Two Spring-Coupled Masses

Consider a mechanical system consisting of two identical masses $m$ which 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 14. 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

$$m\,\ddot{x}_1 = -k\,x_1 +k\,(x_2-x_1),$$ (142)

$$m\,\ddot{x}_2 = -k\,(x_2-x_1) +k\,(-x_2).$$ (143)

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 14: Two degree of freedom mass-spring system.

Equations (142)-(143) can be rewritten in the form

$$\ddot{x}_1 = -2\,\omega_0^{\,2}\,x_1+ \omega_0^{\,2}\,x_2,$$ (144)

$$\ddot{x}_2 = \omega_0^{\,2}\,x_1 - 2\,\omega_0^{\,2}\,x_2,$$ (145)

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

$$x_1(t) = \hat{x}_1\,\cos(\omega\,t-\phi),$$ (146)

$$x_2(t) = \hat{x}_2\,\cos(\omega\,t-\phi),$$ (147)

where $\hat{x}_1$, $\hat{x}_2$, and $\phi$ are constants. Equations (144) and (145) yield

$$-\omega^2\,\hat{x}_1\,\cos(\omega\,t-\phi) = \left( -2\,\omega_0^{\,2}\,\hat{x}_1+ \omega_0^{\,2}\,\hat{x}_2\right)\,\cos(\omega\,t-\phi),$$ (148)

$$-\omega^2\,\hat{x}_2\,\cos(\omega\,t-\phi) = \left( \omega_0^{\,2}\,\hat{x}_1-2\,\omega_0^{\,2}\,\hat{x}_2\right)\,\cos(\omega\,t-\phi),$$ (149)

or

$$(\hat{\omega}^{\,2}-2)\,\hat{x}_1 +\hat{x}_2 = 0,$$ (150)

$$\hat{x}_1 + (\hat{\omega}^{\,2}-2)\,\hat{x}_2 = 0,$$ (151)

where $\hat{\omega}=\omega/\omega_0$. Note that by searching for a solution of the form (146)-(147) we have effectively converted the system of two coupled linear differential equations (144)-(145) into the much simpler system of two coupled linear algebraic equations (150)-(151). The latter equations have the trivial solutions $\hat{x}_1=\hat{x}_2=0$, but also yield

$$\frac{\hat{x}_1}{\hat{x}_2}=-\frac{1}{(\hat{\omega}^{\,2}-2)} =-( \hat{\omega}^{\,2}-2).$$ (152)

Hence, the condition for a nontrivial solution is

$$(\hat{\omega}^{\,2}-2)\,(\hat{\omega}^{\,2}-2)-1 = 0.$$ (153)

In fact, if we write Equations (150)-(151) in the form of a homogenous (i.e., with a null right-hand side) $2\times 2$ matrix equation, so that

$$\left( \begin{array}{cc} \hat{\omega}^{\,2}-2 & 1\\ [0.5ex] ... ...) = \left( \begin{array}{c} 0\\ [0.5ex] 0 \end{array}\right) ,$$ (154)

then it becomes apparent that the criterion (153) can also be obtained by setting the determinant of the associated $2\times 2$ matrix to zero.

Equation (153) can be rewritten

$$\hat{\omega}^{\,4} - 4\,\hat{\omega}^2+3 = (\hat{\omega}^{\,2}-1)\,(\hat{\omega}^{\,2}-3)=0.$$ (155)

It follows that

$$\hat{\omega} = 1 \mbox{~or~~} \sqrt{3}.$$ (156)

Here, we have neglected the two negative frequency roots of (155)--i.e., $\hat{\omega}=-1$ and $\hat{\omega}=-\sqrt{3}$--since a negative frequency oscillation is equivalent to an oscillation with an equal and opposite positive frequency, and an equal and opposite phase: i.e., $\cos(\omega\,t-\phi)\equiv \cos(-\omega\,t+\phi)$. It is thus apparent that the dynamical system pictured in Figure 142 has two unique frequencies of oscillation: i.e., $\omega=\omega_0$ and $\omega=\sqrt{3}\,\omega_0$. These are called the normal frequencies of the system. Since 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 easily be deduced from Equation (152). Thus, for $\omega=\omega_0$ (i.e., $\hat{\omega}=1$), we get $\hat{x}_1=\hat{x}_2$, so that

$$x_1(t) = \hat{\eta}_1\,\cos(\omega_0\,t-\phi_1),$$ (157)

