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 ) 33sin(3 ) 3cos(3 )

t tW 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 ty t t dt t dt

tt dt t t dt

t

tt 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 tt t t t

tt t

Cont….

• General solution

1 2

cos(3 )( ) cos(3 ) sin(3 ) ln sec(3 ) tan(3 )

3

ty 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

ty t e

y t t

Solution

• First, need to divide out by a t.

• The Wronskian for the fundamental set of solution is

1 11y y y t

t t

1( 1)

1

t

t t t

t

e tW e e t te

e

Cont…

• The particular solution is:

• General solution is:

2

( 1) ( )( ) ( 1)

( 1) ( 1)

( ( 2)) ( 1)

2 2

tt

p t t

t t

t t

t t e tY 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 2ty 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

tx e( )k kx y

tx e

Example Find the general solution

Solution:

• Taking the transformation the equation reduces to:

2 32 2 , 0x y xy y x x

tx e

33 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 tF 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( ) tb t e3te 3

1

tS 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( )

2y 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:

𝐿𝑑𝐼

𝑑𝑡+ 𝑅𝐼 +

𝑄

𝐶= 𝐸(𝑡)

Since 𝐼 =𝑑𝑄

𝑑𝑡, equation becomes

𝐿𝑑2𝑄

𝑑𝑡2 + 𝑅𝑑𝑄

𝑑𝑡+

1

𝐶𝑄 = 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

𝐿𝑑2𝐼

𝑑𝑡2+ 𝑅

𝑑𝐼

𝑑𝑡+

𝐼

𝐶𝐼 = 𝐸′(𝑡)

Question: Find the charge and current at time 𝑡 in the circuit of figure if 𝑅 = 40Ω, 𝐿 = 1𝐻, 𝐶 = 16 × 10−4𝐹, 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), 𝑑2𝑄

𝑑𝑡2 + 40𝑑𝑄

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 𝑄𝐶 𝑄𝐶 𝑡 = 𝑒−20𝑡(𝑐1 cos 15𝑡 + 𝑐2 sin 15𝑡) For the method of undetermined coefficients we try particular solution, 𝑄𝑃 𝑄𝑃 𝑡 = 𝐴 cos 10𝑡 + 𝐵 sin 10𝑡

𝑄𝑃′ 𝑡 = −10𝐴 sin 10𝑡 + 10𝐵 cos 10𝑡

𝑄𝑃′′ 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

𝑄𝑃 𝑡 =1

697(84 cos 10𝑡 + 64 sin 10𝑡)

General solution is 𝑄 𝑡 = 𝑄𝐶 𝑡 + 𝑄𝑃 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 + 𝑐2𝑒−15

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 + 𝑐2𝑒−1

5𝑠 + 160𝑠 (b)

2) 𝑦𝑐 • Let 𝑦𝑐 = 𝑥 𝑠 Given 𝑥 0 = −100

𝑐1 + 𝑐2𝑒−1

5𝑠 + 160 = −100

𝑐1 + 𝑐2𝑒−15(0) + 160 = −100

𝑐1 + 𝑐2 = −100 (1) Given 𝑥′ 0 = 0

𝑥 𝑠 = 𝑐1 + 𝑐2𝑒−15𝑠 + 160𝑠

𝑥′ 𝑠 = −1

5𝑐2𝑒−

1

5𝑠 + 160

−1

5𝑐2𝑒−

15 0 + 160 = 0

−1

5𝑐2 = −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 + 𝑐2𝑒−1

5𝑠 + 160𝑠

𝑥(𝑠) = −900 + 800𝑒−1

5𝑠 + 160𝑠

Let 𝑠 = 𝑡

𝑥(𝑡) = −900 + 800𝑒−1

5𝑡 + 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−15

𝑡 + 160𝑡 = 100

800𝑒−15

𝑡 + 160𝑡 = 100 + 900

800𝑒−1

5𝑡 + 160𝑡 = 1000

𝑒−1

5𝑡 +

160

800𝑡 =

1000

800

𝑒−1

5𝑡 +

1

5𝑡 =

5

4

𝑒−15

𝑡 =5

4−

1

5𝑡

• ln both side to eliminate exponent;

ln 𝑒−15

𝑡 = ln(5

4−

1

5𝑡 )

