MAP 2302 In-Class Activity #2 - Fall 2011
(1) Recall that according to Newton’s Second Law of motion,
F = ma.
But we know that
a=
dv
dt
and
dx
.
dt
Thus, Newton’s Second Law of motion can be written as
v=
d2 x
.
dt2
According to Hooke’s Law, when an object attached to a spring is displaced a
small distance from its resting position (its equilibrium position), the restoring
force is proportional to the displacement, but in the opposite direction as the
displacing force. In other words,
F =m
F = −kx
where k is the spring constant which determines the strength of the restoring
force. Thus, the equation of motion for am object attached to a spring sliding
along a frictionless surface is
d2 x
k
d2 x
=
−kx
or
+ x = 0.
2
2
dt
dt
m
(a) Suppose a spring is stretched one unit to the right and is at rest when it
is released; i.e., x(0) = 1 and x0 (0) = 0. Determine the solution to the
differential equation in terms of k and m under these conditions.
m
(b) The frequency, f , with which a spring oscillates is the reciprocal of the period,
T ; i.e., f = T1 . Find the frequency with which the spring in the previous part
oscillates. What happens to the frequency as m increases? Is this what you
would expect? Why or why not?
(c) Now suppose the object slides along a surface with friction. Recall that at
low speeds, friction is proportional to velocity. Thus, the force due to friction
is
dx
Ff = −λv = −λ .
dt
In this case the equation of motion becomes
dx
d2 x
λ dx
k
d2 x
=
−kx
−
λ
or
+
+ x = 0.
2
2
dt
dt
dt
m dt
m
Suppose that m = 2, λ = 4, and k = 202. Determine the solution to the
differential equation using the same initial conditions as previously.
m
(d) Determine the eventual fate of the object’s motion; that is, determine
lim x(t).
t→∞
(e) Graph the solution from t = 0 to t = 6. Note: If you do not have a graphing
calculator, you can go to the website www.wolframalpha.com which is free.
The command to graph is
plot 0 your f unction here0 , t = 0 to t = 6, y = −1 to 1
(2) Do problems 1 - 5 in the project on the following pages.
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Project4
Bungee Jumping
To the Instructor:
You might want to wait until Chapter 5 is covered to assign
http://www.idahoparks.org//Park_Pictures/Malad_Gorge/index.htm
this project.
Figure 1 The falls of the Malad River. Note
the pedestrian bridge
Suppose that you have no sense. Suppose that you are standing on a
bridge above the Malad River Canyon* and that you plan to jump off
that bridge. See Figure 1. You have no suicide wish. Instead, you plan
to attach a bungee cord to your feet, to dive gracefully into the void,
and to be pulled back gently by the cord before you hit the river that is
174 feet below. You have brought a number of different cords to affix to
your feet, including several standard bungee cords, a climbing rope, and
a steel cable. You need to choose the stiffness and length of a cord to
avoid the unpleasantness associated with an unexpected water landing.
You are undaunted by this task, because you know math!
Each of the cords you have brought will be tied off so as to be
100 feet long when hanging from the bridge. Call the position at the
bottom of the cord 0, and measure the position of your feet below that
“natural length” as x(t), where x increases as you go down and is
a function of time t. See Figure 2. Then at the time you jump,
x(0) 100, and if your 6-foot frame hits the water head first, then at
that time x(t) 174 100 6 68.
You know that the acceleration due to gravity is a constant g, so the
force pulling downwards on your body is mg. You know that when you
leap from the bridge, air resistance will increase proportionally to your
speed, providing a force in the opposite direction to your motion of
about v, where is a constant and v is your velocity. Finally, you
know that Hooke’s law describing the action of springs says that the
bungee cord will eventually exert a force on you proportional to your
distance past the natural length of the cord. Thus you know that the
force of the cord pulling you back from destruction can be expressed as
b(x) 100 ft
x ( t)
0,kx,
x0
x 0.
(1)
The number k 0 in (1) is called the spring constant and is where the stiffness of the
cord you use influences the equation. For example, if you used the steel cable, then k
would be very large, giving a tremendous stopping force very suddenly as you passed
the natural length of the cable. This could lead to discomfort, injury, or even a Darwin
award. You want to choose the cord with a value of k large enough to stop you above
or just touching the water but not too suddenly. Consequently, you are interested in
finding the distance you fall below the natural length of the cord as a function of the
spring constant. To do that, you must solve the differential equation that we have
derived in words above: The net force mx on your body is given by
mx mg b(x) x.
