Assignment grouping 2(bungee jumping) (edit)

SMN 3043
ORDINARY DIFFERENTIAL EQUATIONS
ASSIGNMENT 3
BUNGEE JUMPING
NAME MATRIC NO.
NOORUL ASMA BINTI ABDUL SAMAT D20092036907
NUR HAFIZAH IZZATI BINTI MD RASIP D20111048854
NORFAEZAH BINTI HAMZAH D20111048860
NOR ATIQAH FATIHAH BINTI ABDULLAH D20111048872
NORIZAN BINTI NORDIN D20111048874

1.0 TWO METHODS TO SOLVE HIGHER
ORDER ODE
1) Variation of Parameters
2) Euler-Cauchy ODE

Variation of Parameters
Consider the differential equation,
Assume that y1(t) and y2(t) are a fundamental
set of solutions for
Then a particular solution to the
nonhomogeneous differential equation is,

Example 1
• Find a general solution to the following
differential equation
• The differential equation the actually be
solving is
2 18 6tan(3 )y y t 
9 3tan(3 )y y t 

Cont…
• Complementary solution is:
• So, we have
• The Wronskian of these two function is
1 2( ) cos(3 ) sin(3 )cy t c t c t 
1
2
( ) cos(3 )
sin(3 )
y t t
y t


2 2cos(3 ) sin(3 )
3cos (3 ) 3sin (3 ) 3
3sin(3 ) 3cos(3 )
t t
W t t
t t
   


Cont….
• Particular solution:
2
2
3sin(3 )tan(3 ) 3cos(3 )tan(3 )
( ) cos(3 ) sin(3 )
3 3
sin (3 )
cos(3 ) sin(3 ) sin(3 )
cos(3 )
1 cos (3 )
cos(3 ) sin(3 ) sin(3 )
cos(3 )
cos(3 ) sec(3 ) cos(3 ) sin(3 ) sin(3 )
p
t t t t
y t t dt t dt
t
t dt t t dt
t
t
t dt t t dt
t
t t t dt t t dt
   
  

  
   
 
 
 
 
 
   
cos(3 ) sin(3 )
ln sec(3 ) tan(3 ) sin(3 ) cos(3 )
3 3
cos(3 )
ln sec(3 ) tan(3 )
3
t t
t t t t
t
t t
   
  

Cont….
• General solution
1 2
cos(3 )
( ) cos(3 ) sin(3 ) ln sec(3 ) tan(3 )
3
t
y t c t c t t t   

Example 2
Find the general solution to
Given that:
Form a fundamental set of solutions for the
homogeneous differential equation
2
( 1)ty t y y t    
1
2
( )
( ) 1
t
y t e
y t t

 

Solution
• First, need to divide out by a t.
• The Wronskian for the fundamental set of
solution is
1 1
1y y y t
t t
      
 
1
( 1)
1
t
t t t
t
e t
W e e t te
e

     

Cont…
• The particular solution is:
• General solution is:
2
( 1) ( )
( ) ( 1)
( 1) ( 1)
( ( 2)) ( 1)
2 2
t
t
p t t
t t
t t
t t e t
Y t e dt t dt
te te
e t e dt t dt
e e t t t
t t



   
 
   
    
   
 
 
2
1 2( ) ( 1) 2 2t
y t c e c t t t     

Cauchy-Euler
• The Cauchy-Euler equation has the form
• Where are constants.
• Each term contains
• The transformation reduces the equation
to a linear ODE with constant coefficient in the
variable t. Notice that we assume x>0, and t=ln x.
• Using chain rule, since y function of t through x.
( ) 1 ( 1)
0 1 1...... ( )n n n n
n na x y a x y a xy a y b x 

    
t
x e
( )k k
x y
t
x e

Example
Find the general solution
Solution:
• Taking the transformation the equation
reduces to:
2 3
2 2 , 0x y xy y x x    
t
x e
3
3 2 t
t ty y y e   

