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Control Engineering

Tutorial #2

Dynamics of Mechanical

Systems

GUC Faculty of Engineering and Material Science

Department of Mechatronics

Control Engineering ENME 503

Dr. Ayman Ali El-Badawy





Tutorial #2 Dynamics of Mechanical Systems

As far of now, we know how to model mechanical systems and get

the equations of motion.

System equation

of motion

Modeling

Mathematical

ModelPhysical Dynamic

system

In this course, we are interested in

• Linear time invariant systems.

• The equations of motion are ODE’s.

• If not linear, we linearize.

Free response of undamped systems

Undamped system “spring mass system”

System equation of motion

For this system, we define









1Problem

Plot the solution of a linear spring-mass system with frequency ωn = 2 rad/s,

x0 = 1 mm, and v0= 2.34 mm/s, for at least two periods.

Solution:

• The equation of motion of this system can be written as

0

0

0

2

xx

xm

kx

kxxm

n

Remember

• Substituting with the given natural frequency, we get the

following DE.

04 xx

The given initial conditions are

smmx

mmx

/34.2)0(

1)0(

Remember that the equation of

motion, together with the initial values,

is called “the initial value problem”.





04 xx smmx

mmx

/34.2)0(

1)0(

• To solve this initial value problem, we assume the solution

taetx )(

• Plugging this solution into the DE, we get

2

04

04

2,1

2

2

j

aeea tt

DE Characteristic equation

tjtj eaeatx 2

2

2

1)( DE General Solution

• From the lecture, this reduces to

)2sin()2cos()(21

tAtAtx

• Constants A1 and A2 are obtained from the initial conditions





smmx

mmx

/34.2)0(

1)0(

1

21

21

1

)0sin()0cos(1

)2sin()2cos()(

A

AA

tAtAtx

2

2

21

17.1

)0cos(2)0sin(234.2

)2cos(2)2sin(2)(

A

A

tAtAtx

Final Solution

)2sin(17.1)2cos()( tttx

Nice solution, but we still can simplify it more…





)2sin(17.1)2cos()( tttx

• From the lecture, we have

• So, the final answer can be written as;

)71.02sin(54.1)( ttx

04 xx

smmx

mmx

/34.2)0(

1)0(

)71.02sin(54.1)( ttx

The problem The solution





Free response of undamped systems (The General case)

• For any undamped system, we have

Free response solution

Where

Free response plot





• For the given problem, the final answer can be obtained directly from the

general solution

04 xxsmmx

mmx

/34.2)0(

1)0(

2n

)71.02sin(54.1)( ttx

Substituting by the given values, we get

A = 1.54 mm and Φ= 40 deg = 0.71 rad.





)7.02sin(54.1)( ttx

Finally, the solution is given by

The time period of this solution is T = 2π/ωn = 3.14 sec. i.e. the solution will be

repeated each 3.14 sec.

Plotting this function using MATLAB, we use the following code.





The resulting plot:

By examining the

resulting plot, we

find that it starts

from x = 1 mm

which is the initial

displacement

given, and the

slope at this point

is equal to the

initial velocity.





Free response of damped systems

Damped system “spring-mass-damper system”


For this system we define

Undamped Natural Frequency

Damping Ratio

nm

c

2

02

0

2

xxx

xm

kx

m

cx

nn

Damping Ratio

The free response of the system is determined depending on the value of

the damping ratio

nm

c

2

1

1

1

Underdamped system, which means that the system will

oscillate until it reaches the equilibrium position

Critically damped system, which means that the system

will return to the equilibrium position without oscillation

Overdamped system, which means that the system will

return to the equilibrium position without oscillation

Note that the critically damped system is faster than overdamped system when

it returns back to the equilibrium position.

For this system, we define wd









2Problem

Solve the following initial value problems.

0)0(,1)0(,035 xxxxx

1)0(,3)0(,025.1 yyyyy

4)0(,1)0(,025.23 wwwww

Solution:

