4
0, x 0
DSolve x t
4x t
x 0
0 ,x t ,t
x t
2 Cos 2 t
Out[851]=
2,
solves
x
x0
4x
0
4, x 0
and
0
x
x0
4x
0
2, x 0
0
naming the results de3 and de4, respectively. We graph the solutions
on the interval 0, Π with Plot in Figure 5-4. Note how we use Map to
extract the formula for each solution from de2, de3, and de4.
In[852]:= toplot Map # 1, 1, 2 &, de2, de3, de4
Out[852]= Cos 2 t , 4 Cos 2 t , 2 Cos 2 t
In[853]:= Plot Evaluate toplot , t, 0, Π ,
PlotStyle
GrayLevel 0 , GrayLevel 0.3 ,
GrayLevel 0.6
We see that the initial position affects only the amplitude of the function (and direction in the case of the negative initial position). The mass
passes through the equilibrium position (x 0) at the same time in all
three cases.
For (b), we need to solve the initial-value problem
x
x0
4x
0
0, x 0
Β
for Β
1, 4, 2. In this case, we deﬁne a procedure d that, given Β,
returns the solution to the initial-value problem.
5.1 Harmonic Motion
331
2
1
1
2
3
4
5
6
-1
-2
Figure 5-5 Simple harmonic motion: varying the initial velocity
In[854]:= d Β
Module
,
DSolve x t
x 0
0, x 0
4x t
0,
Β ,x t ,t
We then use Map to apply d to the list of numbers {1,4,-2} and
name the resulting output solutions. (Note that the same result is
obtained by using the keyboard shortcut for Map, /@, and entering
solutions=d/@{1,4,-2}.)
In[855]:= solutions
Out[855]=
x t
x t
Map d, 1, 4, 2
1
Sin 2 t
,
2
2 Sin 2 t
, x t
Sin 2 t
We see that solutions consists of three lists. For example, the solution
to the initial-value problem when Β
2 is contained in the third list in
solutions. We now extract the formula for the solution with Part
([[...]]).
In[856]:= solutions
Out[856]=
3, 1 , 1 , 2
Sin 2 t
All three solutions are graphed together on 0, 2Π with Plot in
Figure 5-5.
In[857]:= Plot Evaluate x t /.solutions , t, 0, 2Π ,
PlotStyle
GrayLevel 0 , GrayLevel 0.3 ,
GrayLevel 0.6
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Chapter 5 Applications of Higher-Order Differential Equations
Notice that varying the initial velocity affects the amplitude (and
direction in the case of the negative initial velocity) of each function.
The mass passes through the equilibrium position at the same time in
all three cases.
5.1.2 Damped Motion
Equation (5.1) disregards all retarding forces acting on the motion of the mass and
a more realistic model which takes these forces into account is needed.
Studies in mechanics reveal that resistive forces due to damping are functions
c dx/ dt, FR
c dx/ dt 3 , or
of the velocity of the motion. Hence, for c > 0, FR
FR c sgn dx/ dt , where
dx
sgn
dt
1, dx/ dt > 0
0, dx/ dt
0
1, dx/ dt < 0
are typically used to represent the damping force. Incorporating damping into
c dx/ dt, the displacement function, x t , is
equation (5.1) and assuming that FR
found by solving the initial-value problem
dx
d2x
kx 0
c
dt 2
dt
dx
x0
Α,
0
Β.
dt
m
This calculation is identical to
those followed in Chapter 4
for second-order linear
homogeneous equations with
constant coefﬁcients.
(5.2)
From our experience with second-order ordinary differential equations with constant coefﬁcients in Chapter 4, the solutions to initial-value problems of this type
greatly depend on the values of m, k, and c.
Suppose we assume that solutions of the differential equation have the form
rert and x
r2 ert , we have by substitution into the differxt
ert . Because x
2 kt
rt
rt
ential equation mr e
cre
ke
0, so ert mr2 cr k
0. The solutions to the
characteristic equation are
c2 4mk
c
.
r
2a
Hence, the solution depends on the value of the quantity c2 4mk. In fact, problems
of this type are characterized by the value of c2 4mk as follows.
1. c2 4mk > 0. This situation is said to be overdamped because the damping
coefﬁcient c is large in comparison to the spring constant k.