Normal Modes & Harmonics on a String

Pluck a guitar string and it does not make a noise — it makes a note. Pluck it harder and the note is louder but not higher; pin it down at the twelfth fret and the note jumps up exactly one octave. A drum, by contrast, makes a thud with no clear pitch at all. The difference is the whole subject of this article: a string fixed at both ends is not free to vibrate at just any frequency. It is allowed a strict, evenly spaced menu of frequencies — a fundamental and its whole-number multiples — and nothing in between. The previous article in this chapter built the standing wave: send a wave down a string, let it reflect off a fixed end, and the incoming and outgoing waves add to a pattern that stands still, with nodes that never move and antinodes that swing hardest. Here we pin both ends down and ask the decisive question: which standing waves survive? A fixed end must be a node, so only those patterns that place a node at each end can live on the string. That single boundary rule — node here, node there — quantises the string. Fit one loop, two loops, three loops between the ends, never one-and-a-half, and out drops the harmonic series

f_n = n\,\frac{v}{2L} = n f_1, \qquad n = 1, 2, 3, \dots

with the wave speed $v = \sqrt{T/\mu}$ carried over from earlier in the chapter. This is why a string is a musical object and a slammed door is not, why halving a length raises the pitch an octave, and why tightening a peg sharpens a note — but only as the square root of the tension. Every one of those facts is one line of algebra away.

Key takeaways

A string fixed at both ends must have a node at each end, so it can only carry standing waves that fit a whole number of half-wavelengths into its length $L$ .
That boundary rule gives $\lambda_n = 2L/n$ and $f_n = n\,v/(2L) = n f_1$ : a fundamental $f_1$ and every whole-number multiple — the harmonic series, with no gaps.
The wave speed is $v = \sqrt{T/\mu}$ (quoted from earlier in the chapter), where $\mu$ is the mass per unit length in kg per metre.
The $n$ -th mode has $n$ antinodes and $n+1$ nodes; adjacent nodes are $\lambda_n/2 = L/n$ apart.
Mersenne's three laws fall straight out: $f \propto 1/L$ , $f \propto \sqrt{T}$ , $f \propto 1/\sqrt{\mu}$ . So double the tension and the pitch rises by only $\sqrt{2}$ , not $\times 2$ .
Halving the length raises the pitch by exactly one octave ( $\times 2$ ) — the physics behind every fret and every fingered note.
Key idea: a string fixed at both ends is a frequency filter — it rings only at $f_1, 2f_1, 3f_1, \dots$ because only those standing waves place a node at each clamped end, and the spacing of that menu is set entirely by $L$ , $T$ and $\mu$ .

The core insight

Clamping both ends of a string forces a node at each end, which admits only the standing waves that fit a whole number of half-wavelengths into the length; the surviving modes have frequencies $f_n = n\,v/(2L)$ with $v = \sqrt{T/\mu}$ , an evenly spaced harmonic series whose fundamental and spacing are fixed by the length, tension and linear mass density alone.

Where this lives today

Every stringed instrument is equation (5) made playable. A piano spans its range by all three of Mersenne's dials at once: the strings lengthen and thicken toward the bass (raising $L$ and $\mu$ ) and are tuned by tension pegs (setting $T$ ), so a grand piano's longest bass string is over two metres while its top treble string is a few centimetres. A guitarist transposing up an octave frets at the twelfth fret — the exact midpoint — halving $L$ ; a violinist's vibrato is a fast, tiny wobble of $L$ heard as a wavering pitch. Beyond music, the same boundary-condition mathematics sets the resonances of a suspension-bridge deck, the vibration modes of a turbine blade, and the standing electron waves in a quantum well — engineers avoid a bridge's natural $f_n$ the way a luthier seeks a string's. The harmonic series you hear when a string is plucked is the audible face of one of the deepest ideas in physics: a bounded system has a discrete spectrum.

