Harmonics and overtones — on strings and in pipes: Worked examples

The tools we will reuse (earned before use)

Definition The only symbols on this page

$L$ — the length of the string or air column, in metres. Picture a ruler laid along it.
$v$ — the speed the wave travels down that medium, in metres per second. On a string it comes from tension (see Speed of a wave on a string); in air it is the speed of sound.
$λ$ (the Greek letter "lambda") — the wavelength: the length of one complete repeat of the wave shape, in metres. Picture the distance from one wave-crest to the next crest.
$f$ — the frequency: how many full wobbles happen each second, measured in hertz (Hz).
$n$ — a plain counting number $1, 2, 3, \dots$ that labels which allowed pattern we mean.
$T$ — the tension in the string: how hard it is pulled taut, measured in newtons (N). Picture two hands stretching the string. (Used only in Ex 8.)
$μ$ (the Greek letter "mu") — the linear mass density: how many kilograms of string sit in each metre of its length, in kg/m. A thick heavy string has large $μ$ . (Used only in Ex 8.)
$r$ — the radius of a pipe's opening, in metres. Picture the distance from the centre of the hole to its rim. (Used only in Ex 9.)

The scenario matrix

Here is the complete list of classes of situation this topic can throw at you. Every worked example below is tagged with the cell it fills, so you can check the whole grid gets covered.

Formula Every case this topic contains

Cell	System / condition	What makes it tricky	Example
A	String, both ends fixed — all harmonics	baseline node–node	Ex 1
B	Pipe open both ends — all harmonics	antinode–antinode, same math	Ex 2
C	Pipe closed one end — odd only	overtone ≠ harmonic number	Ex 3
D	Degenerate: shortest possible fit ( $n = 1$ )	is $n = 1$ really the smallest?	Ex 4
E	Zero / limiting input ( $v \to 0$ , $L \to \infty$ )	what happens at the edges	Ex 5
F	Real-world word problem	translate words → $L, v$	Ex 6
G	Exam twist: "shared harmonic?" trap	odd-vs-all numbering	Ex 7
H	Inverse problem: given $f$ , find $L$ or $v$	run the formula backward	Ex 8
I	End correction (real pipe)	$L_{eff} = L + 0.6 r$	Ex 9

We now fill every cell.

Ex 1 — Cell A · String fixed at both ends

Figure — Harmonics and overtones — on strings and in pipes

Figure s01 — A string clamped at both ends. The red curve is the fundamental standing wave: a single hump (antinode) in the middle with a node at each black wall. The double-headed arrow marks that the whole length equals half a wavelength, $L = λ /2$ .

Worked example Statement

A steel string of length $L = 0.80$ m carries transverse waves at $v = 480$ m/s. It is clamped at both ends. Find the fundamental $f_{1}$ and the third harmonic $f_{3}$ .

Forecast (guess first!): Will $f_{3}$ be exactly triple $f_{1}$ , or something messier? Jot a number.

Step 1 — draw the shortest fit. Why this step? Both ends are clamped (nodes), so the simplest standing wave is one hump between two nodes. Look at Figure s01: trace the red curve with your finger — it starts pinned at the left black wall, rises to a single bulge, and comes back to zero at the right wall. Count the zero-crossings that touch a wall: exactly two. That single hump spans a node-to-node distance, which the picture rule tells us is $λ /2$ . Therefore $L = \frac{λ _{1}}{2} \Rightarrow λ_{1} = 2 L = 1.60 m .$

Step 2 — build the general pattern (WHY $f_{n} = n f_{1}$ ). Why this step? We must justify the ladder, not just assert it. To make the $n$ -th mode, keep the two end-nodes and pack in more humps. Each extra hump adds one more node-to-node stretch of $λ /2$ , so $n$ humps fill a length $n \cdot λ_{n} /2$ . Since that must equal $L$ , $L = n \frac{λ _{n}}{2} \Rightarrow λ_{n} = \frac{2 L}{n}, n = 1, 2, 3, \dots$ Feed each $λ_{n}$ through the master relation: $f_{n} = \frac{v}{λ _{n}} = \frac{n v}{2 L} = n f_{1}, f_{1} = \frac{v}{2 L} .$ So every integer multiple is allowed — that is the whole reason $f_{n} = n f_{1}$ .