Solution
• Corresponding homogeneous equation:
• Characteristic equation:
• Fundamental set of solution:
3 2 0t ty y y   
2
1 2
3 2 ( 2)( 1) 0
2, 1
r r r r
r r
     
 
 2
,t t
F e e

Cont….
• Complementary solution:
• Non-homogeneous term is
• The UC set of is
• The candidate for particular solution is:
• Computing the derivatives
2
1 2( ) t t
cy t c e c e 
3
( ) t
b t e
3t
e  3
1
t
S e
3
( ) t
py t Ae
3
3
( ) 3
( ) 9
t
t
yp x Ae
yp x Ae
 
 

Cont…
• Substituting into the equation
• General solution:
3 3 3 3
3
2 3
1 2
9 9 2
1
2
1
( )
2
1
( )
2
t t t t
t
p
t t t
Ae Ae Ae e
A
y t e
y t c e c e e
  


  
2 3
1 2
1
( )
2
y x c x c x x  

2.0 Application of Higher Order
Differential Equations
Electric Circuit
Q = Charge
Current, I =
𝑑𝑄
𝑑𝑡
L = Inductance
R = Resistance
1
𝐶
= Elastance
𝐸 𝑡 = Electromotive Force
C = Capacitor

In the figure 1, it contains an electromotive force E (supplied by
battery or generator) a resister, an inductor, L , and a capacitor in
series C.
If the charge on the capacitor at time, t is 𝑄 = 𝑄 𝑡 the current
is the root of change of Q with respect to 𝑡: 𝐼 =
𝑑𝑄
𝑑𝑡
.
Kirchhoff’s voltage law says that the sum of these voltage drops is
equal to supplied voltage:
𝐿
𝑑𝐼
𝑑𝑡
+ 𝑅𝐼 +
𝑄
𝐶
= 𝐸(𝑡)
Since 𝐼 =
𝑑𝑄
𝑑𝑡
, equation becomes
𝐿
𝑑2 𝑄
𝑑𝑡2 + 𝑅
𝑑𝑄
𝑑𝑡
+
1
𝐶
𝑄 = 𝐸(𝑡) Equation 1

The charge, 𝑄0 and current, 𝐼0 are known at time 0, then we have
initial conditions,
𝑄 0 = 𝑄0 therefore, 𝑄′
0 = 𝐼 𝑜 = 𝐼
Initial value problem can be solved by methods of Additional: Non
homogeneous linear equations.
A differential equation for the current can be obtained by
differentiating Equation 1 with respecting to 𝑡 and remembering
that 𝐼 =
𝑑𝑄
𝑑𝑡
𝐿
𝑑2 𝐼
𝑑𝑡2
+ 𝑅
𝑑𝐼
𝑑𝑡
+
𝐼
𝐶
𝐼 = 𝐸′(𝑡)

Question:
Find the charge and current at time 𝑡 in the circuit of figure if
𝑅 = 40Ω, 𝐿 = 1𝐻, 𝐶 = 16 × 10−4 𝐹, 𝐸 𝑡 = 100 cos 10𝑡 and the
initial charge and current are both 0.
From Equation 1 with given values of 𝐿, 𝑅, 𝐶 and 𝐸(𝑡),
𝑑2 𝑄
𝑑𝑡2 + 40
𝑑𝑄
𝑑𝑡
+ 625𝑄 = 100 cos 𝑡
Auxiliary question is 𝑟2 + 40𝑟 + 625 = 0 with root,
𝑟 =
−40± −900
2
= −20 ± 15𝑖

Solution of complementary equation is 𝑄 𝐶
𝑄 𝐶 𝑡 = 𝑒−20𝑡(𝑐1 cos 15𝑡 + 𝑐2 sin 15𝑡)
For the method of undetermined coefficients we try particular
solution, 𝑄 𝑃
𝑄 𝑃 𝑡 = 𝐴 cos 10𝑡 + 𝐵 sin 10𝑡
𝑄 𝑃
′ 𝑡
= −10𝐴 sin 10𝑡 + 10𝐵 cos 10𝑡
𝑄 𝑃
′′ 𝑡
= −100𝐴 cos 10𝑡 − 100𝐵 sin 10𝑡

