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Chapter 5. Applications of Higher-Order Differential Equations

# 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 .

328

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,

330

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

332

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.

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