3台灣師範大學機電科技學系

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台灣師範大學機電科技學系 C. R. Yang, NTNU MT

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Chapter 3 Harmonically Excited Vibration

3

台灣師範大學機電科技學系

C. R. Yang, NTNU MT

-2-

Learning Objectives

Find the response of undamped and viscously damped single-DOF systems subjected to different types of harmonic force, including base excitation and rotating unbalance

Distinguish between transient, steady-state, and total solutions

Understand the variations of magnification factor and phase angles with the frequency of excitation and the phenomena of resonance and beats

Find the response of systems involving Coulomb, hysteresis, and other types of damping

Identifyself-excited problemsand investigate their stability aspects

Derivetransfer functions of systems governed by linear differential equations with constant coefficients

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Learning Objectives

Solve harmonically excited single-DOF vibration problems using Laplace transforms

Derive frequency transfer functionfrom the general transfer function and represent frequency-response characteristicsusingBode diagrams

Solve harmonically excited vibration response using MATLAB

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Chapter Outline

3.1 Introduction 3.2 Equation of Motion

3.3 Response of an Undamped System Under Harmonic Force 3.4 Response of a Damped System Under Harmonic Force 3.5 Response of a Damped System Under

3.6 Response of a Damped System Under the Harmonic Motion of the Base

3.7 Response of a Damped System Under Rotating Unbalance 3.8 Forced Vibration with Coulomb Damping

3.9 Forced Vibration with Hysteresis Damping 3.10 Forced Motion with Other Types of Damping 3.11 Self-Excitation and Stability Analysis

t

ei

F t F() ₀ ^

(2)

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Chapter Outline

3.12 Transfer-Function Approach 3.13 Solutions Using Laplace Transforms 3.14 Frequency Transfer Functions

-6-

3.1 Introduction

3.1

-7-

3.1 Introduction

•Forced vibrations occurs whenever external energy is always supplied to the system during vibration.

•The external force can be supplied through either an applied force or an imposed displacement excitation, which may be harmonic, nonharmonic but periodic, nonperiodic, or random in nature.

•

The response of a system to a harmonic excitation is called harmonic response.The harmonic excitations is the form:

where F0is the amplitude, is the frequency, and is the phase angle of the harmonic excitation. The value of depends on the value of F(t) at t=0 and is usually taken to be zero.

0 0 0

( )

( ) ⁱ ^t ( ) cos( ) ( ) sin( )

F t F e ^{ }^ or F t F  t or F t F  t

-8-

3.1 Introduction

•Transient response is defined as the response of a dynamic system to suddenly applied nonperiodic excitations.

•Under a harmonic excitation, the response of the system will also be harmonic.

•If the frequency of excitation coincides with the natural frequency of the system, the response will be very large, which is known as resonance and should be avoided.

•The vibration produced by a unbalanced rotating machine, the oscillation of a tall chimney (煙囪) due to vortex (渦流) shedding (旋渦脫落)in a steady wind, and the vertical motion of an automobile on a sinusoidal road surface are examples of harmonically excited vibration.

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3.2 Equation of Motion

3.2

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3.2 Equation of Motion

•From the figure below, the equation of motion using Newton’s Second Law of Motion states that

•The homogeneous solution is obtained from the homogeneous equation(the free vibration)

( ) (3.1) mx  cx  kx  F t

0 (3.2) mx  cx  kx 

A spring-mass-damper system To find the homogeneous solution xh

and particular solution x_p, and general solution as x(t)=xh+xp

As seen in Section 2.6.2, this free vibration dies out with time under each of the three possible conditions of damping (underdamping, critical damping, and overdamping) and under all possible initial conditions

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3.2 Equation of Motion

•The variations of homogeneous and general solutions with time for a typical case are shown in the figure below.

Homogenous and general solutions of Eq. (3.1) for an underdamped case Homogeneous solution: Transient vibration

Particular solution:Steady-state vibration

General solution

(the free-vibration part)

(the force-vibration part)

Xh(t) dies out and x(t) becomes xp(t) after time The rate at which the

transient motion decays depends on the values of the system parameters k, c, and m

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3.3 Response of an Undamped System Under Harmonic Force

3.3

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3.3 Response of an Undamped System Under Harmonic Force

•Consider an undamped systemsubjected to a harmonic force. If a harmonic force acts on the mass m of the system,

•The homogeneous solutionis given by:

( ) 0cos F t F t

) 3 . 3 (

0cos t F kx x

m 



) 4 . 3 ( sin cos

)

