Two Spring-Coupled Masses

Consider a mechanical system consisting of two identical masses $m$ that are free to slide over a frictionless horizontal surface. Suppose that the masses are attached to one another, and to two immovable walls, by means of three identical light horizontal springs of spring constant $k$, as shown in Figure 3.1. The instantaneous state of the system is conveniently specified by the displacements of the left and right masses, $x_1(t)$ and $x_2(t)$, respectively. The extensions of the left, middle, and right springs are $x_1$, $x_2-x_1$, and $-x_2$, respectively, assuming that $x_1=x_2=0$ corresponds to the equilibrium configuration in which the springs are all unextended. The equations of motion of the two masses are thus

$$m\,\ddot{x}_1 = -k\,x_1 +k\,(x_2-x_1),$$ (3.1)

$$m\,\ddot{x}_2 = -k\,(x_2-x_1) +k\,(-x_2).$$ (3.2)

Here, we have made use of the fact that a mass attached to the left end of a spring of extension $x$ and spring constant $k$ experiences a horizontal force $+k\,x$, whereas a mass attached to the right end of the same spring experiences an equal and opposite force $-k\,x$.

Figure 3.1: Two-degree-of-freedom mass-spring system.

Equations (3.1)–(3.2) can be rewritten in the form

$$\ddot{x}_1 = -2\,\omega_0^{\,2}\,x_1+ \omega_0^{\,2}\,x_2,$$ (3.3)

$$\ddot{x}_2 = \omega_0^{\,2}\,x_1 - 2\,\omega_0^{\,2}\,x_2,$$ (3.4)

where $\omega_0=\sqrt{k/m}$. Let us search for a solution in which the two masses oscillate in phase with one another at the same angular frequency, $\omega$. In other words,

$$x_1(t) = \hat{x}_1\,\cos(\omega\,t-\phi),$$ (3.5)

$$x_2(t) =\hat{x}_2\,\cos(\omega\,t-\phi),$$ (3.6)

where $\hat{x}_1$, $\hat{x}_2$, and $\phi$ are constants. Equations (3.3) and (3.4) yield

$$-\omega^{\,2}\,\hat{x}_1\,\cos(\omega\,t-\phi) =\left( -2\,\omega_0^{\,2}\,\hat{x}_1+ \omega_0^{\,2}\,\hat{x}_2\right)\,\cos(\omega\,t-\phi),$$ (3.7)

$$-\omega^{\,2}\,\hat{x}_2\,\cos(\omega\,t-\phi) =\left( \omega_0^{\,2}\,\hat{x}_1-2\,\omega_0^{\,2}\,\hat{x}_2\right)\,\cos(\omega\,t-\phi),$$ (3.8)

or (after cancelling the common factor $\cos(\omega\,t-\phi)$, dividing through by $\omega_0$, and rearranging),

$$(\hat{\omega}^{\,2}-2)\,\hat{x}_1 +\hat{x}_2 =0,$$ (3.9)

$$\hat{x}_1 + (\hat{\omega}^{\,2}-2)\,\hat{x}_2 =0,$$ (3.10)

where $\hat{\omega}=\omega/\omega_0$. Note that, by searching for a solution of the form (3.5)–(3.6), we have effectively converted the system of two coupled linear differential equations (3.3)–(3.4) into the much simpler system of two coupled linear algebraic equations (3.9)–(3.10). The latter equations have the trivial solutions $\hat{x}_1=\hat{x}_2=0$, but also yield

$$\frac{\hat{x}_1}{\hat{x}_2}=-\frac{1}{(\hat{\omega}^{\,2}-2)} =-( \hat{\omega}^{\,2}-2).$$ (3.11)

Hence, the condition for a nontrivial solution is

$$(\hat{\omega}^{\,2}-2)\,(\hat{\omega}^{\,2}-2)-1 = 0.$$ (3.12)

In fact, if we write Equations (3.9)–(3.10) in the form of a homogenous (i.e., with a null right-hand side) $2\times 2$ matrix equation, so that

$$\left( \begin{array}{cc} \hat{\omega}^{\,2}-2, & 1\\ [0.5ex] ... ...t) = \left( \begin{array}{c} 0\\ [0.5ex] 0 \end{array}\right) ,$$ (3.13)

then it is apparent that the criterion (3.12) can also be obtained by setting the determinant of the associated matrix to zero (Riley 1974).