(2)
Here mg is your weight, 160 pounds, and x is the rate of change of your position
below the equilibrium with respect to time — that is, your velocity. The constant for air resistance depends on a number of things, including whether you wear your
Figure 2 The fall from the
bridge. As depicted here, x(t) 0
*The Malad River Canyon is located in the Malad Gorge State Park near Hagerman, Idaho.
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Project 4 BUNGEE JUMPING
skin-tight pink Spandex or your skater shorts and XXL T-shirt, but you know that
the value today is about 1.
Differential equation (2) is nonlinear, but inside it are two linear equations
struggling to get out. You know how to solve such equations from your work in
Chapters 4 and 5. When x 0, the equation is mx mg x, while after you
pass the natural length of the cord it is mx mg kx x. You solve each
equation separately and then piece the solutions together when x(t) 0.
PROBLEM 1. Solve the equation mx x mg for x(t), given that
you step off the bridge — that is, no jumping, no diving! “Stepping off”
means that the initial conditions are x(0) 100, x(0) 0. Use mg 160,
1, and g 32.
PROBLEM 2. Use the solution from Problem 1 to compute the length
of time you free-fall (that is, the time it takes to go the natural length of the
cord: 100 feet).
PROBLEM 3. Compute the derivative of the solution you found in
Problem 1 and evaluate it at the time you found in Problem 2. You have
found your downward speed when you pass the point where the cord starts
to pull.
Problem 1 has given you an expression for your position t seconds after you
step off the bridge, before the bungee cord starts to pull you back. Notice that it does
not depend on the value of k. When you pass the natural length of the bungee cord, it
does start to pull back, so the differential equation changes. Let t1 denote the time
you computed in Problem 2, and v1 denote the speed you calculated in Problem 3.
PROBLEM 4. Solve the initial-value problem
mx x kx mg,
x(t 1) 0,
x(t 1) v1.
For now you may use the value k 14, but eventually you will need to replace
this number with the values of k for the cords you brought. The solution x(t)
represents your position below the natural length of the cord after it starts to
pull back.
Now you have an expression for your position as the cord pulls on your body.
All you have to do is find the time t2 at which you stop going down. When you stop
going down, your velocity is zero — that is, x(t2 ) 0.
PROBLEM 5. Compute the derivative of the expression you found in
Problem 4 and solve for the value of t where the derivative is zero. Denote
this time as t2. Be careful that the time you compute is greater than t1 — there
are several times when your motion stops at the top and bottom of your
bounces! After you find t2, substitute it back into the solution you found in
Problem 4 to find your lowest position.
PROBLEM 6 (CAS). You have brought a soft bungee cord with k 8.5,
a stiffer cord with k 10.7, and a climbing rope for which k 16.4. Which, if
any, of these cords can you use safely under the given conditions?
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Project 4 BUNGEE JUMPING
193
As you see, knowing a little bit of math is a dangerous thing. The assumption
that the drag due to air resistance is linear applies only for low speeds. By the time
you swoop past the natural length of the cord, that approximation is only wishful
thinking, so your actual mileage may vary. Moreover, springs behave nonlinearly
in large oscillations, so Hooke’s law is only an approximation. Do not trust your
life to an approximation made by a man who has been dead for two hundred years.
Leave bungee jumping to the professionals.
Still Curious?
PROBLEM 7. You have a bungee cord for which you have not determined the spring constant k. To do so, you suspend a weight of 10 pounds
from the end of the 100-foot cord, causing it to stretch 1.2 feet. What is the
value of k for this cord?
PROBLEM 8 (CAS). What would happen if your 220-pound friend uses
the bungee cord whose spring constant is k 10.7?
PROBLEM 9 (CAS). If your heavy friend wants to jump anyway, then
how short should you make the cord so that he does not get wet?
PROBLEM 10 (CAS). Graph the solutions you found in Problems 1
and 2 on the same coordinate axes. Explain the differences.
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Project 4
Bungee Jumping
1. Rewrite the equation in the form
x"
1, and mg = 160 with g
We have f3
homogeneous equation is ,A2 +!,A
dUELto gravity
=
f3 x' - g = O.
m
<
32 implies m = 5, so the characteristic polynomial for the 0, with roots given by ,A
a solution of form xp(t)
=
0 and ,A
!. The forcing function 160t, so we conclude that the formal solution mgt
"~
IS
x(t)
-100 so A
Now x(O)
B
=
B
+ 160t. o implies
-100, while x'(O)
160. Thus A
A/5
=
800 and B = -900, so x(t) = 800e- t / 5
900 + 160t. -
2. We simply need to solve for the time at which x(t)
800e- t / 5
-
O. In other words, we require
900
160t
O.