Step 3 — plug in numbers. Why this step? We now turn the general formulas into the two specific pitches the question asks for. $f_{1} = \frac{v}{2 L} = \frac{480}{1.60} = 300 Hz, f_{3} = 3 f_{1} = 900 Hz .$

Verify: Units: $\frac{m/s}{m} = 1/s = Hz$ ✓. Sanity: three humps must fit in the same length, so each hump is a third as long, so $λ$ is a third and $f$ is triple — matches.

Ex 2 — Cell B · Pipe open at both ends

Figure s02 — A pipe open at both ends (thin black tube walls). The red curve is the fundamental: an antinode at each open end and one node in the middle. The double-headed arrow shows $L = λ /2$ , exactly the same spacing as the clamped string.

Worked example Statement

A flute-like pipe open at both ends is $L = 0.34$ m long; sound speed $v = 340$ m/s. Find $f_{1}$ and the first overtone.

Forecast: Same formula as the string, or different because it's air not string?

Step 1 — draw the shortest fit. Why? Both ends open $\Rightarrow$ antinodes. Look at Figure s02: the red curve is biggest at both ends (the open mouths, where air swings freely) and crosses zero once in the middle — that single central node is the only pinned point. The shortest standing wave with an antinode at each end is therefore antinode–node–antinode. Antinode to the next antinode is also $λ /2$ , so the spacing is identical to the string: $L = \frac{λ _{1}}{2} \Rightarrow λ_{1} = 2 L = 0.68 m .$

Step 2 — build the general pattern. Why this step? Same node-counting argument as Ex 1: adding each extra half-wave keeps antinodes at both open ends, so $n$ half-waves fill the length, $L = n \frac{λ _{n}}{2} \Rightarrow λ_{n} = \frac{2 L}{n}, f_{n} = \frac{n v}{2 L} = n f_{1} .$ Every integer is allowed, exactly as for the string.

Step 3 — convert to frequency. Why this step? We have the allowed wavelength but the question asks for a pitch, and the master relation $f = v / λ$ is the only bridge from a length we can draw to a frequency we can hear — so we apply it here. $f_{1} = \frac{v}{λ _{1}} = \frac{340}{0.68} = 500 Hz .$

Step 4 — first overtone. Why? "Overtone" = next allowed frequency above $f_{1}$ . All harmonics exist here, so the next one is $f_{2} = 2 f_{1} = 1000$ Hz. The 1st overtone IS the 2nd harmonic.

Verify: Same $λ = 2 L$ math as the string confirms the parent's claim that string and open pipe share formulas. Units check as before ✓.

Ex 3 — Cell C · Pipe closed at one end (odd only)

Figure s03 — A pipe closed at the left (thick black wall) and open at the right. The red curve is the fundamental: a node at the closed wall growing to a full antinode at the open end. The double-headed arrow shows only a quarter wavelength fits, $L = λ /4$ — a long wave, hence a low pitch.

Worked example Statement

Close one end of that same $L = 0.34$ m pipe ( $v = 340$ m/s). Find $f_{1}$ and its first overtone.

Forecast: Will $f_{1}$ go up, down, or stay the same when we plug one end?

Step 1 — draw the shortest fit. Why? Closed end = node, open end = antinode. Look at Figure s03: notice the red curve is pinned flat against the thick left wall (a node) but swings to its fullest at the open right mouth (an antinode). Follow the double-headed arrow — the red curve only completes a quarter of a full wave shape across the tube: $L = \frac{λ _{1}}{4} \Rightarrow λ_{1} = 4 L = 1.36 m .$ Notice $λ_{1}$ is twice as long as the open pipe's — a longer wave means a lower pitch.