Before we solve, we find the natural frequency and the damping ratio for

each system just to relate the behavior of the response to the damping ratio

1,25.23

447.0,25.12

44.1,31

n

n

n

system

system

system

Underdamped system

Critically damped

Overdamped system





0)0(,1)0(,035 xxxxx

• Assume the solution

taetx )(

• The characteristic equation

0352

• The roots of the Characteristic

equation are

3.4,7.0 21

tt eaeatx 3.4

2

7.0

1)(

• The General solution is

• Invoking the initial conditions

gives the final solution as

tt eetx 3.47.0 2.02.1)(

)(tx

t

tt eetx 3.47.0 2.02.1)(





1)0(,3)0(,025.1 yyyyy


taety )(

• The characteristic equation

025.12


equation are

jj 5.0,5.0 21

)()( 21

5.0 tjjtt eaeaety

• The General solution is

• As in problem 1, this solution

can be recast in the form

))sin()cos(()( 21

5.0 tAtAety t



))sin(5.2)cos(3()( 5.0 ttety t

)(ty

t





4)0(,1)0(,025.23 wwwww


taetw )(• The characteristic equation

025.232


equation are

5.1,5.1 21

tt teaeatw 5.1

2

5.1

1)(

• Since the roots are repeated

here, the General solution is

given by



tt teetw 5.15.1 5.2)(

)(tw

t

tt teetw 5.15.1 5.2)(





A spring-mass-damper system has mass of 150 kg, stiffness of 1500 N/m and

damping coefficient of 200 kg/s. Calculate the undamped natural frequency, the

damping ratio and the damped natural frequency. Is the system overdamped,

underdamped, or critically damped? Does the solution oscillate?

3Problem

sradm

kn /16.3

21.02

nm

c

sec/02.31 2 radnd

m = 150 kg, k = 1500 N/m, c = 200 kg/s

The undamped natural frequency

The damping ratio

The damped natural frequency

It is obvious from the damping ratio that the system is underdamped because

< 1, and consequently the system will oscillate.

Solution:





The system of problem 3 is given an initial velocity of 10 mm/s and an initial

displacement of -5 mm. find the form of the response and plot it for as long as it

takes to die out.

Solution:

4Problem

)154.102.3sin(47.5)( 6636.0 tetx t

Now, we evaluate A and Φ

A = 5.47 mm

and Φ = -66.13 deg = -1.154 rad.

Finally the solution has the form:





Now we plot the solution using MATLAB using the following code






where

Here, the solution has the form

Forced Response of damped systems





5Problem

For the shown system, find the particular solution for the given inputs

)(25.1 tfyyy

Solution:

)Re(10)2cos(10)()

)Im(5)sin(5)()

2

2

1

ti

it

ettfb

ettfa

)sin(525.1

)(25.1 1

tyyy

tfyyy

Here, we define the operator D

2

22,

dt

dD

dt

dD

iteyDD

tyDD

5~)25.1(

)sin(5)25.1(

2

2

)(DP

a)





iteyDD 5~)25.1( 2

)(DP

25.1)( 2 iiP

25.1

5~2

ii

ey

it

p

)25.0()06.1(

5~

25.0

5~

ie

y

i

ey

it

p

it

p

)]sin())(cos(25.0[(72.4~ titiyp

)]sin()sin(25.0

)cos()cos(25.0[72.4~

tti

tityp

)Im(5)sin(5)(1

itettf

Since the input is given by

Then the Particular solution will be

)~Im( pp yy

)]sin(25.0)cos([72.4 ttyp





)2cos(1025.1

)(25.1 2

tyyy

tfyyy

tieyDD

tyDD

22

2

10~)25.1(

)2cos(10)25.1(

)(DP

b)

25.12)2()( 2 iiP

25.12)2(

10~2

2

ii

ey

ti

p

)275.2()56.11(

10~

275.2

10~

2

2

ie

y

i

ey

ti

p

ti

p

)]2sin()2)(cos(275.2[(87.0~ titiyp

)]2sin(2)2sin(75.2

)2cos(2)2cos(75.2[87.0~

tti

tityp

))Re(10)2cos(10)( 2

1

tiettf

Since the input is given by

Then the Particular solution will be

)~Re( pp yy

)]2sin(2)2cos(75.2[87.0 ttyp





6Problem

For the shown system, find its total solution

1)0(,3)0(),sin(525.1 yytyyy

Solution:

The total solution is simply the superposition of the homogeneous solution shown

in problem 2 and the particular solution shown in problem 5

))sin()cos(()( 21

5.0

shomogeneou tAtAety t

)]sin(25.0)cos([72.4)( ttty particular

)]sin(25.0)cos([72.4))sin()cos((

)()()(

21

5.0

shomogeneou

tttAtAe

tytyty

t

particular

Then, invoke the initial conditions to get the constants values A1 and A2.





A 100-kg mass is suspended by a spring of stiffness mN /1030 3

with a viscous-damping constant of 1000 N.s/m. The mass is initially

at rest and in equilibrium. Calculate the steady-state displacement

amplitude and phase if the mass is exited by a harmonic cosine force

of 80N at 3 Hz.

7Problem

mN /1030 3

sradfn /84.182

)84.18cos(80 tF

M =100 kg, k =

Mass is initially at rest and in equilibrium means that the initial

displacement and velocity are equal to zero.

Exciting force natural frequency

The exciting force is

To find the solution of the system we have first to know the type of damping

in the system.

, and c = 1000 N.s/m

Solution:

F





sradm

kn /32.17

288.02

nm

c

The undamped natural frequency

The damping ratio

The value of the damping ratio shows that the system is underdamped and

the solution is given by

Now we can calculate X, and Θ

X = 0.04 m Steady state displacement amplitude.

Θ = 1.856 rad Steady state phase.





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