ch. 3: forced vibration of 1-dof systempioneer.netserv.chula.ac.th/~pphongsa/vibration/ch3.pdf ·...

Ch. 3: Forced Vibration of 1-DOF System

3.0 Outline

Harmonic ExcitationFrequency Response FunctionApplicationPeriodic ExcitationNon-periodic ExcitationImpulse ResponseArbitrary Excitation

3.0 Outline


3.1 Harmonic Excitation

Force input function of the harmonic excitation is theharmonic function, i.e. functions of sines and cosines. This type of excitation is common to many systeminvolving rotating and reciprocating motion. Moreover,many other forces can be represented as an infiniteseries of harmonic functions. By the principle ofsuperposition, the response is the sum of the individual harmonic response.

It is more convenient to use the frequency domaintechnique in solving the harmonic excitation problems.This is because the response to differentexcitation frequencies can be seen in one graph.




( )( )

0

20 0 0

0

20

Let us focus on the particular solution ofcos

normalize the equation of motion2 cos , /

Re

solve for from 2

and the solution

n n

i t

i tn n

mx cx kx F t

x x x f t f F m

f t f e

z t z z z f e

ω

ω

ω

ζω ω ω

ζω ω

+ + =

+ + = =

⎡ ⎤= ⎣ ⎦∴ + + =

( ) ( ) ( )

( ) ( ) ( )( ) ( )

( )

2 20

is the real part of ; Re

Assume the solution to have the same form as the forcing function same frequency as the input w/ different mag. and phase

2

i t

i t i tn n

z t x t z t

z t Z i e

i Z i e f e

fZ i

ω

ω ω

ω

ω ζωω ω ω

ω

= ⎡ ⎤⎣ ⎦

=

− + + =

=( )

( )

20 0

22 2

02

/2 1 / 2 /

1 / 2 /

n

n n n n

n n

fi i

Fk i

ωω ω ζωω ω ω ζω ω

ω ω ζω ω

=− + − +

=⎡ ⎤− +⎣ ⎦



( ) ( )

( )

( ) ( )

( ) ( ) ( )

( )( ) ( )

002

02

2

0

2 22

12

1 2

Re , /1 2

1If is the frequency response1 2

cos

1where magnitude1 2

2tan phas1

i t i t

i tn

i

Fz t e H i F ek r i r

Fx t e rk r i r

H i H i ek r i r

x t F H i t

H ik r r

rr

ω ω

ω

θ

ωζ

ω ωζ

ω ωζ

ω ω θ

ωζ

ζθ −

= =⎡ ⎤− +⎣ ⎦

⎡ ⎤⎢ ⎥∴ = =

⎡ ⎤− +⎢ ⎥⎣ ⎦⎣ ⎦

= =⎡ ⎤− +⎣ ⎦

∴ = +

= =− +

−= =

−

( ) ( )

e

The system modulates the harmonic input by

the magnitude and phase H i H iω ω



( ) ( )( ) ( ) ( ) ( )( )

1 2

0

1

total response homogeneous soln. particular soln.Recall the homogeneous solution of the underdamped system

cos or sin cos

cos cos

or

n n

n

n

t th d h d d

td

t

x Ce t x e A t A t

x t Ce t F H i t

x t e A

ζω ζω

ζω

ζω

ω φ ω ω

ω φ ω ω θ

− −

−

−

= +

= − = +

∴ = − + +

= ( ) ( ) ( )2 0

1 2

sin cos cos

The initial conditions will be used to determine , or ,They will be different from those of free responsebecause the transient term now is partly due to the excitatio

d dt A t F H i t

C A A

ω ω ω ω θ

φ

+ + +

n force

and partly due to the initial conditions



Ex. 1 Compute and plot the response of a spring-masssystem to a force of magnitude 23 N, drivingfrequency of twice the natural frequency and i.c.given by x0 = 0 m and v0 = 0.2 m/s. The massof the system is 10 kg and the spring stiffnessis 1000 N/m.



( )

( ) ( )( )( )

( )

32 2

31 2

31 2

2

/ 1000 /10 10 rad/s/ 2 0

2 10 20 rad/s1 1 0.333 10

1 2 1000 1 2

sin cos 23 0.333 10 cos

cos sin 23 0.333 10 sin

i.c. 0 0 23 0.333 10

n

n

n n

n n n n

k mc m

H ik r i r

x t A t A t t

x t A t A t t

x A

ωζ ωω

ωζ

ω ω ω

ω ω ω ω ω ω

−

−

−

= = =

= =

= × =

= = = − ×⎡ ⎤− + × −⎣ ⎦

= + − × ×

= − + × × ×

= = − × ×

( )( ) ( )

3 32

1 1

3

, 7.667 10

0 0.2 10 , 0.02

0.02sin10 7.667 10 cos10 cos 20 m

A

x A A

x t t t t

− −

−

= ×

= = × =

∴ = + × −



ωresponse finally becomes ω, and in phase



ωresponse finally becomes ω, and out of phase



F0ωnt/(2k)

( ) ( )

( ) 0

In case of 0 and , the guess solution of the form

cos sin is invalid. This is becauseit has the same form as the homogeneous solution.

