mathematical modeling & simulation, math...

Continuous Mathematical Models: Second Order Models – Mechanical & Electrical Systems

Mathematical Modeling &

Simulation, MATH 463

Dr. Alam Z M Barkat Ali First Order Models, MATH 463 1

Instuctor: Dr. Samy MZIOUslides from Dr. Alam Z M Barkat Ali

Second Order ODE Based Models• A linear model based on a second order ODE has the following

general form:

where a(t), b(t), c(t), f(t) are known functions and , areconstants. The function f(t) is called input, driving function, orforcing function.

• The model is homogenous if f(t) = 0, otherwise it is non-homogeneous. A solution y(t) of the differential equation thatsatisfies initial conditions is called output or response of the systemrepresented by the model.

• Second order models have many important applications in practicalworld. We will consider some very simple dynamical systems inwhich mathematical model is a second order ordinary differentialequation, with a(t), b(t), c(t) being constants.

---------------------------------------------------------------------------------------Dr. Alam Z M Barkat Ali First Order Models, MATH 463 2

)0( ,)0( ),()()()(2

2

yytfytcdt

dytb

dt

ydta

Spring-Mass System: Hooke’s Law• Suppose a flexible spring of length is

suspended vertically from a rigid supportand a body of mass m is attached to its freeend. The spring stretches by an amount s &attains an equilibrium position.

• According to Hooke’s law the spring exertsa restoring force Fr on the body, which isopposite to the direction of stretch & isproportional to the amount of elongation,i.e.

Fr s Fr = - k s

where k is constant of proportionality,called spring constant.

• At equilibrium position weight mg of thebody will balance the restoring force, i.e.

mg = k s mg – k s =0Dr. Alam Z M Barkat Ali First Order Models, MATH 463 3

------------------------------------------------------------------------------------------------------------

l

Spring-Mass System: Free undamped Motion• Forces acting on the body are Fr & its weight

mg. If the body attached to the spring isnow displaced by an amount from theequilibrium position, restoring force of thespring becomes

• We assume that, other than Fr , there are noresistive forces acting on the system. Thus,total force F on the body is:

• By the Newton’s second law we can write

Hence, equation of motion of the system is

Dr. Alam Z M Barkat Ali 4First Order Models, MATH 463

y

).( sykFr

kykyksmgyskmgF )(

2

2

dt

ydmF

ydt

xdy

m

k

dt

ydyk

dt

ydm 2

2

2

2

2

2

2

mkydt

yd/ , 22

2

2

-------------------------------------------------------------------------------------------------------------

Example 1: Free Undamped System• A body weighing 2 pounds (lbs) is attached with a spring and

stretches the spring by 6 inches. At time t=0 the body is releasedfrom a point 8 inches below the equilibrium position with an upwardvelocity of 4/3 feet per second.

Set up a mathematical model to find displacement of the body fromequilibrium position at time t.

Solve the model analytically and find a formula for the displacementat any time t.

Draw simulation block diagram, use it to simulate the model andanalyze simulation result.

---------------------------------------------------------------------------------------


Example 1: Mathematical Model

• Suppose that y(t) denotes distance of the body from equilibriumposition. Then

But, stretch s = 6 inches = ½ feet, and Weight = 2 lbs

mg = 2 m=2/g = 2/32 = 1/16 slug. Thus, by Hooke’s law:

mg = k s 2 = k 1/2 k = 4. Thus 2 = k/m = 4/(1/16)=64

Since, the body is released from 8 inches below the equilibriumposition with an upward velocity of 4/3 feet per second. Therefore,

y(0) = 8 inches = 2/3 feet and y’(0) = - 4/3 feet/second,downward direction is positive upward is negative. Thus, requiredmathematical is:

--------------------------------------------------------------------------------------Dr. Alam Z M Barkat Ali First Order Models, MATH 463 6

mkydt

yd/ , 22

2

2

064 64 2

2

2

2

ydt

ydy

dt

yd

3

4)0(,

3

2)0(,0 64

2

2

yyydt

yd

Example 1: Analytical Solution

•

------------------------------------------------------------------------Dr. Alam Z M Barkat Ali First Order Models, MATH 463 7

