Standing Waves – Standing Waves

What is the relationship between the frequency spectrum of the sound wave and the vibration on the string? Let us first consider only the fundamental tone f₀.

Plucking causes the string to vibrate. This vibration travels through the air at the speed of sound. Let us consider two points: point A in the air, and point B on the string. In the air, there is a sound wave that is propagating. The fundamental tone has the frequency f₀, which corresponds to a wheel with a rotation frequency f₀. Its spatial and temporal course is visualised here.

The situation at point B on the string is different, because the vibration is reflected back and forth at the two end points of the string. There are thus two waves which travel in opposite directions and overlap, forming a standing wave. The frequency of this standing wave is also f₀. There are two nodes at the edges, where the string does not move. An antinode forms in the middle. This standing wave on the string, with the length l, is equivalent to one-half a wavelength on the string. This is the vibration behind the fundamental tone with the frequency f₀.

When the frequency doubles, the wavelength is halved. This happens because the product of frequency and wavelength is always equal to the wave propagation rate on the string. That means there are now two antinodes on the vibrating string. There is also one additional node, which is located precisely in the middle of the string.

The wavelength is reduced again when the frequency is tripled. There is exactly one additional antinode on the vibrating string, and, correspondingly, there is also another node.

Let’s sum it up. Standing waves are formed when multiples of one-half wavelength fit within the length of the vibrating string. In the simplest scenario, there are only two nodes at the edges, and one antinode, which is also the fundamental tone. The first overtone has one node more than the fundamental tone; the second has two nodes more, and so on. Doubling the frequency halves the wavelength relative to the fundamental tone. Tripling the frequency reduces the wavelength to one third. It goes on like this for all the multiples of the frequency of the fundamental tone.

Let us now return to the question of unambiguity. Is this spectrum a unique fingerprint for the vibrating string with the length l? Well, it might seem to be the case at first, since there is an apparent correlation between the frequency and the string length l, such that l=(n·λ)/2.

Let us consider two different vibrating strings having different lengths and different wave propagation rates, C₁ and C₂. At first, the longer string sounds lower. That means, we have an equidistant spectrum again, but the fundamental tone f₀ has moved down a little. All multiples of f₀, or overtones, have moved down with it.

However, if we increase the string tension of the longer string, thus adjusting the wave propagation rate C₂ such that the ratio of wave propagation rate and length is equal for both vibrating strings, the result will be an identical spectrum. That is, the fundamental tone is equal, as are the overtones, even though the strings have various lengths.

Of course! You can, for example, tune a guitar so that two different strings will have the same fundamental tone.

If just one vibrating string does not allow us to draw clear conclusions on the instrument used, it will certainly not be the case with more complex instruments. I am sorry, Bob!