Dynamics cheat sheet

4 Steady state solutions to single degree freedom system
4.1 Summary:
( )
my ′′ + cy ′ + ky = Re ˆF e
iϖt
{ }
x = Re ˆXe
iϖt
ˆX = ˆF D (r, ζ)
k
1
D (r, ζ) =
(1 − r 2 ) + 2iζr
{ }
ˆF
x = Re
k |D (r, ζ)| ei(ϖt−θ)
θ = tan −1
2ζr
1 − r 2
4.2 Details
( )
Let load be harmonic and represented in general as Re ˆF e
iϖt
where ˆF is the complex amplitude of the force.
Hence system is represented by
( )
my ′′ + cy ′ + ky = Re ˆF e
iϖt
( ) ˆF
y ′′ + 2ζω n y ′ + ωny 2 = Re
m eiϖt
)
)
)
Let y = Re
(Ŷ e
iϖt
Hence y ′ = Re
(iϖŶ eiϖt , y ′′ = Re
(−ϖ 2 Ŷ e iϖt , therefore the differential equation
becomes
(
Re −ϖ 2 Ŷ e iϖt) (
+ 2ζω n Re iϖŶ eiϖt) + ωn 2 Re
(Ŷ ) ( )
e
iϖt ˆF
= Re
m eiϖt
Dividing numerator and denominator ω 2 n gives
Where r = ϖ ω n
, hence the response is
Therefore, the phase of the response is
y = Re
( )
arg (y) = arg ˆF
( ˆF
−ϖ 2 + 2ζω n iϖ + ωn) 2 Ŷ =
m
ˆF
m
Ŷ =
(−ϖ 2 + 2ζω n iϖ + ωn)
2
Ŷ = ˆF 1
k (1 − r 2 ) + i2ζr
( )
ˆF 1
k (1 − r 2 ) + i2ζr eiϖt
( 2ζr
− tan −1 (1 − r 2 )
20
)
+ ϖt