(t C₁ t C₂ t

x_h  _n  _n

2 /

)1

/ (k m

n



where is the natural frequency

( mx  cx  kx  0 (3.2))

-14-

3.3 Response of an Undamped System Under Harmonic Force

•Because the exciting force and particular solution is harmonic and has the same frequency, we can assume a solution in the form:

Thus, the total solution of Eq. (3.3) is where X is the Max. amplitude of x_p(t)

) 5 . 3 ( cos

)

(t X t

x_p  

0 0

2 2

2

3 6

1 1

/ ^st ( . )

n

F F k

X k m m

k



  



  

    

  k

stF₀/



where denotes the static deflection

0

1 2 2

( ) _h( ) _p( ) cos _n sin _n F cos (3.7)

x t x t x t C t C t t

k m

  

     



代入(3.3)式，等號兩邊做係數比較求出X (待定係數法)

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3.3 Response of an Undamped System Under Harmonic Force

•Using initial conditions

Hence

The Max. amplitude in Eq. (3.6) can be expressed as ) 8 . 3 (

, ₂ ⁰

2 0 0 1

n

C x m k x F

C  

 

 



) 9 . 3 ( cos

sin cos

) (

2 0

0 2

0 0

m t k

F

x t m t

k x F t

x n

n n

 



 





 



 







 





 

 

2

1 (3.10)

st 1

n

X

 



  

  

 

0 0

( ) ( ) x t x and x t x

The ratio of the dynamic to the static amplitude of the motion is called magnification factor, amplification factor, or amplitude ratio

/ _st X 

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3.3 Response of an Undamped System Under Harmonic Force

•The quantity is called the magnification factor, amplification factor, or amplitude ratio.

•The variation of the amplitude ratiowith the frequency ratiois shown in the figure.

•The response of the system can be identified to be of three types.

X/st

Magnification factor of an undamped system

1 0 1

2 1

3 1

.

n



 





2

1

st 1

n

X

 



  

  

 

Case 1

Case 2 Case 3

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3.3 Response of an Undamped System Under Harmonic Force

•Case 1:

When 0 < < 1, the denominator in Eq.(3.10) is positive and the response is given by Eq.(3.5) without change. The harmonic response of the system is in phase with external force, shown in figure.

/ _n

 

Harmonic response when 0/_n1

2

1 (3.10)

st 1

n

X

 



  

  

 

) 5 . 3 ( cos

)

(t X t

x_p  

相角差=0^o

-18-

3.3 Response of an Undamped System Under Harmonic Force

•Case 2:

When > 1, the denominator in Eq.(3.10) is negative and the steady-state solution can be expressed as

where the amplitude is redefined as

The response of the system to a harmonic force of very high frequency is close to zero

/ _n

 

) 11 . 3 ( cos )

(t X t

x_p  

) 12 . 3 ( 1

2

 



 





n

X st





Harmonic response when /_n1

相角差=180^o (   _n , X0)

cos( )

X  t

 

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3.3 Response of an Undamped System Under Harmonic Force

•Case 3:

When , the amplitude X given by Eq.(3.10) or (3.12) becomes infinite. The forcing frequencyis equal to the natural frequency nof the system is called resonance.

The total response if the system at resonance is rewrite Eq. (3.9) as / n 1

  

0 0 0

0 2 2

0 0

2 0

0 2

0

( ) cos sin cos (3.9)

cos sin (cos cos )

cos cos

cos sin

1 ( )

n n

n

st

n n

n

n n

n

F x F

x t x t t t

k m k m

x t x t t t

x t x

m

t F k

t t

  

    



 

 

 

 

  

 

   

     

 

 

  

 

  

     

 

    

 

  

   

  (3.13)

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與時間成正比關係

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-21-

Harmonic response when

  /

_n

 1 ( . ., i e  

_n

)

0

( ) 0cos sin sin (3.15)

2

st n

n n n

n

t x t x t x t   t

  

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3.3 Response of an Undamped System Under Harmonic Force

•Total Response

The total response of the system, Eq.(3.7) or Eq.(3.9), can also be expressed as

( ) cos( ) 2cos ; for 1 (3.16)

1 ⁿ

n

st

n

x t A t t 

  



   



 

 

 

1 2

0

( ) cos _n sin _n 2cos (3.7)

x t C t C F t

t k m

  

  

 

•Or

And

( ) cos( ) 2cos ; for 1 (3.17)

1

st

n n

n

x t A t  t



 

 _  

 

   

 

  

( ) cos( ) cos cos( ) 2 cos

1

n n

st

n

x t A t  X t A t  t



 

 

  

 

 

   

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3.3 Response of an Undamped System Under Harmonic Force

( ) cos( ) 2cos ; for 1 (3.16)

1 ⁿ

n

st

n

x t A  t t 

  



   



 

 

 

-24-

( ) cos( ) 2cos ; for 1 (3.17)

1

st

n n

n

x t A t  t



 

 _  

 

   

 

  

The complete motion can be expressed as the sum of two cosine curves of different frequencies.