The physical picture

A standing wave on a string is the sum of two identical travelling waves running in opposite directions — one you sent, one that bounced back off a fixed end. Add them with the sum-to-product identity and the travelling character cancels, leaving a shape that stays put and merely breathes up and down:

y(x,t) = 2A\,\sin(kx)\,\cos(\omega t) \qquad (1)

where: $A$ — amplitude of each travelling wave; $k = 2\pi/\lambda$ — the wave number; $\omega = 2\pi f$ — the angular frequency; $x$ — position along the string. The important thing is that space and time have separated: $\sin(kx)$ fixes the shape, $\cos(\omega t)$ sets the timing. Points where $\sin(kx) = 0$ never move — those are the nodes. Points where $\sin(kx) = \pm 1$ swing hardest — the antinodes. The string does not travel anywhere; it stands and oscillates in place. Now impose the physics of the clamps. Both ends are tied down, so the string cannot move there: $y = 0$ at $x = 0$ and at $x = L$ , for all time. This is a boundary condition, and it is the entire reason a string is choosy about its frequencies. At $x = 0$ , equation (1) already gives $y = 0$ because $\sin 0 = 0$ — good, that end is free of charge. At $x = L$ we need $\sin(kL) = 0$ , which is true only when $kL$ is a whole number of $\pi$ :

kL = n\pi \quad\Rightarrow\quad \frac{2\pi}{\lambda_n}L = n\pi \quad\Rightarrow\quad \lambda_n = \frac{2L}{n}, \qquad n = 1, 2, 3, \dots \qquad (2)

Read equation (2) as geometry, because that is all it is: a node sits at each end, so the length must hold a whole number of half-loops. One half-wavelength gives the fundamental; two gives the next mode; three the next. You physically cannot fit one-and-a-half loops between two nodes, which is why the in-between frequencies simply do not exist on the string.

Figure

The first three modes of a string fixed at both ends, with nodes marked as dots and the half wavelength count and frequency shown for each

Turn the wavelengths into frequencies with the one imported fact, $v = f\lambda$ . The wave speed on the string comes from the tension pulling it taut and the mass it has to drag along:

v = \sqrt{\frac{T}{\mu}} \qquad (3)

where: $T$ — the tension in newtons; $\mu = m/L$ — the linear mass density, mass per unit length, in kg per metre. We do not re-derive equation (3) here — it is the wave-speed-on-a-string result established earlier in this chapter — but you can see it must be right on units alone: tension has units of $\mathrm{kg\,m\,s^{-2}}$ and $\mu$ has units of $\mathrm{kg\,m^{-1}}$ , so $\sqrt{T/\mu}$ has units of $\mathrm{m\,s^{-1}}$ , a speed, and it is the only combination that does. Stiffer pull, faster wave; heavier string, slower wave.

The harmonic series

Combine equation (2) with $f = v/\lambda$ and the whole spectrum appears:

f_n = \frac{v}{\lambda_n} = \frac{v}{2L/n} = n\,\frac{v}{2L} = n f_1, \qquad f_1 = \frac{v}{2L} \qquad (4)

where: $f_1$ — the fundamental frequency (also called the first harmonic); $f_n$ — the $n$ -th harmonic. This is the punchline of the article. The allowed frequencies are $f_1, 2f_1, 3f_1, 4f_1, \dots$ — a perfectly even ladder, each rung a whole-number multiple of the bottom one. There are no missing rungs: a string carries every harmonic (that "missing even harmonics" story belongs to the closed pipe, next article). This even, gap-free ladder is exactly what the ear recognises as a single musical pitch — the fundamental sets the note, the higher harmonics colour its tone. Put numbers to it. Take a light steel wire with $\mu = 1.0\ \mathrm{g\,m^{-1}} = 1.0 \times 10^{-3}\ \mathrm{kg\,m^{-1}}$ under $T = 90\ \mathrm{N}$ of tension. First the speed, and it comes out clean:

v = \sqrt{\frac{90}{1.0\times10^{-3}}} = \sqrt{90000} = 300.0\ \mathrm{m\,s^{-1}}.