Equation (3.12) can be rewritten

$$\hat{\omega}^{\,4} - 4\,\hat{\omega}^2+3 = (\hat{\omega}^{\,2}-1)\,(\hat{\omega}^{\,2}-3)=0.$$ (3.14)

It follows that

$$\hat{\omega} = 1 \sqrt{3}.$$ (3.15)

Here, we have neglected the two negative frequency roots of (3.14)—that is, $\hat{\omega}=-1$ and $\hat{\omega}=-\sqrt{3}$—because a negative frequency oscillation is equivalent to an oscillation with an equal and opposite positive frequency, and an equal and opposite phase. In other words, $\cos(\omega\,t-\phi)\equiv \cos(-\omega\,t+\phi)$. It is thus apparent that the dynamical system pictured in Figure 3.1 has two unique frequencies of oscillation: namely, $\omega=\omega_0$ and $\omega=\sqrt{3}\,\omega_0$. These are called the normal frequencies of the system. Because the system possesses two degrees of freedom (i.e., two independent coordinates are needed to specify its instantaneous configuration), it is not entirely surprising that it possesses two normal frequencies. In fact, it is a general rule that a dynamical system with $N$ degrees of freedom possesses $N$ normal frequencies.

The patterns of motion associated with the two normal frequencies can be deduced from Equation (3.11). Thus, for $\omega=\omega_0$ (i.e., $\hat{\omega}=1$), we get $\hat{x}_1=\hat{x}_2$, so that

$$x_1(t) = \hat{\eta}_1\,\cos(\omega_0\,t-\phi_1),$$ (3.16)

$$x_2(t) = \hat{\eta}_1\,\cos(\omega_0\,t-\phi_1),$$ (3.17)

where $\hat{\eta}_1$ and $\phi_1$ are arbitrary constants. This first pattern of motion corresponds to the two masses executing simple harmonic oscillation with the same amplitude and phase. Such an oscillation does not stretch the middle spring. On the other hand, for $\omega=\sqrt{3}\,\omega_0$ (i.e., $\hat{\omega}=\sqrt{3}$), we get $\hat{x}_1=-\hat{x}_2$, so that

$$x_1(t) = \hat{\eta}_2\,\cos\left(\sqrt{3}\,\omega_0\,t-\phi_2\right),$$ (3.18)

$$x_2(t) =-\hat{\eta}_2\,\cos\left(\sqrt{3}\,\omega_0\,t-\phi_2\right),$$ (3.19)

where $\hat{\eta}_2$ and $\phi_2$ are arbitrary constants. This second pattern of motion corresponds to the two masses executing simple harmonic oscillation with the same amplitude but in anti-phase; that is, with a phase shift of $\pi$ radians. Such oscillations do stretch the middle spring, implying that the restoring force associated with similar amplitude displacements is greater for the second pattern of motion than for the first. This accounts for the higher oscillation frequency in the second case. (The inertia is the same in both cases, so the oscillation frequency is proportional to the square root of the restoring force associated with similar amplitude displacements.) The two distinctive patterns of motion that we have found are called the normal modes of oscillation of the system. Incidentally, it is a general rule that a dynamical system possessing $N$ degrees of freedom has $N$ unique normal modes of oscillation.

The most general motion of the system is a linear combination of the two normal modes. This immediately follows because Equations (3.1) and (3.2) are linear equations. [In other words, if $x_1(t)$ and $x_2(t)$ are solutions then so are $\alpha\,x_1(t)$ and $\alpha\,x_2(t)$, where $\alpha$ is an arbitrary constant.] Thus, we can write

$$x_1(t) = \hat{\eta}_1\,\cos(\omega_0\,t-\phi_1)+ \hat{\eta}_2\,\cos\left(\sqrt{3}\,\omega_0\,t-\phi_2\right),$$ (3.20)

$$x_2(t) = \hat{\eta}_1\,\cos(\omega_0\,t-\phi_1)-\hat{ \eta}_2\,\cos\left(\sqrt{3}\,\omega_0\,t-\phi_2\right).$$ (3.21)

We can be sure that this represents the most general solution to Equations (3.1) and (3.2) because it contains four arbitrary constants; namely, $\hat{\eta}_1$, $\phi_1$, $\hat{\eta}_2$, and $\phi_2$. (In general, we expect the solution of a second-order ordinary differential equation to contain two arbitrary constants. It, thus, follows that the solution of a system of two coupled, second-order, ordinary differential equations should contain four arbitrary constants.) These constants are determined by the initial conditions.