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-25-

3.3 Response of an Undamped System Under Harmonic Force

•Beating Phenomenon

If the forcing frequency is close to, but not exactly equal to, the natural frequency of the system, beating may occur.

The phenomenon of beating can be expressed as Eq. (3.18) from Eq.

(3.9) under the initial conditions of :

0 0 0

0 2 2

0

2 2

0

2 2

(cos

( ) cos sin cos (3.9)

( )

(3.18 cos )

(2 sin sin )

2 )

2

n

n n

n

t t

F x F

x t x t t t

k m k m

F m

  

 

   

 

   

        

 



 



0 0 0

x x 

-26- 0

2 2

0

( )

( ) (2 ) (3.18)

/ (

s

3.22)

si in

n 2 2

s n 2 s ni i

n

F m n

x t

F t

t

m t

 t



 



 



 



 

  







 ₂ ₂

2 (3.19) 2 (3.20) (3.19) (3.20)

4

n n n

n

 

  

  

  



  

 



  

數值小、週期大數值大、週期小

波的振幅

波的振幅

the period of beating

-27-

3.3 Response of an Undamped System Under Harmonic Force

•Beating Phenomenon

The time between the points of zero amplitude or the points of maximum amplitude is called the period of beatingand is given by

The frequency of beating defined as







b

 2 

n



2 2 2

(3.23) 2

2 2 1

b

b n

  

    



     



2 (3.19)

n n

 

  



  

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3.3 Response of an Undamped System Under Harmonic Force

Example 3.1 Plate Supporting a Pump

A reciprocating pump, having a mass of 68kg, is mounted at the middle of a steel plate of thickness 1cm, width 50cm, and length 250cm, clamped along two edges as shown in Figure. During operation of the pump, the plate is subjected to a harmonic force, F(t) = 220 cos 62.832t N. Find the amplitude of vibration of the plate.

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-29-

3.3 Response of an Undamped System Under Harmonic Force

Example 3.1 Plate Supporting a Pump Solution

The plate can be modeled as a fixed-fixed beam having Young’s modulus (E) = 200GPa, length = 250cm, and area moment of inertia,

The bending stiffness of the beam is given by:

2 2 3 9 4

1(50 10 )(1 10 ) 41.667 10 m

I12  ^  ^   ^

(E.1) N/m

82 . 102400 )

10 250 (

) 10 667 . 41 )(

10 200 )(

192 ( 192

3 2

9 9

3 



 

 _ ^

l k EI

a=50 cm

b=1 cm plate

1 3

I12ab

-30-

3.3 Response of an Undamped System Under Harmonic Force

Example 3.1 Plate Supporting a Pump Solution

Thus,

0

2 2

220 102400.82 68(62.832) 0.00132487 m 1.32487 mm (E.2) X F

k m

 

 

   

0

2

( . ) 3 6 X F

k m 

 

0 220 68

102400.82 / 62.832 /

F N

m kg

k N m

rad s





The negative sign indicates that the response x(t) of the plate is out of phasewith the excitation F(t)

-31-

3.4 Response of a Damped System Under Harmonic Force

3.4

-32-

3.4 Response of Damped System Under Harmonic Force



²



cos

 

sin

 

0cos ( .3 26) X^ km

  

t c

  

t ^F



t

•If the forcing function is given by

•The equation of motion becomes

•The particular solution of Eq. (3.24) is

•By substituting Eq.(3.25) into Eq.(3.24)

•Using trigonometric relations,

( ) 0cos F t F t

0  3 24

   cos ( . ) mx cx kx F t

3 25 ( ) cos( ) ( . ) x tp X  t

cos( ) cos cos sin sin

sin( ) sin cos cos sin

t t t

     

  

  

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-33-

•Equating the coefficients of cost and sint on both sides of the resulting equation, we obtain

 

2

0 2

cos sin

sin cos 0 (3.27)

