Two friends play the same note on two harmoniums at a bhajan evening. One harmonium is a little out of tune. The note does not sound steady. It swells and fades, wah, wah, wah, a few times a second. When one friend presses a slightly different key, the throbbing becomes faster. Neither harmonium throbs when it plays alone.
Interference in time
When we superpose two waves of the same frequency, we get a stable interference pattern in space. The chapter Interference and Reflection of Sound showed such a pattern: loud places and quiet places that stay where they are. Now superpose two waves with slightly different frequencies. The result is a pattern that changes in time.
At one point in space, the amplitude of the sound grows and fades in a steady rhythm. This periodic variation in the intensity of sound is called beats. The beats are the throbbing of the two harmoniums.
When two sound waves of slightly different frequencies are superposed at a point, the loudness at that point rises and falls periodically. This periodic variation in the intensity of the sound is called beats.
Two runners on a circular track
Imagine two runners on a circular track, one slightly faster than the other. At the start they run side by side. This is the picture of two waves in phase, which gives a loud sound. The faster runner slowly pulls ahead of the slower one.
After a while, the two runners are on opposite sides of the track. This is the picture of two waves out of phase, which gives a quiet sound. When the faster runner has lapped the slower one, they are side by side again, and the sound is loud again. The rate at which they lap each other corresponds to the beat frequency.
Read the lap counter of Fig 1.1 at the end of the run. In \(10\,\text{s}\), A has run \(4\) laps and B has run \(5\). B has gained exactly one lap, so the two meet once in \(10\,\text{s}\). The rate of meeting is \(0.5 - 0.4 = 0.1\) per second. It is the difference of the two rates, not either rate alone.
The phase difference grows with time
Two sound waves of slightly different frequencies behave in the same way. At one point, they produce two disturbances whose phase difference depends on time. So at some instants the two waves are in phase, and at other instants they are out of phase. Fig 1.2 shows this.
In Fig 1.2, \(s_1\) completes \(10\) cycles across the figure and \(s_2\) completes \(12\). At the left edge, the middle and the right edge, a crest meets a crest. At one quarter and three quarters of the width, a crest meets a trough. So the second wave gains one full cycle on the first between each pair of in-phase instants.
A tuning fork of \(440\,\text{Hz}\) and another of \(441\,\text{Hz}\) are sounded together. What does a listener hear?
The mathematics of beats
Let us represent two sound waves of equal amplitude \(A\) and slightly different frequencies, \(f\) and \(f + \Delta f\). Here \(\Delta f\) is the small difference between the two frequencies. At a fixed point in space, the displacements of the two waves can be written as
\[ s_1 = A\cos(2\pi f t) \quad \text{and} \quad s_2 = A\cos\big(2\pi (f + \Delta f) t\big) \]
By the principle of superposition, the resultant displacement is \(s = s_1 + s_2\). The derivation below turns this sum into a product.
\[ s = A\cos(2\pi f t) + A\cos\big(2\pi (f + \Delta f) t\big) \]
Why. the principle of superposition: at one point the displacements of the two waves add.
\[ \cos a + \cos b = 2\cos\left(\dfrac{a - b}{2}\right)\cos\left(\dfrac{a + b}{2}\right) \]
Why. the sum-to-product identity of trigonometry. Take \(a = 2\pi (f + \Delta f) t\) and \(b = 2\pi f t\).
\[ \dfrac{a - b}{2} = 2\pi \dfrac{\Delta f}{2} t, \qquad \dfrac{a + b}{2} = 2\pi\left(f + \dfrac{\Delta f}{2}\right) t \]
Why. half the difference and half the sum of the two phases.
\[ s(t) = \left[2A\cos\left(2\pi \dfrac{\Delta f}{2} t\right)\right] \cos\left(2\pi\left(f + \dfrac{\Delta f}{2}\right) t\right) \]
Read it. put the two half-angles into the identity and multiply by the common amplitude \(A\).
Visualizing the resultant wave
The equation we derived describes the wave in Fig 1.3. It can be read as a high-frequency wave whose amplitude is changed slowly by a much slower wave.
The equation has two parts. Each part does a different job:
- Rapid oscillation. The term \(\cos\big(2\pi (f + \Delta f/2) t\big)\) is the fast vibration at the average frequency. This is the main pitch you perceive.
- Amplitude envelope. The term in the brackets, \(\big[2A\cos(2\pi \tfrac{\Delta f}{2} t)\big]\), acts as a slowly changing amplitude. It creates the loud-soft-loud pulsing. This term is called the amplitude envelope.
