Physics Models on Spring-mass Systems-Mathematical Modeling and Simulation-Lecture Slides, Slides of Mathematical Modeling and Simulation

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Lecture Slides

on

Modeling and Simulation

Lecture: Second Order Linear & Homogeneous Models

Models : Physics models on Spring-mass Systems

Second Order Linear Homogeneous Models

• Models that are based on second order ordinary differential

equations have the following general form:

• This equation is linear.
• If G ( t ) = 0 for all t , then the equation is called homogeneous.

Otherwise the equation is nonhomogeneous.

P( t )yQ(t)yR(t)y G(t)

Let us consider Models with Constant Coefficients

• Two solutions of this equation are
• The general solution is of the following form:

y^  y  0

t t y t e y t e

 1 (^ ) ^ , 2 ( )^ 

t t y t ce c e

 ( )  1  2

0

2 y ^ woy

So we have a second order linear ODE based model:

Let us choose m = k = 1 and then we get wo =

Example 1: second order homogeneous model

• Now let us consider the following initial conditions for the model
• We have already found a general solution of the form

y^  y  0 , y( 0 ) 3 , y( 0 ) 1

t t

y t ce c e



Example 1: second order homogeneous model

• Our model and solution are
• Using MATLAB/Simulink, let us simulate

this system.

• Graphs of simulations are given here. The

graph on the right suggests that both initial conditions are satisfied.

0.0 0.2 0.4 0.6 0.8 1.

0

1

2

3

4

5

6

y(t)

time

ode45 solver in simulink was used.

t t y(t ) e e

y y , y( ) , y( )

   

    

2

0 0 3 0 1

simout To Workspace

Scope

1/s

Integrator

1/s

Integrator

0 1 2 3 4 5

0

50

100

150

200

250

300

y(t)

time

ode45 solver in simulink was used.

Short time behavior

long time behavior

Example 1: second order homogeneous model

Characteristic Equation in Second Order Models

To solve the 2nd^ order equation with constant coefficients,

ay^ bycy  0 ,

we begin by assuming a solution of the form:

y = exp (r t).

General Solution for such models

• Using the quadratic formula on the characteristic equation

we obtain two solutions, r 1 and r 2.

• There are three possible results:
  • The roots r 1 , r 2 are real and r 1  r 2.
  • The roots r 1 , r 2 are real and r 1 = r 2.
  • The roots r 1 , r 2 are complex.

Here in first case, we will assume r 1 , r 2 are real and r 1  r 2.

Then the general solution has the form:

2 ar  brc 

r t r t

y t ce c e

1 2

a

b b ac r 2

4

2    

Initial Conditions in 2

nd

Order Models

• For the initial value problem

we use the general solution

together with the initial conditions to find c 1 and c 2. That

is,

• Since we are assuming r 1  r 2 , it follows that a solution

of the form y = ert^ to the above initial value problem will

always exist, for any set of initial conditions.

ay bycy  0 , y(t 0 ) y 0 , y(t 0 ) y 0 ,

10 2 0 1 0 2 0

1 0 2 0

1 2

0 1 0 2 1 2

0 0 2 1 11 2 2 0

1 2 0 ,

rt r t rt r t

rt r t

e r r

y r y e c r r

y y r c c re c r e y

c e c e y  



   

    

 



  

 

r t r t y t ce c e

1 2 ( ) 1  2

y^  5 y 6 y  0 , y  0  2 , y  0  3

y^  5 y 6 y  0

m

k wo 

2

m

C damping coefficient

Let us put k = 6, m = 1 and C = 5 for a particular system. Then

The mathematical model is

Let us also assume initial conditions as y(0) = 2 , y’(0) = 3.

Then model becomes

0

2 y ^ ywoy 

Example 2: A spring with restoring effects

• Finally the model equation becomes:
• Then
• Factoring yields two solutions, r 1 = -2 and r 2 = -
• The general solution has the form:
• Using initial conditions:
• Thus particular solution is

y^  5 y 6 y  0 , y  0  2 , y  0  3

( ) 5 6 0  2  3  0

2 y t  e  r  r   r r 

rt

t t y t ce c e

3 2

2 ( ) 1

   

9 , 7 2 3 3

2 1 2 1 2

1 2    





  

  c c c c

c c

t t y t e e

2 3 ( ) 9 7

   

Example 2: A spring with restoring effects

Let us also now find the maximum value attained by the solution.

ln( 7 / 6 )

2 3

2 3

2 3

 

 

 

y

t

t

e

e e

y t e e

y t e e

t

t t

set t t

t t

0 1 2 3 4 5

0.

0.

0.

1.

1.

2.

2.

y(t)

time

ode45 solver in simulink was used.

Example 2: A spring with restoring effects

2 nd^ Order Models: Complex Roots of Characteristic Equation

• Recall the model equation:

where a , b and c are constants.

• Assuming an exponential solution leads to characteristic equation:
• Quadratic formula (or factoring) yields two solutions, r 1 & r 2 :
• If b^2 – 4 ac < 0, then complex roots: r 1 =  + i , r 2 =  - i 
• Thus

ay^ bycy  0

2 y t  e  ar brc

rt

a

b b ac r 2

4

2    

 i t  i t

y t e y t e

    

1 (^ )^ , 2 ( )

• Our two solutions thus far are complex-valued functions:
• We would prefer to have real-valued solutions, since our differential

equation has real coefficients.

• To achieve this, recall that linear combinations of solutions are themselves

solutions:

• Ignoring constants, we obtain the two solutions

y t e t ie t

y t e t ie t

t t

t t

 

 

( ) cos sin

( ) cos sin

2

1

y t y t ie t

y t y t e t

t

t









( ) ( ) 2 sin

( ) ( ) 2 cos

1 2

1 2

 

 

y t e t y t e t

t t  

  3 (^ )  cos , 4 ( )  sin

2nd Order Models: Complex Roots of Characteristic Equation

• Consider a spring of mass m (1) which tries to restore to its original state

when force is applied and its displacement is proportional to the applied force but opposite in direction. The proportionality constant (k) is 1 for this system.

• It has also damping effect due to friction of air and surface. The damping

force is proportional to the velocity of the mass. The proportionality constant is 0.1.

• Let us setup the mathematical model for such a system:

force y force ky

dt

dy damping force C

dt

dy ky C dt

d y Net force  m    2

2

Example 3: A Spring Revisited Again