X k m c F

X k m c

   

   

 

   

 

 

     

 

1 2 2

2

2 2 2 1/ 2

0 0

2

0 0

22 2

0

2

2 1/

sin t

sin cos 0 tan

cos

(3 [

.28

an ( )

[ (

( ) ]

sin

) ] )

k m c c

k m

if A k m c

F

k m A X F

k m c

k m c c

A

F F

c k m

F

A k m c

A A

 

 

 

      

 

   

   

 



      



  

 

 



 



 



3 25 ( ) cos( ) ( . )

x tp X t just find X and

 



 

The particular solution

=Steady-state response

(3.29)

-34-

3.4 Response of Damped System Under Harmonic Force

•The figure shows typical plots of the forcing function and steady- state response.

(a) Graphical representation (b) Vectorial representation for Eq.(3.26)

  ^ ^ ^ ^

     

2

0 2

0

3 26

cos sin cos ( . )

cos cos sin cos

X k m t c t F t

kX t m X t c X t F t

      

        

     

 

     

 t

 t  t

-35-

3.4 Response of Damped System Under Harmonic Force

•Dividing both the numerator and denominator of Eq. (3.28) by k and making the following substitutions

m; k

n



2 2 ; 2 n;

c n

c c c c

c m mk m

 

    

0; k F

st



n

r 

 

   

 

0 0

1 2 1 2

2 2

2 2 2 2 2 2

1 2 1 2

2 2

2 2 2 2

2 2 2

1 2 1 2

2 2

2 2 2

2 2

3 28

4

1 1

2

1

4

1

/ /

/ ( . )

( )

st st

n

st st

n

s

n

t

F F k

X

k m c k m c

c

k

m

k k k

r r

X

mk

   

 

   



 

 











 

       

   

   

 

     

     

     

   

 

      

    

 

 





 

1 2 1 2

2 2

2 2 2 2 2

2 2

1 3 30

4 1 2

/ / ( . )

( )

n n

r r

   

 

      

    

  

  Frequency ratio:

Undamped natural frequency:

Deflection under the static force:

-36-

1 1

2 2

1 1

2 2

1 1

2

2 2

3 29 1

2 2

1 1

1 1 3 31

tan tan

( )

t

( . )

an ta

tan tan

n

( . )

n n

c c k

k m m k

m

mk k k

m k m k

r r

 

  



 

 







 



 

   

      

   

 

 

       

 

 

   

    

 

 

   



m; k

n



2 2 ; 2 _n;

c n

c c c c

c m mk m

 

    

0; k F

st



n

r 

 

Frequency ratio:

Undamped natural frequency:

Deflection under the static force:

(10)

-37-

3.4 Response of Damped System Under Harmonic Force

•The following characteristics of the magnification factor (M=X/_st) can be noted from the figure as follows:

 

2 1 2

2 2 2

2 2

2 1 2

2 2

1 1 4

1 3 30

1 2

/

/ ( . )

( )

st

n n

X

r r

   

 



    

  

 

   

2

1 (3.10)

st 1

n

X

 



  

  

  0 if  

-38-

3.4 Response of Damped System Under Harmonic Force

1. For an undamped system , Eq.(3.30) reduces to Eq.(3.10), and as .

2. Any amount of damping reduces the magnification factor (M) for all values of the forcing frequency.

3. For any specified value of r, a higher value of damping reduces the value of M.

4. In the degenerate (退化的) case of a constant force (when r = 0), the value of M = 1.

) 0 (  



M  r  1 (   0)

-39-

3.4 Response of Damped System Under Harmonic Force

5. The reduction in M in the presence of damping is very significant at or near resonance.

6. The amplitude of forced vibration becomes smaller with increasing values of the forcing frequency (that is, as ).

7. For , the maximum value of Moccurs when

which can be seen to be lower than the undamped natural frequency nand the damped frequency

0 M  r  

0 1

2 0.71



  

2 2

1 2 or _n 1 2 (3.32)

n

r

    

  



 

1 2

d n

  

(see Problem 3.32)

-40-

3.4 Response of Damped System Under Harmonic Force

8. The maximum value of X (when ) is given by:

and the value of X at by

9. For when r = 0. For , the graph of M monotonically decreases with increasing values of r.

1 2 2

r  



²



² ² ^{1/ 2} ^max ²

1 (3.30)

1 (2

1 (3.

