Beats & Interference

A piano tuner sits down with one tuning fork and no electronics. She strikes the fork, plays the string, and listens. What she hears is not two notes. It is one note that throbs: loud, soft, loud, soft, a couple of times a second. She turns the pin a little. The throbbing slows. She turns it further until the throbbing stops, and the string is in tune to better than a tenth of a hertz.

She is using the same physics you met in the last lesson. Two waves arrive at one place, and their displacements add. In standing waves the two waves had the same frequency and travelled opposite ways, and the result was fixed in space: nodes that never move. Here the two waves have slightly different frequencies, or they come from two places instead of one. Loosen either of those conditions and the pattern starts to move, or spreads out across a room. That is all beats and two-source interference are.

Two frequencies at one point

Put a microphone at one spot and let two tones reach it. Take them equal in amplitude and in step at $t = 0$, so the pressure at the microphone is

$y(t) = A\cos(2\pi f_1 t) + A\cos(2\pi f_2 t)$

Here $A$ is the amplitude each wave would have alone, $f_1$ and $f_2$ are the two frequencies in hertz, and $t$ is time in seconds.

Nothing about that sum looks like throbbing yet. To see the throb, rewrite it. The identity you need is the sum-to-product rule, $\cos X + \cos Y = 2\cos\!\left(\frac{X-Y}{2}\right)\cos\!\left(\frac{X+Y}{2}\right)$. Put $X = 2\pi f_1 t$ and $Y = 2\pi f_2 t$:

Read the two factors separately, because they do completely different jobs. The right-hand factor oscillates at the average frequency $(f_1+f_2)/2$. If the tones are 440 Hz and 443 Hz, that is 441.5 Hz: a fast wiggle, and it is the pitch you actually hear. The left-hand factor oscillates at $(f_1-f_2)/2$, which for those numbers is 1.5 Hz. That is far too slow to be a pitch. Instead it acts as a slowly changing amplitude for the fast wiggle: an envelope.

Now the step where nearly everyone loses a factor of two. The envelope has frequency $|f_1-f_2|/2$. But the ear does not hear the sign of the envelope; it hears loudness, which tracks $|{\cos}|$. A cosine reaches an extreme twice per cycle, once at $+1$ and once at $-1$, and both give maximum loudness. So the loudness cycles happen twice as often as the envelope does, and the beat frequency is

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

Not half of it. The absolute value matters too: a beat rate tells you the size of the gap between the frequencies, never which one is higher.

Using beats to tune

Beats are the most sensitive measurement in acoustics. Your ear cannot tell 440 Hz from 441 Hz played one after the other. Played together, they beat once a second, and you can count that in your sleep.

The catch is the absolute value. Three beats per second against a 440 Hz fork means the string is at 437 Hz or 443 Hz, and the sound alone will not say which. You need a second experiment. Tighten the string a little: tightening raises pitch, because a stretched string carries waves at $v=\sqrt{T/\mu}$ and its fundamental is $f = v/2L$, so $f \propto \sqrt{T}$. If the beats speed up, you were already sharp. If they slow down, you were flat.

That square root is worth working with numerically, because problems lean on it.

Example 1: two percent more tension. A string sounds exactly 440 Hz. Its tension is raised by 2.00%, then it is sounded against a 440 Hz fork. What beat rate do you hear?

Frequency goes as $\sqrt{T}$, so multiplying $T$ by 1.0200 multiplies $f$ by $\sqrt{1.0200} = 1.009950$. The new frequency is

$f' = 440 \times 1.009950 = 444.378\ \text{Hz}$

and the beat rate is $|444.378 - 440| = 4.378$ Hz. A 2% change in tension gave roughly a 1% change in frequency (the square root halves it) and 1% of 440 is about 4.4.

Run that backwards and you get the tuner's question. Suppose the string beats at 3.0 Hz against the fork and you know it is flat, so it sits at 437.0 Hz. To reach 440 you need the frequency to grow by a factor $440/437$, which means the tension must grow by the square of that:

$\frac{T'}{T} = \left(\frac{440}{437}\right)^{2} = (1.006865)^2 = 1.013777$

so tighten by 1.378%.

Beats do not care where the frequency difference comes from. Two organ pipes of slightly different length, two identical pipes in air at different temperatures, a string against a fork: the recipe is always the same. Work out each frequency on its own, to four or five figures, then subtract.

Example 2: two organ pipes. Open pipes of length 0.500 m and 0.510 m each sound their fundamental, $f = v/2L$, with $v = 343$ m/s. What beat frequency do you hear?

