1
Waves in Different Forms
You should know how to use the formulas
amplitude−phasef orm
C1 cos(ωt) + C2 sin(ωt) =
zq
}|
C12
+
C22
{
cos(ωt + φ)
C2
where φ = tan−1 −
C1
iφ iωt
Re( ke
| {ze } ) = k cos(ωt + φ)
(1)
(2)
complexif ied
2
Second Order, Transient, Steady State and Forced Solutions
2.1
No Damping (δ = 0) and Forcing Freqency (Ω) 6= Natural Frequency
(ω).
A sinusoidally forced MSD differential equation (DE) is
ẍ + 2δ ẋ + ω 2 x = A cos(Ωt).
(3)
As a special case, when A = 1 (amplitude of forcing function) and δ = 0 (no damping), the forced
MSD DE eq. (3) becomes,
ẍ + ω 2 x = cos(Ωt)
(Ω 6= ω)
(4)
Using the complexification method, we showed in class that the general (forced) solution to this
special case can be written as
x(t) = HomogenousOscillation + P articularOscillation,
where
Homogenous
=
P articular
=
√
√
C1 cos( ω 2 t) + C2 sin( ω 2 t)
1
cos(Ωt)
(Ω 6= ω).
ω 2 − Ω2
when 0=δ 2 < ω 2
Note that the function
χ(ω, Ω) =
ω2
1
− Ω2
(5)
is called an amplitude response curve. We should observe that this general solution x(t) applies only
when Ω 6= ω. You should be able to derive this formula and use it to solve the assigned homework
problems.
2.1.1
Beating (Ω close but 6= ω)
You should be able to find a specific solution (i.e. by finding C1 , C2 ) by applying initial conditions
if given. For example, when x(0) = 0 and ẋ(0) = 0 when showed in class that the general solution
becomes a specific solution
x(t) =
cos(ωt)
cos(Ωt)
− 2
ω 2 − Ω2
ω − Ω2

(6)
slow−oscillations

z }|
{
−2 
Ω+ω
Ω−ω 


= 2
t · sin
t .
 sin
ω − Ω2 
2
2

|
{z
}
(7)
f ast−oscillations
You should be able to verify that eq. (6) and (7) are equal using sin(A ± B) = sin(A) cos(B) ∓
sin(B) cos(A). The advantage of eq. (7) is that, when Ω is chosen near but not equal some given
1
value of ω) we were are able to analyze the oscillations as a competition between the fast–frequency
and the beating (slow)–frequency pieces with frequencies given respectively by
1 Ω+ω
f ast − f requency =
,
(8)
2π
2
1 Ω−ω
beating − f requency =
.
(9)
2π
2
The periods of each of these pieces is, of course, T = period = 1/f requency.
2.2
No Damping (δ = 0) and Resonance (Ω = ω)
We have hanging over our heads the fact that we don’t know the solution to eq. (4) when Ω = ω.
Finding the general solution to eq. (4) when Ω = ω is shown in our book and can be summarized
as: (a) complexify F (t) = cos(ωt) as Fc (t) = eiωt , (b) choose xp (t) = α · t · Fc (t) (the t because
this is what we do in this Ω = ω case) and substitute into the complexified DE ẍ + ω 2 x = cos(ωt)
i
and therefore taking the real part of
which is ẍ + ω 2 x = eiωt (c) solve for α which will be α = − 2ω
i
1
iωt
xp = − 2ω te gives Re(xp (t)) = 2ω t sin(ωt).
HM: For homework you should follow these above steps to see if you get Ω = ω
particular solution.
In summary, when Ω = ω, A = 1, and δ = 0 then eq. (3) reduces
ẍ + ω 2 x = cos(ωt)
(Ω = ω)
(10)
and the complexification method outlined above give us a general solution
x(t) = HomogenousOscillation + U nboundedP articularOscillation,
where
2.3
√
√
C1 cos( ω 2 t) + C2 sin( ω 2 t)
Homogenous
=
U nboundedOscillation
=
1
t sin(ωt)
2ω
when 0=δ 2 < ω 2
(Ω = ω).
Damping (δ 6= 0)
When we reintroduce damping (δ 6= 0), the general solution to eq. (3) can be written as
x(t) = T ransient + SteadyStateOscillation,
(11)
where
√


2
√
2
2
2
δ −ω )t
C1 e(−δ+√δ −ω )t + C2 e(−δ− √
−δt
2
2
T ransient =
e (C1 cos( ω − δ t) + C2 sin( ω 2 − δ 2 t)

C1 e−δt + C2 te−δt
SteadyState
= χ(δ, ω, A, Ω) cos(Ωt + φ)
when δ 2 > ω 2
when δ 2 < ω 2
when δ 2 = ω 2
(Ω 6= ω)
where χ is called the Amplitude Response Curve given by
χ(δ, ω, A, Ω) = p
A
(ω 2 − Ω2 )2 + 4δ 2 Ω2
,
and φ is a phase angle given by
−1
φ = tan
−2δΩ
ω 2 − Ω2
.
You should be able to show that the solution eq. (11) reduces to the solution eq. (5) when δ = 0
and A = 1.
2