)8sin(6

1)8cos(

3

2)( ttty

Example 1: Block diagram


3

4)0(,

3

2)0(,64064

2

2

2

2

yyydt

ydy

dt

yd

------------------------------------------------------------------------------------------------------------

M. Al-Shedokhi

Sticky Note

هذي الواي دبل داش يصير لها تكامل وتطلع لنا واي داش

M. Al-Shedokhi

Sticky Note

هذي واي داش تكامل تطلع لي الواي

Example 1: Simulation Result

•


From this figure we see simple harmonic motion about equilibrium position

in the absence of any resistive force and oscillation do not approach to zero.

Spring-Mass System: Free damped Motion

• Free undamped motion is somewhatunrealistic because in reality there will beresisting force acting on the mass due tosurrounding medium, e.g. it may be due to airresistance or due to a mechanical deviceimparting resistive force.

• This resistive force is called damping force,

denoted by Fd and is considered to be equal

to some power of the instantaneous velocity,which we choose to be 1. If y(t) denotesdisplacement from equilibrium position, at atime t, of the mass attached to a verticalspring. Then

where C is constant of proportionality.


dt

dyCF

dt

dyF dd

----------------------------------------------------------------------------------------------------------


• Therefore, forces acting on the body attached with the spring are:

1. Weight of the body = mg,

2. Restoring force Fr = - k(s + y), Minus indicates force is resistive.

3. Damping Force = - C dy/d t, Minus indicates force is resistive.

• Thus, total force F acting on the body attached to spring is thefollowing:

F = mg – k(s + y) - C dy/d t = mg – k s – k y – C dy/d t

• However, we know the following from the Newton’s second law of motion: F = m a = m d2y/dt2

---------------------------------------------------------------------------------------



• Hence, making use of the last two equations, we can write thefollowing

where 2= C/m and 2 = k/m.

• If the mass attached to spring is released from a point at a distanceof units from equilibrium position with a velocity units persecond. Then, y(0)= and y’(0)=. Hence, equation of motion ofthe system is the following:


02 2

2

2

2

2

ydt

dy

dt

yd

dt

dy

m

Cy

m

k

dt

yd

)0( ,)0( ,02 2

2

2

yyydt

dy

dt

yd

dt

dyCky

dt

dyCkyksmg

dt

ydm

2

2


•

• For solving this equation analytically, we put y(t) = emt. Then,auxiliary equation (AE) is m2 +2m +2=0. Solving this eq. we get

• If

1. 2 > 2 , The system is over - damped.

2. 2 = 2 , The system is critically damped.

3. 2 < 2 , The system is under - damped.

----------------------------------------------------------------------------------------Dr. Alam Z M Barkat Ali First Order Models, MATH 463 13

)0( ,)0( ,02 2

2

2

yyydt

dy

dt

yd

222222

2

22

2

442

m

Example 2: Free (Over) damped motion• Suppose a body weighing 32 lbs stretches a spring by 32 feet.

Assuming that a damping force numerically equal to 3 times theinstantaneous velocity acts on the system. At t = 0 the mass isreleased from a point 10 feet below the equilibrium position fromrest.




---------------------------------------------------------------------------------------




But, stretch s = 32 feet, and Weight of the body = 32 lbs.

mg = 32 m=32/g = 32/32 = 1 slug. Thus, by Hooke’s law:

mg = k s 32 = k 32 k = 1. Thus 2 = k/m = 1/1 = 1.Damping Force = - 3(instantaneous vel.) C = 3 2=C/m = 3

Since, the body is released from 10 feet below the equilibriumposition from rest. Therefore, y(0) = 10 feet and y’(0) = 0feet/second, downward direction is positive upward is negative.Thus, required math model is:


mkmCydt

dy

dt

yd/ ,/2 ,2 22

2

2

03 3 2

2

2

2

ydt

dy

dt

ydy

dt

dy

dt

yd

0)0(,10)0(,0 32

2

yyydt

dy

dt

yd


•



•


0)0(,10)0( , 30 32

2

2

2

yyydt

dy

dt

ydy

dt

dy

dt

yd


•


We see that the initial displacement (downward) of 10 feet decreases and

becomes zero ultimately. This is expected because of the presence of

resistive forces and no driving force.

