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CHAPTER 5

Modeling with Differential Equations

Definition: Modeling is a translation of a physicalphenomenon into a set of equations which describes it.

 SECTION 5.1.1: FREE UNDAMPED MOTION 

• Hook’s Law: k whereks F    −=  is spring constant.The restoring force exerted by the spring in the oppositedirection of stretch is proportional to the amount ofstretch.

• ewton Law: ∑==

n

ii

 F  F 1

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• Equilibrium condition:Weight must be equal torestoring force. mg = ks where k is spring constant.mg - ks =

• !et" mass #ibrate free$ !ree Motion. %f mass isdisplaced by an amount x then the restoring force will

 be &'   s xk    + . (ssume there are no retarding forces.(ccording to )ewton !aw" * = ma implies

m

k where

xdt

xdlyequi#alentor x

m

k 

dt

xdkx

dt

xdm

kxkxmgkskxdt

xdmmg&xs'k 

dt

xdm

=

=+=+⇒−=

⇒−=+−=+−−=⇒++−=

2

2

2

2

2

2

2

2

2

2

2

2

00

0

ω 

ω 

ω ω  imxdt

xd"   ±=⇒=+ 21

2

2

2

0  

Differential equation

•  %s known as differential equation of freeundamped motion.

•  %ts solution &tsinctcosc'e&t'x +1t β β α  +=   t ct ct  x   ω ω  sincos&' +1   += "

after the initial conditions are used to determine +1  " cc  is called equation of motion with period ω 

π +=T   "

Time in seconds taken by the mass to complete a cycle&and frequency π 

ω 

+

1==

T  f    'number of cycles completed in

each second&• %f" interested in amplitude and phase angle# equi#alent

form of equation of motion is  &sin'&'   φ ω    +=   t  At  x  

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• where $mplitude  +++1   cc A   +=  and %hase angle φ   is related

as +

1tanc

c=φ 

t ct c

t  A

c At 

 A

c At  At  A

t  At  At  At  x

ω ω 

ω ω ω φ ω φ 

φ ω φ ω φ ω 

sincos

sincossin&cos'cos&sin'

sincoscossin&sin'&'

+1

+1

+

=+=+

=+=+=

•  )ote: ,are should be taken while e#aluating phaseangle by considering sign associated with c1 and c+ andthe rele#ant appropriate quadrant.

Definitions:

1&(mplitude ' ( &: The maximum distance that anobect mo#es from its equilibrium position.

 ( simple harmonic oscillator mo#es back and forth between the two positions of maximum displacement"

at x = ( and x = - ( .+&eriod 'T&: The time that it takes for an oscillator toexecute one complete cycle of its motion. %f it startsat t = at x = (" then it gets back to x = ( after onefull period at t = T .

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/&*requency 'f&: The number of cycles 'or oscillations&the obect completes per unit time.

&emark  0y definition" f = 1T

The unit of frequency is usually taken to be 1 23 = 1cycle per second.

'imple Harmonic (scillator: (ny obect that oscillates abostable equilibrium position and experiences a restoring forceapproximately described by 2ooke4s law.

5xamples of simple harmonic oscillators include: a massattached to a spring" a car stuck in a ditch being 66rockedout44 and a pendulum.

ote: The negati#e sign in 2ooke4s law ensures that theforce is always opposite to the direction of thedisplacement and therefore back towards the equilibrium

 position 'i.e. a restoring force&.

7. ( +-pounds weight stretches a spring 8 inches. Theweight is released from rest 8 inches below theequilibrium point.

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9ol. 2ooks !aw: The restoring force is proportional toits total stretch.

;&+

1'+4sin"

+

18"

/+

+=⇒===   k k  Law s Hook  g u ft inches slugsm

t ct ct  x

imm x x xdt 

 xd 

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2ooks !aw: The restoring force is proportional to itstotal stretch. +&1

/+'8;4sin"+

/+

8;=⇒===   k k  Law s Hook  g u slugsm

t ct ct  x

imm x x x

dt 

 xd 

1sin1cos&'

111

+1

++

+

+

+=

⇒±=⇒=+⇒=+′′⇒=+ω 

 )ow" initial conditions$ >&'"/+

&'   =′−

=   x x

&?+@.1sin'8

>1sin

+

11cos

/

+&'"

+

1&cos'1&sin'1>&'

/

+&' ++11

−=+−

=

=⇒+−==′=−

=

t t t t  x Hence

ccc xand c x

'b& *ind amplitude and period of motion.$mplitude  +++1   cc A   += =

8

>

;

1

?

