Harmonics in strings and air columns
Standing waves on strings fixed at both ends and in open and closed pipes, the fundamental and higher harmonics, the formulas for their frequencies, why instruments sound different, and the resonance-tube method for measuring the speed of sound.
Pluck the open A string on a guitar and it vibrates 110 times a second. Press the string at the twelfth fret, exactly halfway along, and the note jumps an octave to 220 Hz. Touch the string lightly at that same point without pressing, and you get a pure, bell-like 220 Hz "harmonic" while the whole string is still moving. Every one of these numbers falls out of one idea: a standing wave must fit the length it lives on.
Standing waves on a string
When a string fixed at both ends is plucked, waves travel to each end, reflect and overlap, and a standing wave forms (page 3D). The fixed ends cannot move, so each end must be a node. Only patterns that fit a whole number of half-wavelengths into the length $L$ are possible:
$L = \dfrac{n\lambda}{2} \quad \Rightarrow \quad \lambda_n = \dfrac{2L}{n}, \quad n = 1, 2, 3, \ldots$
Using $f = v/\lambda$, the allowed frequencies are
$f_n = \dfrac{nv}{2L}$
The lowest frequency, $n = 1$, is the fundamental (first harmonic). The others are whole-number multiples of it and are called the second, third, ... harmonics. A plucked string vibrates in many harmonics at once.
The wave speed $v$ on a string depends on the tension and on the mass per unit length: tighter and lighter strings carry waves faster. So a string's pitch can be raised by
- shortening it (pressing a fret): $f \propto 1/L$,
- tightening it (turning the tuning peg): larger $v$,
- using a thinner, lighter string: larger $v$.
(The exact relationship $v = \sqrt{T/\mu}$, with tension $T$ and mass per unit length $\mu$, is beyond the syllabus, but the trends are not.)
A guitar string is 0.65 m long between the nut and the bridge, and its fundamental is 220 Hz. (a) Find the wave speed on the string. (b) Find the frequency of the third harmonic. (c) Where must the string be pressed to play 330 Hz?
Standing waves in air columns
Blow across the top of a bottle or into a flute and the air inside resonates. Sound waves travel up and down the tube and reflect at both ends, even at an open end, and a standing wave forms in the air. It is usual to describe these waves by the displacement of the air:
- At a closed end the air cannot move: a displacement node.
- At an open end the air moves freely: a displacement antinode.
Pipe open at both ends (flute, many organ pipes): an antinode at each end, just like the string's nodes, so
$L = \dfrac{n\lambda}{2}, \qquad f_n = \dfrac{nv}{2L}, \qquad n = 1, 2, 3, \ldots$
Pipe closed at one end (clarinet roughly, a bottle, the resonance tube below): a node at the closed end and an antinode at the open end. The shortest fit is a quarter wavelength, and after that only odd numbers of quarter wavelengths fit:
$L = \dfrac{n\lambda}{4}, \qquad f_n = \dfrac{nv}{4L}, \qquad n = 1, 3, 5, \ldots$
So a closed pipe has a fundamental half that of an open pipe of the same length, and it is missing all the even harmonics. That missing set is part of why a clarinet sounds "hollow" next to a flute.
The antinode at an open end actually sits slightly outside the tube. For a tube of radius $r$, the effective length is about $L + 0.6r$; this end correction matters in accurate experiments.
An organ pipe is 0.50 m long. The speed of sound is 343 m/s. Find the first two possible frequencies if the pipe is (a) open at both ends, (b) closed at one end.
Why instruments sound different
A flute and a violin can play the same note (the same fundamental) but still sound different. The difference, called timbre or tone quality, comes from the mix of harmonics: which ones are present and how strong each is. The waveform on an oscilloscope is the superposition of all of them, so each instrument has its own shape. The yidaki (didjeridu), which you met in Year 9, is a pipe closed at one end by the player's lips, which is why its overtones are mostly odd harmonics; players shape them with their mouth and voice.
Measuring the speed of sound with a resonance tube
With a 384 Hz fork, resonances are heard at 0.212 m and 0.659 m. Find the speed of sound and the end correction.
- In String mode at the start (L = 0.65 m, v = 286 m/s, n = 1) the readouts show λ = 1.300 m and f1 = 220.0 Hz. Set n = 3: f3 = 660.0 Hz. Count the nodes.
- In Open pipe mode (L = 0.50 m, 20 °C, v = 343 m/s) the fundamental is 343.0 Hz. Switch to Closed pipe: 171.5 Hz. Why can't you select n = 2 there?
- Raise the air temperature to 40 °C. What happens to every pipe frequency, and why does an orchestra's wind section go sharp in a hot hall?
- In Resonance tube mode with the 512 Hz fork, slowly increase L and find the two lengths where the loudness meter peaks (about 0.155 m and 0.490 m). Calculate $v = 2f(L_2 - L_1)$.

Test yourself
A string 0.80 m long vibrates in its third harmonic. The wavelength of the waves on it is
$\lambda_n = 2L/n = 1.6/3 = 0.53$ m.
A pipe closed at one end has a fundamental of 150 Hz. Which frequency could it NOT produce?
A closed pipe has only odd harmonics: 150, 450, 750, 1050 Hz ... 300 Hz would be the second harmonic, which does not exist.
A pipe open at both ends and a pipe closed at one end have the same length. The ratio of the open pipe's fundamental to the closed pipe's fundamental is
Open: $v/2L$; closed: $v/4L$. Ratio 2 : 1.
A violinist tightens a string. Which row is correct for the string's fundamental?
The length still fixes $\lambda = 2L$. Greater tension gives greater wave speed, so $f = v/\lambda$ rises.
Label the pipe diagram.
Order these from lowest to highest fundamental frequency (all 1.0 m long, same air).
A string fixed at both ends has a at each end. Its fundamental wavelength is the length of the string. The open end of a pipe is a displacement . A pipe closed at one end produces only harmonics. Raising the temperature of the air in a pipe increases the of sound and so raises the pitch.
A student uses a 440 Hz tuning fork with a resonance tube closed at the bottom by water. The first two resonances are at 0.183 m and 0.573 m. (a) Calculate the speed of sound. (b) Calculate the end correction. (c) Evaluate the student's method of using only these two readings, and suggest how the experiment could be improved.
(a) $L_2 - L_1 = \lambda/2 = 0.573 - 0.183 = 0.390$ m, so $\lambda = 0.780$ m and $v = f\lambda = 440 \times 0.780 = 343\ \text{m s}^{-1}$.
(b) $\lambda/4 = 0.195$ m $= L_1 + e$, so $e = 0.195 - 0.183 = 0.012$ m.
(c) Using the difference between two resonances is a good choice, because the unknown end correction cancels, which improves accuracy. However, each length depends on judging by ear where the sound is loudest, which carries a random error of perhaps a few millimetres, and with only one pair of readings from one fork there is no way to estimate or reduce that error. Reliability would improve by approaching each resonance from both above and below and averaging, and by repeating with several forks of different frequency. Plotting $L_1$ against $1/f$ gives a straight line of gradient $v/4$, using all the data at once. The air temperature should also be recorded, since the speed of sound depends on it.
Coaching Centre