Dynamics cheat sheet

Hence at t = 0 the phase of the response will be
( ) ( ) 2ζr
arg (y) = arg ˆF − tan −1 (1 − r 2 )
So when ˆF
( )
is real, the phase of the response is simply − tan −1 2ζr
(1−r 2 )
4.3 Undamped case
When ζ = 0 the above becomes
For real force this becomes
∣
The magnitude ∣Ŷ ∣ = F k
4.4 damped cases
ζ > 0
1
(1−r 2 )
( ˆF
y = Re
k
∣ ˆF
∣
=
k
)
1
(1 − r 2 ) eiϖt
1
(
(1 − r 2 ) cos
y = F k
and phase zero.
∣
∣Ŷ
y = Re
( ))
ϖt + arg ˆF
1
(1 − r 2 cos (ϖt)
)
( )
ˆF 1
k (1 − r 2 ) + i2ζr eiϖt
∣ ˆF
∣ 1
∣ = √
k
(1 − r 2 ) 2 + (2ζr) 2
) (
arg
(Ŷ = φ = arg ˆF
Hence for real force and at t = 0 the phase of displacement is
( ) 2ζr
− tan −1 1 − r 2
)
− tan −1 ( 2ζr
1 − r 2 )
+ ϖt
lag behind the load.
When r < 1 then φ goes from 0 to −90 0 Therefore phase of displacement is 0 to −90 0 behind force. The
minus sign at the front was added since the complex number is in the denominator. Hence the response will
always be lagging in phase relative for load.
For r > 1
Now 1 − r 2 is negative, hence the phase will be from −90 ◦ to −180 ◦
When r = 1
( ) ˆF 1
y = Re
k i2ζ eiϖt
∣ ∣
∣Ŷ
)
arg
(Ŷ
∣ ˆF
∣
∣ =
k
= −90 ◦
1
2ζ
Now phase is −90 ◦ 21