Suppose the middle spring is very weak compared to the outer two, $l\ll k$. This condition, called weak coupling, affects $\text{Ω}$ but not $\omega$. We can write

We see that $\text{Ω}$ is only very slightly larger than $\omega$ if $l\ll k$.

Let’s define the average frequency ${\omega }_0$ and the difference frequency $\epsilon$ as

The weak-coupling condition, for which $\text{Ω}$ is only slightly larger than $\omega$, means that ${\omega }_0$ is almost exactly the same as $\omega$ while $\epsilon$ is close to zero. If we invert these definitions, the two frequencies that characterize the solutions u and v are

Suppose we establish initial conditions with $A=B=C/2$ and ${\varphi }_{\text{a}}={\varphi }_b=0$. If we use the trigonometric identity $\text{cos}(\alpha \pm \beta )=\text{cos}\alpha co\text{s}\beta \mp \text{s}\text{i}\text{n}\alpha \text{s}\text{i}\text{n}\beta$, we can write Equations 15.67 as

It looks like we’ve made things more complicated, but if we now solve for $x_1$ and $x_2$, as before, we find

The factors of 2 we included in the amplitudes of Equations 15.67 have allowed us to avoid needing factors of $\frac{1}{2}$ in Equations 15.75. This solution has a simple interpretation, seen most easily in the graphs of FIGURE 15.30.

FIGURE 15.30 Weakly coupled oscillators transfer energy back and forth between the two masses.

Because ${\omega }_0\approx \omega$, the $\text{cos}{\omega }_{0}t$ and $\text{sin}{\omega }_{0}t$ terms indicate a fast oscillation at essentially the uncoupled angular frequency $\omega =\sqrt{k\text{/}m}$. In contrast, the terms $\text{cos}\text{ε}t$ and $\text{sin}\text{ε}t$ oscillate very slowly because, for weak coupling, $\text{ε}\approx 0$. We’ve grouped these terms with the amplitude C because, together, they describe a slowly oscillating amplitude. The terms in parentheses in Equations 15.75 are a slowly varying envelope of the rapid oscillations in exactly the same way that an exponentially decaying amplitude is the envelope of a damped oscillation, as you saw in Figure 15.22.

The initial conditions corresponding to Equations 15.75 are that mass 1 is displaced by distance C and then released, while mass 2 is at rest at its equilibrium position. For a while, the motion is almost entirely an oscillation of mass 1. But over time, due to the weak coupling of the middle spring, mass 2 begins to oscillate—at the same frequency—and the amplitude of mass 1 decreases. Eventually, when $\text{ε}t=\pi /2\text{rad,}$ mass 1 has temporarily stopped while mass 2 oscillates with amplitude C. Then the process reverses, with the oscillations returning to mass 1.

What we see is an energy transfer back and forth between the two masses and their attached springs. The kinetic energy of the system is passed back and forth between the masses as first one, then the other, has all or nearly all of the oscillatory motion. This system of two weakly coupled oscillators is an excellent model of more complex systems that exhibit the same pattern of energy transferred back and forth between different parts of the system. And, as you’ll learn in Chapter 17, this fast oscillation with a slowly oscillating amplitude is closely related to the phenomena of beats between two waves that have slightly different frequencies.

CHAPTER 15 CHALLENGE EXAMPLE A swinging pendulum

A pendulum consists of a massless, rigid rod with a mass at one end. The other end is pivoted on a frictionless pivot so that the rod can rotate in a complete circle. The pendulum is inverted, with the mass directly above the pivot point, then released. The speed of the mass as it passes through the lowest point is $5.0\text{m}\text{/}\text{s}$. If the pendulum later undergoes small-amplitude oscillations at the bottom of the arc, what will its frequency be?

MODEL This is a simple pendulum because the rod is massless. However, our analysis of a pendulum used the small-angle approximation. It applies only to the small-amplitude oscillations at the end, not to the pendulum swinging down from the inverted position. Fortunately, energy is conserved throughout, so we can analyze the big swing using conservation of mechanical energy.

VISUALIZE FIGURE 15.31 is a pictorial representation of the pendulum swinging down from the inverted position. The pendulum length is $L$, so the initial height is $2L$.

FIGURE 15.31 Before-and-after pictorial representation of the pendulum swinging down from an inverted position.

FIGURE 15.31 Full Alternative Text

SOLVE The frequency of a simple pendulum is $f=\sqrt{g\text{/}L}\text{/}2\pi$. We’re not given $L$, but we can find it by analyzing the pendulum’s swing down from an inverted position. Mechanical energy is conserved, and the only potential energy is gravitational potential energy. Conservation of mechanical energy $K_1+U_{\text{G}1}=K_0+U_{\text{G}0}$, with $U_{\text{G}}=\text{m}gy$, is

$$\frac{1}{2}m{v}_1{}^2+mg{y}_1=\frac{1}{2}m{v}_0{}^2+mg{y}_0$$

The mass cancels, which is good since we don’t know it, and two terms are zero. Thus

$$\frac{1}{2}{v}_1{}^2=g\left(2L\right)=2gL$$

Solving for $L$, we find

$$L={\displaystyle \frac{v_1{}^2}{4g}}={\displaystyle \frac{(5.0\text{m}\text{/}\text{s}{)}^2}{4\left(9.80\text{m}\text{/}{\text{s}}^2\right)}}=0.638\text{m}$$

Now we can calculate the frequency:

$$f={\displaystyle \frac{1}{2\pi }}\sqrt{{\displaystyle \frac{g}{L}}}={\displaystyle \frac{1}{2\pi }}\sqrt{{\displaystyle \frac{9.80\text{m}\text{/}{\text{s}}^2}{0.638\text{m}}}}=0.62\text{Hz}$$

REVIEW The frequency corresponds to a period of about $1.5\text{s}$, which seems reasonable.