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Next: Exercises Up: Coupled Oscillations Previous: Two Coupled LC Circuits


Three Spring-Coupled Masses

Consider a generalized version of the mechanical system discussed in Section 4.1 that consists of three identical masses $ m$ which slide over a frictionless horizontal surface, and are connected by identical light horizontal springs of spring constant $ k$ . As before, the outermost masses are attached to immovable walls by springs of spring constant $ k$ . The instantaneous configuration of the system is specified by the horizontal displacements of the three masses from their equilibrium positions: namely, $ x_1(t)$ , $ x_2(t)$ , and $ x_3(t)$ . 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

$\displaystyle m\,\ddot{x}_1$ $\displaystyle = -k\,x_1 +k\,(x_2-x_1),$ (202)
$\displaystyle m\,\ddot{x}_2$ $\displaystyle =-k\,(x_2-x_1)+k\,(x_3-x_2),$ (203)
$\displaystyle m\,\ddot{x}_3$ $\displaystyle =-k\,(x_3-x_2)+k\,(-x_3).$ (204)

These equations can be rewritten

$\displaystyle \ddot{x}_1$ $\displaystyle = -2\,\omega_0^{\,2}\,x_1+ \omega_0^{\,2}\,x_2,$ (205)
$\displaystyle \ddot{x}_2$ $\displaystyle =\omega_0^{\,2}\,x_1-2\,\omega_0^{\,2}\,x_2+\omega_0^{\,2}\,x_3,$ (206)
$\displaystyle \ddot{x}_3$ $\displaystyle =\omega_0^{\,2}\,x_2-2\,\omega_0^{\,2}\,x_3,$ (207)

where $ \omega_0=\sqrt{k/m}$ . Let us search for a normal mode solution of the form

$\displaystyle x_1(t)$ $\displaystyle = \hat{x}_1\,\cos(\omega\,t-\phi),$ (208)
$\displaystyle x_2(t)$ $\displaystyle =\hat{x}_2\,\cos(\omega\,t-\phi),$ (209)
$\displaystyle x_3(t)$ $\displaystyle =\hat{x}_3\,\cos(\omega\,t-\phi).$ (210)

Equations (205)-(210) can be combined to give the $ 3\times 3$ homogeneous matrix equation

$\displaystyle \left(\begin{array}{ccc} \hat{\omega}^{\,2}-2 & 1 & 0\\ [0.5ex] 1...
...ay}\right) =\left(\begin{array}{c}0\\ [0.5ex] 0 \\ [0.5ex] 0\end{array}\right),$ (211)

where $ \hat{\omega}=\omega/\omega_0$ . The normal frequencies are determined by setting the determinant of the matrix to zero: that is,

$\displaystyle (\hat{\omega}^{ 2}-2)\left[(\hat{\omega}^{ 2}-2)^2-1\right]-(\hat{\omega}^{ 2}-2)=0,$ (212)

or

$\displaystyle (\hat{\omega}^{ 2}-2) \left[\hat{\omega}^{ 2}-2-\sqrt{2}\right] \left[ \hat{\omega}^{ 2}-2+\sqrt{2}\right]=0.$ (213)

Thus, the normal frequencies are $ \hat{\omega}= \sqrt{2}\left(1-1/\sqrt{2}\right)^{1/2}$ , $ \sqrt{2}$ , and $ \sqrt{2}\left(1+1/\sqrt{2}\right)^{1/2}$ . According to the first and third rows of Equation (211),

$\displaystyle \hat{x}_1:\hat{x}_2:\hat{x}_3 :: 1:2- \hat{\omega}^{\,2}:1,$ (214)

provided $ \hat{\omega}^{\,2}\neq 2$ . According to the second row,

$\displaystyle \hat{x}_1:\hat{x}_2:\hat{x}_3 :: -1:0:1$ (215)

