chapter 2 response to harmonic excitation

© D. J. Inman1/49 Mechanical Engineering at Virginia Tech

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)=F0cost

• is the driving frequency• F0 is the magnitude of the applied force• We take c = 0 to start with

Mk

xDisplacement

F=F0cost


Equations of motion• Solution is the sum of

homogenous and particular solution

• The particular solution assumes form of forcing function (physically the input wins):xp (t) X cos(t)

0

20

00

( ) ( ) cos( )

( ) ( ) cos( )

where ,

n

n

mx t kx t F t

x t x t f tF kf m m

Figure 2.1


Substitute particular solution into the equation of motion:

2

2 20

02 2

cos cos cos

solving yields:

p n px x

n

n

X t X t f tfX

Thus the particular solution has the form:

xp (t) f0

n2 2 cos(t)


Add particular and homogeneous solutions to get general solution:

01 2 2 2

homogeneous

1 2

( )

sin cos cos

and are constants of integration.

particular

n nn

x t

fA t A t t

A A


0 01 2 2 02 2 2 2

02 0 2 2

01 2 1 02 2

01

0

(0) sin 0 cos0 cos0

(0) ( cos0 sin 0) sin 0

( )

n n

n

n nn

n

f fx A A A x

fA x

fx A A A v

vA

vx t

0 00 2 2 2 2sin cos cosn n

n n n

f ft x t t

Apply the initial conditions to evaluate the constants

(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 n = 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, = n +5 v0=0.1m/s and x0= -0.02 m.

Note the obvious presence of two harmonic signals

Go to code demo

0 2 4 6 8 10-0.05

0

0.05

Time (sec)

Dis

plac

emen

t (x)


What happens when is near n?

When the drive frequency and natural frequency are close a beating phenomena occurs

0 5 10 15 20 25 30-1

-0.5

0

0.5

1

Time (sec)

Dis

plac

emen

t (x)

02 2

2sin

2n

n

ft

Larger amplitude

02 2

2( ) sin sin (2.13)

2 2n n

n

fx t t t


What happens when is n?

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 5 10 15 20 25 30

-5

0

5

Time (sec)

Dis

plac

emen

t (x)

xp (t) tX sin(t)

substitute into eq. and solve for X

X f0

2

grows with out bound

01 2( ) sin cos sin( )

2f

x t A t A t t t


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, =2n.

)20cos10(cos10667.710sin02.0)(

:yields then (2.11)Equation

m 109667.7s/rad )20(10

N/kg 3.2

m 02.0rad/s 10m/s 2.0 N/kg, 3.2

kg 10N 23

rad/s 202 rad/s, 10kg 10N/m 1000

3

3222222

0

00

ttttx

f

vmFf

mk

n

n

nn


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.

02 2

02 2

0.1 m

02 2 2

( ) cos 20 cos for zero initial conditions

2trig identity ( ) sin sin2 2

2 2(20 / )0.1 0.12000 (20 )

0.45 kg

nn

n n

n

n

fx t t t

fx t t t

f m

mm


Example 2.1.3 Design a rectangular mount for a security camera.

Compute l 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 ( ) cosEImx x t F t

From strength of materials: I bh3

12

Thus the frequency expression is: 3 32

3 3

312 14n

Ebh Ebhm m

Here we are interested computing l that will make the amplitude less then 0.01m:

2 202 2

02 2

2 202 2

2( ) 0.01 , for 02 0.01

2( ) 0.01, for 0

nn

nn

n

faf

fb


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

0.01 2 f0

n2 2 2 f0 0.01 2 0.01 n

2 0.01 2 2 f0 0.01Ebh3

4ml 3

l 3 0.01Ebh3

4m(0.01 2 2 f0 )0.321 l 0.6848 m

Case (b):

2 f0

n2 2 0.01 2 f0 0.01 n

2 0.01 2 2 f0 0.01 2 0.01Ebh3

4ml 3

l 3 0.01Ebh3

4m(2 f0 0.01 2 )0.191 l 0.576 m


Remembering the constraint that the length must be at least 0.5 m, (a) and (b) yield