Stretch a length $L = 0.75\ \mathrm{m}$ of it between two bridges and the fundamental is

f_1 = \frac{v}{2L} = \frac{300.0}{2 \times 0.75} = \frac{300.0}{1.5} = 200.0\ \mathrm{Hz},

so the string rings at $f_1 = 200\ \mathrm{Hz}$ , $f_2 = 400\ \mathrm{Hz}$ , $f_3 = 600\ \mathrm{Hz}$ , and so on up the ladder. In the third mode ( $n = 3$ ) the string carries three half-loops, so it has $3$ antinodes and $4$ nodes, of which $2$ are the internal ones you can actually see standing still in the middle. Adjacent nodes sit $\lambda_3/2 = L/3 = 0.75/3 = 0.25\ \mathrm{m}$ apart — touch the string lightly a quarter of the way along and you force a node there, killing every mode that does not already have one at that spot, which is exactly how a guitarist plays a "harmonic".

Mersenne's three laws

Equation (4) with $v = \sqrt{T/\mu}$ substituted in is really three laws in one, discovered empirically by Marin Mersenne in 1636, half a century before anyone could derive them:

f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}} \qquad (5)

where: every symbol you can change on an instrument appears once. Read off the three proportionalities, each with the other two held fixed: $f_1 \propto 1/L$ (shorten the string, sharpen the note), $f_1 \propto \sqrt{T}$ (tighten it, sharpen it — but as a square root), and $f_1 \propto 1/\sqrt{\mu}$ (a heavier string sounds lower, which is why a bass string is thick and often wound with wire). The square roots are where marks are won and lost. To raise the pitch a full octave by tension alone you would need to quadruple the tension, because $f \propto \sqrt{T}$ and $\sqrt{4} = 2$ ; merely doubling the tension raises the pitch by only $\sqrt{2} = 1.414$ , about a musical fifth, not an octave. The length law is cleaner and needs no square root, which is why instruments are played by fingering, not by re-tensioning on the fly. Halve the length and you exactly double the frequency:

f_1(L/2) = \frac{v}{2(L/2)} = \frac{v}{L} = 2 \times \frac{v}{2L} = 2 f_1.

For our wire, pressing it to half its length, $0.375\ \mathrm{m}$ , turns $200\ \mathrm{Hz}$ into $400\ \mathrm{Hz}$ — one octave up, the interval every guitarist gets by fretting at the twelfth fret, the exact midpoint of the string.

Figure

A sonometer bench with a wire over two movable bridges stretched by a hanging mass, showing the tension equals mass times g and the fundamental loop between the bridges

The bench in the figure is the sonometer, the standard school apparatus for testing every one of these laws. A wire runs over two movable bridges — the spacing between them is the vibrating length $L$ — and is stretched by a mass $M$ hung over a pulley, so the tension is simply the weight $T = Mg$ . Take $M = 9.0\ \mathrm{kg}$ with $g = 9.8\ \mathrm{m\,s^{-2}}$ (the CBSE and NEET value), and the tension is $T = 9.0 \times 9.8 = 88.2\ \mathrm{N}$ . On the same $\mu = 1.0 \times 10^{-3}\ \mathrm{kg\,m^{-1}}$ wire that gives $v = \sqrt{88.2/10^{-3}} = \sqrt{88200} = 297.0\ \mathrm{m\,s^{-1}}$ , and with the bridges set at $L = 0.50\ \mathrm{m}$ the fundamental is $f_1 = 297.0/(2 \times 0.50) = 297.0\ \mathrm{Hz}$ . Change any one dial and equation (5) predicts the new pitch exactly — that is the point of the whole instrument.

Derivation AWhy a string sounds only the harmonic series

The goal here is to show, from the boundary conditions alone, that a string clamped at both ends is forced to vibrate at the frequencies $f_n = n\,v/(2L)$ and at no others — that the harmonic series is not a musical convention but a geometric inevitability. The only fact we import from outside is the wave speed $v = \sqrt{T/\mu}$ ; everything else is the shape of a standing wave and the physics of a clamp. The strategy is to write the general standing wave, kill it at both ends, and read off which wavelengths survive.

