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Physics 325
Lecture 14
Superposition & Fourier Series
We now investigate driven oscillations with nonsinusoidal driving forces.
We’ll start
with a simple superposition of two sinusoidal driving forces with different frequencies,
and then show how to generalize this result to an arbitrary driving force.
We take a driving force to be
( )
11
2
cos
cos
Ft
F
t F
t
2
ω
=+
(14.1)
and the equation of motion, which follows from Equation (13.2), is
2
12
01
2c
o
s
c
FF
2
o
s
x
xx
t
mm
t
β
ωω
++=
+
±±
±
(14.2)
This equation, as before is linear and therefore the sum of solutions is also a solution.
This inspires us to write two equations with solutions
( ) ( )
and
x
tx
t
:
2
1
0
1
1
2
2
22
0
2
F
2
o
s
o
s
x
t
m
F
x
m
t
±
±
We can add these two equations to find
()
2
2
0
1
2
2
2
cos
cos
dd
x
x
x
x
t
dt
dt
m
m
t
++
+=
+
Therefore,
() ()
x
t
+
is a solution to Equation (14.2).
We’ve demonstrated this for a
superposition of two terms, but it obviously works for an arbitrary number of terms.
Let
() () ()
1
cos
N
nn
n
n
n
F
tF
t
F
t
F t
=
=
∑
=
(14.3)
Then for each
n
we can write
2
0
2
n
nnn
F
c
o
s
n
x
m
t
±
(14.4)
and
1
N
n
n
x
=
=
∑
t
(14.5)
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View Full Documentsatisfies the differential equation
( )
2
0
2
Ft
xxx
m
βω
++=
±±
±
(14.6)
This is the
principle of superposition
, and it works because the differential equation
(14.6) is linear.
In solving the problem of a sinusoidal driving force in Lecture 13, we have already found
solutions to the differential equation (14.4).
Therefore we can make the following
statement:
If the driving force
F(t)
has the form
( ) ( )
cos
nn
n
F
t
n
ω
φ
=−
∑
(14.7)
and the oscillator system obeys Equation (14.6), then the solution for the resulting steady
state motion (long times) is
()
(
2
2
22
0
1
cos
2
n
pn
F
xt
t
m
)
n
n
δ
ωω
β
−+
∑
−
(14.8)
with
1
2
0
2
tan
n
n
n
2
−
=
−
(14.9)
Fourier’s Theorem
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 Fall '08
 Staff
 Physics, mechanics, Force

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