HW#7—Solutions
—Phys410—Fall 2011
Prof. Ted Jacobson
Room 4115, (301)4056020
www.physics.umd.edu/grt/taj/410a/
[email protected]
11.32 (CO
2
vibrations
)
(a)
L
=
1
2
m
( ˙
x
2
1
+ ˙
x
2
3
) +
1
2
M
˙
x
2
2

1
2
k
[(
x
1

x
2
)
2
+ (
x
3

x
2
)
2
].
m
¨
x
1
=

k
(
x
1

x
2
),
m
¨
x
2
=
k
(
x
1
+
x
3

2
x
2
),
m
¨
x
3
=

k
(
x
3

x
2
). These take the
matrix form
M
¨
x
=

Kx
, with
x
= (
x
1
, x
2
, x
3
)
T
, and
M
=
m
0
0
0
M
0
0
0
m
,
K
=
k
1

1
0

1
2

1
0

1
1
.
Now let’s adopt units with
k
= 1 and
m
= 1. Then
K

ω
2
M
=
1

ω
2

1
0

1
2

ω
2
M

1
0

1
1

ω
2
and det(
K

ω
2
M
) =
ω
2
(1

ω
2
)(
Mω
2

(2 +
M
)), so the normal mode frequencies
are
ω
= 0
,
1
,
p
1 + 2
/M
.
(b & c) For
ω
= 0 we have
x
1
=
x
2
=
x
3
. This is the translational mode. For
ω
= 1
we have
x
2
= 0 and
x
3
=

x
1
. This is the asymmetric mode. Note the center of mass
remains at rest.
These two are easily guessed.
For
ω
=
p
1 + 2
/M
, we can insert
the value of
ω
2
into
K

ω
2
M
and find the zero eigenvector.
This yields
x
1
=
x
3
and
x
2
=

(2
/M
)
x
1
, so the two ends move identically and the middle one moves
oppositely with relative amplitude 2
a
. This has the highest frequency, because for a
given displacement of one of the end atoms, the spring is deformed more because of
the opposite motion of the middle atom, so the effective spring constant for that atom
is larger.
(d) For CO
2
,
M
=
M/m
= 3
/
4, so the frequency ratio is
p
1 + 2
/M
=
p
1 + 8
/
3 =
p
11
/
3 = 1
.
91. The measured frequency of the asymmetric mode (oxygens move op
posite) is listed as 2349 in units of cm

1
, while that of the symmetric mode (oxygens
move in same direction) is 1388. The ratio is 1.69, which is smaller than the above
result by 12%. [The website writes 1288 in the text and 1388 in the diagram. Other
web resources I’ve found list other numbers for one or both of these frequencies...]
(e) The actual frequency ratio is lower than in the model. Maybe this is because the
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 Fall '11
 Jacobson
 Physics, mechanics, Mass, uniform rod, normal mode frequencies

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