M. Al-Shedokhi

Sticky Note

قيم موجبة يعني نزل .. ومافيه تردد يعني نزول باتجاه واحد يعني ينزل لين يثبت

Example 3: Free (Critically) damped motion

• Suppose a body weighing 8 lbs stretches a spring by 2 feet.Assuming that a damping force numerically equal to 2 times theinstantaneous velocity acts on the system. At t = 0 the body isreleased from equilibrium position with an upward velocity of 3 feetper second.




---------------------------------------------------------------------------------------




But, stretch s = 2 feet, and Weight of the body = 8 lbs.


mg = k s 8 = k 2 k = 4. Thus 2 = k/m = 4/(1/4) = 16.Damping Force = - 2(instantaneous vel.) C = 2 2=C/m = 8.

Since, the body is released the equilibrium position with an upwardvelocity of 3 feet / second. Therefore, y(0) = 0 and y’(0) = -3,Note: downward direction is positive & upward is negative. Thus,required mathematical model is:


mkmCydt

dy

dt

yd/ ,/2 ,2 22

2

2

0168 168 2

2

2

2

ydt

dy

dt

ydy

dt

dy

dt

yd

3)0(,0)0( ,0 1682

2

yyydt

dy

dt

yd


•

----------------------------------------------------------------------Dr. Alam Z M Barkat Ali First Order Models, MATH 463 21


•

-------------------------------------------------------------------------Dr. Alam Z M Barkat Ali First Order Models, MATH 463 22

3)0(,0)0( , 1680 1682

2

2

2

yyydt

dy

dt

ydy

dt

dy

dt

yd


•

-------------------------------------------------------------------------Dr. Alam Z M Barkat Ali First Order Models, MATH 463 23

We see that the mass displacement first increases and goes to a peak value

above the equilibrium position and then starts coming back to equilibrium

position.

M. Al-Shedokhi

Sticky Note

هنا المنحنى نزل ، لكن قيم سالبة .. بالواقع الجسم طلع فوق وانضغط ثم نزل ( قيم موجبة ) ثم ثبت

Example 4: Free (Under) damped Motion

• Suppose a body weighing 16 lbs is attached to a 5 feet long spring.At equilibrium position the spring measures 8.2 feet. Assume thatthe surrounding medium offers a resistance (damping force) tomotion of the system that is numerically equal to the instantaneousvelocity. At t = 0 the body is released from rest at a point 2 feetabove the equilibrium position.







But, stretch s = 8.2 - 5=3.2 feet, and Weight of the body = 16 lbs.


mg = k s 16 = k 3.2 k = 5, 2 = k/m = 5/(1/2) = 10.Damping Force = - (instantaneous vel.) C = 1 2=C/m = 2.

Since, the body is released from rest at a point 2 feet above theequilibrium position. Therefore, y(0) = -2 and y’(0) = 0, Note:downward direction is positive & upward is negative. Thus, requiredmathematical model is:


mkmCydt

dy

dt

yd/ ,/2 ,2 22

2

2

0102 102 2

2

2

2

ydt

dy

dt

ydy

dt

dy

dt

yd

0)0(,2)0( ,0 1022

2

yyydt

dy

dt

yd


•



•

-----------------------------------------------------------------------Dr. Alam Z M Barkat Ali First Order Models, MATH 463 27

0)0(,2)0( ,0 1020 1022

2

2

2

yyydt

dy

dt

ydy

dt

dy

dt

yd


•

-----------------------------------------------------------------------Dr. Alam Z M Barkat Ali First Order Models, MATH 463 28

From this figure we see that the mass displacement first increases and

goes to a peak value below the equilibrium position and then starts

oscillating with decreasing amplitude.

Spring-Mass System: Driven Motion

• We now consider motion of a spring-masssystem in which an external force f(t) is actingon the mass in addition to damping force andrestoring force. The external force f(t)represents a driving force causing the motion.