;=+  

period ω π +

=T >1

+   π π ==

A2ow many complete cycles will the mass ha#ecompleted at the end of /B second.

CyclesC C  1>>

/   =⇒= π 

π 

'd& (t what time does the mass passes through theequilibrium position heading downward for +nd time.%f = x  and the weight is mo#ing downward for the +nd

time" then  st t  @+1.+?+@.1   =⇒=−   π 'e& (t what time does the mass attain its extreme

displacements on either side of equilibrium position

−−−=++=⇒

+=−⇒=−=′

"+"1"?+@.+&1+'

+?+@.1&?+@.1cos'

/

+>

nnt 

nt t  x If  

π 

π π 

'f& osition of mass at t = / s>?@.&/'   −== x ft

'g& %nstantaneous #elocity at t = / s

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 s ft  x &/'   −==′

'h& (cceleration at t = / s+@+.>?&/'   s  ft  x   ==′′

'i& %nstantaneous #elocity at the times when mass passesthrough equilibrium position.

 s ft t  xand n for nt then x If   /

+>&'"...+"1"&?+@.'

1

1   ±=′=+==   π  '& (t what time

mass is > inches below the equilibrium position

−−−=++=

++==

+"1"&"+?+@.8

>

'1

1

C

&+?+@.8

'1

1

1+

>

nnt 

nt then  ft  x If  

π π 

π π 

'k& (t what time mass is > inches below the equilibrium position heading upward.

−−−=++=

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'E+,-( .).*: !&EE D$M%ED M(,-(

 • There may be a damping force due to #iscous medium or damping de#ice.

•  Damping force i proportiona! to po"er of intantaneo# $e!ocit%. i.e." damping force is

constant multiple of dt dx  .

• With reference to )ewtons !aw" and damping constant β   " the resulting differential

equation of free damped motion isdt 

dxkx

dt 

 xd m   β −−=+

+

• 5qui#alently" + ++

+

+

+

=++⇒−−=   xdt 

dx

dt 

 xd 

dt 

dx

m x

m

k 

dt 

 xd ω λ 

β  where

m

k 

m==+"+   ω 

β λ 

• The auxiliary equation is ++++ +   ω λ λ ω λ    −±−=⇒=++   mmm• The damping factor t e   λ − indicates that displacements become negligible as ∞→t 

+ase ): ++ >−ω λ   • This is o#er-damped case.• Damping coefficient β   E spring constant k.

• 9olution will be t t t  ececet  x++++

+1'&'  ω λ ω λ λ    −−−−

+=

• Fotion will be smooth and non-oscillatory.+ase *: ++ =−ω λ   

This is critically-damped case. A critically damed system con!erges to "ero faster than any other# and withoutoscillating$

(n example of critical damping is the door closer seen on

many hinged doors in public buildings. The recoil mechanisms in most guns are also critically

damped so that they return to their original position" after the recoil due to firing" in the least possible time.

•  9olution will be &'&' +1   t ccet  x  t 

+=  −λ 

• Fotion will be smooth and non-oscillatory.• Fass can pass the equilibrium position only once.

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%hen & ' 1# there is a dou(le root ) *defined a(o!e+# which is real$

( slight decrease in damping force will result oscillatory motion.

 In this case# with only one root )# there is in addition to the solution x*t+ ' e)t  a

 solution x*t+ ' te)t 

• where A and , are determined (y the initial conditions of the system *usually the

initial osition and !elocity of the mass+

+ase /: ++

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whose constant is 18 )m" and the entire system is then submerged in a liquid that imparts adamping force numerically equal to 1 times the instantaneous #elocity. Determine the equationof motion if 'a& The weight is released from rest 1 meter below the equilibrium position.

