Vibrations of stretched strings

Pluck a guitar string and you hear a clean musical note, not a random noise — because the string can only vibrate in a very specific pattern once both ends are clamped down. This article asks a simple question: given a string of length $l$ , tension $T$ and mass per unit length $m$ , what note does it produce, and how does that note change if you tighten the string, shorten it, or swap it for a thicker one? The answer is a single formula, $f_0=\dfrac{1}{2l}\sqrt{T/m}$ , but the real payoff is in the reasoning that gets you there: fixed ends force a standing-wave pattern, the pattern fixes the wavelength, and the wave speed on the string converts that wavelength into a frequency. Along the way you get three "laws of vibration" that a sonometer experiment verifies directly, and that explain everyday facts about guitars, violins and pianos — why pressing a fret raises pitch, why tightening a peg raises pitch, and why bass strings are thick.

The core insight

A string of length $l$ fixed at both ends must have nodes at both ends, forcing its fundamental wavelength to be $\lambda_0=2l$ , which combines with the wave speed $v=\sqrt{T/m}$ to give the fundamental frequency $f_0=\dfrac{1}{2l}\sqrt{T/m}$ .

The core idea

Fixed ends force nodes. A string clamped rigidly at both ends cannot move at either clamp, so both ends must be nodes of any stationary wave that forms on it. A stationary wave pattern of wavelength $\lambda$ can only fit between two fixed nodes if the string's length is a whole number of half-wavelengths:

l = n\frac{\lambda}{2}, \qquad n=1,2,3,\dots

where: $l$ is the string length, $\lambda$ is the wavelength of the standing-wave pattern, and $n$ counts how many half-wavelength "loops" fit along the string.

The fundamental mode. The simplest possible pattern ( $n=1$ ) has just one antinode, at the centre, and nodes only at the two ends. Setting $n=1$ gives the fundamental wavelength:

\lambda_0 = 2l

This is the single most important fact in the topic: the fundamental "loop" is only half a wavelength long, so the wavelength is twice the string length, never equal to it.

Wave speed on the string. A transverse wave travels along the string at a speed set by the tension (the restoring force) and the mass per unit length (the inertia):

v=\sqrt{\frac{T}{m}}

where: $T$ is the tension in the string (newtons) and $m$ is the mass per unit length, i.e. linear mass density (kg/m) — not the total mass of the wire. This relation is taken as given by the syllabus, the way $v=f\lambda$ is taken as given for waves in general.

Only transverse waves survive on a string. A stretched string cannot be usefully compressed along its length the way air can, so it only supports transverse disturbances. This is why $v=\sqrt{T/m}$ describes a transverse stationary wave here, unlike the longitudinal stationary waves in an air column (a closed pipe, for instance).

How a sonometer checks all this. A sonometer is a hollow resonance box carrying a stretched wire over two bridges: one fixed, one movable so the vibrating length $l$ can be changed, with the tension $T$ set by weights hung over a pulley. To test any one law, the other two quantities are held constant while the third is varied, and the wire's note is matched to a tuning fork of known frequency by sliding the movable bridge until the wire resonates — confirmed when a small paper rider balanced on the wire flutters off. Running this three separate ways (varying $l$ , then $T$ , then swapping wires to vary $m$ ) is how the three laws of vibration below were established experimentally, and it is still the standard school demonstration of them.

Figure

A string fixed at both ends showing the fundamental mode with nodes at the two ends and a single antinode at the centre, labelled with wavelength equal to twice the length

Derivation

Derivation AFundamental frequency of a stretched string

The goal is to combine the geometric constraint that fixes the fundamental wavelength of a string clamped at both ends with the physical wave speed set by tension and mass per unit length, arriving at the single testable formula $f_0=\dfrac{1}{2l}\sqrt{T/m}$ that a sonometer experiment can check by ear against a tuning fork.

