The big idea: Pluck a guitar string or blow across a bottle and it only 'sings' at certain special frequencies — its harmonics.
At those frequencies a standing wave fits neatly into the length, and the sound is loud. This is resonance.
The lowest of these is the fundamental (the 1st harmonic); the others are whole-number multiples of it.
Three new words: Node = a point on a standing wave that never moves.
Antinode = a point that swings with the biggest amplitude.
Resonance = when a system is driven at one of its natural frequencies and vibrates strongly.
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What the ends force: A fixed end of a string (or the closed end of a pipe) must be a node.
A free/open end must be an antinode.
The pattern that fits between those ends decides the wavelength.
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Count how many loops (half-wavelengths) fit into the length, and you have the wavelength. Which condition you use depends on the ends.
| Boundary | Ends are… | Wavelength condition | Which harmonics |
|---|---|---|---|
| String, both ends fixed | node — node | λ = 2L ÷ n | all: n = 1, 2, 3, … |
| Pipe open at both ends | antinode — antinode | λ = 2L ÷ n | all: n = 1, 2, 3, … |
| Pipe closed at one end | node — antinode | λ = 4L ÷ n | odd only: n = 1, 3, 5, … |
- wavelength of that harmonic (m)
- length of the string or pipe (m)
- harmonic number (1, 2, 3 … ; for a closed pipe only odd: 1, 3, 5 …)
These two are NOT in the data booklet: You have to know λ = 2L/n and λ = 4L/n — they are not given.
Memory aid: a string or open pipe matches at its ends (both same type), so a half-wave fits → 2L/n. A closed pipe is lopsided (node one end, antinode the other), so a quarter-wave fits → 4L/n, odd harmonics only.
Then turn wavelength into frequency: Once you have the wavelength, the frequency comes from the given wave equation v = fλ (rearranged f = v ÷ λ), where v is the speed of the wave (the speed of sound for a pipe).
- wave speed (m s⁻¹)
- frequency (Hz)
- wavelength (m)
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A guitar string of length 0.65 m is fixed at both ends. A wave travels along it at 260 m s⁻¹. Find the wavelength and frequency of its fundamental (1st harmonic).
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How this is tested — harmonics show up as short calculations and 'read the pattern' questions:
Paper 1A
- A quick calculation — e.g. a pipe closed at one end of a given length and sound speed: find the first two harmonic frequencies.
Paper 2
- Determine a frequency by measuring the spacing of nodes/antinodes (e.g. melted spots in a microwave) and using v = fλ.
The classic trap: A closed pipe has only odd harmonics — its 'second harmonic' is actually n = 3, not n = 2.
Closed pipe — odd harmonics only: For a pipe closed at one end, λ = 4L/n with n = 1, 3, 5, …
So the first harmonic is n = 1 and the next one is n = 3 (there is no n = 2). The frequencies go in the ratio 1 : 3 : 5 …
A pipe of length 0.20 m is closed at one end and open at the other. The speed of sound in the air inside is 340 m s⁻¹. Find the frequencies of its first two harmonics.
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