when $ \hat{\omega}^{\,2} = 2$ . Incidentally, we can only determine the ratios of $ \hat{x}_1$ , $ \hat{x}_2$ , and $ \hat{x}_3$ , rather than the absolute values of these quantities. In other words, only the direction of the vector $ \hat{\bf x}= (\hat{x}_1,\hat{x}_2,\hat{x}_3)$ is well-defined. [This follows because the most general solution, (219), is undetermined to an arbitrary multiplicative constant. That is, if $ {\bf x}(t)=(x_1(t),x_2(t),x_3(t))$ is a solution to the dynamical equations (205)-(207) then so is $ a\,{\bf x}(t)$ , where $ a$ is an arbitrary constant. This, in turn, follows because the dynamical equations are linear.] Let us arbitrarily set the magnitude of $ \hat{\bf x}$ to unity. It follows that the normal mode associated with the normal frequency $ \hat{\omega}_1=\sqrt{2}\,\left(1-1/\sqrt{2}\right)^{1/2}$ is

$\displaystyle \hat{\bf x}_1 = \left(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}\right).$ (216)

Likewise, the normal mode associated with the normal frequency $ \hat{\omega}_2=\sqrt{2}$ is

$\displaystyle \hat{\bf x}_2 = \left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right).$ (217)

Finally, the normal mode associated with the normal frequency $ \hat{\omega}_3=\sqrt{2}\,\left(1+1/\sqrt{2}\right)^{1/2}$ is

$\displaystyle \hat{\bf x}_3 =\left(\frac{1}{2},-\frac{1}{\sqrt{2}}, \frac{1}{2}\right).$ (218)

The vectors $ \hat{\bf x}_1$ , $ \hat{\bf x}_2$ , and $ \hat{\bf x}_3$ are mutually perpendicular. In other words, they are normal vectors. (Hence, the name ``normal'' mode.)

Let $ {\bf x}=(x_1,x_2,x_2)$ . It follows that the most general solution to the problem is

$\displaystyle {\bf x}(t) = \eta_1(t) \hat{\bf x}_1 + \eta_2(t) \hat{\bf x}_2+\eta_3(t) \hat{\bf x}_3,$ (219)

where

$\displaystyle \eta_1(t)$ $\displaystyle =\hat{\eta}_1\,\cos(\hat{\omega}_1\,t-\phi_1),$ (220)
$\displaystyle \eta_2(t)$ $\displaystyle =\hat{\eta}_2\,\cos(\hat{\omega}_2\,t-\phi_2),$ (221)
$\displaystyle \eta_3(t)$ $\displaystyle =\hat{\eta}_3\,\cos(\hat{\omega}_3\,t-\phi_3).$ (222)

Here, $ \hat{\eta}_{1,2,3}$ and $ \phi_{1,2,3}$ are constants. Incidentally, we need to introduce the arbitrary amplitudes $ \hat{\eta}_{1,2,3}$ to make up for the fact that we set the magnitudes of the vectors $ \hat{\bf x}_{1,2,3}$ to unity. Equation (219) yields

$\displaystyle \left(\begin{array}{c}x_1\\ [0.5ex]x_2\\ [0.5ex]x_3\end{array}\ri...
...)\left(\begin{array}{c}\eta_1\\ [0.5ex]\eta_2\\ [0.5ex]\eta_3\end{array}\right)$ (223)

The preceding equation can be inverted by noting that $ \eta_1=\hat{\bf x}_1\cdot \hat{\bf x}$ , et cetera, because $ \hat{\bf x}_1$ , $ \hat{\bf x}_2$ , and $ \hat{\bf x}_3$ are mutually perpendicular unit vectors. Thus, we obtain

$\displaystyle \left(\begin{array}{c}\eta_1\\ [0.5ex]\eta_2\\ [0.5ex]\eta_3\end{...
...ay}\right)\left(\begin{array}{c}x_1\\ [0.5ex]x_2\\ [0.5ex]x_3\end{array}\right)$ (224)

This equation determines the three normal coordinates, $ \eta_1$ , $ \eta_2$ , $ \eta_3$ , in terms of the three conventional coordinates, $ x_1$ , $ x_2$ , $ x_3$ . In general, the normal coordinates are undetermined to arbitrary multiplicative constants.


next up previous
Next: Exercises Up: Coupled Oscillations Previous: Two Coupled LC Circuits
Richard Fitzpatrick 2013-04-08