We use a nonlinear solver like Newton's method to find that the time is approximately tl
=
2.73s.
3. The derivative of our solution is
x'(t)
so x(2.73)
~
=
160( _e- t / 5 + 1),
!I
67.26ft/s. Note that the answer given used a more precise approximation for the answer to Problem 2. Substituting 2.73 in directly gives a number more like 67.3. II
4. This time our characteristic polynomial is ,A2 + ,A/5 + k/5
1
,A = -10
11k
VlOO-'5
and O. This gives complex roots
1
10
I~
11k
VlOO-'5 for k > 1/20. Thus, the formal solution to the homogeneous equation is Ae-t / l0 sin(wt)
Xh(t)
+ Be- t / IO cos(wt) for w = /-1/100 + k/5. The particular solution for this equation.is found using xp(t)
the equation kG = mg, or xp(t)
G
G, giving = mg/k. Now we substitute the initial condition x(2.73)
to find that 221 0
Project 4
Bungee Jumping
Likewise, we take the derivative and use the other initial condition to see that
1 ( X(tl) - -_
160)
.
10
k
Note that X(tl)
any k. When k
=
we-h/10(Acos(tlW)
Bsin(tlw»)
67.26.
O. Thus, we have two simple simultaneous equations to solve for A and B: given
14, then W ~ 1.67, so e-tl/ 10 sin(tlw) ~ -0.75 and e-tI/lO COS(tlW) -0.12. The
system becomes approximately
-0.75A - 0.12B
-0.20wA
=
160
14
0.75wB = 67.26 _ 160
140'
vVe solve this to arrive at a solution given approximately by
x(t)
6.66e- t / 10 sin(wt)
+- 53.70e- t / IO cos(wt)
160
14
Note that all calculations were actually carried out using greater precision than that indicated.
In general, given
al2
=
k we compute w
jk/5 -
0.1 and then compute all =
/10 sin(tlw) and
e- t l/ lO COS(tlW), \Ve then solve the system
"I
anA
al2wA
al2 B
160
k
allwB = 67.26
160
10k
to find the solution to the initial-value problem. There are evidently other ways to pull this together.
For example, Maple can be used to combine all the terms in the second equation instead of using
'1
_I
III
X(tl) = O. Students could thus use this computation to solve many of the remaining problems, but
they are not really expected to do so that's what the CD is for.
5. The derivative of the solution from
Probl~~'4 is x'(t) ~ -90.37e-t / 10 sin(wt)+ 5.76e- t / 1O cos(wt).
We must use a numerical solver to find zeros of this, There is one zero at about t
0:0'38 s, and
others periodically at intervals of 7r/w. Since 0.038s represents a time when you are falling freely,
it is not the solution we want, and neither is the next one at 1.92 s. \Ve use, instead, the next time
that the derivative vanishes, at about t = 3.80s. At that time x(3.80)
~
48.37ft, that is, about 148
feet below the bridge. You had about 168 feet to work with, assuming you didn't want to get wet,
so you have plenty of space to spare.
6. We could use the analytical solutions computed in Problem 4 to do this problem, but that would
be painful. That's why it is marked as a computer problem. vVe use the tool on the CD to plot
solutions for the given k values, and see which of them allow you to hit the water. The tool indicates
that you can probably use any of the bungee cords given with safety, though the one with k = 8.5
222 Project 4
Bungee Jumping
might let you get wet. Figures showing the position of the jumper indicate the relevant path as the
jumper approaches the water.
7. In this case, mg = 10. When the 10 Ib weight comes to rest, we have x, = xl! = 0, so kx = mg.
Since the weight has stretched the cord by 1.2 ft, we conclude that k(1.2)
=
10, or k : : : : 8.3.
8. We use the bungee tool from the CD here again. Changing the weight and k value and clicking "step
off", we see that our friend is at least going to get wet.
horizontal line at the top represents
the water (don't forget that our coordinate system is inverted the positive direction is downward).
Remember that our friend is probably about 6 feet tall, so when the end of the cord scrapes the
water, he is quite wet. Inasmuch as the water is not
deep, he had better not take the chance
on th.!:s jump. He can, however, survive the jump on the cord with k
=
14.
9. We use the CD tool to solve this, since that course is dramatically easier than trying to work
backwards from the ending point to
t~ intitial
condition. When the cord is shortened to about 96
feet (Le. the initial condition is changed'to x(O)
-96), then our friend seems to be able to jump
without getting more than a few hairs wet.
I••••
,t
11111UII
., 223