Substituting into Equation 2, we have
−100𝐴 cos 10𝑡 − 100𝐵 sin 10𝑡 + 40(−10𝐴 sin 10𝑡 +
10𝐵 cos 10𝑡) + 625(𝐴 cos 10𝑡 + 𝐵 sin 10𝑡) = 100 cos 10𝑡
or
(525𝐴 + 400𝐵) cos 10𝑡 + −400𝐴 + 525𝐵 sin 10𝑡 = 100 cos 10𝑡
Equating coefficient, we have
525 + 400𝐵 = 100 and −4000𝐴 + 525𝐵 = 0
Or
21𝐴 + 16𝐵 = 4 and −16𝐴 + 21𝐵 = 0

Solution of this system is,
𝐴 =
84
697
and 𝐵 =
64
697
.
So, particular solution is
𝑄 𝑃 𝑡 =
1
697
(84 cos 10𝑡 + 64 sin 10𝑡)
General solution is
𝑄 𝑡 = 𝑄 𝐶 𝑡 + 𝑄 𝑃 𝑡 = 𝑒−20𝑡 𝑐1 cos 15𝑡 + 𝑐2 sin 15𝑡
+
4
697
(21 cos 10𝑡 + 16 sin 10𝑡)

Imposing the initial condition 𝑄 0 = 0, we get
𝑄 0 = 𝑐1 +
84
697
= 0, 𝑐1 = −
84
697
To impose other initial condition, we differentiate to find the current
𝐼 =
𝑑𝑄
𝑑𝑡
= 𝑒−20𝑡 −20𝑐1 + 15𝑐2 cos 15𝑡 + −15𝑐1 − 20𝑐2 sin 15𝑡
+
40
697
−21 sin 10𝑡 + 16 cos 10𝑡
𝐼 0 = −20𝑐1 + 15𝑐2 +
640
697
= 0, 𝑐2 = −
464
2091
∴ Formula of charge is
𝑄 𝑡 =
4
697
[
𝑒−20𝑡
3
(−63 cos 15𝑡 − 116 sin 15𝑡)
+(21 cos 10𝑡 + 16 sin 10𝑡)] ∎

Expression for current is
𝐼 𝑡 =
1
2091
[ 𝑒−20𝑡 −1920 cos 15𝑡 + 13060 sin 15𝑡
+ 120 −21 sin 10𝑡 + 16 cos 10𝑡 ]∎

PROBLEM 1
QUESTION:
Solve the equation 𝑚𝑥′′
+ 𝛽𝑥′
= 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, 𝛽 = 1, and 𝑔 = 32.

SOLUTION
• We apply the theorem from subtopic 4.3
Homogenous Linear Equation With Constant
Coefficients (Second-Order) by let 𝑦𝑝 as any
particular solution on an interval I.
• To solve a non-homogenous differential equation,
we need to find:
1) 𝑦𝑐 that is a fundamental set of solutions that
form when 𝑦1 and 𝑦2 are linearly independent.
2) then, we find the 𝑦𝑝.

1) 𝑦𝑐
• Since it given the value of 𝑔 and 𝑚𝑔, so the
value of 𝑚 are:
𝑚𝑔 = 160 , 𝑔 = 32
𝑚 =
160
𝑔
=
160
32
𝑚 = 5

𝑚𝑥′′
+ 𝑥′
= 𝑚𝑔
5𝑚2
+ 𝑚 = 160 (a)
5𝑚2
+𝑚 = 0
𝑚 5𝑚 + 1 = 0
𝑚 = 0 , 5𝑚 + 1 = 0
5𝑚 = −1
𝑚 = −
1
5
𝑦𝑐 = 𝑐1 + 𝑐2 𝑒−
1
5
𝑥

