Physics 325, Fall 2006
Solutions to Homework #8
1) This is an exercise in finding the Fourier series of a periodic function.
F
(
t
) =
F
0

sin
ω
f
t

=
1
2
a
0
+
∞
n
=1
a
n
cos(
nω
f
t
) +
b
n
sin(
nω
f
t
)
with
a
n
=
2
τ
τ/
2

τ/
2
F
(
t
) cos(
nω
f
t
)
dt
b
n
=
2
τ
τ/
2

τ/
2
F
(
t
) sin(
nω
f
t
)
dt
The period
τ
= 2
π/ω
f
so
τ/
2 =
π/ω
f
and 2
/τ
=
ω
f
/π
. Actually, this is twice the period,
but it is easier to apply the Fourier series formulas from class if we center the intervals
on zero. Also, it is important to use this defintion of the period because the
ω
f
= 2
π/τ
in the Fourier Series defintion is the same as the
ω
f
in the driving function.
All the
b
n
’s are zero by symmetry (they are the integrals of products of even and odd
functions, which makes an odd function, from

τ/
2 to
τ/
2, so the integrals over the
negative times cancel out the integrals over the positive times).
a
n
=
ω
f
π
π/ω
f

π/ω
f
F
0

sin
ω
f
t

cos(
nω
f
t
)
dt
This is an integral of an even function, and so we can evaluate just the positivetime half
and double it.
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 Fall '09
 Lamb
 Calculus, mechanics, Fourier Series, Work, Cos, Periodic function, Even and odd functions

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