Step 2 — the odd-only ladder. Why this step? Starting from node→antinode, each next mode must add a full half-wave (node→antinode→node→antinode), so only odd quarter-waves fit: $L = (2 n - 1) \frac{λ _{n}}{4} \Rightarrow λ_{n} = \frac{4 L}{2 n - 1}, f_{n} = (2 n - 1) \frac{v}{4 L} = (2 n - 1) f_{1} .$

Step 3 — frequency. Why this step? We now have the allowed wavelength but the question asks for a pitch, so we cross the bridge $f = v / λ$ once more to turn that length into a frequency. $f_{1} = \frac{v}{λ _{1}} = \frac{340}{1.36} = 250 Hz .$ Exactly half the open-pipe answer (500 Hz) — one octave lower, as promised.

Step 4 — first overtone. Why? Allowed pitches are $250, 750, 1250, \dots$ (odd multiples). The first one above $250$ is $f_{1st overtone} = 3 f_{1} = 750 Hz = the 3rd harmonic.$

Verify: $f_{1}^{closed} = 250 = \frac{1}{2} (500) = \frac{1}{2} f_{1}^{open}$ ✓. The even harmonic $500$ Hz simply does not appear in the odd list — matches the parent's "even harmonics forbidden".

Ex 4 — Cell D · Degenerate: is $n = 1$ really the smallest?

Figure s04 — Why $n = 1$ is the smallest closed-pipe mode. The red solid curve is the allowed quarter wave (node at the closed wall, antinode at the open end). The black dotted curve is a shorter shape that would put a bulge at the closed wall — forbidden, because that wall must be a node.

Worked example Statement

A closed pipe has $L = 0.10$ m, $v = 340$ m/s. Could it possibly vibrate with half a quarter-wave — i.e. an even shorter pattern than $n = 1$ ? And what is its actual lowest note?

Forecast: Is there any allowed pattern below the fundamental?

Step 1 — check the boundary rule at the smallest fit. Why? A standing wave MUST place a node at the closed end and an antinode at the open end. Look at Figure s04: compare the two curves at the left wall. The red solid curve sits at zero there (a legal node); the black dotted curve rides up to a bulge there (an illegal antinode at a closed wall). The shortest legal curve that keeps a node at one wall and a bulge at the other is the red quarter wave — you physically cannot draw a shorter one. So $n = 1$ is by definition the smallest; nothing lower exists.

Step 2 — compute that fundamental. Why this step? Having proved the quarter wave is the smallest allowed shape, we now turn its wavelength into the actual lowest pitch with the master relation. $λ_{1} = 4 L = 0.40 m, f_{1} = \frac{v}{λ _{1}} = \frac{340}{0.40} = 850 Hz .$

Verify: The "half-a-quarter-wave" shape (black dotted) would put an antinode at the closed end — which violates the node rule — so it is forbidden. Good: the degenerate case can't sneak below $f_{1}$ . Units ✓.

Ex 5 — Cell E · Zero and limiting inputs

Worked example Statement

Using $f_{1} = v /2 L$ for an open pipe, describe what happens to the pitch as (a) $v \to 0$ , and (b) $L \to \infty$ . Then evaluate the borderline number: an open pipe with $L = 1.00$ m, $v = 340$ m/s.

Forecast: In each limit, does the pitch rise, fall, or vanish?

Step 1 — the $v \to 0$ limit. Why? $f_{1} = v /2 L$ is directly proportional to $v$ . If the wave speed shrinks to zero (imagine a medium where nothing propagates), then $f_{1} = \frac{v}{2 L} v \to 0 0.$ No travelling wave $\Rightarrow$ no standing wave $\Rightarrow$ no note. Physically sensible.