𝑓 𝑥 = 160
𝑦𝑝 = 𝐴
𝑦𝑝
′
= 0
𝑦𝑝
′′
= 0
• When substitute into (a);
5 0 + 0 = 160
0 + 0 = 160 not supposed to get this
answer

• Let 𝑦𝑝 = 𝐴𝑠 , 𝑠 as variable,
𝑦𝑝
′
= 𝐴
𝑦𝑝
′′
= 0
5 0 + 𝐴 = 160
𝐴 = 160
𝑦𝑝 = 160𝑠
𝑦𝑐 = 𝑐1 + 𝑐2 𝑒−
1
5
𝑠
+ 160𝑠 (b)

2) 𝑦𝑐
• Let 𝑦𝑐 = 𝑥 𝑠
Given 𝑥 0 = −100
𝑐1 + 𝑐2 𝑒−
1
5
𝑠
+ 160 = −100
𝑐1 + 𝑐2 𝑒−
1
5
(0)
+ 160 = −100
𝑐1 + 𝑐2 = −100 (1)
Given 𝑥′ 0 = 0
𝑥 𝑠 = 𝑐1 + 𝑐2 𝑒−
1
5 𝑠
+ 160𝑠
𝑥′ 𝑠 = −
1
5
𝑐2 𝑒−
1
5
𝑠
+ 160
−
1
5
𝑐2 𝑒−
1
5 0
+ 160 = 0
−
1
5
𝑐2 = −160
𝑐2 = 800

• After get the value of 𝑐2, substitute the value
into (1)
𝑐1 + 𝑐2 = −100
𝑐1 + 800 = −100
𝑐1 = −900
• Then substitute the value of 𝑐1 and 𝑐2 into (b)
𝑥(𝑠) = 𝑐1 + 𝑐2 𝑒−
1
5
𝑠
+ 160𝑠
𝑥(𝑠) = −900 + 800𝑒−
1
5
𝑠
+ 160𝑠
Let 𝑠 = 𝑡
 𝑥(𝑡) = −900 + 800𝑒−
1
5
𝑡
+ 160𝑡

PROBLEM 2
QUESTION:
Use the solution from Problem 1 to compute the
length of time you free-fall (that is, the time it
takes to go to the natural length of the cord:100
feet)

SOLUTION
• From Problem 1,
𝑥 𝑡 = 100
𝑥 𝑡 = −900 + 800𝑒−
1
5
𝑡
+ 160𝑡 = 100
800𝑒−
1
5
𝑡
+ 160𝑡 = 100 + 900
800𝑒−
1
5
𝑡
+ 160𝑡 = 1000
𝑒−
1
5
𝑡
+
160
800
𝑡 =
1000
800
𝑒−
1
5
𝑡
+
1
5
𝑡 =
5
4

𝑒−
1
5
𝑡
=
5
4
−
1
5
𝑡
• ln both side to eliminate exponent;
ln 𝑒−
1
5
𝑡
= ln(
5
4
−
1
5
𝑡 )
−
1
5
𝑡 = ln(
25
4𝑡
)
−
1
5
𝑡 = ln 25 − ln(4𝑡)

−
1
5
𝑡 + ln(4𝑡) = ln 25
𝑡(−
1
5
+ ln 4) = ln 25
𝑡 =
ln 25
−
1
5
+ln 4
𝑡 = 2.71
 𝑡 = 2.71s

PROBLEM 3
QUESTION:
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 cords starts to pull.

SOLUTION
• From the answer in Problem 1,
𝑥(𝑡) = −900 + 800𝑒−
1
5
𝑡
+ 160𝑡
We need to find it derivatives;
𝑥(𝑡) = −900 + 800𝑒−
1
5
𝑡
+ 160𝑡
𝑥′
𝑡 = −
1
5
(800𝑒−
1
5
𝑡
) + 160
𝑥′
𝑡 = −160𝑒−
1
5
𝑡
+ 160
𝑥′
𝑡 = −160𝑒−
1
5
𝑡
+ 160
 𝑥′
𝑡 = 160(−𝑒−
1
5
𝑡
+ 1)

