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# Three Spring-Coupled Masses

Consider a generalized version of the mechanical system discussed in Section 4.1 that consists of three identical masses which slide over a frictionless horizontal surface, and are connected by identical light horizontal springs of spring constant . As before, the outermost masses are attached to immovable walls by springs of spring constant . The instantaneous configuration of the system is specified by the horizontal displacements of the three masses from their equilibrium positions: namely, , , and . This is manifestly a three degree of freedom system. We, therefore, expect it to possess three independent normal modes of oscillation. Equations (149)-(150) generalize to  (202)  (203)  (204)

These equations can be rewritten  (205)  (206)  (207)

where . Let us search for a normal mode solution of the form  (208)  (209)  (210)

Equations (205)-(210) can be combined to give the homogeneous matrix equation (211)

where . The normal frequencies are determined by setting the determinant of the matrix to zero: that is, (212)

or (213)

Thus, the normal frequencies are , , and . According to the first and third rows of Equation (211), (214)

provided . According to the second row, (215)

when . Incidentally, we can only determine the ratios of , , and , rather than the absolute values of these quantities. In other words, only the direction of the vector is well-defined. [This follows because the most general solution, (219), is undetermined to an arbitrary multiplicative constant. That is, if is a solution to the dynamical equations (205)-(207) then so is , where is an arbitrary constant. This, in turn, follows because the dynamical equations are linear.] Let us arbitrarily set the magnitude of to unity. It follows that the normal mode associated with the normal frequency is (216)

Likewise, the normal mode associated with the normal frequency is (217)

Finally, the normal mode associated with the normal frequency is (218)

The vectors , , and are mutually perpendicular. In other words, they are normal vectors. (Hence, the name normal'' mode.)

Let . It follows that the most general solution to the problem is (219)

where  (220)  (221)  (222)

Here, and are constants. Incidentally, we need to introduce the arbitrary amplitudes to make up for the fact that we set the magnitudes of the vectors to unity. Equation (219) yields (223)

The preceding equation can be inverted by noting that , et cetera, because , , and are mutually perpendicular unit vectors. Thus, we obtain (224)

This equation determines the three normal coordinates, , , , in terms of the three conventional coordinates, , , . In general, the normal coordinates are undetermined to arbitrary multiplicative constants.   Next: Exercises Up: Coupled Oscillations Previous: Two Coupled LC Circuits
Richard Fitzpatrick 2013-04-08