) 2

33) 1

st st

X X

r r

    

 

  

 

 

    

 

 



²



² ² ^{1/ 2}

1 (3.30) 1 . .,

(3.34) (

1 (2 2

1)

) ^s ⁿ

t t

n s

X i e

r r

X r

   

 

  ^

 

   

  

 

 

 

 



, 0

1 2

dM

  dr  1

0.71

  2 

n



(可實驗求得X與，代入求得，反之亦然 )

(11)

-41-

3.4 Response of Damped System Under Harmonic Force

• The following characteristics of the phase angle can be observed from the figure and Eq.(3.31) as follows:

1

2 2

1 2

2 1

2 3 31

1 tan

tan ( . )

n n

r r

 

 

 





 

 

   

 

 

 

     

●

-42-

3.4 Response of Damped System Under Harmonic Force

1.

For an undamped system , Eq.(3.31) shows that the phase angle is 0 for 0 < r < 1 and 180° for r > 1.This implies that the excitation and response are in phasefor 0 < r < 1 and out of phase for r > 1 when .

2. For and 0 < r < 1, the phase angle is given by 0 < Φ < 90°, implying that the response lagsthe excitation.

3. For and r > 1, the phase angle is given by 90° < Φ < 180°, implying that the response leadsthe excitation.

) 0 (

 0



 0



0



-43-

3.4 Response of Damped System Under Harmonic Force

4. For and r = 1, the phase angle is given by Φ = 90°, implying that the phase difference between the excitation and the response is 90°.

5. For and large values of r, the phase angle approaches 180°, implying that the response and the excitation are out of phase.

 0



0



-44-

3.4 Response of Damped System Under Harmonic Force

• Total response: The complete solution is For an underdamped system,

) 35 . 3 ( ) cos(

) cos(

)

(t X0e^^ t0 X t

x d

nt

) 36 . 3 ( 1 ² _n

d  

  

( ) _h( ) _p( ) x t x tx t

270 325

( ) _h _p .( . ) .( . )

xt x x Eq Eq

   

0

1 2 1 2 1 2

2 2 2 2 2 2

2 2 2 2 2

2 2

4 1

1

3 28 3 30

2

/ / /

( ( . )

) )

st st ( .

n n

X F

k m c r r

 

  

  

 

  

          

       

      

1 1 1

2 3 29 2 2 2 3 3

2 2

1 1

tan ( . ) = tan ⁿ tan 1 ( . )

n

c r

k m r

 



 

 ^  ^   ^

 

 

     

         

 

 

1 2

(where_d  _n)

(12)

-45-

0 0 0

0

0 0 0 0

0

0 3 36

( )

cos cos

cos sin sin ( . )

n d

x X X

X X X

t x

x t x

 

     

 



 

    

X₀and ₀can be determined from the initial conditions, , Eq.(3.35) yields

0 0

( 0) ( 0) x t x and x t x

) 35 . 3 ( ) cos(

) cos(

)

(t X₀e^^ t₀ X t

x ⁿ^t _d

The solution of Eq.(3.36) gives X₀and ₀as 解聯立

方程式

(Practice by yourself)

-46-

3.4 Response of Damped System Under Harmonic Force

Example 3.3

Total Response of a System

Find the total responseof a single degree of freedom system with m

=10kg, c = 20 N-s/m, k = 4000 N/m, x₀= 0.01 m, under the following conditions:

a. An external force acts on the system with and .

a. Free vibration with F(t) = 0.

00 x

t F t

F() 0cos N

0100 rad/s F

10



-47-

3.4 Response of Damped System Under Harmonic Force

Example 3.3 Total Response of a System Solution

a. From the given data,

  

   

        

0 4000

10

100 4000

20

2 2 4000 10

2 2

10 20

2 2 2 2

2 2

20 0 025

0 05

1 1 0 05 20 19 974984 0 5

0 025

0 03326

1 2 1 0 05 2 0 5 0 5

rad/s . m

.

. . rad/s

.

. . m (E.1)

. . .

c

n k

n m

F

st k

c c

c km

d n

st

r X

r r









  





  

   

    

  

 

     

0 0

( ) ⁿ^tcos( _d ) cos( ) (3.35)

x t X e^^



t



X

 

t

-48-

3.4 Response of Damped System Under Harmonic Force

Example 3.3

Total Response of a System Solution

We have

Using initial conditions , Eq.(3.36)yield:

(E.2) 814075

. 5 3 . 0 1

5 . 0 05 . 0 tan 2 1

tan¹ 2 ₂ ¹ ₂  



 









 



 





 ^  ^

r

r



0 0

0 01 0 03326 0 997785

0 023186

. cos ( . )( . )

cos . (E.3)

X X



 

  

(E.4) )

814075 . 3 sin(

) 10 )(

03326 . 0 (

sin ) 974984 . 19 ( cos ) 20 )(

05 . 0 (

0 ₀ ₀ ₀ ₀







 X  X 

0 and 01 .

