The ends of a string do not allow just one standing wave. They allow a whole family of them: one loop, two loops, three loops, on and on. Each member of the family is a harmonic, and their frequencies climb in a beautifully simple sequence. This is why a guitar string sounds like a guitar and not like a sine wave — and it is why a flute and a clarinet of the same length play completely different notes.
📘 What you need to know
Harmonics are the only wave patterns the boundary conditions will allow
String fixed at both ends (or pipe open at both ends): λn = 2L/n and fn = nv/2L, with n = 1, 2, 3…
Pipe open at one end: λn = 4L/n and fn = nv/4L, with n = 1, 3, 5… only
On a string, the nth harmonic has n + 1 nodes and n antinodes
In a pipe open at both ends, the nth harmonic has n nodes and n + 1 antinodes
None of these formulas are in the data booklet. You must recall them
The family of allowed patterns
Last page you drew the simplest pattern the ends allow. Now let’s be greedy and add more loops.
Take a string fixed at both ends. Both ends must be nodes. One loop fits. Two loops fit. Three fit. Anything in between would need half a loop hanging off the end, and the fixed end refuses to allow it. So the allowed patterns come in whole loops, and nothing else survives.
Think of it like stepping stones across a river, with the stones fixed in place. You can take one long stride, or two, or three — but you cannot take two and a half. The ends of the string are the riverbanks, and they only accept a whole number of half-wavelengths.
Harmonics on a string (fixed at both ends)
Both ends are nodes. Count the loops — the number of loops is the harmonic number n.
Every extra loop squeezes the wavelength down and pushes the frequency up. The nth harmonic sits at exactly n times the frequency of the first.
Each loop is λ/2, and n loops fill the string:
String fixed at both endsL = n × λ/2 → λn = 2L / nand since v = fλ: fn = nv / 2Ln = 1, 2, 3…
Here L is the length of the string and v is the speed of the travelling waves on it. The first harmonicf1 = v/2L is the lowest note the string can play, and every other harmonic is a whole-number multiple of it.
Harmonics in a pipe open at both ends
Both ends are antinodes now. Remarkably, the wavelengths come out exactly the same as for the string.
The curve is air displacement along the pipe. Both open ends are antinodes, so the pattern is upside-down-ish compared with the string — yet the wavelengths match exactly.
Pipe open at both endsλn = 2L / nfn = nv / 2Ln = 1, 2, 3…identical to the string — but count nodes and antinodes the other way round
Harmonics in a pipe open at one end
This is the odd one out, and examiners know it. One end is an antinode, the other is a node. The simplest pattern is a lonely quarter of a wave.
Now try to fit the “second harmonic”. Two loops would need a node at the open end — not allowed. So the pattern jumps straight to the one with three quarter-waves. Only the odd harmonics can exist.
Each allowed pattern adds two more quarter-waves, never one. That is why the harmonic numbers skip: 1, 3, 5, 7…
Pipe open at one endλn = 4L / nfn = nv / 4Ln = 1, 3, 5… odd numbers only
Why do the odd harmonics matter in real life? A clarinet behaves like a pipe closed at one end, so it plays mostly odd harmonics — that hollow, woody tone. A flute is open at both ends and gets the lot. Same length of tube, completely different voice.
System
Ends
λn
Allowed n
String, both ends fixed
node – node
2L/n
1, 2, 3…
Pipe, both ends open
antinode – antinode
2L/n
1, 2, 3…
Pipe, one end open
antinode – node
4L/n
1, 3, 5…
Look at the ends
gives
Allowed λn
then v = fλ
Allowed fn
🧮 Any harmonics question, in four moves
Check the boundary conditions first. Node–node and antinode–antinode both give 2L/n. Mixed ends give 4L/n.
Decide which n are allowed. All integers, or odd only?
Write fn = nv/2L or nv/4L and substitute.
Rearranging for n? Get n = 2Lf/v, then round down to a whole number.
WE 1
A string of length 0.75 m is fixed at both ends. Waves travel along it at 180 m s⁻¹. Calculate the frequency of the first harmonic, and the wavelength and frequency of the third harmonic.
Step 1 — both ends fixed, so use λn = 2L/nλ₁ = 2 × 0.75 = 1.5 mStep 2 — first harmonic frequencyf₁ = v/λ₁ = 180 / 1.5f₁ = 120 HzStep 3 — third harmonicλ₃ = 2L/3 = 1.5/3 = 0.50 mf₃ = 3f₁ = 3 × 120f₃ = 360 HzOnce you have f₁, every other harmonic is just a multiple of it. No need to redo the whole calculation.
WE 2
A pipe of length 0.60 m is open at one end and closed at the other. The speed of sound in the pipe is 340 m s⁻¹. Determine the frequency of the first harmonic, and the next frequency at which a standing wave can form.
Step 1 — mixed ends, so λn = 4L/nλ₁ = 4 × 0.60 = 2.4 mStep 2 — first harmonicf₁ = 340 / 2.4f₁ = 142 HzStep 3 — the next allowed harmonic is n = 3, not n = 2f₃ = 3 × 141.67f₃ = 425 HzJumping straight from n = 1 to n = 3 is the whole point of this boundary condition. An even harmonic simply cannot fit.
WE 3
A guitar string of length 0.64 m is fixed at both ends and carries waves at 320 m s⁻¹. Determine the highest harmonic that can be heard by a listener whose hearing extends to 4.0 kHz.
Step 1 — write the nth harmonic frequencyfn = nv / 2LStep 2 — rearrange for nn = 2Lfn / v = (2 × 0.64 × 4000) / 320Step 3 — evaluaten = 5120 / 320 = 16the 16th harmonicIf n had come out as 16.7, the answer would still be 16 — you round down, because harmonics only exist at whole numbers.
💡 Top tips
Boundary conditions first, formula second. The ends tell you whether it’s 2L/n or 4L/n.
These formulas are not in the data booklet. Learn them.
Every harmonic is a whole-number multiple of the first: fn = nf1. Find f1 once, then scale.
Solving for n? Round down, never up. There is no 16.7th harmonic.
Sketch first. A quick drawing of the loops beats trying to remember which formula it was.
⚠ Common mistakes
Using 2L/n for a pipe open at one end — it is 4L/n
Calling the next pattern after the first in a closed pipe the “second harmonic”. It’s the third
Swapping the node and antinode counts between strings and open pipes
Thinking v changes between harmonics — the wave speed is fixed by the medium, so only λ and f change
Rounding n up, giving a harmonic the listener can’t actually hear
Forgetting that v for a string is the speed of the waves on the string, not the speed of sound in air
Quick recap: Harmonics are the wave patterns the ends permit. For a string fixed at both ends, and for a pipe open at both ends, λn = 2L/n and fn = nv/2L with n = 1, 2, 3… For a pipe open at one end, λn = 4L/n and fn = nv/4L with n odd only. Always read the ends before you reach for a formula.
One last question hangs over all of this. If a string can only vibrate at its harmonics, what happens when you push it at exactly one of those frequencies? The answer is that the amplitude climbs, and climbs, and climbs. Push a child on a swing at just the right rhythm and you get the same effect. That is resonance, and it’s the next page.
Harmonics not in tune yet?
Book a free meeting and we’ll work through 2L/n versus 4L/n, odd harmonics and past-paper pipe questions together.