1+

1181"18 +

==

⇒===⇒==

m

also

m

k mk 

β λ 

β ω 

&'"1&'

181++

++

+

+

=′=

=++⇒=++

 x xwith

 xdt 

dx

dt 

 xd  x

dt 

dx

dt 

 xd ω λ 

t t ecect  xmmm<

+

+

1

+&'

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7. (fter a mass weighing 1 pounds is attached to a >-feet spring" the spring measure @ feet.This mass is remo#ed and replaced with another mass that weighs < pounds. The entiresystem is placed in a medium that offers a damping force numerically equal to theinstantaneous #elocity.

a& *ind the equation of motion if the mass is initially released from a point I foot below theequilibrium position with a downward #elocity of 1 fts.

  ⇒====⇒= 1.;1/+

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'E+,-( .)./: D-!!E&E,-$L E01$,-( (! D&-2E

  M(,-( 3-,H D$M%-4

The D5 of free damped motiondt 

dxkx

dt 

 xd m   β −−=+

+

 

will be D5 of dri#en motion with dumping if an external force&' x  f    which will cause #ertical oscillation be added to it .

⇒=++⇒+−−= &'&'+

+

+

+

 x f  dt 

dxkx

dt 

 xd m x f  

dt 

dxkx

dt 

 xd m   β β 

&'+ ++

+

 x F  xdt 

dx

dt 

 xd =++   ω λ   where

m

 x f   x F and 

m

k 

m

&'&'"+ + ===   ω 

β λ 

•  )ote that the abo#e is a non-homogeneous equation which can be sol#ed by any of themethod discussed earlier.

Another common mechanical problem arises when a damped harmonic oscillator is

driven by some time-dependent external applied force: thedriven harmonic oscillator.

The most important case is that of a force that oscillates in a sinusoidal manner. If the

driving force is of the form

*'t& = *5cos'Jt K L5&then the differential equation has an exact solution.

The solution has two parts: transient and steady state. The transient portion" which has the samesolution as the damped harmonic oscillator" dies out exponentially and depends on the initialconditions. The steady state portion has an amplitude that remains constant and does not dependon the initial conditions. Thus no mater what initial conditions the oscillator had" it wille#entually acquire beha#ior that is wholly dependent upon the dri#ing force.The amplitude that the oscillator e#entually acquires depends on the relation of the dri#ing

frequency to the natural frequency of the oscillator and on the damping factor. %t is a maximumwhen is a minimum. This occurs when the ratio of the two frequencies is equal toM'1 K N+&.This condition is known as resonance and results in a large amplitude of oscillation. When thedri#ing frequency equals the natural frequency" the amplitude of the steady state portion of thesolution* m

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MO'J+ P J+&+ K ;N+J+Q(s the damping factor approaches 3ero" the steady state amplitude approaches infinity. Thisillustrates the importance of damping in structures susceptible to #ibration such as suspension bridges and steel framed buildings. Without damping" a structure could shake itself to pieces

from a tiny external force with ust the right frequency.( classic demonstration of resonance is the Tacoma )arrows bridge incident. !ong span bridgesare now all designed with open cross section struts to dissipate some of the wind4s force. Theformer aramount ,ommunications building on ,olumbus ,ircle con#erted by Donald Trumpinto luxury apartments in the late 1??s pro#ides a more contemporary example. The base of the building" which is a long narrow prism" was wrapped in scaffolding for the better part of fi#eyears. 0ecause of a structural flaw that allowed the top floors to twist" windows were pryingloose from their casements. ( degree of flexibility is needed in large buildings" but it was fearedthat gale force winds would pop the windows out #iolently" showering the streets below with plate glass. art of the building4s reno#ation included structural repairs to dampen this motion.%n summary" we ha#e seen how a second order linear differential equation" the simple harmonicoscillator" can generate a #ariety of beha#iors. %n the damped harmonic oscillator we sawexponential decay to an equilibrium position with natural periodicity as a limiting case. Thedetermining factor that described the system was the relation between the natural frequency andthe damping factor. %n the dri#en harmonic oscillator we saw transience leading to some steadystate periodicity. The final beha#ior of the system depended on the relation between the dri#ingfrequency and the natural frequency 'and to a lesser extent the damping factor&. The beha#iorsdescribed abo#e are also found in first order nonlinear difference equations$ the quadraticmapping and the related logistic equation. % will re#iew the latter of these and present it in a

manner similar to what has appeared so far.