Step 1. Write the standing wave. Two identical waves travelling in opposite directions along the string, $A\sin(kx - \omega t)$ and $A\sin(kx + \omega t)$ , add by the sum-to-product identity to $y(x,t) = 2A\sin(kx)\cos(\omega t)$ . Space and time separate: $\sin(kx)$ is the frozen shape, $\cos(\omega t)$ is the common up-and-down beat.

Step 2. Apply the left clamp. The string is tied at $x = 0$ , so $y(0, t) = 0$ for all $t$ . Since $\sin(0) = 0$ , this holds automatically — the sine form was the right choice, and it already guarantees a node at the left end.

Step 3. Apply the right clamp, and this is where the quantisation happens. We need $y(L, t) = 2A\sin(kL)\cos(\omega t) = 0$ at all times, so we must have $\sin(kL) = 0$ . The sine vanishes only at whole multiples of $\pi$ , so $kL = n\pi$ with $n = 1, 2, 3, \dots$ A non-whole value of $n$ would leave the right end swinging, which the clamp forbids.

Step 4. Turn the condition into wavelengths and frequencies. With $k = 2\pi/\lambda$ , the rule $kL = n\pi$ becomes $\lambda_n = 2L/n$ . Feed that into $f = v/\lambda$ with $v = \sqrt{T/\mu}$ : $f_n = v/\lambda_n = n v/(2L) = (n/2L)\sqrt{T/\mu}$ . The allowed frequencies are an evenly spaced ladder $f_1, 2f_1, 3f_1, \dots$ , and the string is silent at every frequency between the rungs. Verified: v = √(90 / 0.001) = 300.0 m/s exactly, f₁ = 300.0 / 1.5 = 200.0 Hz, f₂ = 400.0, f₃ = 600.0 Hz, and node spacing L/3 = 0.25 m are recomputed in verify_hs2-modes-on-a-string.py.

Go deeper · Derivation B · 1 minMersenne's laws and the square-root trap

Derivation BMersenne's laws and the square-root trap

Here we extract the three practical laws an instrument-maker actually uses — how the pitch responds to length, tension and thickness — from the single formula $f_1 = (1/2L)\sqrt{T/\mu}$ , and we pin down the number that examiners build their traps around: doubling the tension does not double the pitch. The goal is to make the $\sqrt{T}$ dependence concrete with the locked numbers, so that "double the tension" and "octave up" never get confused again.

Step 1. Isolate each dependence from $f_1 = (1/2L)\sqrt{T/\mu}$ by holding the other two variables fixed. Length: $f_1 \propto 1/L$ . Tension: $f_1 \propto \sqrt{T}$ . Linear density: $f_1 \propto 1/\sqrt{\mu}$ . These are Mersenne's three laws (1636), each a separate empirical discovery, here unified.

Step 2. Test the tension law on the master wire. Start at $T = 90\ \mathrm{N}$ , $v = 300.0\ \mathrm{m\,s^{-1}}$ , $f_1 = 200.0\ \mathrm{Hz}$ at $L = 0.75\ \mathrm{m}$ . Now double the tension to $180\ \mathrm{N}$ : $v = \sqrt{180/10^{-3}} = \sqrt{180000} = 424.3\ \mathrm{m\,s^{-1}}$ , and $f_1 = 424.3/(2\times0.75) = 282.8\ \mathrm{Hz}$ .

Step 3. Read the ratio. The new frequency over the old is $282.8/200.0 = 1.414 = \sqrt{2}$ , not $2$ . Doubling the tension multiplies the pitch by $\sqrt{2}$ — about a musical fifth — because $f \propto \sqrt{T}$ . To gain a whole octave ( $\times 2$ ) you would have to quadruple the tension, and quadrupling the tension on a real string is how strings snap.

Step 4. Contrast the length law, which has no square root. Halving $L$ from $0.75$ to $0.375\ \mathrm{m}$ gives $f_1 = 300.0/(2\times0.375) = 400.0\ \mathrm{Hz}$ , exactly double — one clean octave. This is why pitch is controlled by fingering (change $L$ ) rather than by re-tuning (change $T$ ) in the middle of a phrase. Verified: v(180 N) = √(180 / 0.001) = 424.3 m/s, f₁ = 282.8 Hz, the ratio 282.8 / 200.0 = 1.414 = √2, and the octave f₁(0.375 m) = 400.0 Hz are recomputed in verify_hs2-modes-on-a-string.py.

