Ch. 3: Forced Vibration of 1-dof system

Ch. 3: Forced Vibration of 1-DOF System

3.0 

Outline

„ Harmonic Excitation

„ Frequency Response Function

„ Applications

„ Periodic Excitation

„ Non-periodic Excitation

3.0 Outline

Ch. 3: Forced Vibration of 1-DOF System

3.1 

Harmonic Excitation

Force input function of the harmonic excitation is the

harmonic function, i.e. functions of sines and cosines.  

This type of excitation is common to many system

involving rotating and reciprocating motion.  Moreover,

many other forces can be represented as an infinite

series of harmonic functions.  By the principle of

superposition, the response is the sum of the individual 

harmonic response.

It is more convenient to use the frequency domain

technique in solving the harmonic excitation problems.

This is because the response to different

excitation frequencies can be seen in one graph.

3.1 Harmonic Excitation

Ch. 3: Forced Vibration of 1-DOF System

3.1 Harmonic Excitation

( )

( )

0

2

0

0

0

0

2

0

Let us focus on the particular solution of

cos

normalize the equation of motion

2

cos

,    

/

Re

solve for 

 from  

2

and the solution 

n

n

i t

i t

n

n

mx

cx

kx

F

t

x

x

x

f

t

f

F

m

f t

f

e

z t

z

z

z

f e

ω

ω

ω

ζω

ω

ω

ζω

ω

+

+

=

+

+

=

=

⎡

⎤

=

⎣

⎦

∴

+

+

=

( ) ( )

( )

( )

( )

(

)

(

)

( )

( )

2

2

0

is the real part of 

;   

Re

Assume the solution to have the same form as the forcing function

  same frequency as the input w/ different mag. and phase

2

i t

i t

i t

n

n

z t

x t

z t

z t

Z i

e

i

Z i

e

f e

f

Z i

ω

ω

ω

ω

ω

ζωω ω

ω

ω

=

⎡

⎤

⎣

⎦

=

−

+

+

=

=

(

)

(

)

2

0

0

2

2

2

0

2

/

2

1

/

2

/

         

1

/

2

/

n

n

n

n

n

n

n

f

i

i

F

k

i

ω

ω

ω

ζωω

ω ω

ζω ω

ω ω

ζω ω

=

−

+

−

+

=

⎡

⎤

−

+

⎣

⎦

Ch. 3: Forced Vibration of 1-DOF System

3.1 Harmonic Excitation

( )

( )

( )

( )

( )

( )

( )

(

)

( )

(

)

(

)

0

0

2

0

2

2

0

2

2

2

1

2

1

2

Re

,  /

1

2

1

If 

 is the frequency response

1

2

cos

1

where  

magnitude

1

2

2

tan

 phas

1

i t

i t

i t

n

i

F

z t

e

H i

F e

k

r

i

r

F

x t

e

r

k

r

i

r

H i

H i

e

k

r

i

r

x t

F H i

t

H i

k

r

r

r

r

ω

ω

ω

θ

ω

ζ

ω ω

ζ

ω

ω

ζ

ω

ω θ

ω

ζ

ζ

θ

−

=

=

⎡

⎤

− +

⎣

⎦

⎡

⎤

⎢

⎥

∴

=

=

⎡

⎤

− +

⎢

⎥

⎣

⎦

⎣

⎦

=

=

⎡

⎤

− +

⎣

⎦

∴

=

+

=

=

−

+

−

=

=

−

( )

( )

e

The system modulates the harmonic input by

the magnitude 

 and phase 

H i

H i

ω

ω

Ch. 3: Forced Vibration of 1-DOF System

3.1 Harmonic Excitation

(

)

(

)

( )

(

)

( )

(

)

( )

1

2

0

1

total response

homogeneous soln. particular soln.

