A string or a pipe can’t vibrate at just any frequency — it has a menu of allowed patterns called harmonics, and the exam expects you to know the whole menu. The good news: every entry comes from one simple idea. Count how many loops fit, and the wavelength and frequency follow.
📘 What you need to know
Harmonics are the only wave patterns (and so the only frequencies) that can form standing waves for a given set of boundary conditions
String fixed at both ends: the nth harmonic has n loops, with wavelength λn = 2L/n and frequency fn = nv/2L
On a string, the nth harmonic has (n + 1) nodes and n antinodes
Pipe open at both ends: same formula λn = 2L/n — but flipped: (n + 1) antinodes and n nodes
Pipe open at one end: only odd harmonics exist — λn = 4L/n with n = 1, 3, 5…
These formulas are not in the data booklet — you have to remember them
What Is a Harmonic?
Last page we saw that the boundaries force a node or antinode at each end, and only certain frequencies produce patterns that fit. Each pattern that fits is called a harmonic.
The simplest possible pattern is the first harmonic (also called the fundamental) — it uses the fewest loops the boundary conditions allow. Raise the driving frequency and, one by one, more complicated patterns appear: the second harmonic, the third, and so on. Each harmonic has its own wavelength and its own natural frequency, and the frequency of every harmonic comes from the same trusty wave equation, v = fλ, where v is the speed of the travelling waves in the medium.
Harmonics on a String Fixed at Both Ends
Both ends are clamped, so both ends must be nodes. The simplest pattern between two nodes is a single loop — two nodes at the ends, one antinode in the middle. One loop is half a wavelength, so for the first harmonic:
The first three harmonics on a string of length L fixed at both ends. Each extra harmonic squeezes in one more loop: n loops means λ = 2L/n. Count the letters: the nth harmonic has n antinodes and (n + 1) nodes.
Spot the pattern: every harmonic just adds one more loop. Since each loop is half a wavelength, n loops fill the string when L = n(λ/2). Rearranged:
Wavelength of the nth harmonic — string fixed at both endsλn = 2L ÷ n (n = 1, 2, 3…)
Then v = fλ turns each wavelength into a natural frequency:
Natural frequenciesfn = nv ÷ 2L = n × f1
So the harmonics of a string are simply whole-number multiples of the fundamental: f1, 2f1, 3f1, and so on.
Pipes Open at Both Ends
Swap the string for an air column with both ends open, and the boundary rules flip: now both ends must be antinodes. The simplest pattern is antinode–node–antinode — half a wavelength again, so all the maths carries straight over: λn = 2L/n and fn = nv/2L, for any whole number n.
Only the head-count swaps around: the nth harmonic of an open–open pipe has (n + 1) antinodes and n nodes — the mirror image of the string.
Pipes Open at One End
This is the odd one out — literally. One end is closed (must be a node) and the other is open (must be an antinode). The smallest pattern that starts on N and ends on A is a half-loop: just one node and one antinode, spanning only a quarter of a wavelength. So for the fundamental:
First harmonic — pipe open at one endL = λ/4 → λ1 = 4L
Now here’s the twist. To reach the next allowed pattern, you must add a whole extra loop (a node and an antinode) so the ends still finish on N and A. That jumps the pipe straight from 1 quarter-wavelength to 3 quarter-wavelengths — skipping what would have been the second harmonic entirely. The next pattern after that holds 5 quarter-wavelengths, and so on:
A pipe open at one end can only hold an odd number of quarter-wavelengths: 1, 3, then 5. That’s why only odd harmonics exist here — there is no second or fourth harmonic in this pipe.
Wavelength of the nth harmonic — pipe open at one endλn = 4L ÷ n (n = 1, 3, 5… odd only)
🧭 Solving any harmonics problem
Check the boundary conditions first — they decide which formula you’re allowed to use
Pick the wavelength formula: string fixed both ends or pipe open both ends → λn = 2L/n; pipe open one end → λn = 4L/n with odd n
Sketch if unsure — draw N/A at the ends, fill in loops, and count
Convert with the wave equation — v = fλ links every wavelength to its natural frequency
Quick recap: strings and open–open pipes take every harmonic (λn = 2L/n, fn = nf1); a pipe open at one end takes odd harmonics only (λn = 4L/n, n = 1, 3, 5…).
WE 1
A pipe of length 0.60 m is open at one end and closed at the other. The speed of sound in air is 340 m s⁻¹.
(a) Calculate the frequency of the first harmonic.
(b) State, with a reason, the frequency of the next harmonic that can form in this pipe.
Part (a)
Open at one end, so λ₁ = 4L
λ₁ = 4 × 0.60 = 2.4 mf₁ = v/λ₁ = 340 ÷ 2.4f₁ ≈ 142 HzPart (b)
Only odd harmonics fit this boundary condition, so the next one is the third harmonic
f₃ = 3 × f₁ = 3 × 141.7f₃ = 425 HzNo second harmonic exists here — the pattern jumps straight from n = 1 to n = 3.
WE 2
Transverse waves travel at 300 m s⁻¹ along a stretched wire of length 80 cm, fixed at both ends. Determine the highest harmonic of the wire that a person who can hear frequencies up to 15 kHz would be able to detect.
Set up
L = 0.80 m, v = 300 m s⁻¹, maximum audible frequency fₙ = 15 000 Hz
Rearrange the harmonic frequency formulafₙ = nv/2L ⇒ n = 2Lfₙ/vSubstituten = (2 × 0.80 × 15 000) ÷ 300n = 80 — the 80th harmonicSanity check: the fundamental is 300 ÷ (2 × 0.80) = 187.5 Hz, and 80 × 187.5 Hz = 15 kHz exactly.
💡 Top tips
Neither wavelength formula is in the data booklet — memorise both, or re-derive them by sketching the fundamental and counting loops
Harmonics of a string are exact multiples of the fundamental: fn = n × f1 — often the fastest route in a calculation
The v in fn = nv/2L is the speed of the travelling waves in the medium — the standing wave pattern itself doesn’t move
Quick head-count: string → n antinodes, (n + 1) nodes; open–open pipe → the other way round
⚠ Common mistakes
Using λn = 2L/n for a pipe open at one end — that pipe needs 4L/n with odd n
Calling the pattern after the fundamental in a one-open-end pipe the “second harmonic” — it’s the third (n jumps 1 → 3 → 5)
Writing λ1 = L for a string’s first harmonic — one loop is only half a wavelength, so λ1 = 2L
Diving into the formula without checking the boundary conditions — always identify the ends first
Up next: resonance — what happens when you drive a system at exactly one of its natural frequencies, and why that made soldiers break step on bridges.
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