$f_1 = \frac{343}{2(0.500)} = \frac{343}{1.000} = 343.000\ \text{Hz}$

$f_2 = \frac{343}{2(0.510)} = \frac{343}{1.020} = 336.275\ \text{Hz}$

The beat frequency is $343.000 - 336.275 = 6.725$ Hz. Notice that rounding $f_2$ to 336 Hz would have given 7 Hz, a 4% error in the answer from a 0.08% error in an intermediate. Beat problems subtract two nearly equal numbers, so carry digits you would normally throw away.

The same idea in space

Now hold the frequency fixed and move yourself instead. Two speakers wired to the same amplifier emit exactly the same tone, in step with each other. They sit at different places, so the sound from one has further to travel to your ear than the sound from the other.

Call the two path lengths $r_1$ and $r_2$, and their difference $\Delta r = r_1 - r_2$. Travelling one extra wavelength puts a wave one full cycle behind, which is $2\pi$ radians of phase. Travelling half a wavelength puts it $\pi$ behind. Phase lag is proportional to extra distance, one full $2\pi$ per $\lambda$, so

$\Delta\phi = \frac{2\pi}{\lambda}\,\Delta r$

That single line is the whole of two-source interference, and it is worth saying in words: phase difference is path difference measured in wavelengths, times $2\pi$.

Two waves reinforce when they are a whole number of cycles apart and cancel when they are a half-odd number of cycles apart. Written in path difference, with $m = 0, 1, 2, \dots$:

$\Delta r = m\lambda \quad \text{(loud)}, \qquad \Delta r = \left(m + \tfrac{1}{2}\right)\lambda \quad \text{(quiet)}$

And $\lambda = v/f$, with $v = 343$ m/s for sound in ordinary air, converts between the two languages whenever a problem gives you a frequency and wants a distance.

Example 3: walking to the first quiet spot. Two speakers 3.00 m apart on a wall are driven in phase at 500 Hz. You start on the perpendicular bisector, on a line 4.00 m from the wall and parallel to it, and walk. How far until the first place where the sound drops out?

Start with the wavelength, because every condition is written in wavelengths:

$\lambda = \frac{v}{f} = \frac{343}{500} = 0.6860\ \text{m}$

The first minimum is the smallest path difference that cancels, $\Delta r = \lambda/2 = 0.3430$ m.

Now set up the geometry. Put the speakers at $y = +1.50$ m and $y = -1.50$ m on the wall, and let $y$ be how far you have walked from the bisector. Your distances to the two speakers are

$r_1 = \sqrt{4.00^2 + (y+1.50)^2}, \qquad r_2 = \sqrt{4.00^2 + (y-1.50)^2}$

and you want $r_1 - r_2 = 0.3430$ m. There is no tidy way to unpick those square roots, so close in on $y$ numerically. At $y = 0$ the paths are equal and the difference is zero, so walk outward and watch it grow.

At $y = 0.400$ m: $r_1 = \sqrt{16.00 + 1.90^2} = \sqrt{19.610} = 4.42832$ m and $r_2 = \sqrt{16.00 + 1.10^2} = \sqrt{17.210} = 4.14849$ m, so $\Delta r = 0.27983$ m. Too small.

At $y = 0.500$ m: $r_1 = \sqrt{16.00 + 4.00} = 4.47214$ m and $r_2 = \sqrt{16.00 + 1.00} = 4.12311$ m, so $\Delta r = 0.34903$ m. Slightly too big.

The target 0.34303 sits between them, so interpolate straight across:

$y \approx 0.400 + 0.100 \times \frac{0.34303 - 0.27983}{0.34903 - 0.27983} = 0.400 + 0.100(0.9133) = 0.4913\ \text{m}$

Check it: $r_1 = \sqrt{16 + 1.9913^2} = 4.46825$, $r_2 = \sqrt{16 + 1.0087^2} = 4.12522$, difference 0.34303 m. That is the half wavelength you wanted, so the first quiet spot is 0.491 m from the centre line.

Two habits from that problem carry everywhere. First, convert to a wavelength before you touch the geometry. Second, if the algebra fights you, bracket the answer with two guesses and interpolate: that is a legitimate method, not a retreat.

When the sources are not in step

Everything above assumed the two sources start together. They need not. Wire one speaker with its terminals reversed and it pushes when the other pulls: the sources carry a built-in phase difference of $\phi_0 = \pi$.

Do not memorise a new pair of conditions for that case. Write down the total phase difference from every cause and then ask what makes it a multiple of $2\pi$:

$\Delta\phi_{\text{total}} = \phi_0 + \frac{2\pi}{\lambda}\,\Delta r$

With $\phi_0 = \pi$, the loud and quiet conditions simply trade places. $\Delta r = 0$ on the perpendicular bisector now gives $\Delta\phi_{\text{total}} = \pi$, so the point that would have been loudest is the quietest one in the room. The first constructive point needs the path difference to supply another $\pi$, that is $\Delta r = \lambda/2$.

