Dynamic response to harmonic excitation

© Eng. Vib, 3rd Ed.1/49 @ProfAdhikari, #EG260

Chapter 2 Response to Harmonic Excitation

Introduces the important concept of resonance

© Eng. Vib, 3rd Ed.2/49 College of Engineering

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

Mk

xDisplacement

F=F0cosωt


Equations of motion

• Solution is the sum of homogenous and particular solution

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

( ) cos( )px t X t

Figure 2.1


Substitute particular solution into the equation of motion:

Thus the particular solution has the form:

02 2

( ) cos( )pn

fx t t

( ) cospx t X t


Add particular and homogeneous solutions to get general solution:


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 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

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 ω is near ωn?

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

0

2 2

2( ) sin sin (2.13)

2 2n n

n

fx t t t

Called beating


What happens when ω is ωn?

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

pla

cem

ent (

x)

0

( ) sin( )

substitute into eq. and solve for

2

px t tX t

X

fX


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 =2 ω n.


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.


Example 2.1.3 Design a rectangular mount for a security camera.

Compute l > 0.5 m so that the mount keeps the camera from vibrating more then 0.01 m of maximum amplitude under a windload of 15 N at 10 Hz. The mass of the camera is 3 kg.

0.02 x 0.02 m in cross section


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:

Thus the frequency expression is:

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 0

20.01

2( ) 0.01, for 0

nn

nn

n

fa

f

fb


Case (a) (assume aluminum for the material):3

2 2 200 02 2 3

33

20

20.01 2 0.01 0.01 0.01 2 0.01

4

0.01 0.321 0.6848 m4 (0.01 2 )

nn

f Ebhf f

m

Ebh

m f

Case (b):3

2 2 200 02 2 3

33

20

20.01 2 0.01 0.01 2 0.01 0.01

4

0.01 0.191 0.576 m4 (2 0.01 )

nn

f Ebhf f

m

Ebh

m f


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

0.5 0.576, or > 0.6848 m

Less material is usually desired, so chose case

b, say l = 0.55 m. To check, note that

3

22 23

320 1742 0

12n

Ebh

m

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 less then m. 3(2.7 10 )(0.55)(0.02)(0.02)

0.594 kg

m bh


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

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

fx t t

Solving for the homogenous solution and evaluating the constants yields

0 0 00 2 2 2 2

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

v f fx t x t t t


Section 2.2 Harmonic Excitation of Damped Systems

Extending resonance and response calculation to damped systems


2.2 Harmonic excitation of damped systems

M

kx

Displacement

c

F=F0cosωt


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.


2 2

0

2 2

2 20

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

Substitute into the equations of motion


Write as a matrix equation:

2 20

2 2

( ) 2

02 ( )sn n

sn n

A f

B

2 20

2 2 2 2

( )

( ) (2 )n

sn n

fA

02 2 2 2

2

( ) (2 )n

sn n

fB

Solving for As and Bs:


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.


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.


Example 2.2.1: ωn = 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 A and for forced and unforced case:

0.1

0.133, 0.013

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

X

x t Ae t t

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

Using the equations on slide 23:

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 0.665sin 0.013


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


12

2tan1r

r

Magnitude:

Non dimensionalForm:

Phase:

Frequency ratio:

Proceeding with ignoring the transient

0

2 2 2 2( ) (2 )n n

fX

2

2 2 20 0

1

(1 ) (2 )nXXk

F f r r

n

r

(2.39)

(2.40)


Magnitude plot

0 0.5 1 1.5 2-20

-10

0

10

20

30

40

r

X (

dB

)

2 2 2

1

(1 ) (2 )X

r r• 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

Fig 2.7


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 20

0.5

1

1.5

2

2.5

3

3.5

r

Ph

ase

(rad

)

=0.01 =0.1 =0.3 =0.5 =1

12

2tan1r

r

Fig 2.7


2 2 20

2peak

20 max

10

(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

0 0.2 0.4 0.6 0.80

5

10

15

20

25

30

ζ

0 0.2 0.4 0.6 0.80

0.2

0.4

0.6

0.8

1

rpeak

(dB)0F

Xk

• 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

Fig 2.9ζ


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 2 alternatives:a frequency response approacha transform approach

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


2.3.2 Complex response method

Real part of this complex solution corresponds to the physical solution

(2.47)

(2.48)


Note: These are all complex functions

Choose complex exponential as a solution

(2.49)

(2.50)

(2.51)

(2.52)


0

2 2 2

12

( )0

2 2 2

( ) ( )

tan

( )( ) ( )

j

j tp

FX e

k m c

c

k m

Fx t e

k m c

Using complex arithmetic:

Has real part = to previous solution

(2.53)

(2.54)

(2.55)


Comments:

• 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


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

X(s) L (x(t)) x(t)

0

e stdt


Take the transform of the equation of motion:

Now solve algebraic equation in s for X(s)

0

2 2 2( )

( )( )

F sX 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 sH s

F s ms cs k

H jk m c j

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:

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:

10 2 2 2

2 2

1( )

1 1 1( ) sin sin , n n

n n

t a F Ls Js k

a F kt 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

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


sin dt tan 1 xod

vo nxo

0 00 0x 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

Dynamic response to harmonic excitation

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