Let and be the displacements of the first and second masses,
respectively, from the equilibrium state. It follows that the
extensions of the left-hand, middle, and right-hand springs are
, , and , respectively. The kinetic energy of the system takes the
form

(820) |

A comparison of Equations (819) and (821) with the standard
forms (775) and (780) yields the following
expressions for the mass matrix, , and the force matrix, :

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

(825) |

(826) |

(827) |

The two roots of the above equation are

(828) | |||

(829) |

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

Now, the first row of Equation (824) gives

(832) |

(833) |

According to Equations (830)-(831) and (834)-(835), our two degree of freedom system possesses two normal modes. The first mode oscillates at the frequency , and is a purely *symmetric*
mode: *i.e.*, . Note that such a mode does not stretch
the middle spring. Hence, is independent of . In fact, is simply the characteristic oscillation frequency of a mass on the end of a spring of spring constant . The second mode
oscillates at the frequency , and is a purely
*anti-symmetric* mode: *i.e.*, . Since such a mode
stretches the middle spring, the second mode experiences a greater restoring force than the first, and hence has a higher oscillation frequency: *i.e.*,
.

Note, finally, from Equations (809) and (822), that the normal
coordinates of the system are:

(836) | |||

(837) |

When expressed in terms of these normal coordinates, the kinetic and potential energies of the system reduce to

(838) | |||

(839) |

respectively.