Biomedical Engineering Reference
In-Depth Information
m
1
m
2
Figure 3.4
Diatomic molecule
the equilibrium length of the spring is
R
e
and the length of the extended spring at some
given time is
R
=
x
2
−
x
1
The spring extension is therefore
R
e
Considering atom 1, the spring exerts a force of
k
s
(
x
2
−
x
2
−
x
1
−
x
1
−
R
e
) and so, according to
Newton's second law
m
1
d
2
x
1
d
t
2
=
k
s
(
x
2
−
x
1
−
R
e
)
=
k
s
(
R
−
R
e
)
(3.10)
As far as atom2 is concerned, the extended spring exerts a force of magnitude
k
s
(
x
2
−
x
1
−
R
e
)
in the direction of decreasing
x
2
and so
m
2
d
2
x
2
d
t
2
=−
k
s
(
x
2
−
x
1
−
R
e
)
=−
k
s
(
R
−
R
e
)
(3.11)
After a little rearrangement we find
d
2
R
d
t
2
=−
k
s
m
2
k
s
m
1
−
−
−
(
R
R
e
)
(
R
R
e
)
k
s
1
(
R
1
m
2
=−
m
1
+
−
R
e
)
(3.12)
We now define a quantity μ, called the
reduced mass
,by
1
μ
=
1
m
1
+
1
m
2
and so we have
μ
d
2
R
d
t
2
=−
k
s
(
R
−
R
e
)
(3.13)
which is identical to Equation (3.2) already derived for a single particle of mass μ on
a spring.
The general solution is therefore
A
sin
k
s
B
cos
k
s
μ
t
μ
t
R
=
R
e
+
+
(3.14)
=
√
k
s
/μ.
and the angular frequency is ω