0

₀

0

 x 

x 

(13)

-49-

3.4 Response of Damped System Under Harmonic Force

Substituting Eq.(E.3) into (E.4),

Hence,

(E.5) 002268

. 0 sin ₀

0  

X

(E.7) 586765

. 5

0978176 . cos 0 tan sin

0

0 0

0 0 0







  X X



⁽ ^cos ⁾ ⁽ 0^sin 0⁾²



¹^/² ⁰^.⁰²³²⁹⁷ ^(E.6)

2 0 0

0 X   X  

X

-50-

3.4 Response of Damped System Under Harmonic Force

b. For free vibration, the total response is

Using the initial conditions,

(E.9) 010012 . 974984 0

. 19

01 . 0 20 05 . 01 0 . 0

2 / 2 1 2

2 / 2 1

2 0 0

0 











 



 



  



 















 







d nx x

X 



0 0

( ) ⁿ^tcos( d ) (E.8) x t X e^^ t

1 0 0 1

0

0 05 20

2 865984 19 974984

tan tan . . (E.10)

.

n

d

x x

x

 



     

     

0 and 01 .

0

₀

0

 x 

x 

(See Eqs.(2.73) and (2.75)))

-51-

3.4 Response of Damped System Under Harmonic Force

• Quality factor and bandwidth:

For small values of damping ,

From figure, R1and R2, where the amplification factor falls to , are called half-power pointsbecause the power absorbed (W) by the damper, responding harmonically at a given frequency, is proportional to the square of the

amplitude (See Eq.(2.94))

0 05

(  . )

) 38 . 3 ( 2

1

max

X Q X

st n st



 



 







 





 



 _ _

The value of the amplitude ratio at resonance is called as quality factor or Q factor of the system

) 39 . 3

2

( X c W   



2 Q

2 max

1 (3.33)

2 1

st

X

  

 

  

  

-52-

2

0

(  )

1 (3.38)

2 Q

^

2 2 2

2 2 (1r) (2r) 

3台灣師範大學機電科技學系

Chapter 3

Harmonically Excited Vibration

3

Learning Objectives

Learning Objectives

Chapter Outline

Chapter Outline

3.1

Introduction

3.1

3.1 Introduction

•

3.1 Introduction

3.2

Equation of Motion

3.2

3.2 Equation of Motion

( ) (3.1) mx  cx  kx  F t

0 (3.2) mx  cx  kx 

3.2 Equation of Motion

3.3 Response of an Undamped System Under Harmonic Force

3.3

3.3 Response of an Undamped System Under Harmonic Force



( mx  cx  kx  0 (3.2))

3.3 Response of an Undamped System Under Harmonic Force

代入(3.3)式，等號兩邊做係數比較 求出X (待定係數法)

3.3 Response of an Undamped System Under Harmonic Force

3.3 Response of an Undamped System Under Harmonic Force

3.3 Response of an Undamped System Under Harmonic Force

 

3.3 Response of an Undamped System Under Harmonic Force

3.3 Response of an Undamped System Under Harmonic Force

與時間成正比關係

  /

 1 ( . ., i e  

)

3.3 Response of an Undamped System Under Harmonic Force

3.3 Response of an Undamped System Under Harmonic Force

3.3 Response of an Undamped System Under Harmonic Force

數值小、週期大 數值大、週期小

3.3 Response of an Undamped System Under Harmonic Force









 2 



2 2 2

(3.23) 2

2

2 1

  

    



     



3.3 Response of an Undamped System Under Harmonic Force

3.3 Response of an Undamped System Under Harmonic Force

a=50 cm

b=1 cm plate

3.3 Response of an Undamped System Under Harmonic Force

( . ) 3 6 X F

k m 

 

3.4 Response of a Damped System Under Harmonic Force

3.4

3.4 Response of Damped System Under Harmonic Force





 

 

  

  



cos( ) cos cos sin sin

sin( ) sin cos cos sin

t t t

t t t

代入(3.3)式，等號兩邊做係數比較求出X (待定係數法)

數值小、週期大數值大、週期小

  ^ ^ ^ ^