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Dept. of Mechanical Eng., KAISTDept. of Mechanical Eng., KAISTMAE 351MAE 351:: Wk 4

CHAPTER 3CHAPTER 3

Harmonically Excited Harmonically Excited VibrationVibration

m

ck

xF(t)


3.1. General3.1. General• System subject to external dynamic forces

– Forcing function, Exciting function• Harmonic excitation → Harmonic response

Nonperiodic excitation → Transient response• Harmonic excitations

• Governing equation(nonhomogeneous equation)

General sol’n: x(t) = xh(t) + xp(t)where xh(t) = homogeneous sol’n of

xp(t) = particular sol’n• xh(t) : exponentially decaying (initial transient) vibration

xp(t) : steady-state vibration

( )

( )0

0

0

( )( ) cos( ) sin( )

i tF t F eF t F tF t F t

ω φ

ω φω φ

+=

= +

= +

( )mx cx kx F t+ + =

0mx cx kx+ + =

m

ck

xF(t)


Homogeneous, Particular, & General Solutions(Underdamped 1-DOF Vibration)


3.2. Superposition Principle3.2. Superposition Principle

x(t) = output (response); F(t) = input (excitation)Introduce a linear differential operator

Then, G = ‘Black-box’ of the 2nd order system

( )mx cx kx F t+ + =2

2 ( )d x dxm c kx F tdt dt

+ + =

2

2d dG m c kdt dt

≡ + +

[ ]( ) ( )G x t F t=

or

Consider two excitations and responses:F1(t)=G[x1(t)], F2(t)=G[x2(t)]

Next, considerF3(t)=c1F1(t) + c2F2(t) : linear combination of F1(t) & F2(t)

c1,c2 : known constants

If x3(t) = c1x1(t) +c2x2(t), the system is “linear”: otherwise “nonlinear”G[x3(t)] = G[c1x1(t) + c2x2(t)] = c1G[x1(t)] + c2G[x2(t)]

= c1F1(t) + c2F2(t) = F3(t) ; principle of superposition

F(t)G

x(t)


3.3. Response of an 3.3. Response of an UndampedUndamped System System under Harmonic Forceunder Harmonic Force

For

F(t) : harmonic →→ xp(t): harmonicwith ω with ω

Assume xp(t) = Xcos ωt, then

cosomx kx F tω+ =

1 2

0,( ) cos sinh n n

n

mx kxx t C t C t

km

ω ω

ω

+ == +

=

20( cos cos ) cosX m t k t F tω ω ω ω− + =

02

FXk mω

=−

m

k

xF(t)

; Amplitude of particularsolution

Q: What is the physical meaning of ‘particular solution’?


x(t) = total solution

=

Then,

01 2cos sin cosn n

FC t C t tk m

ω ω ωω2+ +

−

1 2

0 0

0 01 0 22

Two I.C.'s of (0)& (0) ,Let (0) & (0)

;n

x x C Cx x x x

F xC x Ck mω ω

⇒= =

= − =−

0 00 2

02

( ) cos sin

cos

n nn

F xx t x t tk mF t

k m

ω ωω ω

ωω

= − + − + −

HomogeneoussolutionRecall: free vib.

Recall:In free vib. case:

01 0 2;

n

xC x Cω

= =

General Solution


From 002

2 2

0

n

and

1 11

1

where static deflection,

= frequency ratio ,

Dynamic ampl.Static ampl.

magnification factoror amplification factoror amplification ratio

st

st

n

st

st

FX F kk m

Xr

Fk

r

X

δω

δ ωω

δ

ωω

δ

= =−

⇒ = ≡−

−

= =

≡

=

=

Look!X infinity as r 1

“RESONANCE”


A. CharacteristicsA. Characteristics③②①

①: 0 < ω /ωn < 1②: ω /ωn > 1③: ω /ωn = 1

①① F(t) & xp(t) are in phase

②② out-of phase ( r > 1 )xp(t) = -Xcosωt

( 0 < r < 1 )


③③ (r=1)X →∞ ; Resonance ω=ωn

1 2 2

00 2

2

2

00

( ) cos sin cos1 ( )

1cos sin (cos cos )1 ( )

cos cos ( sin ) 1lim lim sin (L'Hospital's rule)1 ( ) 22

( ) cos sin

n n

stn n

n

n n st nn n

nn n

n

n

sn n

n

x t C t C t t

xx t t t t

t t t t t t

xx t x t t

ω ω ω ω

δω ω ωω ω

ω ω δ ω ωω ω ω

ω ω ω ω ωωω ωω

δω ωω

→ →

= + +− /

= + + −− /

− −= =

− / −

⇒ = + + sin2t n

nt tω ω


B. Total ResponseB. Total Response

2

2 2 1 21 2

1

( ) cos( ) cos1 ( )

where ; tan

stn

n

x t A t t

CA C CC

δω φ ωω ω

φ −

= − ±− /

= + =

+ for ω/ωn < 1- for ω/ωn > 1


C. BeatingC. Beating① Linear superposition of individual vibrations:

1 1 1 2 2 2exp[ ( )] exp[ ( )]x A i t A i tω φ ω φ= + + +

Nonperiodic (aperiodic) oscillations in generalWhen ω2 = ω1 + ∆ω, then

1 2 1 1( ) ( )1 2[ ]i i t i t i tx Ae A e e Aeφ φ ω ω ω φ+∆ += − ≡

2 2 1/ 21 2 1 2[ 2 cos( )]A A A A A tφ φ ω1 2= + + − −∆

1 1 1 2 2

1 1 2 2

sin sin( )tans cos( )

A A tAco A t

φ φ ωφφ φ ω

− + + ∆= + + ∆

where

; Approximately simple harmonic vibrationwith slowly varying ampl. A and phase φ


Consider a special case of A1 = A2 and φ1 = φ2 = 0Then,

1/ 21 1

1

1 2

(2 2cos ) 0 2sintan

1 cosBeat frequency b

A A t A Att

f f f f

ωωφω

ω

−

= + ∆ ≤ ≤

∆ = + ∆ ≡ = ∆ = − = ∆

② (Forcing freq., ω) ≈ (Natural freq., ωn (system))

0 0

02

02

0,( / )( ) (cos cos )

( / ) 2sin sin2

nn

n n

n

When x xF mx t t t

F m t t

ω ωω ω

ω ω ω ωω ω

2

2

= =

= −−

+ −= ⋅

− 2


As ωn - ω = ∆ω ≡ 2ε (ε : very small),ω +ωn ≈ 2ω & ωn

2 - ω2 = (ωn + ω)(ωn - ω) ≈ 4εωThen,

0 /( ) sin sin2F mx t t tε ωεω

≈

variable amplitude2 2 2period of beating

2

bn

b n

π π πτε ω ω ω

ω ε ω ω ω

≡ = = =2 − ∆

= = − = ∆

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