Chapter 2 Response to Harmonic Excitation

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

• We take c = 0 to start with

M

F=F0cosωt

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 k

f m m



 



  

 

 

Equations of motion

( ) cos( )

x 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( )

x t

 X  t

2

2 2

0

0

2 2

cos cos cos

solving yields:

p n p

x x

n

n

X t X t f t

X f



    

 

  

 

Thus the particular solution has the form:

0

2 2

( ) cos( )

p

n

x t f 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

A t A t f t

A A

  

 



 

 (2.8)

Apply the initial conditions to evaluate the constants

0 0

1 2 2 2 2 2 2 0

0

2 0 2 2

0

1 2 2 2 1 0

0 1

0

(0) sin 0 cos 0 cos 0

(0) ( cos 0 sin 0) sin 0

( )

n n

n

n n

n

n

f f

x A A A x

A x f

x A A f A v

A v

x t v

   

 

 

 





     

 

  



    



  

 sin _{n 0 2} ⁰ ₂ cos _{n 2} ⁰ ₂ cos

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 ω_n = 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 = ω_n +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 ω

?

0

2 2

( ) 2 sin sin (2.13)

2 2

n n

n

x t f  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 ω

?

0

( ) sin( )

substitute into eq. and solve for

2 x tp tX t

X X f









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) = A₁sinw^t + A₂ cosw^t + f₀

2w ^{t sin(}w^t)

grows with out bound

Example 2.1.1: Compute and plot the response for

=0,v

=0.2 m/s, F=23 N, ω =2ω

.

0 0

0 3

2 2 2 2 2 2

1000 N/m

10 rad/s, 2 20 rad/s 10 kg

23 N 0.2 m/s

2.3 N/kg, 0.02 m

10 kg 10 rad/s

2.3 N/kg

7.9667 10 m (10 20 ) rad / s

Equation (2.11) then yields:

( ) 0.02 sin10 7.96

n n

n

n

k m

v f F

m f

x t t

  



 



    

    

   

 

  67 10 (cos10 ^3 t cos 20 )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

trig identity ( ) 2 sin sin

2 2

2 2(20 / )

0.1 0.1

2000 (20 ) 0.405 kg

n n

n n

n

n

x t f t t

x t f 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:

From strength of materials:

3

12 I  bh

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 2

0

2 2

0

2 2

2 2

0

2 2

( ) 0.01 2 , for 0

2 0.01

( ) 2 0.01, for 0

n n

n

n n

a f f

b f

 

 

   

 

    

 

  

    

 



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

3

2 2 2

0

0 0

2 2 3

3 3

2

0

0.01 2 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 2

0

0 0

2 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

  

 



      



    



(2.7 10 )(0.22)(0.01)(0.01)3

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 2

2 2

3

3 20 1609 0

n 12

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 ( ) _n ( ) sin (2.18) mx t  kx t  F t x t  x t  f t

The particular solution then becomes a sine:

( ) sin (2.19) x tp  X t

Substitution of (2.19) into (2.18) yields: ^{p( ) 2 0 2 sin}

n

x t f t

  



Solving for the homogenous solution and evaluating the constants yields

0 0 0

0 2 2 2 2

( ) cos _n sin _n sin (2.25)

n n n n

v f f

x t x  t   t t

     

 

      

Section 2.2 Harmonic Excitation of Damped Systems

Extending resonance and response calculation to damped

systems

0 2

0

now includes a phase shift

( ) ( ) ( ) cos

( ) 2 ( ) ( ) cos

( ) cos ( )

n n

p

mx t cx t kx t F t

x t x t x t f t

x t X t



  

 

  

  

 

2.2 Harmonic excitation of damped systems

M

x k ^c

Displacement

F=F₀cosωt

Let x_p have the form:

2 2 1

2 2

( ) cos sin

, tan

sin cos

cos sin

p s s

s

s s

s

p s s

p s s

x t A t B t

X A B B

A

x A t B t

x A t B t

 



   

   



 

 

    

 

  

  

Note that we are using the rectangular form, but we could use one of the other forms of the solution.

