3.1 Harmonic Excitations

1

3.1 Harmonic Excitations

3.1.2 Damped SDOF

Consider a damped system subjected to a harmonic force.

If a force acts on the mass m of the system,

t e

A t

e A t

x_h( )  ₁ ^nt sin_d  ₂ ^nt cos_d ) cos(

) ( )

( )

(t cx t kx t F₀ t x

m     

The homogeneous solution is given by:

or

) sin(

)

(t  Ae^  t 

x_h ^nt _d

12





_{d n} where the damped (natural) frequency is

)

0 cos( t F

F  

3.1 Harmonic Excitations

3.1.2 Damped SDOF

t X

t

x_p( )  cos

Since there is damping, assuming the solution in the form is invalid. We have to assume this instead:

t X

t X

t

x_p( )  ₁sin  ₂ cos

We can solve for X¹ and X², but 1) it will take a long time,

2) if the force F is changed to a sine function, we have to redo all the steps.

Thus, it is recommended to use complex representation for solving for the response.

3

3.1 Harmonic Excitations

3.1.2 Damped SDOF

By complex representation of excitation, if then

Let where

Let

Substituting into EOM, and solve for Z,

)

0 cos( t F

F  

] Re[

)

(t F₀ e^{i t}

F  ^

)]

( Re[

)

(t z t

x 

t i n

n e

m t F

z t

z t

z( ) 2 ( ) ² ( )  ^{0 }



t i

p

t Ze

z ( ) 

^

t

ei

F t

kz t

z c t

z

m( ) ( )  ( )  ₀ ^



^{2 2 0} n i ²n



m Z F













 

3.1 Harmonic Excitations

3.1.2 Damped SDOF

n

r 

 

Thus,

where

Define a complex function H(ω) as

 

^_^











  ^it

p e

i r r

k x F

2 ) 1

Re ( ₂ ⁰



^{r r i}



k Z F







 

2 ) 1

( ²

0

] 2

) 1

[(

) 1

( 2

i r r

H k







 





^{i t}



p F He

x  Re ₀ ^

)

0 cos( 

 F H t x_p

) 1 (

H   _ Im[H()] _ 2r

5

3.1 Harmonic Excitations

3.1.2 Damped SDOF

) (

1 )

1 ( ) 1

( _{2 2}



 

 

 k r k m

H

Note that H(ω) is also

] )

[(

1 ]

2 ) 1

[(

) 1

( 2 2

i c m

k i

r r

H k









 







 



2 2

2) ( )

( ) 1

(     

c m

k

H 2

1

1 tan

)]

( Re[

)]

( tan Im[







 



 

 ^{ }

m k

c H

H

Please prove the above by yourself.

If there is NO damping



or 180

0 ) 

( 1 )

1 ( ) 1

( 2   2

 

 k r k m

H

3.1 Harmonic Excitations

3.1.2 Damped SDOF

Thus, the total response is

The constants A¹ and A² can be determined from the initial conditions. Note that this case

) cos(

) ( )

cos(

) sin(

)

(t  A₁e^  t  A₂e^  t  F₀ H  t 

x ^t _d ^t _d

)

0 cos( t F

F  

7

Find the total response of a SDOF system with m = 10 kg, c = 20 N-s/m, k = 4000 N/m, x⁰ = 0.01 m, v⁰ = 0 m/s under

an external force acts on the system with and .

t F

t

F( )  ₀ cos rad/s

10

 ^{F0 100 N}

3.1 Harmonic Excitations

3.1.3 Frequency Response Function (FRF)

FRF = H(ω) = specifies how the system responds to harmonic excitation of various driving frequencies.

1. FRF depends on EOM = system + excitation

2. FRF is complex and is a function of ω and system’s properties

3. FRF is not unique only because of the constant in front, usually it is defined to be dimensionless

()

t H

ei

F₀ ^ F0 H() e^{i(t)}

Input FRF Output

9

FRF plots tell you how the magnitude and

phase angle of the

system’s FRF vary with the excitation

frequency

This is an example of a FRF plots of

Note: phase shift (θ) is -90 deg when r = 1 for any

value of damping.

3.1 Harmonic Excitations

3.1.4 FRF Plots

() H

) 2 (

) 1

( ) 1

( ₂   







 

  k H

i r H r

) cos(

) ( )

( )

(t cx t kx t F₀ t x

m     

To make FRF dimensionless, introduce H’(ω) where

3.1 Harmonic Excitations

3.1.4 FRF Plots

Notes:

1. Peak of |H’(ω)| occurs at where 2. If , no peak is present

3. More damped, smaller |H’(ω)|

4. When r→0, |H’(ω)| →1

() H

i r H r







 

 

2 ) 1

( ) 1

( ₂

2 2

1 



r 2

 1



2

 1

