The kinetic energy of the molecule is written

Clearly, we have a three degree of freedom dynamical system. However, we can reduce this to a two degree of freedom system by only considering

(842) |

Eliminating from Equations (840) and (841), we obtain

(844) |

(845) |

A comparison of the above expressions with the standard
forms (775) and (780) yields the following
expressions for the mass matrix, , and the force matrix, :

(846) | |||

(847) |

Now, the equation of motion of the system takes the form [see Equation (786)]

where is the column vector of the and values. The solubility condition for the above equation is

(849) |

(850) |

The two roots of the above equation are

(851) | |||

(852) |

The fact that the roots are negative implies that both normal modes are indeed

Equation (848) can now be solved, subject to the normalization condition (799), to give the two eigenvectors:

Thus, we conclude from Equations (843) and (853)-(856) that our model molecule possesses two normal modes of oscillation. The first mode oscillates at the frequency , and
is an *anti-symmetric* mode in which and .
In other words, in this mode of oscillation, the two end atoms move in opposite
directions whilst the central atom remains stationary. The second mode oscillates at the frequency , and is a mixed symmetry mode in which
but
. In other words, in this mode of oscillation, the
two end atoms move in the same direction whilst the central atom moves
in the opposite direction.

Finally, it is easily demonstrated that the normal coordinates of the system
are

(857) | |||

(858) |

When expressed in terms of these coordinates, and reduce to

(859) | |||

(860) |

respectively.