−1

5𝑡 = ln(

25

4𝑡)

−1

5𝑡 = ln 25 − ln(4𝑡)

−1

5𝑡 + 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−15

𝑡 + 160𝑡

We need to find it derivatives;

𝑥(𝑡) = −900 + 800𝑒−15

𝑡 + 160𝑡

𝑥′ 𝑡 = −1

5(800𝑒−

1

5𝑡) + 160

𝑥′ 𝑡 = −160𝑒−1

5𝑡 + 160

𝑥′ 𝑡 = −160𝑒−1

5𝑡 + 160

𝑥′ 𝑡 = 160(−𝑒−1

5𝑡 + 1)

• Substitute the value of 𝑡 found in Problem 2 into the derivatives that already calculate before;

𝑡 = 2.71s

𝑥′ 𝑡 = 160(−𝑒−15

𝑡 + 1)

𝑥′ 2.71 = 160(−𝑒−15

(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

32𝑥′′ +

1

160𝑥′ +

14

160𝑥 = 1

160

32𝑥′′ + 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 cos279

10𝑡 + 𝑐2 sin

279

10𝑡

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 cos279

10𝑡 + 𝑐2 sin

279

10𝑡 +

80

7

We know, 𝑥 𝑡1 = 0 and 𝑡1 = 2.71

𝑥 2.71 = 𝑒−2.7110 𝑐1 cos

279(2.71)

10+ 𝑐2 sin

279(2.71)

10𝑡 +

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

10𝑐1 sin

279𝑡

10+

279

10𝑐2 cos

279𝑡

10

−1

10𝑒−

𝑡10 𝑐1 cos

279𝑡

10+ 𝑐2 sin

279𝑡

10𝑡

𝑥′ 2.71

= 𝑒−2.7110 −

279

10𝑐1 sin

279 × 2.71

10+

279

10𝑐2 cos

279 × 2.71

10

−1

10𝑒−

𝑡10 𝑐1 cos

279 × 2.71

10+ 𝑐2 sin

279 × 2.71

10𝑡 = 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 cos279

10𝑡 + 5.43746 sin

279

10𝑡 +

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−

110

𝑡 𝑐1 cos 𝛽 𝑡 + 𝑐2 sin 𝛽𝑡 +80

7

𝑥′ 𝑡 = 𝑒−1

10𝑡 −𝛽𝑐1 sin 𝛽𝑡 + 𝛽𝑐2 cos 𝛽𝑡

−1

10𝑒−

110 𝑐1 cos 𝛽 𝑡 + 𝑐2 sin 𝛽𝑡

We have, 𝑥′ 𝑡2 = 0

𝑒−1

10𝑡2 −𝛽𝑐1 sin 𝛽𝑡2 + 𝛽𝑐2 cos 𝑡2 −1

10𝑒−

110𝑡2 𝑐1 cos 𝛽 𝑡2 + 𝑐2 sin 𝛽𝑡2 = 0

𝑒−1

10𝑡 −𝛽𝑐1 sin 𝛽𝑡2 + 𝛽𝑐2 cos 𝑡2 =1

10𝑒−

110𝑡2 𝑐1 cos 𝛽 𝑡2 + 𝑐2 sin 𝛽𝑡2

−𝛽𝑐1 sin 𝛽𝑡2 + 𝛽𝑐2 cos 𝑡2 =1

10𝑐1 cos 𝛽 𝑡2 + 𝑐2 sin 𝛽𝑡2

−𝛽𝑐1 −1

10𝑐2 sin 𝛽𝑡2 + 𝛽𝑐2 −

1

10𝑐1 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

10𝑡 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.

• Figure shows the position of the jumper.

• 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.

• Figure shows the position of the jumper.

• 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/