The correct particular solution is

ni t

np

x t X i e A t B t

Fx t

ω

ζ ω ω

ω ω ω

ω

= =

= = +

= sin .2 n

t tk

ω


Beat when the driving frequency is close to natural freq.


( ) ( )0 00 2 2

2 2 20 0 1 0 0

2 20

The total solution can be arranged in the form

sin cos cos cos

2 sin tan sin sin2 2

If the system is at rest in

n n nn n

n n n nn

n n

v fx t t x t t t

x v x ft t tv

ω ω ω ωω ω ω

ω ω ω ω ω ωωω ω ω

−

= + + −−

+ ⎛ ⎞ − +⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟− ⎝ ⎠ ⎝ ⎠⎝ ⎠

( ) 02 2

02 2

the beginning,2 sin sin

2 2

The response oscillates with frequency inside2

2the slowly oscillated envelope sin2

The beat frequency is

n n

n

n

n

n

n

fx t t t

f t

ω ω ω ωω ω

ω ω

ω ωω ω

ω ω

− +⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟− ⎝ ⎠ ⎝ ⎠+

−⎛ ⎞⎜ ⎟− ⎝ ⎠

∴ −


Beat when the driving frequency is close to natural freq.



3.2 Frequency Response Function


( ) ( )2

The core of the particular solution to the harmonic function is1 ; frequency response function

1 2

It specifies how the system responds to harmonic excitation.As a standard, we normalize the

H ik r i r

ωζ

=− +

( ) 2

frequency response function1 and then study how it varies as the

1 2excitation frequency and system parameters , vary.It is indeed more convenient since we already normalized the frequen

n

G ir i r

ωζω ζ ω

=− +

( )( )

cy;/ . So we can now study its variation to and .

For the fixed damping ratio, we plot with varies.

has both magnitude and phase magnitude and phase plot.

Then we repeatedly evaluate

nr rG i r

G i

G i

ω ω ζω

ω

=

⇒

( ) by varying .ω ζ



Frequency response plot(Bode diagram)

( )( ) ( )2 22

1

1 2H i

r rω

ζ=

− +

12

2tan1

rrζθ − −⎛ ⎞= ⎜ ⎟−⎝ ⎠


Resonance is defined to be the vibration response atω=ωn, regardless whether the damping ratio is zero.At this point, the phase shift of the response is –π/2.

The resonant frequency will give the peak amplitude for the response only when ζ=0. For ,the peak amplitude will be at , slightly before ωn.For , there is no peak but the max. value of theoutput is equal to the input for the dc signal (of course, for this normalized transfer function).


21 2nω ω ζ= −0 1/ 2ζ< <

1/ 2ζ ≥



Ex. 2 Consider the pivoted mechanism with k=4x103 N/m,l1=0.05 m, l2=0.07 m, l=0.10 m, and m=40 kg.The mass of the beam is 40kg which is pivotedat point O and assumed to be rigid. Calculate cso that the damping ratio of the system is 0.2.Also determine the amplitude of vibration of thesteady-state response if a 10 N force is appliedto the mass at a frequency of 10 rad/s.



( ) ( )

( )

12 2 1 1

2 212 1

2

12 2

0.1 10cos10 0.5 0.0049 59.050.004915.37, 0.2 , 627.3 Ns/m

2 0.5

O O

nn

l lM I Fl mgl Mg c l l k l l

l l l lml M M

t cc c

θ θ θ θ θ

θ

θ θ θ

ω ζω

ω

−⎛ ⎞⎡ ⎤= − − − −⎜ ⎟⎣ ⎦ ⎝ ⎠⎡ ⎤+ −⎛ ⎞= + +⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦× = + +

= = = =× ×

=

∑

( ) ( )( )

10, 0.65061 0.02677 24.268

59.05 0.5767 0.26

0.02677 cos 10 0.424ss

r

H ii

t

ω

θ

=

= = − °+

= −



Ex. 3 A foot pedal for a musical instrument is modeledas in the figure. With k=2000 kg/s2, c=25 kg/s,m=25 kg, and F(t)=50cos2πt N, compute thesteady-state response assuming the system startsfrom rest. Use the small angle approximation.



( ) ( )

( ) ( )

2

2

0.15 0.05 0.05 0.05 0.1 0.15

5 1003.75 50cos 2 , positive CW6 3

Find the parameters2.98, 0.0373, 2 , 2.108

1 0.0087 177.41 2

since 0, the transie

O O

n

M I F k c m

t

r

H ik r i r

θ θ θ θ

θ θ θ π

ω ζ ω π

ωζ

ζ

⎡ ⎤= × − × − × = ×⎣ ⎦

+ + =

= = = =

= = − °− +

≠

∑

( ) ( ) ( )0

nt response will die out

cos 0.435cos 2 3.096ss F H i t tθ ω ω θ π= + = −

ch. 3: forced vibration of 1-dof systempioneer.netserv.chula.ac.th/~pphongsa/vibration/ch3.pdf ·...

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