7. ( 18-pound weight stretches a spring

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/

1/cos11+/sin/cos   ==⇒=+′+′′⇒+=   , At  x x xt  ,t  A x     

&/sin/'cos/

1t t  x

    +=∴

&+

;@

sin+

;@

cos'&' +1+

t ct ce x xt  x

t 

 c  +=+=

−

&/sin/'cos/

1t t  ++

;@/

8;"/;&'+&' +1   −=−=⇒=′=   cc xand  x

*inally &+

;@sin;@/

8;

+

;@cos/

;'&' + t t e x xt  x

t 

 c   −−=+=

−

&/sin/'cos/

1t t  ++

7. ( mass of 1 slug" when attached to a spring" stretches it + feet and then comes to rest inthe equilibrium position. 9tarting at t = " an external force equal to with

t t   f   ;sin

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Dri6en Motion 3ithout Damping7..&E'($+E

!₀ is a constant periodic force# if 89 then &esonance occurs;

&esonance

•

-n physics# resonance is the tendency of a system to oscillate at ma

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• Soltage drop across inductor =+

+

dt 

-d  L

dt 

di L   = . Soltage drop across resistor =

dt 

d- . .i  = .

Soltage drop across capacitor =C 

-

• Hirchhoffs 9econd !aw : %mpressed Soltage 5't& on a closed loop = 9um of #oltage drop in

the loop

  &'1

+

+

t  / -C dt 

d- .

dt 

-d  L   =++⇒  a non-homogeneous D5.

7. *ind the charge on the capacitor in an !R, series circuit when h L;

1= " R=+Ω "

 f  C /

1= " 5't& = S" q'& = ; ," and i'& = (. %s the charge on the capacitor e#er

equal 3ero.

9ol.1+

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7. *ind the charge on the capacitor and the current in an !R, series circuit. *ind the maximumcharge on the capacitor.

h L 1= " R=1Ω "   f  C  ;.= " 5't& = / S" q'& = ," and i'& = + (.

9ol.+>1

/+>11&'1

+

+

+

+

+

=++

⇒=++⇒=++⇒

mm

-

dt 

d-

dt 

-d t  / -

C dt 

d- .

dt 

-d  L

1+.;.11+.&'

;.11+.+&'&'

1+.&'

1+."

"&'>">

>>

+1

>+

>1

>+

>1

++−=⇒

=−=⇒=′=

++=

⇒=⇒=

+=⇒=⇒

−−

−−

−−

t t 

t t 

  

t t c

teet -

cand c-and -conditioninitial 

tecect -

 A A-assume- For 

tecect -m

t t  teedt 

d-t i >> @+&'   −− −== . Faximum charge will be when

/>1@+&'>>

=⇒=−==

  −−

t teedt 

d-t i

  t t 

11' &/>1'>&/>1'> ≈++−=  −− tee-

7. *ind steady-state charge and steady-state current in an !R, series circuit when h L 1= "R=+Ω "   f  C  +>.= " 5't& = >cost S.

9ol: &'1

+

+

≠=++   .witht  / -C dt 

d- .

dt 

-d  L For 

&'t -c  is called transient solution.%f &'t  /   is periodic or constant then &'t -  is called steady-state solution.

 )ow" t -dt d-

dt 

-d cos>;++

+

=++

1/1"1/1>

cos>sin&+/'cos&+/'

cos>;+"

sincos

cossinsincos&'

==

⇒=−++

⇒=+′+′′

−−=′′

+−=′⇒+=

 , A

t t  A ,t  , A

t --- Hence

t  ,t  A-

t  ,t  A-t  ,t  At - 

9teady-state charge ist t t i

iscurrent  state0teadyand t t t -

 

 

cos1/

1sin

1/

1>&'

sin1/

1cos

1/

1>&'

+−=

−+=

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'E+,-( .*: L-E$& E01$,-('

DE!LE+,-( (! @E$M:

$A-' (! 'BMME,&B DE!LE+,-(CEL$',-+ +1&2E

• 0ending moment F'x& at a point x along the beam is related to the load per unit length w'x&as &'

+

+

 xwdx

  d =

•   κ  /I  x 1    =&'  where 5 is Boung’s modulus of elasticity" % is moment of inertia of cross-section of the beam and product 5% is known as fle

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Embedded at left but free at right end:

&'&'&"'&'

&"'&'&"'&'

is force shear  L yismoment (ending  L y

isdeflectionof   sloe ydeflectionno y

=′′′=′′

=′=

  x = x = !