• Substitute the value of 𝑡 found in Problem 2
into the derivatives that already calculate
before;
𝑡 = 2.71s
𝑥′
𝑡 = 160(−𝑒−
1
5
𝑡
+ 1)
𝑥′
2.71 = 160(−𝑒−
1
5
(2.71)
+ 1)
𝑥′
2.71 = 66.9465
 𝑥′
2.71 = 66.9465 ft/s

PROBLEM 4
Solve the initial-value problem
𝑚𝑥′′
+ 𝛽𝑥′
+ 𝑘𝑥 = 𝑚𝑔, 𝑥 𝑡1 = 0, 𝑥′
𝑡1 = 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.

SOLUTION
Let 𝑡1 = 2.71 and 𝑣1 = 66.95
Have, 𝑚𝑔 = 160, 𝛽 = 1 and 𝑔 = 32, 𝑘 = 14
𝑚𝑥′′ + 𝛽𝑥′ + 𝑘𝑥 = 𝑚𝑔
1
32
𝑥′′ +
1
160
𝑥′ +
14
160
𝑥 = 1
160
32
𝑥′′ + 𝑥′ + 14𝑥 = 160
5𝑥′′
+ 𝑥′
+ 14𝑥 = 160

Solve for 𝑦𝑐
Auxillary Equation: 5𝑚2 + 𝑚 + 14 = 0
𝑚 =
−1± 1−4(5)(14)
2(5)
𝑚 =
−1± −279
10
𝑚 =
−1±𝑖 279
10
Compare with 𝑚 =∝ ±𝛽𝑖
We have, ∝= −
1
10
and 𝛽 =
279
10
∴ 𝑦𝑐 = 𝑒−
1
10 𝑐1 cos
279
10
𝑡 + 𝑐2 sin
279
10
𝑡

Solve for 𝑦𝑝,
𝑦𝑝 = 𝐴
𝑦𝑝
′ = 0
𝑦𝑝
′′ = 0
5 0 + 0 + 14𝐴 = 160
𝐴 =
160
14
=
80
7
∴ 𝑦𝑝 =
80
7
The solution for is 𝑥 𝑡 = 𝑦𝑐 + 𝑦𝑝
So, 𝑥 𝑡 = 𝑒−
1
10 𝑐1 cos
279
10
𝑡 + 𝑐2 sin
279
10
𝑡 +
80
7

We know, 𝑥 𝑡1 = 0 and 𝑡1 = 2.71
𝑥 2.71 = 𝑒−
2.71
10 𝑐1 cos
279(2.71)
10
+ 𝑐2 sin
279(2.71)
10
𝑡 +
80
7
= 0
0.76262 −0.18473𝑐1 − 0.98279𝑐2 +
80
7
= 0
0.14088𝑐1 + 0.74950𝑐2 =
80
7
𝑐1 = 81.12274 − 5.32013𝑐2

And we have, 𝑥′ 𝑡1 = 𝑣1, 𝑡1 = 2.71 and 𝑣1 =66.95
𝑥′
𝑡 = 𝑒−
𝑡
10 −
279
10
𝑐1 sin
279𝑡
10
+
279
10
𝑐2 cos
279𝑡
10
−
1
10
𝑒−
𝑡
10 𝑐1 cos
279𝑡
10
+ 𝑐2 sin
279𝑡
10
𝑡
𝑥′
2.71
= 𝑒−
2.71
10 −
279
10
𝑐1 sin
279 × 2.71
10
+
279
10
𝑐2 cos
279 × 2.71
10
−
1
10
𝑒−
𝑡
10 𝑐1 cos
279 × 2.71
10
+ 𝑐2 sin
279 × 2.71
10
𝑡 = 66.95
1.26599𝑐1 − 0.16036𝑐2 = 66.95