Limiting cases

case	result	meaning
$n = 1$	$\lambda_1 = 2L$ , $f_1 = v/2L$	the fundamental: one half-loop, longest wavelength, lowest note
tension $\to 4T$	$f_1 \to 2f_1$	quadrupling tension raises the pitch one octave (because $\sqrt{4} = 2$ )
tension $\to 2T$	$f_1 \to \sqrt{2}\,f_1 = 1.414 f_1$	doubling tension raises it only a musical fifth, not an octave
length $\to L/2$	$f_1 \to 2f_1$	halving the length is a clean octave — no square root
$\mu \to 4\mu$	$f_1 \to f_1/2$	a string four times heavier per metre sounds an octave lower
$n$ -th mode	$n$ antinodes, $n+1$ nodes, spacing $L/n$	touch a node point and you select that harmonic

Common trapDoubling the tension does not double the pitch

The single most expensive misconception on this topic is treating $f \propto T$ when the truth is $f \propto \sqrt{T}$ . Double the tension and the pitch rises by $\sqrt{2} = 1.414$ (about $41\%$ ), not by $\times 2$ ; to raise it a full octave you must quadruple the tension. The square root also hides in the mass: it is $1/\sqrt{\mu}$ , not $1/\mu$ , so a string four times heavier per metre sounds only half the frequency, one octave down. And $\mu$ must be in kilograms per metre: a wire quoted as $1\ \mathrm{g\,m^{-1}}$ is $1 \times 10^{-3}\ \mathrm{kg\,m^{-1}}$ , and forgetting the factor of a thousand puts $\sqrt{1000} = 31.6$ into your speed — the classic factor-of-thirty blunder that turns $300\ \mathrm{m\,s^{-1}}$ into $9.5\ \mathrm{m\,s^{-1}}$ or $9500\ \mathrm{m\,s^{-1}}$ . Before you square-root anything, write $\mu$ with its unit beside it and check it reads kg per metre.

Worked examples

Board level — the sonometer wire (CBSE and ICSE pattern)

A wire of linear mass density $1.0\ \mathrm{g\,m^{-1}}$ is stretched between two bridges $0.75\ \mathrm{m}$ apart under a tension of $90\ \mathrm{N}$ . Find the wave speed, the fundamental frequency, and the frequencies of the second and third harmonics.

Step 1. Convert before touching a formula — this is where the marks are lost. $\mu = 1.0\ \mathrm{g\,m^{-1}} = 1.0 \times 10^{-3}\ \mathrm{kg\,m^{-1}}$ . Tension and length are already SI: $T = 90\ \mathrm{N}$ , $L = 0.75\ \mathrm{m}$ .

Step 2. Wave speed from equation (3): $v = \sqrt{T/\mu} = \sqrt{90/(1.0\times10^{-3})} = \sqrt{90000} = \boxed{300.0\ \mathrm{m\,s^{-1}}}$ .

Step 3. Fundamental from equation (4): $f_1 = v/(2L) = 300.0/(2 \times 0.75) = \boxed{200.0\ \mathrm{Hz}}$ .

Step 4. The harmonics are whole-number multiples: $f_2 = 2f_1 = \boxed{400.0\ \mathrm{Hz}}$ and $f_3 = 3f_1 = \boxed{600.0\ \mathrm{Hz}}$ . Sense-check: the ladder is evenly spaced by $200\ \mathrm{Hz}$ , as a fixed-fixed string must be. Had the numbers come out unevenly spaced, you would have made a slip.

Board level — retuning by tension and by length (ISC and A-Level pattern)

The wire above sounds $200\ \mathrm{Hz}$ . (a) The tension is doubled to $180\ \mathrm{N}$ ; find the new fundamental. (b) Instead, keeping the original tension, the vibrating length is halved to $0.375\ \mathrm{m}$ ; find the new fundamental. Comment on the two results.

