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Massachusetts
Institute
of
Technology
Physics 8.03 Spring 2004
Final Exam Solutions
Thursday, December 16, 2004
Solution for Problem 1
−
Coupled Oscillators
(a)
If
m
A
=
∞
,
then
ω
A
=
0
and
ω
B
=
k
+
k
.
m
B
(b)
If
k
=
0,
then
m
A
,m
B
and
k
±
form
an
isolate
system
and
the
center
of
mass
of
this
system
doesn’t
move.
Therefore,
m
A
x
A
=
−
m
B
x
B
(1)
The
force
applied
on
m
A
is
given
by
m
A
¨
x
A
=
−
k
±
(
x
A
−
x
B
)
(
2
)
Solving
them
gives
±
¨
x
A
+
x
A
k
±
²
m
A
+
m
B
m
A
m
B
³
=0
⇒
ω
=
k
±
M
(3)
where
M
=
m
A
m
B
m
A
+
m
B
is
called
the
“reduced
mass”.
(c)
If
k
±
=
0,
then
ω
A
=
k
m
A
and
ω
B
=
k
m
B
.
(d)
For
the
general
situation,
the
coupled
equations
of
motion
are
m
A
x
A
=
−
kx
A
+
k
±
(
x
B
−
x
A
)
(
4
)
¨
m
B
x
B
=
−
kx
B
−
k
±
(
x
B
−
x
A
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²
³
³
³
³
³
(e)
Assume
that
x
A
=
A
cos
ωt
and
x
B
=
B
cos
ωt
.
The
equations
of
motions
become
k
+
k
±
k
±
−
ω
2
A
+
A
−
B
=0
(6)
m
A
m
A
k
+
k
±
k
±
−
ω
2
B
+
B
−
A
=0
(7)
m
B
m
B
rewrite
them
with
A
and
B
as
the
variables
±
k
+
k
±
²
k
±
A ω
2
−
+
B
=0
(8)
m
A
m
A
±
k
+
k
±
²
k
±
A
+
ω
2
−
B
=0
(9)
m
B
m
B
for
A
and
B
have
nonzero
root,
the
determinant
should
be
equal
to
zero,
that
is
±
k
+
k
±
²±
²
k
±
2
k
+
k
±
ω
2
−
ω
2
−
−
=
0
(10)
m
A
m
A
m
A
m
B
±
1
²
1
ω
4
−
ω
2
(
k
+
k
±
)
+
+
(
k
+
k
±
)
2
−
k
±
2
=
0
(11)
m
A
m
B
m
A
m
B
The
solutions
for
ω
are
´
2
ω
2
p
1
p
=
±
p
2
−
4
q
=
p
±
−
q
(12)
1
,
2
2
2
2
4
1
where
p
=(
k
+
k
±
)
m
A
+
1
and
q
=
(
k
+
k
)
2
−
k
2
.
m
B
m
A
m
B
Solution for Problem 2
−
Dispersive String
(a)
The
phase
velocity
is
ω
T
v
p
=
=
+
αk
2
(13)
k
µ
(b)
The
group
velocity
is
T
+2
αk
2
dω
T
αk
2
µ
v
g
=
=
+
αk
2
+
µ
=
µ
(14)
dk
µ
T
+
αk
2
T
+
αk
2
µ
µ
(c)
There
are
three
normal
oscillation
modes
of
the
string
nπ
2
π
The
wave
numbers
k
n
=
L
and
the
frequencies
ω
n
=
v
p
n
k
n
=
v
p
n
.
Therefore
λ
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 Physics, Mass

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