Substitute value 𝑐1 into the equation
1.26599(81.12274 − 5.32013𝑐2) − 0.16036𝑐2 = 66.95
𝑐2 = 5.43746
And substitute back into 𝑐1, we get,
𝑐1 = 52.19475
Hence,
𝑥 𝑡 = 𝑒−
1
10 52.19475 cos
279
10
𝑡 + 5.43746 sin
279
10
𝑡 +
80
7
∎

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

SOLUTION
𝑥 𝑡 = 𝑒−
1
10
𝑡
𝑐1 cos 𝛽 𝑡 + 𝑐2 sin 𝛽𝑡 +
80
7
𝑥′ 𝑡 = 𝑒−
1
10 𝑡
−𝛽𝑐1 sin 𝛽𝑡 + 𝛽𝑐2 cos 𝛽𝑡
−
1
10
𝑒−
1
10 𝑐1 cos 𝛽 𝑡 + 𝑐2 sin 𝛽𝑡
We have, 𝑥′ 𝑡2 = 0
𝑒−
1
10 𝑡2
−𝛽𝑐1 sin 𝛽𝑡2 + 𝛽𝑐2 cos 𝑡2 −
1
10
𝑒−
1
10 𝑡2
𝑐1 cos 𝛽 𝑡2 + 𝑐2 sin 𝛽𝑡2 = 0
𝑒−
1
10 𝑡
−𝛽𝑐1 sin 𝛽𝑡2 + 𝛽𝑐2 cos 𝑡2 =
1
10
𝑒−
1
10 𝑡2
𝑐1 cos 𝛽 𝑡2 + 𝑐2 sin 𝛽𝑡2

−𝛽𝑐1 sin 𝛽𝑡2 + 𝛽𝑐2 cos 𝑡2 =
1
10
𝑐1 cos 𝛽 𝑡2 + 𝑐2 sin 𝛽𝑡2
−𝛽𝑐1 −
1
10
𝑐2 sin 𝛽𝑡2 + 𝛽𝑐2 −
1
10
𝑐1 cos 𝛽 𝑡2 = 0
Substitute values 𝑐1, 𝑐2 and 𝛽
−87.72617 sin 𝛽𝑡 + 3.86287 cos 𝛽𝑡 = 0
−87.72617 sin 𝛽𝑡2 = −3.86287 cos 𝛽𝑡2
sin 𝛽𝑡2
cos 𝛽𝑡2
= 0.04403

tan 𝛽𝑡2 = 0.04403
𝛽𝑡2 = tan−1 0.04403
𝛽𝑡2 = 0.04400
Substitute value of 𝛽
𝑡2 = 0.02634𝑠 ∎

Find the lowest position,
𝑥 𝑡 = 𝑒−
1
10 𝑡
𝑐1 cos 𝛽 𝑡 + 𝑐2 sin 𝛽𝑡 +
80
7
Substitute value 𝑡2 = 0.02634
So, 𝑥 0.02634 = 63.67417ft ∎

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?

Solution
• Problem 6 has been marked as computer
problem. Therefore, we use the tool to plot
solutions for the given k values (k – spring
constant).

# Note about the tool
• The horizontal line at the top represents the
water.
• Note that, the coordinate system is inverted
– the positive direction is downwards.
• And remember that the jumper is 6 feet tall.
• The value of k only can be set in positive value
only.
• Click the Play button to have the bungee jumper
“step-off” the bridge at x(0) = -100

• Given soft bungee cord, k = 8.5 (set k = 8.6) and from
the problem 1, we know that w = 160
• Figure shows the position of the jumper.

• From the figure, we can see that the river
is 250 feet below and x (0) = -100.
• When we click the Play button, it shows
that the jumper does not get wet.
• Then, from the graph, it shows that value
of x is about 68

• Given stiffer cord, k = 10.7 (k = 10.8) and from the
problem 1, we know that w = 160.
• The animation
of the tool shows
that the jumper
also does not get
wet.
• Then, from the
graph, it shows
that value of x
is about 58

• Given climbing rope, k = 16.4 and from the problem 1,
we know that w = 160.
• For this climbing
rope, it shows that
the jumper also
does not get wet.
• And from the
graph, x is about
48.