0.5 l 0.576, or l > 0.6848 m

Less material is usually desired, so chose case a, say l = 0.55 m. To check, note that

n

2 2 3Ebh3

12ml 3 20 21742 0

Thus the case a condition is met.Next check the mass of the designed beam to insure it does not change the frequency. Note

it is much less then m.

m lbh3

(2.7 103)(0.55)(0.02)(0.02)3

2.376 10 4 kg


A harmonic force may also be represented by sine or a complex exponential. How does thischange the solution?

20 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:

xp (t) X sint (2.19)

Substitution of (2.19) into (2.18) yields:0

2 2( ) sinpn

fx t t

Solving for the homogenous solution and evaluating the constants yields

x(t) x0 cos nt

v0

n

n

f0

n2 2

sin nt

f0

n2 2 sint (2.25)


Section 2.2 Harmonic Excitation of Damped Systems

Extending resonance and response calculation to damped systems


2.2 Harmonic excitation of damped systems

0

20

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 tx t X t

Mkx

Displacement

c

F=F0cost


2 2 1

2 2

( ) cos sin

, tan

sin cos

cos sin

p s s

ss s

s

p s s

p s s

x t A t B t

BX A BA

x A t B t

x A t B t

Let xp have the form:

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


0)()2(

)2()(

/2,0for ly Specifical time.allfor 0sin2

cos)2(

22

022

22

022

snsn

snsn

snsns

snsns

BA

fBA

ttBAB

tfABA

Substitute into the equations of motion


Write as a matrix equation:

2 20

2 2

( ) 202 ( )

sn n

sn n

A fB

2 20

2 2 2 2

( )( ) (2 )

ns

n n

fA

Bs 2n f0

(n2 2 )2 (2 n )2

Solving for As and Bs:


102 22 2 2 2

2( ) cos( tan )( ) (2 )

np

nn n

X

fx t t

Substitute the values of As and Bs into xp:

Add homogeneous and particular to get total solution:

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

particular or homogeneous or transient solutionsteady state solution

( ) sin( ) cos( )ntdx t Ae t X t


Things to notice about damped forced response• If = 0, undamped equations result• Steady state solution prevails for large t• Often we ignore the transient term (how

large is , 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.


Example 2.2.1: n = 10 rad/s, = 5 rad/s, = 0.01, F0= 1000 N, m = 100 kg, and the initial conditions x0 = 0.05 m and v0 = 0. Compare A and f for forced and unforced case:

X 0.133, 0.013

x(t) Ae 0.1t sin(9.99t ) 0.133cos(5t 0.013)

v(t) 0.01Ae 0.1t sin(9.999t )

9.999Ae 0.1t cos(9.999t ) 0.665sin(5t 0.013)applying the intial conditions :A 0.519 (0.05), 0.875 (1.561)

Using the equations on slide 6:

Differentiating yields:

The numbers in ( ) are those obtained by incorrectly using the

freeresponse values

x(0) 0.05 Asin 0.133cos( 0.013)v(0) 0 0.01Asin 9.999A cos 0.665sin0.013


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


tan 1 2r1 r2

Magnitude:

Non dimensionalForm:

Phase:

Frequency ratio:

Proceeding with ignoring the transient

02 2 2 2( ) (2 )n n

fX

2

2 2 20 0

1

(1 ) (2 )nXXk

F f r r

r n

(2.39)

(2.40)


Magnitude plot

0 0.5 1 1.5 2-20

-10

0

10

20

30

40

r

X (d

B)

=0.01 =0.1 =0.3 =0.5 =1

X 1

(1 r2 )2 (2r)2• Resonance is close to r = 1

• For = 0, r =1 defines resonance

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

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

Fig 2.7


Phase plot

• Resonance occurs at = /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 radians