Substitute into the equations of motion

 

2 2

0

2 2

2 2

0

2 2

( 2 ) cos

2 sin 0

for all time. Specifically for 0, 2 /

( ) (2 )

( 2 ) ( ) 0

s n s n s

s n s n s

n s n s

n s n s

A B A f t

B A B t

t

A B f

A B

    

    

 

   

   

   

    

 

  

   

Write as a matrix equation:

2 2

0

2 2

( ) 2

2 ( ) 0

n n s

n n s

A f

B

   

   

      

            

 

Solving for A_s and B_s:

2 2

0

2 2 2 2

( )

( ) (2 )

n s

n n

A   f

   

 

 

0

2 2 2 2

2

( ) (2 )

n s

n n

B   f

   

  

Substitute the values of A

and B

into x

:

0 1

2 2

2 2 2 2

( ) cos( tan 2 )

( ) (2 )

n p

n n n

X

x t f t



  

 

   

  

      

Add homogeneous and particular to get total solution:

particular or homogeneous or transient solution

steady state solution

( ) ^nt sin( _d ) cos( )

x t  Ae^





 

Note: that A and Φ will not have the same values as in Ch 1, for the free response. Also as t gets large, transient dies out.

Things to notice about damped forced response

• If ζ = 0, undamped equations result

• Steady state solution prevails for large t

• Often we ignore the transient term (how large is ζ, how long is t?)

• Coefficients of transient terms (constants of integration) are effected by the initial conditions AND the forcing function

• For underdamped systems at resonance the, amplitude is finite.

(0) 0.05

sin 0.133cos( 0.013) (0) 0 0.01 sin

9.999 cos 0.665sin 0.013 x

A

v A

A









  

  

 

Example 2.2.3: ω_n = 10 rad/s, ω = 5 rad/s, ζ = 0.01, F₀= 1000 N, m = 100 kg, and the initial conditions x₀ = 0.05 m and v₀ = 0.

Compare amplitude A and phase Φ for forced and unforced case:

Using the equations on slide 23:

0.1

0.133, 0.013

( ) ^t sin(9.99 ) 0.133cos(5 0.013) X

x t Ae t t







  

   

Differentiating yields:

0.1

0.1

( ) 0.01 sin(9.999 ) 9.999 cos(9.999 ) 0.665sin(5 0.013)

applying the intial conditions :

0.083 (0.05), 1.55 (1.561)

t

t

v t Ae t

Ae t

t

A











  

 

 

  

Proceeding with ignoring the transient

• Always check to make sure the transient is not significant

• For example, transients are very important in earthquakes

• However, in many machine applications transients may be ignored

2

2 2 2

0 0

1

(1 ) (2 )

X n

Xk

F f r r



  

 

Proceeding with ignoring the transient

Magnitude: ₍ ^{2 2 2}₎ ⁰ _{(2 )}²

n n

X f

   

   (2.39)

Frequency ratio:

n

r 

 

Non dimensional Form:

(2.40) Phase: ^{  tan1}





2 2 2

1

(1 ) (2 )

X

r  r

  

0 0.5

1

1.5

2

Magnitude plot

• Resonance is close to r = 1

• For ζ = 0, r =1 defines resonance

• As ζ grows resonance moves r <1, and X decreases

• The exact value of r, can be found from differentiating the magnitude

 

1

2

tan 2

1 r

r

  ^ 



Phase plot

• Resonance occurs at Φ = π/2

• The phase changes more rapidly when the damping is small

• From low to high values of r the phase always changes by 180⁰ or π radians

0 0.5 1 1.5 2

0 0.5 1 1.5 2 2.5 3 3.5

r Phase (rad)



 =0.1

 =0.3

 ^=0.5

 ⁼¹

Fig 2.7

2 2 2 0

2 peak

0 max 2

1 0

(1 ) (2 )

1 2 1 1/ 2

1 2 1

d Xk d

dr F dr r r

r

Xk F



 

 

 

 

 

  

     

   

    

 

  

  

Example 2.2.3 Compute max peak by differentiating:

(2.41)

(2.42)

Effect of Damping on Peak Value

• The top plot shows how the peak value becomes very large when the

damping level is small

• The lower plot shows how the frequency at which the peak value occurs reduces with increased damping

• Note that the peak value is only defined for values

ζ<0.707

0 0.2 0.4 0.6 0.8

0 5 10 15 20 25 30

ζ

0 0.2 0.4 0.6 0.8

0 0.2 0.4 0.6 0.8 1

rpeak

(dB) F0

Xk

Fig 2.9

ζ

Section 2.3 Alternative Representations

• A variety methods for solving differential equations

• So far, we used the method of undetermined coefficients

• Now we look at 3 alternatives:

a geometric approach

a frequency response approach a transform approach

• These also give us some insight and additional useful tools.

