We have already seen how Lagrangian dynamics applies to simple mechanical systems, e.g.

- Lagrangian Dynamics Revisited (1)
- We now examine a more advanced application, namely in terms of modeling a
*linear triatomic molecule*. We sketch the basic model system below:

Kinetic energy:

T = ½ ( m
x1’ ^{2}
- M x2**’**** **** ^{2}** + m x3’

^{2})

Potential energy:

V = ½ k[ (x2 – x1) – b] ^{2}
+
½ k[ (x3 – x2) – b]^{ 2}

**T** =

½
[(m….0…..0)

[(0….M…..0)

[(0….0… ..m)

And make use here of the refined position parameter in terms of displacements h_{ i}:

h_{ i }** ** = x _{i
}** **– x o
_{i }** **

So: h_{
1 }** _{ } ** = x

_{1 }

**– x o**

_{1}

x _{02}_{
}** **– x o
_{1} = x _{03
}** **– x o_{2} = b

** **Then the potential energy is:

V = ½ k[ (h 2 – h 1) ^{2}
+
½ k[ (h 3 – h 2) ^{2} ]** **

** **And the kinetic energy:

T = ½ ( m
h 1’ ^{2} - M h 2**’**** **** ^{2}** + m h 3’

^{2})

Rewriting V:

V = ½ k[
(2 h 2** **^{2} + h 1 ^{2}
+ h 3 ^{2}
- 2 h 1 h 2** ** - 2 h 3 h 2 )]

In tensor form:

**V** =

½
[(k…
.-k…..0)

[(-k….2k…. -k)

[(0….-k…
.. k)

__T__** ** = **h** **×**__
T __**×**** ****h**

By Lagrange’s eqns.(Tensor form)

__T __**×**** ****h **+ ** **__V
__**×**** ****h
=
** 0 ** **

Assume a solution of form:
exp (iw t)

w** **^{2}** ** __T
__**×**** ****h
**+ ** **__V
__**×**** ****h
=
** 0 ** **

And the determinant of coefficients must = 0 for non-trivial
soln.

è | w** **^{2}** T + V ** |** ****= ** 0

è

Secular determinant:

(k - w** **^{2}**
m ** …-k… . .0)

(-k …..2 k
- w** **^{2}**
m ..** -k)

(0….
….-k….. k - w** **^{2}**
m**)

*three solutions *for the angular frequency:

w1 = 0, w2 = Ö(k/ m), w3 = Ö{ k/ m (1 + 2m/ M)}

For amplitudes, three solutions are obtained:

h 1 = A1 exp (iw t)

__Suggested Problem__:

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