Note that \(2\pi \tfrac{\Delta f}{2} t\) is the same as \(\pi\,\Delta f\,t\). So the envelope in Fig 1.3 is written \(\pm 2A\cos(\pi\,\Delta f\,t)\). Both forms are the same function.
Two waves of equal amplitude have frequencies \(256\,\text{Hz}\) and \(260\,\text{Hz}\). What is the frequency of the cosine inside the amplitude envelope?
Perceived loudness and beat frequency
The human ear perceives loudness from the intensity of the sound. The intensity is proportional to the square of the amplitude. So we hear the greatest loudness when the amplitude envelope is at its maximum, positive or negative. A beat is one cycle of loudness, from maximum to minimum to maximum. It occurs as the intensity rises and falls.
Now count. The amplitude term \(\cos(2\pi \tfrac{\Delta f}{2} t)\) goes through one full cycle in a time \(T_{\text{env}} = 1/(\Delta f/2)\). But the ear hears a loud moment when the amplitude is greatest, that is, when the cosine is \(+1\) or \(-1\). This happens twice in each cycle of the envelope, as the humps of Fig 1.4 show. Therefore the frequency of the beats is twice the frequency of the envelope.
\[ f_{\text{envelope}} = \dfrac{\Delta f}{2} \]
Why. the envelope is \(2A\cos(2\pi \tfrac{\Delta f}{2} t)\), a cosine of frequency \(\Delta f/2\).
\[ f_{\text{beat}} = 2 \times f_{\text{envelope}} \]
Why. the loudness follows the square of the envelope, which peaks at \(+2A\) and again at \(-2A\) in every envelope cycle.
\[ f_{\text{beat}} = 2 \times \dfrac{\Delta f}{2} = \Delta f \]
Read it. substitute the envelope frequency.
These are the two key formulas for beats. The main tone you hear has a frequency equal to the average of the two source frequencies. The frequency of the pulsations of loudness is the absolute difference of the two source frequencies. We write the difference with a modulus, \(|f_1 - f_2|\), because a number of beats per second is never negative.
The envelope has frequency \(\Delta f/2\), but the beat frequency is \(\Delta f\). Do not halve the difference. Loudness does not care about the sign of the amplitude, so a crest of the envelope at \(-2A\) is as loud as one at \(+2A\).
Two tuning forks of \(256\,\text{Hz}\) and \(260\,\text{Hz}\) are sounded together. How many beats does a listener hear in one second?
Tones of \(440\,\text{Hz}\) and \(444\,\text{Hz}\) of equal amplitude sound together. What pitch does the listener hear inside the beats?
A \(512\,\text{Hz}\) fork and a \(508\,\text{Hz}\) fork are sounded together. What is the time between two moments of greatest loudness, in seconds?
The amplitude envelope \(2A\cos(2\pi \tfrac{\Delta f}{2} t)\) has a frequency of \(\Delta f/2\). Why does the ear hear \(\Delta f\) beats per second and not \(\Delta f/2\)?
The ear responds to intensity, which is proportional to the square of the amplitude. In one cycle of the envelope the amplitude reaches \(+2A\) once and \(-2A\) once. Both give the same greatest loudness. So there are two loud moments per envelope cycle, and the beat frequency is \(2 \times \Delta f/2 = \Delta f\).
Around 1700, the French mathematician Joseph Sauveur wanted the actual number of vibrations per second of a musical note. No instrument could count so fast. Sauveur sounded two organ pipes whose pitches differed by a known musical interval, so the ratio of their frequencies was known. He counted the slow beats per second, which gave the difference of the two frequencies. The ratio and the difference together fixed each frequency. Sauveur also gave the science of sound its name, acoustics.
Now return to the two harmoniums. Their two reeds sound nearly equal frequencies. At the ear the two waves drift in and out of phase, so the loudness swells and fades. The number of throbs in one second is the difference of the two frequencies. When one friend presses a key further from the other's note, the difference grows, and the throbs come faster.
- Taking the beat frequency as half the difference, because the envelope has frequency \(\Delta f/2\).The loudness peaks at \(+2A\) and at \(-2A\), twice per envelope cycle. So \(f_{\text{beat}} = \Delta f\).
- Taking the pitch heard as one of the two source frequencies.The fast factor of the resultant vibrates at the average, \((f_1 + f_2)/2\).
- Expecting beats from two very different frequencies, such as \(440\,\text{Hz}\) and \(660\,\text{Hz}\).Beats are heard only when \(\Delta f\) is small, a few per second. A large difference is heard as two notes.