Step 1. Doubling the tension (part a). The speed scales as $\sqrt{T}$ : $v = \sqrt{180/(1.0\times10^{-3})} = \sqrt{180000} = 424.3\ \mathrm{m\,s^{-1}}$ , so $f_1 = 424.3/(2\times0.75) = \boxed{282.8\ \mathrm{Hz}}$ .

Step 2. Read the ratio: $282.8/200.0 = 1.414 = \sqrt{2}$ . Doubling the tension raised the pitch by $\sqrt 2$ , about a musical fifth — not an octave. To double the frequency you would need four times the tension.

Step 3. Halving the length (part b). Now $f_1 = v/(2L) = 300.0/(2\times0.375) = \boxed{400.0\ \mathrm{Hz}}$ , exactly double the original.

Step 4. The contrast is the lesson. Length has no square root, so halving it is a clean octave; tension does have a square root, so doubling it falls short of an octave. This is precisely why a guitarist changes notes by fretting (length) rather than by retensioning mid-phrase.

Competitive level — find the wave speed from a played note (JEE-Main and NEET pattern)

A guitar's open A2 string has a vibrating length of $0.65\ \mathrm{m}$ and sounds a fundamental of $110\ \mathrm{Hz}$ . (a) What is the wave speed on the string? (b) If the string's linear mass density is $6.0\ \mathrm{g\,m^{-1}}$ , what tension is it under? Take the tabulated A2 pitch as exact.

Step 1. Invert the fundamental formula. From $f_1 = v/(2L)$ , the speed is $v = 2Lf_1 = 2 \times 0.65 \times 110 = \boxed{143.0\ \mathrm{m\,s^{-1}}}$ . No tension or density is needed for the speed — the played note already contains it.

Step 2. Now use $v = \sqrt{T/\mu}$ backwards for the tension: $T = \mu v^2$ . Convert the density first: $\mu = 6.0\ \mathrm{g\,m^{-1}} = 6.0 \times 10^{-3}\ \mathrm{kg\,m^{-1}}$ .

Step 3. Substitute: $T = 6.0\times10^{-3} \times (143.0)^2 = 6.0\times10^{-3} \times 20449 = \boxed{122.7\ \mathrm{N}}$ . A plausible tension for a wound bass-side guitar string, which is a good sanity check.

Step 4. Notice the shape of the shortcut: whenever you are given the frequency a string actually sounds, the fastest route to the speed is $v = 2Lf_1$ , and only then do you reach for $\sqrt{T/\mu}$ if the tension is asked. Computing $v$ from $\sqrt{T/\mu}$ first, when the note is handed to you, is wasted work. Timing note: about 60 seconds. The trap is doing it the long way — students who reach for $\sqrt{T/\mu}$ when the frequency is already given have missed that $v = 2Lf_1$ is one line. Learn to spot which data the question hands you.

Exam traps

$\mu$ in kilograms per metre, always. A wire quoted as " $1\ \mathrm{g\,m^{-1}}$ " or " $2\ \mathrm{g\,cm^{-1}}$ " must be converted before it touches $\sqrt{T/\mu}$ . The g-to-kg slip is a factor of a thousand under a square root, i.e. $\sqrt{1000} = 31.6$ in the speed — the single most common wrong answer on this topic. Write the unit next to the number and check it reads kg per metre before you square-root.

$f \propto \sqrt{T}$ , never $f \propto T$ . Doubling the tension multiplies the pitch by $\sqrt 2 = 1.414$ , not by $2$ . Quadrupling gives the octave. NEET and JEE distractor lists always include the " $\times 2$ " answer for a doubled tension — it is there to catch exactly this error.

Count nodes and antinodes correctly. The $n$ -th mode has $n$ antinodes but $n+1$ nodes, of which $n-1$ are internal (visible in the middle). "Number of nodes = number of antinodes" is wrong for a fixed-fixed string, and questions asking for the number of internal nodes want $n - 1$ .