Recall the homogeneous solution of the underdamped system

cos

 or 

sin

cos

cos

cos

or 

n

n

n

n

t

t

h

d

h

d

d

t

d

t

x

Ce

t

x

e

A

t

A

t

x t

Ce

t

F H i

t

x t

e

A

ζω

ζω

ζω

ζω

ω

φ

ω

ω

ω

φ

ω

ω θ

−

−

−

−

=

+

=

−

=

+

∴

=

−

+

+

=

(

)

( )

(

)

2

0

1

2

sin

cos

cos

The initial conditions will be used to determine  ,  or  ,

They will be different from those of free response

because the transient term now is partly due to the excitatio

d

d

t

A

t

F H i

t

C

A A

ω

ω

ω

ω θ

φ

+

+

+

n force

and partly due to the initial conditions

Ch. 3: Forced Vibration of 1-DOF System

3.1 Harmonic Excitation

Ex. 1  Compute and plot the response of a spring-mass

system to a force of magnitude 23 N, driving

frequency of twice the natural frequency and i.c.

given by x

0

= 0 m and v

0

= 0.2 m/s.  The mass

of the system is 10 kg and the spring stiffness

is 1000 N/m.

Ch. 3: Forced Vibration of 1-DOF System

3.1 Harmonic Excitation

(

)

( )

(

)

( )

( )

( )

3

2

2

3

1

2

3

1

2

2

/

1000 /10

10 rad/s

/ 2

0

2 10

20  rad/s

1

1

0.333 10

1

2

1000

1 2

sin

cos

23 0.333 10 cos

cos

sin

23 0.333 10 sin

i.c.  

0

0

23 0.333 10

n

n

n

n

n

n

n

n

k m

c

m

H i

k

r

i

r

x t

A

t

A

t

t

x t

A

t

A

t

t

x

A

ω

ζ

ω

ω

ω

ζ

ω

ω

ω

ω

ω

ω

ω

ω

ω

−

−

−

=

=

=

=

=

= ×

=

=

=

= −

×

⎡

⎤

− +

× −

⎣

⎦

=

+

−

×

×

=

−

+ × ×

×

= =

−

×

×

( )

( )

(

)

3

3

2

1

1

3

,   

7.667 10

0

0.2 10

,   

0.02

0.02 sin10

7.667 10

cos10

cos 20  m

A

x

A

A

x t

t

t

t

−

−

−

=

×

=

= ×

=

∴

=

+

×

−

Ch. 3: Forced Vibration of 1-DOF System

3.1 Harmonic Excitation

Ch. 3: Forced Vibration of 1-DOF System

3.1 Harmonic Excitation

Ex. 2  Find the total response of a SDOF system with

m = 10 kg, c = 20 Ns/m, k = 4000 N/m, x

0

= 0.01 m,

v

0

= 0 m/s under an external force F(t) = 100cos10t.

Ch. 3: Forced Vibration of 1-DOF System

3.1 Harmonic Excitation

(

)

( )

(

)

( )

( )

(

)

(

)

( )

(

)

( )

2

2

0

2

1

2

1

/

20  rad/s

/ 2

0.05

/

0.5

1

1

332.6

6

0.0666

1

2

4000

1 0.5

2 0.05 0.5

cos

33.26

3cos 10

0.0666

sin

cos

,  1

19.975 

rad/s

s

n

n

n

n

n

p

t

h

d

d

d

n

t

k m

c

m

r

H i

E

k

r

i

r

i

x

t

F H i

t

E

t

x t

e

A

t

A

t

x t

e

A

ζω

ζω

ω

ζ

ω

ω ω

ω

ζ

ω

ω θ

ω

ω

ω

ω

ζ

−

−

=

=

=

=

=

=

=

=

=

−

−

⎡

⎤

− +

× −

+ ×

×

⎣

⎦

=

+

=

−

−

=

+

=

−

=

=

(

)

( )

(

)

( )

(

)

(

)

( ) (

)

2

0

1

2

1

2

0

in

cos

cos

sin

cos

cos

sin

          

sin

n

n

d

d

t

t

n

d

d

d

d

d

d

t

A

t

F H i

t

x t

e

A

t

A

t

e

A

t

A

t

F H i

t

ζω

ζω

ω

ω

ω

ω θ

ζω

ω

ω

ω

ω

ω

ω

ω

ω

ω θ

−

−

+

+

+

= −

+

+

−

−

+

Ch. 3: Forced Vibration of 1-DOF System

3.1 Harmonic Excitation

Ch. 3: Forced Vibration of 1-DOF System

3.1 Harmonic Excitation

ω

response

finally becomes ω, and in phase

Ch. 3: Forced Vibration of 1-DOF System

3.1 Harmonic Excitation

ω

response

finally becomes ω, and out of phase

Ch. 3: Forced Vibration of 1-DOF System

3.1 Harmonic Excitation

F

0

ω

n

t/(2k)

( )

( )

( )

0

In case of 

0 and 

,  the guess solution of the form

cos

sin

 is invalid.  This is because

it has the same form as the homogeneous solution.

