A musician tunes a sitar string against a tuning fork of \(256\,\text{Hz}\). With the string a little slack, the two notes together do not sound steady. The sound swells and fades, wah, wah, wah, a few times a second. As the musician tightens the string, the throbbing slows. When the string is exactly in tune, the throbbing stops and one steady note remains.
Two arms that turn at nearly the same rate
Give each note its phasor. The two arms now turn at slightly different rates, \(2\pi f_1\) and \(2\pi f_2\). The angle between them is no longer fixed. It grows steadily, by \(2\pi(f_2 - f_1)\) in each second. So the triangle of the two arms and their resultant keeps changing shape. When the arms line up, the resultant has its largest length \(a_1 + a_2\). When they point opposite ways, it has its smallest length \(|a_1 - a_2|\). For two equal arms the resultant swings from \(2A\) down to zero.
Now watch the two arms from a platform that turns with the first arm. On this platform the first arm stands still. The second arm turns slowly, at the difference of the two rates, \(\Delta f\) turns in each second. Every time it comes round to line up with the first arm, the sum is loud. Every time it points the opposite way, the sum is quiet.
When two sinusoids of slightly different frequencies are added, the amplitude of the sum rises and falls periodically. For sound, the loudness rises and falls. These periodic rises and falls are called beats, and the number of them in one second is the beat frequency.
The sum as a product
The phasor picture shows why beats happen. The sum-to-product identity writes the sum exactly. The derivation proves the identity from the two sum formulas of trigonometry, and then uses it.
\[ \cos(p + q) = \cos p\cos q - \sin p\sin q \]
Why. the formula for the cosine of a sum.
\[ \cos(p - q) = \cos p\cos q + \sin p\sin q \]
Why. the same formula with \(q\) replaced by \(-q\).
\[ \cos(p + q) + \cos(p - q) = 2\cos p\cos q \]
Why. add steps 1 and 2; the sine terms cancel.
\[ \cos a + \cos b = 2\cos\dfrac{a + b}{2}\cos\dfrac{a - b}{2} \]
Why. put \(p + q = a\) and \(p - q = b\), so \(p = (a + b)/2\) and \(q = (a - b)/2\).
\[ y = 2A\cos(\pi\,\Delta f\,t)\cos(2\pi\bar{f}\,t) \]
Read it. put \(a = 2\pi f_1t\) and \(b = 2\pi f_2t\), multiply by \(A\), and use that the cosine is even.
The product has two factors, and each does its own job. The factor \(\cos(2\pi\bar{f}t)\) is a fast oscillation at the mean frequency. For sound, this is the pitch you hear. The factor \(2A\cos(\pi\Delta f\,t)\) changes slowly and acts as a changing amplitude. Its outline, drawn dashed in Fig 4.2, is called the envelope.
Why the beat frequency is the whole difference
The envelope \(2A\cos(\pi\Delta f\,t)\) repeats after a time \(2/\Delta f\). So the envelope itself has the frequency \(\Delta f/2\). The ear, however, cannot hear the sign of the envelope. A large negative amplitude is just as loud as a large positive one. In one cycle the envelope reaches its largest size twice, once at \(+2A\) and once at \(-2A\). So the loudness peaks twice per cycle of the envelope, and the beat frequency is \(2 \times \Delta f/2 = \Delta f\).
The phasor picture of Fig 4.1 gives the same answer more directly. The second arm laps the first \(\Delta f\) times in each second, and each lap brings one moment of lining up. One lining up is one loud moment. So the loud moments come \(\Delta f\) times a second.
The envelope reaches zero only when the two amplitudes are equal. With unequal amplitudes the loudness swings between \((a_1 + a_2)^2\) and \((a_1 - a_2)^2\) and never falls silent. The picture of one tone that swells and fades also needs \(\Delta f\) to be small. Up to about \(10\) beats a second the ear counts them. At larger differences the sound becomes rough, and further on the ear hears two separate notes.
Two tuning forks of \(256\,\text{Hz}\) and \(260\,\text{Hz}\) are sounded together. How many times a second does the loudness reach its peak?
Two sources of \(440\,\text{Hz}\) and \(444\,\text{Hz}\) sound together. What is the frequency of the tone heard, the pitch of the beating sound, in Hz?
Two notes of \(300\,\text{Hz}\) and \(303\,\text{Hz}\) are drawn as phasors, and we watch them from a frame that turns with the first. How does the second phasor move in that frame?
Beats as a tuning tool
A beat frequency is the size of a difference, \(|f_1 - f_2|\). So a known frequency and a counted beat rate leave two candidates for the unknown one. A second observation decides between them. Wax on the prongs of a fork adds mass and lowers its frequency. Tightening a string raises its frequency. If the change brings the unknown nearer the known frequency, the beats slow down.
This is the worked example of Beats and the Doppler Effect, lesson 2, done with the tool. A tuning fork A of frequency \(256\,\text{Hz}\) gives \(4\) beats per second with a fork B of unknown frequency. When a little wax is put on the prongs of B, the beats fall to \(2\) per second. Find the frequency of B before it was waxed.
Known: \(f_A = 256\,\text{Hz}\), \(f_{beat} = 4\,\text{Hz}\) before waxing, and \(2\,\text{Hz}\) after. Waxing adds mass to B, so \(f_B\) decreases.
From \(|f_A - f_B| = 4\), the two candidates are \(f_B = 256 + 4 = 260\,\text{Hz}\) and \(f_B = 256 - 4 = 252\,\text{Hz}\).
If \(f_B = 260\,\text{Hz}\), a slight lowering brings B closer to \(256\,\text{Hz}\), and the beats fall. If \(f_B = 252\,\text{Hz}\), a slight lowering takes B further away, and the beats rise. The beats fell.
So \(f_B = 260\,\text{Hz}\) before waxing. After waxing it is about \(258\,\text{Hz}\), which gives the observed \(2\) beats per second.
A sitar string sounded with a \(384\,\text{Hz}\) fork gives \(3\) beats per second. When the string is tightened slightly, the beats fall to \(2\) per second. What was the frequency of the string before it was tightened?
Two sinusoids of nearby frequencies \(f_1\) and \(f_2\) are added. Which statements are true?
Beats and the Doppler Effect uses this identity for beats as interference in time, the beat frequency and the tuning of forks and strings. The same product appears in Superposition, Damped and Forced Oscillations, where two SHMs of slightly different periods drift in and out of step. Radio engineers use the same product when they mix two nearby frequencies to get a low difference frequency.
- Taking the envelope's frequency, \(\Delta f/2\), as the beat frequency.The loudness peaks twice in every cycle of the envelope. The beat frequency is \(\Delta f\).
- Giving one answer for an unknown frequency from one beat count.A beat count gives \(|f_1 - f_2|\). Write both candidates, then use the wax or the tension to decide.
- Expecting silence between beats when the two sounds are unequal.The quietest sound has the amplitude \(|a_1 - a_2|\). It is silent only for equal amplitudes.