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: i.e., $x_1(t)$, $x_2(t)$, and $x_3(t)$. Clearly, this is a three degree of freedom system. We, therefore, expect it to possesses three independent normal modes of oscillation. Equations (142)-(143) generalize to

$$m\,\ddot{x}_1 = -k\,x_1 +k\,(x_2-x_1),$$ (195)

$$m\,\ddot{x}_2 = -k\,(x_2-x_1)+k\,(x_3-x_2),$$ (196)

$$m\,\ddot{x}_3 = -k\,(x_3-x_2)+k\,(-x_3).$$ (197)

These equations can be rewritten

$$\ddot{x}_1 = -2\,\omega_0^{\,2}\,x_1+ \omega_0^{\,2}\,x_2,$$ (198)

$$\ddot{x}_2 = \omega_0^{\,2}\,x_1-2\,\omega_0^{\,2}\,x_2+\omega_0^{\,2}\,x_3,$$ (199)

$$\ddot{x}_3 = \omega_0^{\,2}\,x_2-2\,\omega_0^{\,2}\,x_3,$$ (200)

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

$$x_1(t) = \hat{x}_1\,\cos(\omega\,t-\phi),$$ (201)

$$x_2(t) = \hat{x}_2\,\cos(\omega\,t-\phi),$$ (202)

$$x_3(t) = \hat{x}_3\,\cos(\omega\,t-\phi).$$ (203)

Equations (198)-(203) can be combined to give the $3\times 3$ homogeneous matrix equation

$$\left(\begin{array}{ccc} \hat{\omega}^{\,2}-2 & 1 & 0\\ [0.5... ...(\begin{array}{c}0\\ [0.5ex] 0 \\ [0.5ex] 0\end{array}\right),$$ (204)

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

$$(\hat{\omega}^{\,2}-2)\left[(\hat{\omega}^{\,2}-2)^2-1\right]-(\hat{\omega}^{\,2}-2)=0,$$ (205)

or

$$(\hat{\omega}^{\,2}-2)\,\left[\hat{\omega}^{\,2}-2-\sqrt{2}\right]\,\left[ \hat{\omega}^{\,2}-2+\sqrt{2}\right]=0.$$ (206)

Thus, the normal frequencies are $\hat{\omega}= \sqrt{2}\,(1-1/\sqrt{2})^{1/2}$, $\sqrt{2}$, and $\sqrt{2}\,(1+1/\sqrt{2})^{1/2}$. According to the first and third rows of Equation (204),

$$\hat{x}_1:\hat{x}_2:\hat{x}_3 :: 1:2- \hat{\omega}^{\,2}:1,$$ (207)

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

$$\hat{x}_1:\hat{x}_2:\hat{x}_3 :: -1:0:1$$ (208)

when $\hat{\omega}^{\,2} = 2$. Note that 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, (212), 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 (198)-(200) 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 $\hat{\omega}_1=\sqrt{2}\,(1-1/\sqrt{2})^{1/2}$ is

$$\hat{\bf x}_1 = \left(\frac{1}{2},\,\frac{1}{\sqrt{2}},\,\frac{1}{2}\right).$$ (209)

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

$$\hat{\bf x}_2 = \left(-\frac{1}{\sqrt{2}},\,0,\,\frac{1}{\sqrt{2}}\right).$$ (210)

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

$$\hat{\bf x}_3 =\left(\frac{1}{2},-\frac{1}{\sqrt{2}},\,\frac{1}{2}\right).$$ (211)

Note that the vectors $\hat{\bf x}_1$, $\hat{\bf x}_2$, and $\hat{\bf x}_3$ are mutually perpendicular: i.e., 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

$${\bf x}(t) = \eta_1(t)\,\hat{\bf x}_1 + \eta_2(t)\,\hat{\bf x}_2+\eta_3(t)\,\hat{\bf x}_3,$$ (212)

where

$$\eta_1(t) = \hat{\eta}_1\,\cos(\hat{\omega}_1\,t-\phi_1),$$ (213)

$$\eta_2(t) = \hat{\eta}_2\,\cos(\hat{\omega}_2\,t-\phi_2),$$ (214)

$$\eta_3(t) = \hat{\eta}_3\,\cos(\hat{\omega}_3\,t-\phi_3).$$ (215)

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 arbitrarily set the magnitudes of the vectors $\hat{\bf x}_{1,2,3}$ to unity. Equation (212) yields

$$\left(\begin{array}{c}x_1\\ [0.5ex]x_2\\ [0.5ex]x_3\end{arra... ...ay}{c}\eta_1\\ [0.5ex]\eta_2\\ [0.5ex]\eta_3\end{array}\right)$$ (216)

The above equation can easily be inverted by noting that the matrix is unitary: i.e., its transpose is equal to its inverse. Thus, we obtain

$$\left(\begin{array}{c}\eta_1\\ [0.5ex]\eta_2\\ [0.5ex]\eta_3... ...begin{array}{c}x_1\\ [0.5ex]x_2\\ [0.5ex]x_3\end{array}\right)$$ (217)

This equation determines the three normal coordinates, $\eta_1$, $\eta_2$, $\eta_3$, in terms of the three conventional coordinates, $x_1$, $x_2$, $x_3$. Note that, in general, the normal coordinates are undetermined to arbitrary multiplicative constants. Incidentally, the above matrix equation can also be obtained directly from (212), which yields $\eta_{1,2,3}={\bf x}\cdot\hat{\bf x}_{1,2,3}$ (since $\hat{\bf x}_{1,2,3}$ are mutually perpendicular unit vectors).