The correct particular solution is 

n

i t

n

p

x t

X i

e

A

t

B

t

F

x

t

ω

ζ

ω ω

ω

ω

ω

ω

=

=

=

=

+

=

sin

.

2

n

t

t

k

ω

Ch. 3: Forced Vibration of 1-DOF System

Beat  

when the driving frequency is close to natural freq.

3.1 Harmonic Excitation

( )

(

)

0

0

0

2

2

2

2

2

0

0

1

0

0

2

2

0

The total solution can be arranged in the form

sin

cos

cos

cos

2

      

sin

tan

sin

sin

2

2

If the system is at rest in 

n

n

n

n

n

n

n

n

n

n

n

n

v

f

x t

t

x

t

t

t

x

v

x

f

t

t

t

v

ω

ω

ω

ω

ω

ω

ω

ω

ω

ω ω

ω ω

ω

ω

ω

ω

−

=

+

+

−

−

+

⎛

⎞

−

+

⎛

⎞

⎛

⎞

=

+

+

⎜

⎟

⎜

⎟

⎜

⎟

−

⎝

⎠

⎝

⎠

⎝

⎠

( )

0

2

2

0

2

2

the beginning,

2

sin

sin

2

2

The response oscillates with frequency 

 inside

2

2

the slowly oscillated envelope 

sin

2

The beat frequency is 

n

n

n

n

n

n

n

f

x t

t

t

f

t

ω ω

ω ω

ω

ω

ω ω

ω ω

ω

ω

ω ω

−

+

⎛

⎞

⎛

⎞

=

⎜

⎟

⎜

⎟

−

⎝

⎠

⎝

⎠

+

−

⎛

⎞

⎜

⎟

−

⎝

⎠

∴

−

Ch. 3: Forced Vibration of 1-DOF System

Beat  

when the driving frequency is close to natural freq.

3.1 Harmonic Excitation

Ch. 3: Forced Vibration of 1-DOF System

3.2 

Frequency Response Function

3.2 Frequency Response Function

( )

(

)

2

The core of the particular solution to the harmonic function is

1

;  frequency response function

1

2

It specifies how the system responds to harmonic excitation.

As a standard, we normalize the 

H i

k

r

i

r

ω

ζ

=

− +

( )

2

frequency response function

1

 and then study how it varies as the

1

2

excitation frequency   and system parameters  ,

 vary.

It is indeed more convenient since we already normalized the frequen

n

G i

r

i

r

ω

ζ

ω

ζ ω

=

− +

( )

( )

cy;

/

.  So we can now study its variation to   and  .

For the fixed damping ratio, we plot 

 with   varies.

 has both magnitude and phase

magnitude and phase plot.

Then we repeatedly evaluate 

n

r

r

G i

r

G i

G i

ω ω

ζ

ω

ω

=

⇒

( )

 by varying  .

ω

ζ

Ch. 3: Forced Vibration of 1-DOF System

3.2 Frequency Response Function

Frequency response plot

(Bode diagram)

( )

(

)

(

)

2

2

2

1

1

2

H i

r

r

ω

ζ

=

−

+

1

2

2

tan

1

r

r

ζ

θ

−

−

⎛

⎞

=

⎜

⎟

−

⎝

⎠

Ch. 3: Forced Vibration of 1-DOF System

Resonance is defined to be the vibration response at

ω=ω

n

, regardless whether the damping ratio is zero.

At this point, the phase shift of the response is –π/2.

The resonant frequency will give the peak amplitude for 

the response only when ζ=0.  For             ,the peak 

amplitude will be at                         , slightly before ω

n

.

For               , there is no peak but the max. value of the

output is equal to the input for the dc signal (of course,