The fundamental is the first harmonic. A string fixed at both ends carries every harmonic $1, 2, 3, \dots$ with no gaps. Do not confuse it with the closed pipe (next article), which skips its even harmonics. Also mind the overtone naming: the first overtone is the second harmonic ( $2f_1$ ), the second overtone is the third harmonic, and so on — overtone number is one less than harmonic number.

Which $g$ on a sonometer? The tension is $T = Mg$ , so state your $g$ : CBSE and NEET use $9.8\ \mathrm{m\,s^{-2}}$ , A-Level uses $9.81$ , some IGCSE material uses $10$ . A hanging $9.0\ \mathrm{kg}$ gives $88.2\ \mathrm{N}$ at $9.8$ but $90\ \mathrm{N}$ at $10$ — a real difference in the final frequency, so never mix values mid-solution.

Origins · 1 minJacques Charles François Sturm and how this was discovered

Jacques Charles François Sturm

1803–1855 · Geneva & Paris

Charles-François Sturm was a Geneva-born mathematician who, with Joseph Liouville in 1836–37, created the theory that now bears both their names: Sturm–Liouville theory, the study of which vibration patterns a bounded system is allowed to have. That is not a footnote to this article — it is this article, generalised. The clamped string's demand that only certain wavelengths survive is the simplest Sturm–Liouville problem, and its evenly spaced spectrum $f_n = n f_1$ is that theory's cleanest example. Sturm did not build sonometers or study music; he was a pure mathematician who gave physics the machinery that decides which normal modes any bounded object may ring at. (He also, with Colladon in 1826, made the first good measurement of the speed of sound in water — a story for the speed-of-sound article.) His theorem on counting a function's zeros is still taught today.

History

1636

— Marin Mersenne, in Harmonie universelle, publishes the three laws of the vibrating string — pitch varies as one over the length, as the square root of the tension, and inversely as the square root of the thickness — measured empirically, before any theory existed.

1638

— Galileo, in Two New Sciences, ties pitch to frequency of vibration and describes sympathetic resonance between strings tuned alike.

1715

— Brook Taylor gives the first mathematical treatment of the fundamental mode of a vibrating string, deriving its frequency from the string's shape.

1747

— Jean d'Alembert writes down the wave equation for the string and its travelling-wave solution, the calculus that underlies the picture used here.

1837

— Charles-François Sturm and Joseph Liouville publish the eigenvalue theory that explains, in full generality, why a bounded string is allowed only a discrete set of normal modes.

Going further

Texts. NCERT Physics Class 11, Ch. 15 ("Waves"), the sections on standing waves and normal modes of a stretched string · H. C. Verma, Concepts of Physics Vol. 1, "Wave Motion and Waves on a String" · Halliday, Resnick and Walker, "Waves — I", standing waves on a string · Cambridge A-Level 9702 topic 8 (stationary waves).

Provenance. All results standard. Recompute: $v = \sqrt{90/10^{-3}} = 300.0\ \mathrm{m\,s^{-1}}$ ; $f_1 = 300.0/1.5 = 200.0$ , $f_2 = 400.0$ , $f_3 = 600.0\ \mathrm{Hz}$ ; node spacing $L/3 = 0.25\ \mathrm{m}$ ; doubled tension $v = \sqrt{180/10^{-3}} = 424.3\ \mathrm{m\,s^{-1}}$ , $f_1 = 282.8\ \mathrm{Hz}$ , ratio $282.8/200.0 = 1.414 = \sqrt 2$ ; halved length $f_1 = 400.0\ \mathrm{Hz}$ ; sonometer $T = 9.0 \times 9.8 = 88.2\ \mathrm{N}$ , $v = \sqrt{88200} = 297.0\ \mathrm{m\,s^{-1}}$ , $f_1(L=0.5) = 297.0\ \mathrm{Hz}$ ; guitar $v = 2\times0.65\times110 = 143.0\ \mathrm{m\,s^{-1}}$ , $T = 6.0\times10^{-3}\times143.0^2 = 122.7\ \mathrm{N}$ .

Prerequisites. Periodic Motion and the Definition of SHM — every point on the string executes SHM, and the standing wave is SHM shared across a length. Leads to. Standing Waves in Air Columns: Open and Closed Pipes — the same boundary-condition logic for a pipe · ARCHETYPE: Sonometer and String-Harmonic Problems — the exam drill for every variant of equation (5).

