Chapter 2 Response to Harmonic Excitation

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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 )= F 0 cos w t w is the driving frequency F 0 is the magnitude of the applied force

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### Chapter 2 Response to Harmonic Excitation

Introduces the important concept of resonance

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2.1 Harmonic Excitation of Undamped Systems
• Consider the usual spring mass damper system with applied force F(t)=F0coswt
• w is the driving frequency
• F0 is the magnitude of the applied force

Displacement

x

F=F0cost

k

M

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Equations of motion

Figure 2.1

• Solution is the sum of homogenous and particular solution
• The particular solution assumes form of forcing function (physically the input wins):

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Substitute particular solution into the equation of motion:

Thus the particular solution has the form:

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Apply the initial conditions to evaluate the constants

(2.11)

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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 wn = w because it produces division by 0.
• If forcing frequency is close to natural frequency the amplitude of particular solution is very large

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Response for m=100 kg, k=1000 N/m, F=100 N, w = wn +5

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

0.05

0

Displacement (x)

-0.05

0

2

4

6

8

10

Time (sec)

Note the obvious presence of two harmonic signals

Go to code demo

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What happens when w is near wn?

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

1

0.5

0

Displacement (x)

Larger amplitude

-0.5

-1

0

5

10

15

20

25

30

Time (sec)

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5

Displacement (x)

0

-5

0

5

10

15

20

25

30

Time (sec)

What happens when w is wn?

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!

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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, w=2wn.

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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.

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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.

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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:

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

Case (b):

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

Less material is usually desired, so chose case

a, say l = 0.55 m.

To check, note that

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.

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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:

Substitution of (2.19) into (2.18) yields:

Solving for the homogenous solution and evaluating the constants yields

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### Section 2.2 Harmonic Excitation of Damped Systems

Extending resonance and response calculation to damped systems

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Displacement

x

k

F=F0coswt

M

c

2.2 Harmonic excitation of damped systems

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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.

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Substitute into the equations of motion

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Write as a matrix equation:

Solving for As and Bs:

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Substitute the values of As and Bs into xp:

Add homogeneous and particular to get total solution:

Note: that A and f will not have the same values as in Ch 1,

for the free response. Also as t gets large, transient dies out.

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

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Example 2.2.1: wn = 10 rad/s, w = 5 rad/s, z = 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:

Using the equations on slide 6:

Differentiating yields:

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

response values

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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

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Proceeding with ignoring the transient

Magnitude:

(2.39)

Frequency ratio:

Non dimensional

Form:

(2.40)

Phase:

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40

z

=0.01

z

=0.1

30

z

=0.3

z

=0.5

20

z

=1

X (dB)

10

0

-10

-20

0

0.5

1

1.5

2

r

Magnitude plot
• Resonance is close to r = 1
• For z = 0, r =1 defines resonance
• As z grows resonance moves r <1, and X decreases
• The exact value of r, can be found from differentiating the magnitude

Fig 2.7

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3.5

z

=0.01

3

z

=0.1

z

=0.3

2.5

z

=0.5

z

=1

2

1.5

1

0.5

0

0

0.5

1

1.5

2

r

Phase plot
• Resonance occurs at f = p/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 p radians

Fig 2.7

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Example 2.2.3 Compute max peak by differentiating:

(2.41)

(2.42)

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Effect of Damping on Peak Value

30

• 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 z<0.707

25

20

15

10

5

0

0

0.2

0.4

0.6

0.8

z

1

0.8

0.6

rpeak

0.4

0.2

0

0

0.2

0.4

0.6

0.8

z

Fig 2.9

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### 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.

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2.3.1 Geometric Approach
• Position, velocity and acceleration phase shifted each by p/2
• Therefore write each as a vector
• Compute X in terms of F0 via vector addition

Im

C

D

C

E

q

A

B

q

B

Re

A

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Using vector addition on the diagram:

At resonance:

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2.3.2 Complex response method

(2.47)

(2.48)

Real part of this complex solution corresponds to the physical solution

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Choose complex exponential as a solution

(2.49)

(2.50)

(2.51)

(2.52)

Note: These are all complex functions

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Using complex arithmetic:

(2.53)

(2.54)

(2.55)

Has real part = to previous solution

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• 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

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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

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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

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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

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Take the transform of the equation of motion:

Now solve algebraic equation in s for X(s)

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

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Transfer Function Method

With zero initial conditions:

The transfer

function

(2.59)

(2.60)

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

Taking the Laplace transform:

Taking the inverse Laplace transform:

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Notes on Phase for Homogeneous and Particular Solutions
• Equation (2.37) gives the full solution for a harmonically driven underdamped SDOF oscillator to beHow do we interpret these phase angles?Why is one added and the other subtracted?

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Non-Zero initial conditions

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Zero initial displacement

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Zero initial velocity

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Phase on Particular Solution
• Simple “atan” gives -π/2 < f < π/2
• Four-quadrant “atan2” gives 0 < f < π

F0cos(wt)

f

Xcos(wt-f)

f

2zwnw

2zwnw

f

wt

wn2-w2

wn2-w2

w>wn

w<wn

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