2.3.1 Geometric Approach

• Position, velocity and acceleration phase shifted each by π/2

• Therefore write each as a vector

• Compute X in terms of F₀ via vector addition

Re Im

) cos(wt -q kX

A B

C

D

E kX

X cw

X mw²

F0

C

A B

X m

k )

( - w²

X cw

F0





Using vector addition on the diagram:

2 2 2 2 2 2

0

0

2 2 2

( ) ( )

( ) ( )

F k m X c X

X F

k m c

 

 

  

  

At resonance:

,

2

X F

c

 

  

real part imaginary part

0

harmonic input

cos ( sin )

( ) ( ) ( )

j t

j t

Ae A t A t j

mx t cx t kx t F e





 

 

  

2.3.2 Complex response method

(2.47)

(2.48)

Real part of this complex solution corresponds to the physical solution

2

0

0 2 0

2

the frequency response function

( )

( )

( )

( ) ( )

( ) 1

( ) ( )

j t p

j t j t

x t Xe

m cj k Xe F e

X F H j F

k m c j

H j k m c j



 

 

  

  



   

 

 

  

Choose complex exponential as a solution

(2.49) (2.50)

(2.51)

(2.52)

Note: These are all complex functions

0

2 2 2

1

2

( )

0

2 2 2

( ) ( )

tan

( )

( ) ( )

j

j t p

X F e

k m c

c k m

x t F e

k m c



 

 

 



 







  

 

     

  

Using complex arithmetic:

(2.53)

(2.54)

(2.55)

Has real part = to previous solution

Comments:

• Label x-axis Re(e^jωt) and y-axis Im(e^jωt) results in the graphical approach

• It is the real part of this complex solution that is physical

• The approach is useful in more complicated problems

Example 2.3.1: Use the frequency response approach to compute the particular solution of an undamped system

 

 

2

0 0

2 2

0 0

2 2

( ) ( ) ( ) ( )

Let ( )

j t j t

n j t

p

j t j t

n

n

mx t kx t F e x t x t f e

x t Xe

Xe f e X f

 



 



 

 

    



   

 



2.3.3 Transfer Function Method The Laplace Transform

• Changes ODE into algebraic equation

• Solve algebraic equation then compute the inverse transform

• Rule and table based in many cases

• Is used extensively in control analysis to examine the response

• Related to the frequency response function

The Laplace Transform approach:

• See appendix B and section 3.4 for details

• Transforms the time variable into an algebraic, complex variable

• Transforms differential equations into an algebraic equation

• Related to the frequency response method

0

( ) ( ( )) ( ) ^st

X s x t x t e dt



 ^L 



Take the transform of the equation of motion:

0

2 0

2 2

cos

( ) ( )

mx cx kx F t ms cs k X s F s

s





   

  



Now solve algebraic equation in s for X(s)

0

2 2 2

( ) ( )( )

X s F s

ms cs k s 

   

To get the time response this must be “inverse

transformed”

2

2

2

( ) ( ) ( )

( ) 1

( ) ( ) ( ) 1

frequency response function ms cs k X s F s

X s H s

F s ms cs k

H j  k m c j

 

   

 

 

  



Transfer Function Method With zero initial conditions:

(2.59)

(2.60)

The transfer function

Example 2.3.2 Compute forced response of the suspension system shown using the Laplace transform

Summing moments about the shaft:

( ) ( ) 0 sin

J







Taking the Laplace transform

  

2

0 2 2

0 2 2 2

( ) ( ) ( ) 1

Js X s kX s aF s X s a F

s Js k





 

 



 

 

Taking the inverse Laplace transform:

  

1

0 2 2 2

2 2

( ) 1

1 1 1

( ) sin sin _n , _n

n n

t a F L

s Js k

a F k

t t t

J J

 



    

   

  

 

    

 

      

Notes on Phase for Homogeneous and Particular Solutions

 Equation (2.37) gives the full solution for a harmonically driven underdamped SDOF oscillator to be

   

homogeneous solution particular solution

( )

sin

cos

x t  Ae

 t    X   t 

How do we interpret these phase angles?