Step 2 — the $L \to \infty$ limit. Why? $L$ is in the denominator, so a huge pipe pushes $f_{1}$ toward zero: infinitely long pipes hum infinitely deep (a rumble below hearing). $f_{1} = \frac{v}{2 L} L \to \infty 0.$

Step 3 — the finite borderline value. Why this step? Limits tell us the edges; now we anchor the middle with one concrete, checkable number to make sure the formula behaves sensibly between the extremes. $f_{1} = \frac{340}{2 ( 1.00 )} = 170 Hz .$

Verify: Both limits push to $0$ Hz — consistent, because either killing the speed or stretching the length lengthens/loses the wave. The $170$ Hz number is a low, gentle pitch — plausible for a metre-long open tube. Units ✓.

Ex 6 — Cell F · Real-world word problem

Worked example Statement

A didgeridoo player has a straight wooden tube open at both ends, $L = 1.50$ m. On a warm day the speed of sound is $346$ m/s. What is the drone (fundamental) pitch, and what is the second harmonic she can overblow to?

Forecast: Above or below middle-ish (~250 Hz)?

Step 1 — translate words to symbols. Why? "Open both ends" $\Rightarrow$ antinode–antinode $\Rightarrow$ use $f_{1} = v /2 L$ with all harmonics allowed. That is the entire physics content of the sentence. $f_{1} = \frac{v}{2 L} = \frac{346}{2 ( 1.50 )} = \frac{346}{3.00} \approx 115.33 Hz .$

Step 2 — the overblown note. Why? Overblowing excites the next mode; open pipes allow every integer, so the second harmonic is $f_{2} = 2 f_{1} = \frac{346}{1.50} \approx 230.67 Hz .$

Verify: A didgeridoo drone sits around 60–130 Hz in real life, so $\approx 115$ Hz is right in the pocket ✓. $f_{2}$ is exactly double $f_{1}$ as an open pipe demands.

Ex 7 — Cell G · Exam twist: the "shared harmonic" trap

Worked example Statement

Claim on the exam: "An open pipe and a closed pipe of the same length $L = 0.50$ m ( $v = 340$ m/s) both produce a 340 Hz tone somewhere in their harmonic series." True or false?

Forecast: Sounds reasonable — decide before reading on.

Step 1 — list the OPEN pipe's allowed frequencies. Why? We must check which modes exist, not just plug a formula. Open: $f_{1} = \frac{340}{2 ( 0.50 )} = 340$ Hz, then $680, 1020, \dots$ — 340 Hz is the fundamental, definitely present.

Step 2 — list the CLOSED pipe's allowed frequencies. Why? Closed pipes forbid even harmonics. $f_{1} = \frac{340}{4 ( 0.50 )} = 170$ Hz, then odd multiples: $170, 510, 850, 1190, \dots$

Step 3 — compare. Why this step? The claim is about a shared tone, so the only way to settle it is to ask whether the same number appears in both lists — set membership, not a new formula. Is 340 Hz in the closed list ${170, 510, 850, 1190, \dots}$ ? No — 340 equals $2 \times 170$ , an even multiple of the closed fundamental, which is exactly the family the closed pipe throws away. The closest closed tones straddle it ( $170$ below, $510$ above) but never hit it.

Answer: The shared-tone claim is FALSE. The two pipes share no common tone here: the open pipe's series is ${340, 680, 1020, \dots}$ and the closed pipe's is ${170, 510, 850, \dots}$ , and these lists never overlap (open = even $\times 170$ , closed = odd $\times 170$ ).

Verify: The trap is assuming both pipes fill in every integer. Because the closed list is odd-only, the open pipe's fundamental lands in the gap — precisely the parent note's numbering warning ✓.

Ex 8 — Cell H · Inverse problem (formula run backward)

Worked example Statement

A guitar's open string sounds a fundamental of $f_{1} = 110$ Hz. The string length is $L = 0.64$ m. What wave speed $v$ does that require, and what tension $T$ gives it via > $v=\sqrt{T/\mu}$? (Take linear density $μ = 3.0 \times 1 0^{- 3}$ kg/m.)

