Physics 325
Lecture 13
Driven Oscillations
Now we move to the case of oscillators with driving forces.
We will start with sinusoidal
driving forces and linear velocity-dependent damping.
The equation of motion is
0
cos
mx
bx
kx
F
t
ω
+
+=
±±
±
(13.1)
Here
k
is the spring constant and
is the frequency of the driving force, which can be
different from the
resonant
or natural frequency,
0
km
=
, of the oscillator.
Dividing
Equation (13.1) through by
m
and defining
β=
b/2m
as before, and
A=F
0
/m,
we get
2
0
2c
o
s
x
xx
A t
β
++=
±
(13.2)
The procedure for finding the general solution to this differential equation is to first find
the “complementary” solution, which is the solution to the corresponding homogenous
equation with the right hand side set to zero, and then find the
particular
solution which
actually solves the non-homogenous equation.
This may sound strange, but since it is a
linear equation, we can add solutions.
Adding the solution to the homogenous equation
does not affect the particular solution yielding the answer we want.
We have already solved the complementary equation.
That’s just our damped oscillator
without a driving force.
So the complementary function is
()
22
0
12
t
t
c
xt e
A
e
A
e
βω
−−
−
−
0
t
⎡
⎤
=+
⎢
⎥
⎣
⎦
(13.3)
which is Equation (12.9).
A reasonable guess for the particular solution is
( )
cos
sin
p
x
tB
t
(13.4)
which has the same frequency as the driving frequency.
This makes sense since the
complementary solution dies out at large times since it corresponds to a damped
oscillator.
We substitute Equation (13.4) into Equation (13.2) to get
1
2
01
02
cos
sin
2
sin
2
cos
cos
sin
cos
B
t
B
t
B
t
BtBt
At
ωω
−−−
+
++ =
The terms on the left that have sin
t
must sum to zero, and the terms that have cos
t
must sum to
A
:
0
1
21
0
2
2
20
B
BB
B
ωβ
−+
+ =
A

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*Sign up*And finally
()
22
0
12
00
2
A
A
BB
ωω
βω
ω
ωβω
ωβ
−
==
−+
So the particular solution is not exactly simple:
( )
0
2
2
0
cos
2
sin
2
p
At
A
xt
t
β
=
(13.5)
To understand this solution better, we would like to be able to write it as some sinusoidal
piece with a phase and an amplitude:
( ) ( )
cos
p
xt D
t
δ
=
−
(13.6)
Using our trig identities we expand this as
cos
cos
sin
sin
p
t
D
t
δω
=+
Comparing with Equation (13.5), we can write
0
2
2
0
2
2
0
cos
2
2
sin
2

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