• From above, the value of k for climbing rope is
the largest among the three cord that is k = 16.4.
Followed with stiffer cord that is k = 10.7, while
soft bungee jumping is the smallest that is k =
8.5.
• But we also need to consider the type of cord to
avoid unpleasantness associated with an
unexpected water landing.
• The weakness of using stiffer cord is the cord is
too stiff, then your body will no longer form a
connected set after you hit the end of the cord.
• While, The weakness of using climbing rope for
the bungee jumping is the rope does not have
the spring so it might cause the rope to broke.

• Therefore, the soft bungee cord is the suitable
cord to use for the bungee jumping.
• It is because, when the person jumps,
the cord stretches and the jumper flies
upwards again as the cord recoils, and
continues to oscillate up and down until all
the energy is dissipated.
• Hence, we can conclude that, the position of
the jumper that indicate the relevant path as
the jumper approaches the water is the one
with k = 8.5 that is the soft bungee cord.

PROBLEM 7
You have a bungee cord for which you 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?

SOLUTION
Given weight, mg = 10 and x(t) = 1.2
• We can use the equation of net force :
• We know that,
b(x) = -kx for x> 0  mx’’ = mg – kx - 𝛽𝑥′
Since the question ask the value of k, then
rewrite the equation into :
mx’’ = mg + b(x) - 𝜷𝒙′
kx = mg - 𝜷𝒙′ - mx’’

Then,
x(t) = 1.2,
x’(t) = 0,
x”(t)= 0.
Substitute all the value in the equation,
mx” = mg – kx - 𝜷𝒙′
kx = mg - 𝜷𝒙′ - mx”
k(1.2) = 10 – 0 – 0
k =
𝟏𝟎
𝟏.𝟐
k = 8.3 #
 Therefore, the value of k for this cord is 8.3

Problem 8 (CAS)
What would happen if your 220-pound
friend uses the bungee cord whose
spring constant is k = 10.7

• By using the tool. Set the weight , mg= 220 and k =
10.7 (k = 10.8).
• Click the Play button to “step-off” . We can see
that our friend is at least going to get wet.

• It is because, before the jumping, the
velocity, x’ = 0.
• After the jumping, the velocity change
and the cord stretch into x = 68.
• Remember that our friend is about 6 feet
tall.
• Therefore, we can conclude that at x =
68 it will cause our friend to become wet.

Problem 9
If your heavy friend wants to jump anyway,
then how short should you make the cord
so that he does not get wet?

• By using the tool. Set the weight of our friend into
the highest weight in the too that is 250 and set
first we set k = 10.7 (k = 10.8)
• Click the Play button to “step-off” . We can see that
our friend is going to
get wet.
• It is because the
bigger the weight of
the jumper, the
bigger the stretch
of the cord.
• Hence, when the end
of the cord scrapes the
water, our friend is
going to get wet.

• To overcome him to not become wet, we can
work backwords from the ending point to the
initial condition.
• When the cord is shortened to about 96 feet
(means that the initial condition is changed to
x(0) = -96), then our friend seems to be able to
jump without getting wet.
• Other than that, we can also change the value
of k.
• Change the value of k become larger and the
value of k is about 14. (k = 14).
• Hence, we can see that our friend does not get
wet.

Reference
Book
• Curtis F. Gerald, Patrick O. Wheatley. (2004).
Applied Numerical Analysis (Seventh ed.).
USA: Greg Tobin.
• Richard L. Burden, J. Douglas Faires. (2005).
Numerical Analysis (8th ed.). USA: Bob Pirtle.
• Zill, D. G. (2009). A First Course in Differential
Equations With Modelling Applications (Ninth
ed ed.). USA: Brooks/Cole Cengage Learning.

Website
• http://www.cengage.com/math/book_content
/0495108243_zill/zill_DE/project/final/publish
/tool/tool.html
• http://www.idea.wsu.edu/Bungee/

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