• If y(t) denotes displacement of the mass fromequilibrium position at a time t. Then, theforces acting on the spring – mass system are:

1. Weight of the body = mg,

2. Restoring force Fr = - k(s + y),

3. Damping force = - C dy/d t,

4. An external force = f(t)

• Minus sign simply indicates that the force isresistive. Note: downward direction is positive.

Dr. Alam Z M Barkat Ali First Order Models, MATH 463 29---------------------------------------------------------------------------------------------------------

Spring-Mass System: Driven Motion

• Thus, the total force F acting on the body attached to the spring isthe following:

F = mg – k(s + y) - C (dy/d t) + f(t) = mg – k s – k y – C (dy/d t)

We know by Newton’s second law of motion: F = ma = m d2y/dt2.

Hence we can write

m d2y/dt2 = mg – k s – k y – C (dy/d t) + f(t)= – k y – C dy/d t +f(t)

where 2= C/m and 2 = k/m. If the mass attached to spring isreleased from a point at a distance of units from equilibriumposition with a velocity units per second. Then, y(0)= andy’(0)=. Hence, equation of motion of the system is the following:


m

tftFtFy

dt

dy

dt

yd

m

f(t)

dt

dy

m

Cy

m

k

dt

yd )()( ),(2 2

2

2

2

2

)0( ,)0( ),(2 2

2

2

yytFydt

dy

dt

yd

Example 5: Damped Driven Motion

• Suppose a body weighing 6.4 lbs attached to a spring stretches it by3.2 feet. Assume that damping force, numerically equal to 1.2 timesthe instantaneous velocity acts on the mass. In addition an externalforce f(t)=5 co s ( 4 t ) is being applied to the system. At t = 0 thebody is released from rest at a point 0.5 feet below the equilibriumposition.






• Suppose y(t) denotes distance of the body from equilibrium position.Then

But, f(t) = 5 co s ( 4 t ) , stretch s = 3.2 ft, and weight = 6.4 lbs.

mg = 6.4 m=6.4/g = 6.4/32 = 1/5 slug. Thus, by Hooke’s law:

mg = k s 6.4 = k 3.2 k = 2, 2 = k/m = 2/(1/5) = 10.Damping Force =-1.2(instantaneous vel.) C = 1.2 2=C/m = 6.

Since, the body is released from rest at a point 0.5 feet below theequilibrium position. Therefore, y(0) = -2 and y’(0) = 0, Note:downward direction is positive & upward is negative. Thus, requiredmathematical model is:


m

tftF

m

k

m

CtFy

dt

dy

dt

yd )()( , ,2 ),(2 22

2

2

tm

tftFtFy

dt

dy

dt

ydtFy

dt

dy

dt

yd4cos25

)()( ),(106 )(106

2

2

2

2

0)0(,2

1)0( ,4cos25 106

2

2

yytydt

dy

dt

yd


•



•


0)0(,2

1)0( ,4cos25 1064cos25 106

2

2

2

2

yytydt

dy

dt

ydty

dt

dy

dt

yd



From this figure we see that the system keeps performing sinusoidal motion

until forcing function f(t) = 5 c o s ( 4 t ) turns off, in which case the amplitudes

of the oscillation would diminish and this is what we also expect intuitively.

Electrical Systems: LRC Series Circuits

• Ohm’s Law: The voltage drop across a resistor is proportional to

the current flowing through it (Ohm’s law). The constant ofproportionality is denoted by R and is called resistance, which ismeasured in ohms. If VR denotes voltage drop across a resistor.Then

• Law of Inductance: Experiments have shown that the voltage

drop across an inductor is proportional to the instantaneous timerate of change of the electrical current I. The constant ofproportionality is denoted by L and is called inductance. Theinductance is measured in Henrys and time in seconds. If VLdenotes voltage across the inductor. Then

--------------------------------------------------------------------------------------

IRVIV RR

dt

dILVL


• Law of Capacitors: Experiments have shown that the voltage

drop across a capacitor is proportional to the instantaneous chargeQ on the capacitor. The constant of proportionality is dented by 1/C,where C is called capacitance. The capacitance is measured inFarad and charge in Coulombs. If VC denotes voltage drop acrossthe capacitor. Then

• Kirchhoff's Voltage Law: The algebraic sum of all of the voltage

drops around any closed loop is zero or the applied voltage VE in a

closed loop is equal to the voltage drops across the rest of the loop.

