| T |
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| μ |
The physics behind musical instruments is beautifully simple. The sounds are associated with standing waves, which come from constructive interference between waves traveling in both directions along a string or a tube.
Why do some notes, when played together, sound good, while others do not go together well at all? To answer that requires some understanding of the scale.
Twelve notes make up one octave, covering a factor of two in frequency. Each note on the scale is a factor of 21/12 times the previous one. Let's look at one octave starting at a frequency of 440 Hz, which happens to be an A.
| Note | 2n/12 = ? | Fraction | Frequency (Hz) |
|---|---|---|---|
| A | 20/12 = 1.00 | - | 440 |
| A# | 21/12 = 1.06 | - | 466 |
| B | 22/12 = 1.12 | - | 494 |
| C | 23/12 = 1.19 | 6/5 | 523 |
| C# | 24/12 = 1.26 | 5/4 | 554 |
| D | 25/12 = 1.33 | 4/3 | 587 |
| D# | 26/12 = 1.41 | - | 622 |
| E | 27/12 = 1.50 | 3/2 | 659 |
| F | 28/12 = 1.59 | - | 698 |
| F# | 29/12 = 1.68 | 5/3 | 740 |
| G | 210/12 = 1.78 | - | 784 |
| G# | 211/12 = 1.89 | - | 831 |
| A | 212/12 = 2.00 | 2/1 | 880 |
When notes are played together, such as when a string instrument is played, they sound best to our ears when the frequencies of the sounds are in integer ratios. As shown in the table, this happens for several of the notes on the scale.
A major chord on the guitar, for instance, consists of the notes 4 steps and 7 steps away from the base note. These have frequencies that are about 5/4 and 3/2 times the frequency of the base note, respectively.
We'll look at the guitar in particular, but the same general principles apply to all string instruments.
Tuning a guitar means setting the fundamental frequency of each string to a particular value. Standard tuning has the six strings covering two octaves:
| String | Note |
|---|---|
| Bottom (lightest) | E |
| Second | B |
| Third | G |
| Fourth | D |
| Fifth | A |
| Sixth | E |
The fundamental frequency is given by f = v/λ, where the wavelength is determined by the length of the string and v is given by:
| v | = | ( |
|
) | ½ |
All the strings are under approximately the same tension and they're all the same length. The lighter the string the faster the wave speed and the higher the frequency. Tuning a given string to a precise frequency is done by adjusting its tension.
Pressing down on a particular string shortens its length. This decreases the wavelength, which increases the frequency. On a guitar the correct length for every note on the scale is marked with metal bars called frets. Moving your finger from one fret to the next changes the frequency by one note, which is a factor of 21/12. To increase the frequency by this factor the length must be multiplied by a factor of 2-1/12 = 0.944, which is why the frets gradually get closer together as you move away from the neck of the guitar.
One of the big differences between a string instrument and a wind instrument is that with a wind instrument you don't have as much control over the wave velocity. The wave is moving through a column of air, so the speed is the speed of sound in air, which is temperature dependent.
As with strings, resonance plays an important role for wind instruments. The wavelengths at which resonance occurs are proportional to the length of the instrument, so you can play different notes by changing the length. This is done most often by pressing a key to open a particular hole in the instrument, for instance, but can also be accomplished by changing the length of the instrument itself, as is done in a trombone.
In general the larger the instrument the longer the wavelength of the fundamental and the lower the frequency range of the instrument.
A particular guitar string has a mass of 3.0 grams and a length of 0.75 m. A standing wave on the string has the shape shown in the simulation below - the wave has a frequency of 1200 Hz.
(a) What is the speed of the wave?
(b) What is the tension of the string?
Part (a). The speed of the wave is given by:
v = fλ
From the simulation, it can be seen that he wavelength is 2/3 of the length of the string (1.5λ = L).
λ = 0.50 m and f = 1200 Hz, so:
v = 1200 * 0.5 = 600 m/s.
Part (b). The tension in the string can be found from the relationship:
| v | = | ( |
|
) | ½ |
where μ, the mass per unit length of the string, is
| μ | = |
|
= 0.004 kg/m |
T = μ v2 = 0.004 *600*600 = 1440 N