How loud is a point in between

Maxima and minima are the easy points. Sometimes a problem drops you between them and asks for the amplitude there. Two waves of equal amplitude $A$ and phase difference $\Delta\phi$ add like two arrows of equal length set apart by that angle. Lay them nose to tail and the closing arrow is the resultant. The triangle is isosceles, so the resultant has length

$A_{\text{res}} = 2A\left|\cos\!\left(\frac{\Delta\phi}{2}\right)\right|$

Check it at the ends. At $\Delta\phi = 0$ you get $2A$, full reinforcement. At $\Delta\phi = \pi$ you get zero, full cancellation. Both are what the path rules already said, which is a sign the formula is the general version of them.

Example 4: an odd point with reversed wiring. Two speakers emit 343 Hz but are driven exactly out of phase. Each alone would give displacement amplitude 1.00 mm at your ear, and your two path lengths differ by 0.750 m. What is the resultant amplitude?

The wavelength is $\lambda = 343/343 = 1.000$ m, so the path difference is 0.750 wavelengths. That contributes

$\frac{2\pi}{1.000}(0.750) = 1.5\pi\ \text{rad}$

and the wiring contributes another $\pi$. The total is $\Delta\phi = 1.5\pi + \pi = 2.5\pi$, which is the same as $0.5\pi$ once you drop a whole $2\pi$. Then

$A_{\text{res}} = 2(1.00)\left|\cos\!\left(\frac{0.5\pi}{2}\right)\right| = 2.00\cos(45^\circ) = 2.00(0.7071) = 1.414\ \text{mm}$

Halfway between full reinforcement and silence, which is where a quarter-cycle offset should land you.

A harder one: beats from a single source

Beats need two frequencies, but not two instruments. Any process that shifts one copy of a sound will do, and motion is the usual one. That is the Doppler effect, the subject of the next lesson; take two of its results on loan here. A source moving away from a listener at speed $v_s$ is heard at $f\,v/(v+v_s)$, and moving toward at $f\,v/(v-v_s)$.

Example 5: a buzzer approaching a wall. A 500 Hz buzzer moves toward a flat wall at 3.00 m/s. You stand well behind the buzzer, in line with its motion, and hear both the direct sound and the echo. Take $v = 343$ m/s. What beat frequency do you hear?

Two paths, so two frequencies. Handle them one at a time.

The direct sound comes from a source moving away from you, since the buzzer is heading for the wall on the far side:

$f_{\text{direct}} = 500 \times \frac{343}{343 + 3.00} = 500 \times \frac{343}{346} = 495.665\ \text{Hz}$

The echo takes two steps. First the wall receives sound from a source moving toward it:

$f_{\text{wall}} = 500 \times \frac{343}{343 - 3.00} = 500 \times \frac{343}{340} = 504.412\ \text{Hz}$

Then the wall re-radiates whatever it received. The wall is not moving, and you are not moving, so no further shift happens: the echo reaches you at 504.412 Hz.

Two tones at your ear, so subtract:

$f_{\text{beat}} = 504.412 - 495.665 = 8.747\ \text{Hz}$

The structure is what to take from this. A beat problem never has a trick in the subtraction. The work is always in getting the two frequencies right, and a reflector is best treated as two events: a listener, then a fresh source.

Recognising the technique

Reach for beats when a problem names two sources of nearly the same frequency and asks for a rate, a throb, a count of loud moments, or a time between them. The signatures: "beats per second", "sounded together", "how many beats in 10 seconds", a fork plus anything, two pipes of almost equal length, two sources with one of them moving. Method: compute each frequency separately to five figures, subtract, take the absolute value. If the problem asks which is higher, look for the second piece of information that breaks the tie.

Reach for path difference when there is one frequency and two places, and the question is about where rather than when. The signatures: two speakers, a listener who walks or stands at a named spot, a Quincke tube whose slider moves, "the lowest frequency for which she hears nothing", "how many frequencies below 1000 Hz". Method: find $\Delta r$ from the geometry, find $\lambda = v/f$, and compare. If the unknown is a frequency, invert it: a minimum needs $\Delta r = (m+\frac{1}{2})\lambda$, so $f = (m+\frac{1}{2})v/\Delta r$, and running $m = 0, 1, 2, \dots$ lists every frequency that works and lets you count the ones under a cutoff.

Two more habits close the gap on the harder problems. When a problem specifies how the sources are wired, write $\Delta\phi_{\text{total}} = \phi_0 + 2\pi\Delta r/\lambda$ before anything else, and never start from a memorised $m\lambda$. And when it asks for an amplitude rather than a position, you want $2A|\cos(\Delta\phi/2)|$, which means you need the phase difference itself, not only whether it is loud or quiet.

Both halves of this lesson are one idea. Beats are interference in time; two speakers are interference in space. The next lesson takes the remaining piece (what motion does to a frequency) and makes it exact.