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### Transcript of Chapter 2 Response to Harmonic Excitation 2018. 1. 30.آ  2 2 2 2 22 2 ( ) cos( tan ) ( ) (2 ) n p...

• Chapter 2 Response to Harmonic Excitation

Introduces the important concept of resonance

• 2.1 Harmonic Excitation of Undamped Systems

• Consider the usual spring mass damper system with

applied force F(t)=F0cosωt

• ω is the driving frequency

• F0 is the magnitude of the applied force

M

F=F 0 cosωt

Displacement

x k

• 0

2

0

0 0

( ) ( ) cos( )

( ) ( ) cos( )

where ,

n

n

mx t kx t F t

x t x t f t

F kf m m

 

  

 

 

Equations of motion

( ) cos( )px t X t

• Solution is the sum of

homogenous and

particular solution

• The particular solution

assumes form of forcing

function (physically the

input wins):

• Substitute particular solution into the equation of motion:

( ) cos( )px t X t 2

2 2

0

0

2 2

cos cos cos

solving yields:

p n px x

n

n

X t X t f t

f X

    

 

  

 

Thus the particular solution has the form:

0

2 2 ( ) cos( )p

n

f x t t

  

• Add particular and homogeneous solutions to get general

solution:

0 1 2 2 2

homogeneous

1 2

( )

sin cos cos

and are constants of integration.

particular

n n

n

x t

f A t A t t

A A

    

  

(2.8)

• Apply the initial conditions to evaluate the constants

0 0 1 2 2 02 2 2 2

0 2 0 2 2

0 1 2 1 02 2

0 1

0

(0) sin 0 cos0 cos0

(0) ( cos0 sin 0) sin 0

( )

n n

n

n n

n

n

f f x A A A x

f A x

f x A A A v

v A

v x t

   

 

   

       

   

     

  

 0 00 2 2 2 2sin cos cosn n n n n

f f t x t t  

   

     

  

(2.11)

• Comparison of free and forced response

• Sum of two harmonic terms of different frequency

• Free response has amplitude and phase effected by forcing

function

• Our solution is not defined for ωn = ω because it produces

division by 0.

• If forcing frequency is close to natural frequency the

amplitude of particular solution is very large

• Response for m=100 kg, k=1000 N/m, F=100 N, ω = ωn +5

v0=0.1m/s and x0= -0.02 m.

Note the obvious presence of two harmonic signals

0

Displacement (x)

Time (sec)

• 0

What happens when ω is near ωn?

0

2 2

2 ( ) sin sin (2.13)

2 2

n n

n

f x t t t

   

 

         

    

When the drive frequency and natural frequency

are close a beating phenomena occurs

0

2 2

2 sin

2

n

n

f t

 

 

   

  

Larger

amplitude

Time (sec)

Displacement (x)

• What happens when ω is ωn?

0

( ) sin( )

substitute into eq. and solve for

2

px t tX t

X

f X

When the drive frequency and natural

frequency are the same the amplitude

of the vibration grows without bounds.

This is known as a resonance

condition.

The most important

concept in Chapter 2!

0

2 2

2 sin

2

n

n

f t

 

 

   

  

0 5 1 0

1 5

20

25

x(t) = A1 sinwt + A2 coswt +

f0

2w t sin(wt)

grows with out bound

• Example 2.1.1: Compute and plot the response for

m=10 kg, k=1000 N/m, x0=0,v0=0.2 m/s, F=23 N, ω =2ωn.

0 0

30

2 2 2 2 2 2

10 kg

23 N 0.2 m/s 2.3 N/kg, 0.02 m

2.3 N/kg 7.9667 10 m

(10 20 ) rad / s

Equation (2.11) then yields:

( ) 0.02sin10 7.96

n n

n

n

k

m

vF f

m

f

x t t

  

 

    

    

     

  367 10 (cos10 cos 20 )t t 

• Example 2.1.2 Given zero initial conditions a

harmonic input of 10 Hz with 20 N magnitude and k=

2000 N/m, and measured response amplitude of 0.1m,

compute the mass of the system.

 

 

0

2 2

0

2 2

0.1 m

0

2 2 2

( ) cos 20 cos for zero initial conditions

2 trig identity ( ) sin sin

2 2

2 2(20 / ) 0.1 0.1

2000 (20 )

0.405 kg

n

n

n n

n

n

f x t t t

f x t t t

f m

m

m

   

   

 

  

  

          

    

     

 

• Example 2.1.3 Design a rectangular mount for a

security camera.

0.02 x 0.02 m

in cross section

Compute l > 0.5 m so that the mount keeps the camera from

vibrating more then 0.01 m of maximum amplitude under a

wind load of 15 N at 10 Hz. The mass of the camera is 3 kg.

• Solution: Modeling the mount and camera as a beam

with a tip mass, and the wind as harmonic, the equation

of motion becomes:

03

3 ( ) cos

EI mx x t F t 

From strength of materials:

3

12

bh I 

Thus the frequency expression is: 3 3

2

3 3

3

12 14 n

Ebh Ebh

m m   

Here we are interested computing l that will make the

amplitude less then 0.01m:

2 20

2 2

0

2 2 2 20

2 2

2 ( ) 0.01 , for 0

2 0.01

2 ( ) 0.01, for 0

n

n

n n

n

f a

f

f b

   

   

 

     

       

 

• Case (a) (assume aluminum for the material):

3 2 2 20

0 02 2 3

3 3

2

0

2 0.01 2 0.01 0.01 0.01 2 0.01

4

0.01 0.02 0.272 m 4 (0.01 2 )

n

n

f Ebh f f

m

Ebh

m f

    

        

     

Case (b):

3 2 2 20

0 02 2 3

3 3

2

0

2 0.01 2 0.01 0.01 2 0.01 0.01

4

0.01 0.012 0.229 m 4 (2 0.01 )

n

n

f Ebh f f

m

Ebh

m f

    

       

     

• 3(2.7 10 )(0.22)(0.01)(0.01)

0.149 kg

m bh

 

Remembering the constraint that the length must be at

least 0.2 m, (a) and (b) yield

0.2 0.229, or choose = 0.22 m  To check, note that

  3

22 2

3

3 20 1609 0

12 n

Ebh

m       

• A harmonic force may also be represented by sine or a

complex exponential. How does this

change the solution?

2

0 0( ) ( ) sin or ( ) ( ) sin (2.18)nmx t kx t F t x t x t f t     

The particular solution then becomes a sine:

( ) sin (2.19)px t X t

Substitution of (2.19) into (2.18) yields:

0

2 2 ( ) sinp

n

f x t t

  

Solving for the homogenous solution and evaluating the constants yields

0 0 0 0 2 2 2 2

( ) cos sin sin (2.25)n n n n n n

v f f x t x t t t

   

     

 