Three Spring-Coupled Masses

Consider a generalized version of the mechanical system discussed in Section 3.2 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 (3.1)–(3.2) generalize to give

$$m\,\ddot{x}_1 = -k\,x_1 +k\,(x_2-x_1),$$ (3.52)

$$m\,\ddot{x}_2 =-k\,(x_2-x_1)+k\,(x_3-x_2),$$ (3.53)

$$m\,\ddot{x}_3 =-k\,(x_3-x_2)+k\,(-x_3).$$ (3.54)

These equations can be rewritten

$$\ddot{x}_1 = -2\,\omega_0^{\,2}\,x_1+ \omega_0^{\,2}\,x_2,$$ (3.55)

$$\ddot{x}_2 =\omega_0^{\,2}\,x_1-2\,\omega_0^{\,2}\,x_2+\omega_0^{\,2}\,x_3,$$ (3.56)

$$\ddot{x}_3 =\omega_0^{\,2}\,x_2-2\,\omega_0^{\,2}\,x_3,$$ (3.57)

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

$$x_1(t) = \hat{x}_1\,\cos(\omega\,t-\phi),$$ (3.58)

$$x_2(t) =\hat{x}_2\,\cos(\omega\,t-\phi),$$ (3.59)

$$x_3(t) =\hat{x}_3\,\cos(\omega\,t-\phi).$$ (3.60)

Equations (3.55)–(3.60) can be combined to give the $3\times 3$ homogeneous matrix equation

$$\left(\begin{array}{ccc} \hat{\omega}^{\,2}-2, & 1, & 0\\ [0.5ex]... ...ay}\right) =\left(\begin{array}{c}0\\ [0.5ex] 0 \\ [0.5ex] 0\end{array}\right),$$ (3.61)

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

$$(\hat{\omega}^{\,2}-2)\left[(\hat{\omega}^{\,2}-2)^2-1\right]-(\hat{\omega}^{\,2}-2)=0,$$ (3.62)

or

$$(\hat{\omega}^{\,2}-2)\,\left[\hat{\omega}^{\,2}-2-\sqrt{2}\right]\,\left[ \hat{\omega}^{\,2}-2+\sqrt{2}\right]=0.$$ (3.63)

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 (3.61),

$$\hat{x}_1:\hat{x}_2:\hat{x}_3 :: 1:2- \hat{\omega}^{\,2}:1,$$ (3.64)

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

$$\hat{x}_1:\hat{x}_2:\hat{x}_3 :: -1:0:1$$ (3.65)

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, (3.69), 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 (3.55)–(3.57) then so is $\alpha\,{\bf x}(t)$, where $\alpha$ 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

$$\hat{\bf x}_1 = \left(\frac{1}{2},\,\frac{1}{\sqrt{2}},\,\frac{1}{2}\right).$$ (3.66)

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

$$\hat{\bf x}_2 = \left(-\frac{1}{\sqrt{2}},\,0,\,\frac{1}{\sqrt{2}}\right).$$ (3.67)

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

$$\hat{\bf x}_3 =\left(\frac{1}{2},-\frac{1}{\sqrt{2}},\,\frac{1}{2}\right).$$ (3.68)

Note that 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

$${\bf x}(t) = \eta_1(t)\,\hat{\bf x}_1 + \eta_2(t)\,\hat{\bf x}_2+\eta_3(t)\,\hat{\bf x}_3,$$ (3.69)

where

$$\eta_1(t) =\hat{\eta}_1\,\cos(\omega_1\,t-\phi_1),$$ (3.70)

$$\eta_2(t) =\hat{\eta}_2\,\cos(\omega_2\,t-\phi_2),$$ (3.71)

$$\eta_3(t) =\hat{\eta}_3\,\cos(\omega_3\,t-\phi_3).$$ (3.72)

Here, $\hat{\eta}_{1,2,3}$ and $\phi_{1,2,3}$ are arbitrary 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 (3.69) yields

$$\left(\begin{array}{c}x_1\\ [0.5ex]x_2\\ [0.5ex]x_3\end{array}\ri... ...)\left(\begin{array}{c}\eta_1\\ [0.5ex]\eta_2\\ [0.5ex]\eta_3\end{array}\right)$$ (3.73)

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

$$\left(\begin{array}{c}\eta_1\\ [0.5ex]\eta_2\\ [0.5ex]\eta_3\end{... ...ay}\right)\left(\begin{array}{c}x_1\\ [0.5ex]x_2\\ [0.5ex]x_3\end{array}\right)$$ (3.74)

This equation determines the three normal coordinates, $\eta _1$, $\eta _2$, $\eta_3$, in terms of the three physical coordinates, $x_1$, $x_2$, $x_3$. In general, the normal coordinates are undetermined to arbitrary multiplicative constants.