7. 9ol#e &' xw /Ly i! = . 0eam is of length !" embedded at left and simply supported at rightend. w'x& = w = const " G x G !.

9ol. .&'  /I 

w y xw /Iy   i!i! =⇒=  

*or /;+

/+1

; "   xc xc xcc ym y cc   +++=⇒=

*or ;+;

int"   x /L

w y

 /I 

w yof  egration successi!e y  

i!

     =⇒=

2ence" ;/;+

/+1+;

 x /I 

w xc xc xcc y   ++++=  .

 

&+>/';<&'

;<

>

18

"&'"&'"&'"&'

;/++

;

+

/

+1

 x Lx x L /I 

w x y

 /I 

 Lwcand 

 /I 

 Lwc

cc L y L y y y

+−=⇒

−==

==⇒=′′==′=

'b& Vbtain the graph of deflection cur#e if ! = 1 and w = ;

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&'"&'"   ===+′′   L y y y y   λ   'ol.

+ase ): %f  =λ   then &'&' +1+1   ==⇒==+=⇒=′′   cc L y yand  xcc y yso y = is the tri#ial solution.+ase *: %f λ   then  xc xc y   λ λ  sincos +1   +=

.&sin'

sin

sin&'

sin"&'

+

+

++

++

+1

 solutiontheis L

 xnc y

 L

nn L L If  

 solutiontheis ythencif   Lc L y

 xc ylyconse-uent c y

π 

π λ π λ λ 

λ 

λ 

=

⇒=⇒=⇒=

==⇒=⇒=

==⇒=

 

(,E:

.&sin'

"/"+"1+

++

ionseigenfunct ing corresond called are L

 xn y

and  seigen!aluecalled aren for  L

n

π 

π λ 

=

−−−==

0.  9ol#e" boundary-#alue problem

&;'"&'"   ===+′′

  π λ    y y y y  

'ol.

+ase ): %f  =λ   then &;'&' +1+1   ==⇒==+=⇒=′′   cc y yand  xcc y y

  π 

so y = is the tri#ial solution.

+ase *: %f λ   then  xc xc y   λ λ  sincos +1   +=

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⇒=

⇒=⇒=⇒=

==⇒=⇒=

==⇒=

 solutiontheisnxc y

nn If  

 solutiontheis ythencif  c y

 xc ylyconse-uent c y

&;sin'

18;

;

sin

;

sin&;'

sin"&'

+

+

++

+1

λ π π 

λ π 

λ 

π λ 

π 

λ 

.&;sin'

"/"+"118 +

ionseigenfunct ing corresond theasnx y

and  seigen!alueasn  for n

=

−−−==λ 

0: 9ol#e" boundary-#alue problem&>'"&'"&1'+   ===++′+′′   y y y y y   λ   

'ol.  λ λ    −±−=⇒=+++ 1&1'++ mmm

+ase ): %f  =λ   thensolutiontri6ialtheis5=⇒==

⇒==+=⇒−−=  −−

 ycc

 y yand  xecec ym   x x

&>'&'1"1

+1

+1

+ase *: %f >>sin

>sin&>'

sin"&'

+

++

++>

+1

 solutiontheis xn

e y

n

n

n If  

 solutiontheis ythencif  ce y

 xce ylyconse-uent c y

 x

 x

π 

π 

λ π λ λ 

λ 

λ 

−

−

−

=

⇒−−−==⇒=⇒=

==⇒=⇒=

==⇒=

.&>

sin'

"/"+"1+>

++

ionseigenfunct ing corresond are xn

e y

and  seigen!aluearen for n

 x   π 

π λ 

−=

−−−==

ED