Why is one added and the other subtracted?

 

sin _dt 

tan 1 ^{o d}

o n o

x

v x

 



 



0

0

0

x  v 

Non-Zero initial conditions

Zero initial displacement

Zero initial velocity

Phase on Particular Solution

• Simple “atan” gives -π/2 < Φ < π/2

• Four-quadrant “atan2” gives 0 < Φ < π

2.4 Base Excitation

• Important class of vibration analysis

– Preventing excitations from passing from a vibrating base through its mount into a structure

• Vibration isolation

– Vibrations in your car – Satellite operation

– Disk drives, etc.

) (x y

k  c(x^  y^)

FBD of SDOF Base Excitation

=- ( - )- ( - )=

+ + = + (2.61)

F k x y c x y mx mx cx kx cy ky



System Sketch System FBD

M M

x(t)

k c

base

y(t)

SDOF Base Excitation (cont)

Assume: ( ) sin( ) and plug into Equation(2.61)

+ + = cos( ) + sin( ) (2.63)

harmonic forcing functions

y t Y t

mx cx kx c Y t kY t



  



For a car, ^{  2}_^{  2 V}_

The steady-state solution is just the superposition of the two individual particular solutions (system is linear).

0 0

2 2

+2 + = 2 cos( ) + sin( ) (2.64)

c f s

f

n n n n

x





 







Particular Solution (sine term)

With a sine for the forcing function,

 

 

2

0

0 2 2 2 2

2 2

0 2 2 2 2

+2 + = sin

cos sin sin( )

where

2

( ) 2

( )

( ) 2

n n s

ps s s s s

n s

s

n n

n s

s

n n

x x x f t

x A t B t X t

A f

B f

  

   

 

   

 

   

   

 

 

 

 

Use rectangular form to make it easier to add the cos term

Particular Solution (cos term)

With a cosine for the forcing function, we showed

 

 

2

0

2 2

0

2 2 2 2

0

2 2 2 2

+2 + = cos

cos sin cos( )

where

( )

( ) 2

2

( ) 2

n n c

pc c c c c

n c

c

n n

n c

c

n n

x x x f t

x A t B t X t

A f

B f

  

   

 

   

 

   

   

 

 

  

Magnitude X/Y

Now add the sin and cos terms to get the magnitude of the full particular solution

   

 

2 2 2 2

0 0

2 2

2 2 2 2 2 2

2

0 0

2 2 2 2

(2 )

( ) 2 ( ) 2

where 2 and

1 (2 )

if we define this becomes (2.70)

(1 ) 2

n

c s n

n

n n n n

c n s n

f f

X Y

f Y f Y

r X Y r

r r



 

        

  





 

 

   

 

  

 

 

2 2 2 2

1 (2 )

(2.71)

(1 ) 2

X r

Y r r





 

 

0

2 2

2 sin

2

n n

f   t

 

  

 

  

W h e n

t h e

d r i v e

f r e q

0 0.5 1 1.5 2 2.5 3

-20 -10

0

2 2

2 sin

2

n n

f  t

 

  

 

  

0 10 20 30 40

Frequency ratio r X/Y (dB)

e is close to r = 1

• For ζ = 0, r =1 defines resonance

• As ζ grows

resonance moves r <1,

and X







The relative magnitude plot

of X/Y versus frequency ratio: Called the Displacement Transmissibility

=0.01

=0.1

=0.3

=0.7

Figure 2.13

From the plot of relative Displacement Transmissibility observe that:

• X/Y is called Displacement Transmissibility Ratio

• Potentially severe amplification at resonance

• Attenuation for r > sqrt(2) Isolation Zone

• If r < sqrt(2) transmissibility decreases with damping ratio Amplification Zone

• If r >> 1 then transmissibility increases with damping ratio X_p~2Yζ/r

Next examine the Force Transmitted to the mass as a function of the frequency ratio

2

2 2

( ) ( )

At steady state, ( ) cos( ),

so =- cos( )

T

T

F k x y c x y mx

x t X t

x X t

F m X k r X

 

  



     

 



 

From FBD

x(t) m

FT

k c y(t) base