For instance, suppose that $x_1=a$, $\dot{x}_1=0$, $x_2=0$, and $\dot{x}_2=0$ at $t=0$. It follows, from Equations (3.20) and (3.21), that

$$a =\hat{\eta}_1\,\cos\phi_1 + \hat{\eta}_2\,\cos\phi_2,$$ (3.22)

$$=\hat{\eta}_1\,\sin\phi_1+\sqrt{3}\,\hat{\eta}_2\,\sin\phi_2,$$ 0 (3.23)

$$=\hat{\eta}_1\,\cos\phi_1 -\hat{\eta}_2\,\cos\phi_2,$$ 0 (3.24)

$$=\hat{\eta}_1\,\sin\phi_1-\sqrt{3}\,\hat{\eta}_2\,\sin\phi_2,$$ 0 (3.25)

which implies that $\phi_1=\phi_2=0$ and $\hat{\eta}_1=\hat{\eta}_2=a/2$. Thus, the system evolves in time as

$$x_1(t) =\frac{a}{2}\,\cos(\omega_0\,t)+\frac{a}{2}\,\cos(\sqrt{3}\,\omega_0\,t)= a\,\cos(\omega_-\,t)\,\cos(\omega_+\,t),$$ (3.26)

$$x_2(t) =\frac{a}{2}\,\cos(\omega_0\,t)-\frac{a}{2}\,\cos(\sqrt{3}\,\omega_0\,t)=a\,\sin(\omega_-\,t)\,\sin(\omega_+\,t),$$ (3.27)

where $\omega_\pm = [(\sqrt{3}\pm 1)/2]]\,\omega_0$, and use has been made of the trigonometric identities $\cos a + \cos b\equiv 2\,\cos[(a+b)/2]\,\cos[(a-b)/2]$ and $\cos a - \cos b \equiv -2\,\sin[(a+b)/2]\,\sin[(a-b)/2]$. (See Appendix B.) This evolution is illustrated in Figure 3.2.

Figure 3.2: Time evolution of the physical coordinates of the two-degree-of-freedom mass-spring system pictured in Figure 3.1. The initial conditions are $x_1=a$, $\stackrel{\,\,\mbox{\large.}}{x}_1 = x_2 = \,\stackrel{\,\,\mbox{\large .}}{x}_2=0$. The solid curve corresponds to $x_1$, and the dashed curve to $x_2$. Here, $T_0=2\pi /\omega _0$.

Finally, let us define the so-called normal coordinates, which (in the present case) take the form

$$\eta_1(t) = [x_1(t) + x_2(t)]/2,$$ (3.28)

$$\eta_2(t) = [x_1(t)-x_2(t)]/2.$$ (3.29)

It follows from Equations (3.20) and (3.21) that, in the presence of both normal modes,

$$\eta_1(t) =\hat{\eta}_1\,\cos\left(\omega_0\,t-\phi_1\right),$$ (3.30)

$$\eta_2(t) =\hat{\eta}_2\,\cos\left(\sqrt{3}\,\omega_0\,t-\phi_2\right).$$ (3.31)

Thus, in general, the two normal coordinates oscillate sinusoidally with unique frequencies, unlike the physical coordinates, $x_1(t)$ and $x_2(t)$. See Figures 3.2 and 3.3. This suggests that the equations of motion of the system should look particularly simple when expressed in terms of the normal coordinates. In fact, it can be seen that the sum of Equations (3.3) and (3.4) reduces to

$$\ddot{\eta}_1 = -\omega_0^{\,2}\,\eta_1,$$ (3.32)

whereas the difference gives

$$\ddot{\eta}_2 = -3\,\omega_0^{\,2}\,\eta_2.$$ (3.33)

Thus, when expressed in terms of the normal coordinates, the equations of motion of the system reduce to two uncoupled simple harmonic oscillator equations. The most general solution to Equation (3.32) is (3.30), whereas the most general solution to Equation (3.33) is (3.31). Hence, if we can guess the normal coordinates of a coupled oscillatory system then the determination of the normal modes of oscillation is considerably simplified.

Note that, in general, the normal coordinates of a multiple-degree-of-freedom oscillatory system are linear combinations of the physical coordinates. By definition, the equations of motion of the system become decoupled from one another when expressed in terms of the normal coordinates. In fact, each equation of motion takes the form of a simple harmonic oscillator equation whose characteristic frequency is one of the normal frequencies. Consequently, each normal coordinate executes simple harmonic oscillation at its associated normal frequency.

Figure 3.3: Time evolution of the normal coordinates of the two-degree-of-freedom mass-spring system pictured in Figure 3.1. The initial conditions are $x_1=a$, $\stackrel{\,\,\mbox{\large .}}{x}_1 = x_2 =\, \stackrel{\,\mbox{\large .}}{x}_2=0$. The solid curve corresponds to $\eta _1$, and the dashed curve to $\eta _2$. Here, $T_0=2\pi /\omega _0$.