Thus far we have looked at the superposition of waves that have the same wavelength and frequency. We can also use the principle of superposition to investigate a phenomenon that is easily demonstrated with two sources of slightly different frequency.

If you listen to two sounds with very different frequencies, such as a high note and a low note, you hear two distinct tones. But if the frequency difference is very small, just one or two hertz, then you hear a single tone whose intensity is modulated once or twice every second. That is, the sound goes up and down in volume, loud, soft, loud, soft, …, making a distinctive sound pattern called beats.

Consider two sinusoidal waves traveling along the x-axis with angular frequencies ${\omega }_1=2\pi f_1$ and ${\omega }_2=2\pi f_2$. The two waves are

where the subscripts 1 and 2 indicate that the frequencies, wave numbers, and phase constants of the two waves may be different.

To simplify the analysis, let’s make several assumptions:

1. The two waves have the same amplitude a,

2. A detector, such as your ear, is located at the origin $(x=0)$,

3. The two sources are in phase $({\varphi }_{10}={\varphi }_{20})$, and

4. The source phases happen to be ${\varphi }_{10}={\varphi }_{20}=\pi \text{rad}$.

None of these assumptions is essential to the outcome. All could be otherwise and we would still come to basically the same conclusion, but the mathematics would be far messier. Making these assumptions allows us to emphasize the physics with the least amount of mathematics.

With these assumptions, the two waves as they reach the detector at $x=0$ are

where we’ve used the trigonometric identity $\text{sin}(\pi -\theta )=\text{sin}\theta$. The principle of superposition tells us that the net displacement of the medium at the detector is the sum of the displacements of the individual waves. Thus

Earlier, for interference, we used the trigonometric identity

We can use this identity again to write Equation 17.41 as

where ${\omega }_{\text{avg}}=\frac{1}{2}({\omega }_1+{\omega }_2)$ is the average angular frequency and ${\omega }_{\text{mod}}=\frac{1}{2}|{\omega }_1-{\omega }_2|$ is called the modulation frequency. We’ve used the absolute value because the modulation depends only on the frequency difference between the sources, not on which has the larger frequency.

We are interested in the situation when the two frequencies are very nearly equal: ${\omega }_1\approx {\omega }_2$. In that case, ${\omega }_{\text{avg}}$ hardly differs from either ${\omega }_1$ or ${\omega }_2$ while ${\omega }_{\text{mod}}$ is very near to—but not exactly—zero. When ${\omega }_{\text{mod}}$ is very small, the term $\text{cos}({\omega }_{\text{mod}}t)$ oscillates very slowly. We have grouped it with the 2a term because, together, they provide a slowly changing “amplitude” for the rapid oscillation at frequency ${\omega }_{\text{avg}}$.

FIGURE 17.29 is a history graph of the wave at the detector $(x=0)$. It shows the oscillation of the air against your eardrum at frequency $f_{\text{avg}}={\omega }_{\text{avg}}\text{/}2\pi =\frac{1}{2}(f_1+f_2)$. This oscillation determines the note you hear; it differs little from the two notes at frequencies $f_1$ and $f_2$. We are especially interested in the time-dependent amplitude, shown as a dashed line, that is given by the term $2a\text{cos}\left({\omega }_{\text{mod}}t\right)$. This periodically varying amplitude is called a modulation of the wave, which is where ${\omega }_{\text{mod}}$ gets its name.

FIGURE 17.29 Beats are caused by the superposition of two waves of nearly identical frequency.

As the amplitude rises and falls, the sound alternates as loud, soft, loud, soft, and so on. But that is exactly what you hear when you listen to beats! The alternating loud and soft sounds arise from the two waves being alternately in phase and out of phase, causing constructive and then destructive interference.

Imagine two people walking side by side at just slightly different paces. Initially both of their right feet hit the ground together, but after a while they get out of step. A little bit later they are back in step and the process alternates. The sound waves are doing the same. Initially the crests of each wave, of amplitude a, arrive together at your ear and the net displacement is doubled to 2a. But after a while the two waves, being of slightly different frequency, get out of step and a crest of one arrives with a trough of the other. When this happens, the two waves cancel each other to give a net displacement of zero. This process alternates over and over, loud and soft.

Notice, in Figure 17.29, that the sound intensity rises and falls twice during one cycle of the modulation envelope. Each “loud-soft-loud” is one beat, so the beat frequency $f_{\text{beat}}$, which is the number of beats per second, is twice the modulation frequency $f_{\text{mod}}={\omega }_{\text{mod}}\text{/}2\pi$. From the above definition of ${\omega }_{\text{mod}}$, the beat frequency is

The beat frequency is simply the difference between the two individual frequencies.

EXAMPLE 17.11 Detecting bats with beats

The little brown bat is a common species in North America. It emits echolocation pulses at a frequency of 40 kHz, well above the range of human hearing. To allow researchers to “hear” these bats, the bat detector shown in FIGURE 17.30 combines the bat’s sound wave at frequency $f_1$ with a wave of frequency $f_2$ from a tunable oscillator. The resulting beat frequency is then amplified and sent to a loudspeaker. To what frequency should the tunable oscillator be set to produce an audible beat frequency of 3 kHz?