Forecast: Will $v$ be a few hundred m/s (typical for strings)?

Step 1 — invert $f_{1} = v /2 L$ for $v$ . Why? We are given the pitch and length and want the speed, so algebraically solve for the unknown — multiply both sides by $2 L$ : $f_{1} = \frac{v}{2 L} \Rightarrow v = 2 L f_{1} = 2 (0.64) (110) = 140.8 m/s .$

Step 2 — get the tension. Why? The speed on a string comes from $v = T / μ$ ; to solve for $T$ we square both sides (undoing the square root) and multiply by $μ$ : $v = T / μ \Rightarrow T = μ v^{2} = (3.0 \times 1 0^{- 3}) (140.8)^{2} \approx 59.47 N .$

Verify: $v = 140.8$ m/s is a believable low-E-ish string speed ✓; $T \approx 59.5$ N is a realistic guitar tension (a few kilograms of pull). Units: $kg/m \cdot (m/s)^{2} = kg\cdotm/s^{2} = N$ ✓.

Ex 9 — Cell I · End correction (the real-pipe fix)

Worked example Statement

An open organ pipe has physical length $L = 0.30$ m and radius $r = 0.02$ m, $v = 340$ m/s. Compute the naive fundamental, then the corrected fundamental using $L_{eff} = L + 0.6 r$ per open end (two open ends).

Forecast: Will the correction raise or lower the pitch?

Step 1 — naive value. Why? Baseline to compare against, using the clean open-pipe formula. $f_{1}^{naive} = \frac{v}{2 L} = \frac{340}{0.60} \approx 566.67 Hz .$

Step 2 — add the end corrections. Why? The antinode actually bulges a little outside each open end, so the wave behaves as if the tube were longer. Two open ends means two corrections: $L_{eff} = L + 2 (0.6 r) = 0.30 + 2 (0.6) (0.02) = 0.324 m .$

Step 3 — corrected frequency. Why this step? The whole point of the correction was to fix the pitch, so we feed the larger effective length back into $f_{1} = v /2 L_{eff}$ to get the value a real pipe actually sounds. $f_{1}^{corr} = \frac{v}{2 L _{eff}} = \frac{340}{0.648} \approx 524.69 Hz .$

Verify: A longer effective tube gives a lower pitch, so $524.69 < 566.67$ Hz ✓ — the correction always drops the frequency, exactly as the parent's mistake-box warns.

Active recall

Recall Quick self-test

In Ex 3, why does closing one end drop the fundamental to half? ::: The node–antinode boundary needs only $λ /4$ , so $λ = 4 L$ is double the open pipe's $2 L$ ; double $λ$ halves $f$ . In Ex 7, why isn't 340 Hz in the closed pipe's list? ::: 340 Hz is an even multiple of the closed fundamental 170 Hz, and closed pipes forbid even harmonics. In Ex 5, both limits ( $v \to 0$ , $L \to \infty$ ) send $f_{1}$ to what value? ::: Zero hertz — no note. In Ex 8, how do you get $v$ from $f_{1}$ and $L$ ? ::: Invert $f_{1} = v /2 L$ to $v = 2 L f_{1}$ . In Ex 9, does the end correction raise or lower the pitch? ::: Lowers it, because the effective length is larger.

Connections

Standing waves and superposition — why only certain $λ$ survive at all.
Reflection of waves at boundaries — the node/antinode rules used in every step.
Speed of a wave on a string — supplies $v$ in Ex 1 and Ex 8.
Speed of sound in gases — supplies $v$ in Ex 2, 3, 6, 9.
Beats and resonance — how a driver locks onto these frequencies.
Timbre and Fourier synthesis — why the mix of these overtones defines an instrument.