$$x_2(t) = \hat{\eta}_1\,\cos(\omega_0\,t-\phi_1),$$ (158)

where $\hat{\eta}_1$ and $\phi_1$ are constants. This first pattern of motion corresponds to the two masses executing simple harmonic oscillation with the same amplitude and phase. Note that 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

$$x_1(t) = \hat{\eta}_2\,\cos\left(\sqrt{3}\,\omega_0\,t-\phi_2\right),$$ (159)

$$x_2(t) = -\hat{\eta}_2\,\cos\left(\sqrt{3}\,\omega_0\,t-\phi_2\right),$$ (160)

where $\hat{\eta}_2$ and $\phi_2$ are constants. This second pattern of motion corresponds to the two masses executing simple harmonic oscillation with the same amplitude but in anti-phase: i.e., 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 which 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.

Now, the most general motion of the system is a linear combination of the two normal modes. This immediately follows because Equations (142) and (143) are linear equations. [In other words, if $x_1(t)$ and $x_2(t)$ are solutions then so are $a\,x_1(t)$ and $a\,x_2(t)$, where $a$ is an arbitrary constant.] Thus, we can write

$$x_1(t) = \hat{\eta}_1\,\cos(\omega_0\,t-\phi_1)+ \hat{\eta}_2\,\cos\left(\sqrt{3}\,\omega_0\,t-\phi_2\right),$$ (161)

$$x_2(t) = \hat{\eta}_1\,\cos(\omega_0\,t-\phi_1)-\hat{ \eta}_2\,\cos\left(\sqrt{3}\,\omega_0\,t-\phi_2\right).$$ (162)

Note that we can be sure that this represents the most general solution to Equations (142) and (143) because it contains four arbitrary constants: i.e., $\hat{\eta}_1$, $\phi_1$, $\hat{\eta}_2$, 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 ordinary differential equations should contain four arbitrary constants.) Of course, 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 (161) and (162), that

$$a = \hat{\eta}_1\,\cos\phi_1 + \hat{\eta}_2\,\cos\phi_2,$$ (163)

$$0 = \hat{\eta}_1\,\sin\phi_1+\sqrt{3}\,\hat{\eta}_2\,\sin\phi_2,$$ (164)

$$0 = \hat{\eta}_1\,\cos\phi_1 -\hat{\eta}_2\,\cos\phi_2,$$ (165)

$$0 = \hat{\eta}_1\,\sin\phi_1-\sqrt{3}\,\hat{\eta}_2\,\sin\phi_2,$$ (166)

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

$$x_1(t) = a\,\cos(\omega_-\,t)\,\cos(\omega_+\,t),$$ (167)

$$x_2(t) = a\,\sin(\omega_-\,t)\,\sin(\omega_+\,t),$$ (168)

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 $\cos a - \cos b \equiv -2\,\sin[(a+b)/2]\,\sin[(a-b)/2]$. This evolution is illustrated in Figure 15. [Here, $T_0=2\pi/\omega_0$. The solid curve corresponds to $x_1$, and the dashed curve to $x_2$.]

Figure 15: Coupled oscillations in a two degree of freedom mass-spring system.

Finally, let us define the so-called normal coordinates,

$$\eta_1(t) = [x_1(t) + x_2(t)]/2,$$ (169)

$$\eta_2(t) = [x_1(t)-x_2(t)]/2.$$ (170)

It follows from (161) and (162) that, in the presence of both normal modes,

$$\eta_1(t) = \hat{\eta}_1\,\cos(\omega_0\,t-\phi_1),$$ (171)

$$\eta_2(t) = \hat{\eta}_2\,\cos(\sqrt{3}\,\omega_0\,t-\phi_2).$$ (172)

Thus, in general, the two normal coordinates oscillate sinusoidally with unique frequencies, unlike the regular coordinates, $x_1(t)$ and $x_2(t)$--see Figure 15. 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 is easily seen that the sum of Equations (144) and (145) reduces to

$$\ddot{\eta}_1 = -\omega_0^{\,2}\,\eta_1,$$ (173)

whereas the difference gives

$$\ddot{\eta}_2 = -3\,\omega_0^{\,2}\,\eta_2.$$ (174)

Thus, when expressed in terms of the normal coordinates, the equations of motion of the system reduce to two uncoupled simple harmonic oscillator equations. Of course, the most general solution to Equation (173) is (171), whereas the most general solution to Equation (174) is (172). 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.