0 0.5 1 1.5 20

0.5

1

1.5

2

2.5

3

3.5

r

Phas

e (r

ad)

=0.01 =0.1 =0.3 =0.5 =1

21

12

rr tan

Fig 2.7


2 2 20

2peak

20 max

1 0(1 ) (2 )

1 2 1 1/ 2

1

2 1

d Xk ddr F dr r r

r

XkF

Example 2.2.3 Compute max peak by differentiating:

(2.41)

(2.42)


Effect of Damping on Peak Value

0 0.2 0.4 0.6 0.80

5

10

15

20

25

30

0 0.2 0.4 0.6 0.80

0.2

0.4

0.6

0.8

1

rpeak

(dB)0FXk

• 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 <0.707

Fig 2.9


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 approacha frequency response approacha 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 /2• Therefore write each as a vector • Compute X in terms of F0 via vector addition

Re

Im

)cos( tkXA

B

C

D

E

kX

XcXm 2

0F

C

BAXmk )( 2

Xc0F


2220

2222220

)()(

)()(

cmk

FX

XcXmkF

Using vector addition on the diagram:

At resonance:

2

, X F0

c


2.3.2 Complex response method

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

Real part of this complex solution corresponds to the physical solution

(2.47)

(2.48)


20

002

2

the frequency response function

( )

( )

( )( ) ( )

1( )( ) ( )

j tp

j t j t

x t Xe

m cj k Xe F eFX H j F

k m c j

H jk m c j

Note: These are all complex functions

Choose complex exponential as a solution

(2.49)

(2.50)

(2.51)

(2.52)


02 2 2

12

( )02 2 2

( ) ( )

tan

( )( ) ( )

j

j tp

FX ek m c

ck m

Fx t ek m c

Using complex arithmetic:

Has real part = to previous solution

(2.53)

(2.54)

(2.55)


Comments:• Label x-axis Re(ejwt) and y-axis

Im(ejwt) 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

The equation of motion is written as

20 0

2 20

02 2

( ) ( ) ( ) ( )

Let ( )

j t j tn

j tp

j t j tn

n

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

x t Xe

Xe f e

fX


2.3.3 Transfer Function Method

• 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


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

( ) ( ( )) ( ) stX s x t x t e dt

L


0

2 02 2

cos

( ) ( )

mx cx kx F tF s

ms cs k X ss

Take the transform of the equation of motion:

Now solve algebraic equation in s for X(s)

X(s) F0s

(ms2 cs k)(s2 2 )

To get the time response this must be “inverse transformed”


(ms2 cs k)X(s) F(s) X(s)F(s)

H (s) 1

ms2 cs k

H ( j ) 1

k m 2 c j frequency response function

The transferfunction

With zero initial conditions:Transfer Function Method

(2.59)

(2.60)


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

Summing moments about the shaft:

0( ) ( ) sinJ t k t aF t

Taking the Laplace transform:

20 2 2

0 2 2 2

( ) ( )

1( )

Js X s kX s aFs

X s a Fs Js k

Taking the inverse Laplace transform:

1

0 2 2 2

2 2

1( )

1 1 1( ) sin sin , n nn n

t a F Ls Js k

a F kt t tJ 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

How do we interpret these phase angles?Why is one added and the other subtracted?

homogeneous solution particular solution

( ) sin cosntdx t Ae t X t


sin dt 1tan o d

o n o

xv x

x0 0 v0 0

Non-Zero initial conditions


Zero initial displacement

sin dt 1 1 0tan tan 0o d

o n o o

xv x v

x0 0 v0 0


tdsin

21 1

tan

ono

do

xvx

Zero initial velocity

x0 0 v0 0


Phase on Particular Solutionxp (t) X cos t

• Simple “atan” gives -π/2 < < π/2• Four-quadrant “atan2” gives 0 < < π

F0cos(t)

Xcos(t-)t

2n2n

n2-2 n

2-2

>n <n

22

1 2

n

ntan

chapter 2 response to harmonic excitation

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