Go deeper (UG). The full theory of the vibrating string, with the standing waves emerging as the eigenfunctions of the wave equation, is developed in Normal Modes and Standing Waves on a String in the university mechanics track. It is enrichment — nothing on this page depends on it.

Why can't a string vibrate at just any frequency, the way I can wave a rope at any speed?

A free rope you wave by hand has only one fixed end (your hand controls the other), so it is far less constrained. A string clamped at both ends must sit still at each end for all time, and only a standing wave that already places a node at each end can do that. Those are the patterns that fit a whole number of half-loops into the length — one loop, two loops, three — and their frequencies are $f_1, 2f_1, 3f_1, \dots$ A frequency between two rungs would need a non-whole number of half-loops, which would leave an end swinging, and the clamp forbids that. The boundary conditions, not the plucking, choose the menu.

What is the difference between an overtone and a harmonic? My book uses both.

They count the same modes but start counting from different places. The harmonics are numbered by the multiple: the first harmonic is $f_1$ (the fundamental), the second harmonic is $2f_1$ , the third is $3f_1$ . The overtones are the modes above the fundamental, so the first overtone is the second harmonic ( $2f_1$ ), the second overtone is the third harmonic ( $3f_1$ ), and so on. The rule is simply: overtone number = harmonic number $-1$ . For a string, which carries every harmonic, this is a clean shift. It gets trickier for a closed pipe, which is missing its even harmonics — there the first overtone is the third harmonic.

The formula uses v = √(T/μ) but I have not derived that. Am I allowed to just use it?

Yes — that result was established earlier in this chapter, in the article on wave speed on a string, and quoting it here is exactly right. This article's job is what the boundary conditions do (the harmonic series), not where the wave speed comes from. If you want a sanity check that does not need the derivation, use units: tension is measured in $\mathrm{kg\,m\,s^{-2}}$ and $\mu$ in $\mathrm{kg\,m^{-1}}$ , so $\sqrt{T/\mu}$ has units of $\mathrm{m\,s^{-1}}$ — a speed — and it is the only combination of $T$ and $\mu$ that gives a speed at all. That is why the formula has to have this form even before you prove it.

When I lightly touch a guitar string at its midpoint I get a clear high note. What is happening?

You are forcing a node at the midpoint. Any mode that does not already have a node there is killed, because you are pinning the string still at a point where that mode wants to move. The fundamental has its antinode at the midpoint, so it dies; the second harmonic $2f_1$ has a node exactly there, so it survives — and you hear a note one octave up. Touch at a third of the length and you select the third harmonic, and so on. Guitarists call these "harmonics" or "flageolet tones", and they are a direct, audible demonstration of equation (2): the surviving modes are precisely those whose node pattern already fits where you touched.

Does a thicker string always sound lower than a thin one?

Only if nothing else changes. Pitch goes as $1/\sqrt{\mu}$ , so a string with four times the mass per metre sounds an octave lower — at the same length and tension. That is why bass strings are thick, often a thin core wound with heavy wire to raise $\mu$ without making the string too stiff to bend. But a thick string tightened hard or shortened can easily outpitch a thin, loose, long one: all three of Mersenne's dials act together. On a real instrument the maker balances $L$ , $T$ and $\mu$ so that every string, thick or thin, sits at a comfortable tension while sounding its intended note.

Is this the same physics as a standing wave in a pipe or a microwave oven?

The idea is identical; only the boundary rules change. Any bounded system — a string, an air column, the electromagnetic field in a microwave cavity, an electron in a box — is allowed only the standing waves that satisfy its boundary conditions, and that always produces a discrete set of modes. For a string fixed at both ends the rule is "node at each end", giving $f_n = n\,v/(2L)$ with every harmonic present. An air column has different end rules (open ends are antinodes), which the next article works out and which is why a closed pipe drops its even harmonics. The unifying statement — bounded system, discrete spectrum — is one of the load-bearing ideas of all of physics.