FIGURE 17.30 The operation of a bat detector.

FIGURE 17.30 Full Alternative Text

SOLVE Combining two waves with different frequencies gives a beat frequency

$$f_{\text{beat}}=|f_1-f_2|$$

A beat frequency will be generated at 3 kHz if the oscillator frequency and the bat frequency differ by 3 kHz. An oscillator frequency of either 37 kHz or 43 kHz will work nicely.

REVIEW The electronic circuitry of radios, televisions, and cell phones makes extensive use of mixers to generate difference frequencies.

Beats aren’t limited to sound waves. FIGURE 17.31 shows a graphical example of beats. Two “fences” of slightly different frequencies are superimposed on each other. The difference in the two frequencies is two lines per inch. You can confirm, with a ruler, that the figure has two “beats” per inch, in agreement with Equation 17.43.

FIGURE 17.31 A graphical example of beats.

Beats are important in many other situations. For example, you have probably seen movies where rotating wheels seem to turn slowly backward. Why is this? Suppose the movie camera is shooting at 30 frames per second but the wheel is rotating 32 times per second. The combination of the two produces a “beat” of 2 Hz, meaning that the wheel appears to rotate only twice per second. The same is true if the wheel is rotating 28 times per second, but in this case, where the wheel frequency slightly lags the camera frequency, it appears to rotate backward twice per second!

CHAPTER 17 CHALLENGE EXAMPLE An airplane landing system

Your firm has been hired to design a system that allows airplane pilots to make instrument landings in rain or fog. You’ve decided to place two radio transmitters 50 m apart on either side of the runway. These two transmitters will broadcast the same frequency, but out of phase with each other. This will cause a nodal line to extend straight off the end of the runway. As long as the airplane’s receiver is silent, the pilot knows she’s directly in line with the runway. If she drifts to one side or the other, the radio will pick up a signal and sound a warning beep. To have sufficient accuracy, the first intensity maxima need to be 60 m on either side of the nodal line at a distance of 3.0 km. What frequency should you specify for the transmitters?

MODEL The two transmitters are sources of out-of-phase, circular waves. The overlap of these waves produces an interference pattern.

VISUALIZE For out-of-phase sources, the center line—with zero path-length difference—is a nodal line of maximum destructive interference because the two signals always arrive out of phase. FIGURE 17.32 shows the nodal line, extending straight off the runway, and the first antinodal line—the points of maximum constructive interference—on either side. Comparing this to Figure 17.27, where the two sources were in phase, you can see that the nodal and antinodal lines have been reversed.

FIGURE 17.32 Pictorial representation of the landing system.

FIGURE 17.32 Full Alternative Text

SOLVE Point P, 60 m to the side at a distance of 3000 m, needs to be a point of maximum constructive interference. The distances are

$$\begin{array}{c}{r}_1=\sqrt{(3000\text{m}{)}^2+{(60\text{m}-25\text{m})}^2}=3000.204\text{m}\hfill \\ r_2=\sqrt{(3000\text{m}{)}^2+{(60\text{m}+25\text{m})}^2}=3001.204\text{m}\hfill \end{array}$$

We needed to keep several extra significant figures because we’re looking for the difference between two numbers that are almost the same. The path-length difference at P is

$$\text{Δ}r=r_2-r_1=1.000\text{m}$$

We know, for out-of-phase transmitters, that the phase difference of the sources is $\text{Δ}{\varphi }_0=\pi \text{rad}$. The first maximum will occur where the phase difference between the waves is $\text{Δ}\varphi =1\cdot 2\pi \text{rad}$. Thus the condition that we must satisfy at P is

$$\text{Δ}\varphi =2\pi \text{rad}=2\pi {\displaystyle \frac{\text{Δ}r}{\text{λ}}}+\pi \text{rad}$$

Solving for $\text{λ}$, we find

$$\text{λ}=2\text{Δ}r=2.00\text{m}$$

Consequently, the required frequency is

$$f=\frac{c}{\text{λ}}=\frac{3.00\times {10}^8\text{ m}\text{/}\text{s}}{2.00\text{ m}}=1.50\times {10}^8\text{ Hz}=150\text{ MHz}$$

REVIEW 150 MHz is slightly higher than the frequencies of FM radio $(\approx 100\text{MHz})$ but is well within the radio frequency range. Notice that the condition to be satisfied at P is that the path-length difference must be $\frac{1}{2}\text{λ}$. This makes sense. A path-length difference of ${\displaystyle \frac{1}{2}}\text{λ}$ contributes $\pi \text{rad}$ to the phase difference. When combined with the $\pi \text{rad}$ from the out-of-phase sources, the total phase difference of $2\pi \text{rad}$ creates constructive interference.