For example: If we consider the following LRC series circuit. Then,


QC

VC

1


CLRE VVVV

--------------------------------------------------------------------------------------------------------


• Suppose an E(t) volt battery is connected to the following LRCseries electrical circuit and I(t) denotes current in circuit & q(t)charge on capacitor at a time t.

• By using the Kirchhoff’s voltage, we can write the followingequation:

but the charge q(t) and the current are related by relation:

If initial charge on the capacitor is q0 and electric current in circuitis I0 . Then, q(0) = and q’(0) = serve as the initial conditions.


EECRL VqC

RIdt

dILVVVV

1

dt

dqtI )(

(1) 1

2

2

EVqCdt

dqR

dt

qdL


• Clearly, equation (1) is similar to the one that governs spring-masssystem. Consequently, nomenclature in the analysis of circuits isalso similar.

• If E(t) = 0 and R = 0 , then the electrical vibrations of the circuitare said to be free undamped, and electrical vibrations do notapproach zero.

• If E(t) = 0, R 0. then the circuit is said to be free damped. Forsolving equation (1) we put q(t) = emt . Then, the auxiliary equationis

Lm 2 + Rm + 1/C = 0

Therefore, solution of equation (1) is given by

• Clearly, the factor e-R/(2L) 0 as t, Therefore, q(t) 0 as t


L

CLRRm

2

/42

t

L

CLRct

L

CLRcetq LR

2

/4exp

2

/4exp)(

2

2

2

1

)2/(


• Model

Put , Then

AE is: Lm 2 + Rm + 1/C = 0

• Depending upon the value of discriminant R2 - 4L/C, we say that thecircuit is

1. Over damped if R 2 - 4L/C > 0

2. Critically damped if R 2 - 4L/C = 0

3. Under damped if R 2 - 4L/C < 0


,1

2

2

EVqCdt

dqR

dt

qdL

mteq

L

CLRRm

2

/4

2

Example 6: Underdamed LRC Series Circuit

• A zero volt battery is connected to an LRC series circuit, in whichinductance is L = 0.25 henrys, resistance is R = 10 andcapacitance is C = 0.001 Farad. Initially 4 coulombs of charge ispresent on the capacitor and initial current in the circuit is 0.

Set up a mathematical model to find electric charge on the capacitorat any time t.

Solve the model analytically and find a formula for the elecyriccharge on the capacitor at time t.


---------------------------------------------------------------------------------------


Example 6: Mathematical Model• Suppose q(t) denotes electric charge on the capacitor and I(t) the

electric current in the circuit at any time t. Then, using Ohm’s Law,Law of inductance and law of capacitors, we can write the following

Now, using Kirchhoff’s voltage Law, we can write the following:

Since, initial charge on capacitor is 4 coulombs and initial current inthe circuit is 0 amperes. Thus, Q(0) = 4, I(0) = 0 q’ (0) = 0.Hence, Mathematical Model is the following:


QQC

Vdt

dI

dt

dILVIRIV CLR 001.0

1 ,0.25 ,10

dt

dQIVQ

CRI

dt

dILVVVV EECRL but ,

1

010001025.0 1

2

2

2

2

Qdt

dQ

dt

QdVQ

Cdt

dQR

dt

QdL E

0)0( ,4)0( ,0400040010001025.02

2

2

2

QQQdt

dQ

dt

QdQ

dt

dQ

dt

Qd


•


0)0( ,4)0( ,010001025.02

2

QQQdt

dQ

dt

Qd


•


0)0( ,4)0( ,40004004000402

2

2

2

QQQdt

dQ

dt

QdQ

dt

dQ

dt

Qd

Example 6: Simulation Result/Scope

•


We see that

electric charge

on capacitor

oscillates

initially with

decreasing

amplitude

and becomes

0 after about

0.2 seconds,

in agreement

with analytical

solution.

Notice the

similarity with

graph in

example 4.

Analogy Between electrical systems andmechanical Systems.


•

------------------------------------------------------------------------

Second Order Models Mechanical / Electrical Systems

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