LECTURE 17: FORCED OSCILLATIONS AND RESONANCE
MINGFENG ZHAO
October 21, 2015
Mass-Spring System:
Figure 1. Mass-Spring System
Let x(t) be the displacement of the mass, then
mx00 + cx0 + kx = F (t) .
We are interested in periodic forcing, F (t) = F0 cos(ωt) with ω > 0. So we have
mx00 + cx0 + kx = F0 cos(ωt) .
Rewrite the equation, we have
x00 + 2px0 + ω02 x =
F0
cos(ωt) ,
m
where
c
p=
≥ 0,
2m
r
and ω0 =
1
k
> 0.
m
2
MINGFENG ZHAO
Undamped forced motion and resonance
The differential equation for the undamped forced motion (c = 0) is:
x00 + ω02 x =
The general solution to x00 + ω02 x =
x(t) =
F0
cos(ωt),
m
ω > 0.
F0
cos(ωt) is:
m



 A cos(ω0 t) + B sin(ω0 t) +


 A cos(ω0 t) + B sin(ω0 t) +
F0
cos(ωt), if ω0 6= ω,
− ω2 ]
m[ω02
F0
t sin(ω0 t),
2mω0
.
if ω0 = ω
F0
F0
t and
t, in particular, the
2mω0
2mω0
amplitude of x(t) will go to infinity. This kind of behavior is called resonance or perhaps pure resonance.
In the case that ω0 = ω, when t → ∞, the graph of x(t) oscillates between −
Figure 2. Graph of
1
t sin(πt)
π
Damped forced motion and practical resonance
The differential equation for the damped forced motion (c > 0) is:
x00 + 2px0 + ω02 x =
F0
cos(ωt).
m
The characteristic equation of x00 + 2px0 + ω02 x = 0 is:
r2 + 2pr + ω02 = 0.
Solve r2 + 2pr + ω02 = 0, we have
r1,2 = −p ±
q
p2 − ω02 .
LECTURE 17: FORCED OSCILLATIONS AND RESONANCE
3
The general solution to x00 + 2px0 + ω02 x = 0 is:



Aer1 t + Ber2 t ,
overdamping, that is, p2 − ω02 > 0



Ae−pt + Bte−pt ,
critical damping, that is, p2 − ω02 = 0 .
xc (t) =
q
q




 Ae−pt cos
ω02 − p2 · t + Be−pt sin
ω02 − p2 · t , underdamping, that is, p2 − ω02 < 0
In any case, since p 6= 0, then xc (t) → 0 as t → ∞. Since ωi is not a solution to r2 + 2pr + ω02 = 0, then we can let
F0
xp (t) = D cos(ωt) + E sin(ωt) be a particular solution to x00 + 2px0 + ω02 x =
cos(ωt), then
m
x0p (t)
= −Dω sin(ωt) + Eω cos(ωt)
x00p (t)
= −Dω 2 cos(ωt) − Eω 2 sin(ωt)
x00p (t) + 2px0p (t) + ω02 xp (t)
= −Dω 2 cos(ωt) − Eω 2 sin(ωt) + 2p [−Dω sin(ωt) + Eω cos(ωt)]
+ω02 [D cos(ωt) + E sin(ωt)]
=
−Dω 2 + 2pEω + Dω02 cos(ωt) + −Eω 2 − 2pDω + Eω02 sin(ωt)
=
F0
cos(ωt).
m
Then we have
−Dω 2 + 2pEω + Dω02 =
F0
,
m
and
− Eω 2 − 2pDω + Eω02 = 0.
Solve D and E, we get
D=
F0 [ω02 − ω 2 ]
,
m [(2ωp)2 + (ω02 − ω 2 )2 ]
So a particular solution to x00 + 2px0 + ω02 x =
xp (t) =
and E =
2ωpF0
.
+ (ω02 − ω 2 )2 ]
m [(2ωp)2
F0
cos(ωt) is:
m
2ωpF0
F0 [ω02 − ω 2 ]
· cos(ωt) +
sin(ωt) .
m [(2ωp)2 + (ω02 − ω 2 )2 ]
m [(2ωp)2 + (ω02 − ω 2 )2 ]
The general solution xc to x00 + 2px0 + ω02 x = 0 is called the transient solution, denoted by xtr , the particular solution
F0
xp found in above to x00 + 2px0 + ω02 x =
cos(ωt) is called the steady periodic solution, denoted by xsp , which is a
m
periodic function with frequency ω.
Since xtr (t) → 0 as t → ∞, so for any given initial data, at infinity, the solution will be close the the steady periodic
solution xsp (t).
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MINGFENG ZHAO
Figure 3. Solutions with differential initial data for k = m = F0 = 1, c = 0.7 and ω = 1.1
For the steady periodic solution:
xsp (t) =
F0 [ω02 − ω 2 ]
2ωpF0
· cos(ωt) +
sin(ωt).
2
2
2
2
2
m [(2ωp) + (ω0 − ω ) ]
m [(2ωp) + (ω02 − ω 2 )2 ]
we know that
I. xsp (t) is a periodic function
II. xsp (t) has frequency ω, and the period
2π
.
ω
III. The amplitude of xsp is:
s
C(ω)
=
F0 [ω02 − ω 2 ]
m [(2ωp)2 + (ω02 − ω 2 )2 ]
=
F0
1
·p
.
2
m
(2ωp) + (ω02 − ω 2 )2
2
+
2ωpF0
2
m [(2ωp) + (ω02 − ω 2 )2 ]
Notice that
0
C (ω)
Then
=
1
2(2ωp) · 2p + 2(ω02 − ω 2 ) · (−2ω)
F0
· −
·
3
m
2
[(2ωp)2 + (ω02 − ω 2 )] 2
=
F0
2ω[ω02 − ω 2 − 2p2 ]
·
m [(2ωp)2 + (ω02 − ω 2 )] 23
=
2F0 −ω[ω 2 − (ω02 − 2p2 )]
·
.
m [(2ωp)2 + (ω02 − ω 2 )] 23
2
LECTURE 17: FORCED OSCILLATIONS AND RESONANCE
5
– If ω02 − 2p2 ≤ 0, then C(ω) has the maximum value at ω = 0. But we assume that ω > 0, so C(ω) can not
F0
attain its maximum, that is, C(ω) < C(0) =
for all ω > 0.
mω02
p
– If ω02 − 2p2 > 0, then C(ω) has the maximum value at ω = ω02 − 2p2 , that is,
q
2F0
F0
1
=p
max C(ω) = C( ω02 − 2p2 ) =
·p
ω>0
m
c2 [4k 2 − c2 ]
4p2 [ω02 − p2 ]
In this case, ω =
1
p
4p2 [ω02
− p2 ]
p
ω02 − 2p2 is called the practical resonance frequency, and max C(ω) =
ω>0
F0
·
m
is called the practical resonance amplitude.
Figure 4. Graph of C(ω) for k = m = F0 = 1 with c = 0.4, 0.8 and 1.6
Problems you can do:
Lebl’s Book [2]: All exercises on Page 83.
Braun’s Book [1]: All exercises on Page 172 and Page 173. Read all materials in Section 2.6.
References
[1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
[2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca