Spring-Coupled Masses

Consider the two degree of freedom dynamical system pictured in Figure 37. In this system, two point objects of mass $m$ are free to move in one dimension. Furthermore, the masses are connected together by a spring of spring constant $k$, and are also each attached to fixed supports via springs of spring constant $k'$.

Figure 37: Two spring-coupled masses.

Let $q_1$ and $q_2$ 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 $q_1$, $q_2-q_1$, and $-q_2$, respectively. The kinetic energy of the system takes the form

$$K = \frac{m}{2} \,(\dot{q}_1^{\,2} + \dot{q}_2^{\,2}),$$ (819)

whereas the potential energy is written

$$U= \frac{1}{2}\left[k'\,q_1^{\,2} + k\,(q_2-q_1)^2+ k'\,q_2^{\,2}\right].$$ (820)

The above expression can be rearranged to give

$$U= \frac{1}{2}\left[(k+k')\,q_1^{\,2} -2\,k\,q_1\,q_2 + (k+k')\,q_2^{\,2}\right].$$ (821)

A comparison of Equations (819) and (821) with the standard forms (775) and (780) yields the following expressions for the mass matrix, ${\bf M}$, and the force matrix, ${\bf G}$:

$${\bf M} = \left( \begin{array}{cc} m& 0\\ 0& m\end{array}\right),$$ (822)

$${\bf G} = \left(\begin{array}{cc} -k-k'& k\\ k& -k-k'\end{array}\right).$$ (823)

Now, the equation of motion of the system takes the form [see Equation (786)]

$$({\bf G} -\lambda\,{\bf M})\,{\bf x} = {\bf0},$$ (824)

where ${\bf x}$ is the column vector of the $q_1$ and $q_2$ values. The solubility condition for the above equation is

$$\left\vert{\bf G} - \lambda\,{\bf M}\right\vert = 0,$$ (825)

or

$$\left\vert\begin{array}{cc} -k-k'-\lambda\,m& k\\ [0.5ex] k&-k-k'-\lambda\,m \end{array}\right\vert = 0,$$ (826)

which yields the following quadratic equation for the eigenvalue $\lambda$:

$$m^2\,\lambda^2 + 2\,m\,(k+k')\,\lambda + k'\,(k'+2\,k) = 0.$$ (827)

The two roots of the above equation are

$$\lambda_1 = - \frac{k'}{m},$$ (828)

$$\lambda_2 = - \frac{(2\,k+k')}{m}.$$ (829)

The fact that the roots are negative implies that both normal modes are oscillatory in nature: i.e., the original equilibrium is stable. The characteristic oscillation frequencies of the modes are

$$\omega_1 = \sqrt{-\lambda_1} = \sqrt{\frac{k'}{m}},$$ (830)

$$\omega_2 = \sqrt{-\lambda_2} = \sqrt{\frac{2\,k+k'}{m}}.$$ (831)

Now, the first row of Equation (824) gives

$$\frac{q_1}{q_2}= \frac{k}{k+k'+\lambda\,m}.$$ (832)

Moreover, Equations (799) and (822) yield the following normalization condition for the eigenvectors:

$${\bf x}_k^T\,{\bf x}_k = m^{-1},$$ (833)

for $k=1,2$. It follows that the two eigenvectors are

$${\bf x}_1 = (2\,m)^{-1/2}\,(1,\,\,\,\, 1),$$ (834)

$${\bf x}_2 = (2\,m)^{-1/2}\,(1,-1).$$ (835)

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 $\omega_1$, and is a purely symmetric mode: i.e., $q_1 = q_2$. Note that such a mode does not stretch the middle spring. Hence, $\omega_1$ is independent of $k$. In fact, $\omega_1$ is simply the characteristic oscillation frequency of a mass $m$ on the end of a spring of spring constant $k'$. The second mode oscillates at the frequency $\omega_2$, and is a purely anti-symmetric mode: i.e., $q_1=-q_2$. 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., $\omega_2> \omega_1$.

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

$$\eta_1 = \sqrt{\frac{m}{2}}\,(q_1 + q_2),$$ (836)

$$\eta_2 = \sqrt{\frac{m}{2}}\,(q_1-q_2).$$ (837)

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

$$K = \frac{1}{2}\,(\dot{\eta}_1^{\,2} + \dot{\eta}_2^{\,2}),$$ (838)

$$U = \frac{1}{2}\,(\omega_1^{\,2}\,\eta_1^{\,2} + \omega_2^{\,2}\,\eta_2^{\,2}),$$ (839)

respectively.