Step 1. Fix the wavelength from the boundary condition. Both ends of the string are forced to be nodes, so only stationary-wave patterns with $l=n\lambda/2$ can exist on the string. The fundamental is the simplest such pattern, $n=1$ , giving $\lambda_0=2l$ : the fundamental wavelength is twice the physical length of the string.

Step 2. Bring in the wave speed. Independently of the boundary condition, a transverse wave on this string travels at $v=\sqrt{T/m}$ , fixed entirely by the string's tension $T$ and its mass per unit length $m$ — nothing here depends on how the string is vibrating.

Step 3. Combine with $v=f\lambda$ . Every travelling (and hence every stationary) wave obeys $v=f\lambda$ , so the fundamental frequency is $f_0=v/\lambda_0$ . Substituting $\lambda_0=2l$ from Step 1 and $v=\sqrt{T/m}$ from Step 2 gives

f_0=\frac{v}{\lambda_0}=\frac{1}{2l}\sqrt{\frac{T}{m}}

Step 4. Read off the three laws of vibration. Holding two of $l$ , $T$ , $m$ fixed and varying the third shows $f_0\propto1/l$ (law of length), $f_0\propto\sqrt{T}$ (law of tension), and $f_0\propto1/\sqrt{m}$ (law of mass) — exactly the proportionalities a sonometer verifies by sliding the bridge, changing the hanging weight, or swapping wires, matching resonance against a known tuning fork each time (a small paper rider flutters off the wire at resonance). Each law also has an everyday face: pressing a guitar string at a fret shortens $l$ and raises the pitch (law of length); turning a tuning peg increases $T$ and raises the pitch (law of tension); and choosing a heavier-gauge, larger- $m$ wire for a bass string lowers the pitch at the same $l$ and $T$ (law of mass).

Worked example

A sonometer wire of length $0.6\ \text{m}$ has mass per unit length $4\times10^{-4}\ \text{kg/m}$ under tension $90\ \text{N}$ . Find (a) the wave speed, (b) the fundamental frequency.

Step 1. Wave speed: $v=\sqrt{T/m}=\sqrt{90/(4\times10^{-4})}=\sqrt{2.25\times10^{5}}=474.3\ \text{m/s}$ .

Step 2. Fundamental frequency: $f_0=\dfrac{v}{2l}=\dfrac{474.3}{2\times0.6}=\dfrac{474.3}{1.2}=395.2\ \text{Hz}$ , which matches the direct formula $f_0=\dfrac{1}{2l}\sqrt{T/m}$ .

Answer: $v=474.3\ \text{m/s}$ , $f_0=395.2\ \text{Hz}$ . Notice the units check quietly: $T/m$ has units of $\text{N}\div(\text{kg/m})=\text{m}^2/\text{s}^2$ , so its square root is a speed in m/s, and dividing that speed by a length in metres correctly leaves a frequency in Hz.

Competitive stretch

The same wire's fundamental is $256\ \text{Hz}$ at tension $64\ \text{N}$ (length and mass per unit length unchanged from before). What tension raises it to $384\ \text{Hz}$ ?

Step 1. Since $l$ and $m$ are fixed, only tension changes, so $f_0\propto\sqrt{T}$ , giving $T_2=T_1\left(f_2/f_1\right)^2$ .

Step 2. Substitute: $T_2=64\times(384/256)^2=64\times(1.5)^2=64\times2.25=\mathbf{144\ \text{N}}$ — the tension must rise from $64\ \text{N}$ to $144\ \text{N}$ .

Common traps

Writing $\lambda_0=l$ instead of $\lambda_0=2l$ : the fundamental "loop" between two nodes is only half a wavelength, so the string's length is half the wavelength, not the whole thing.
Using the total mass of the wire in $v=\sqrt{T/m}$ instead of mass per unit length — this throws the answer off by a factor related to $l$ .
Applying a law of vibration (length, tension or mass) while forgetting the other two quantities must be held fixed — "frequency halves when length doubles" only holds at constant $T$ and $m$ .
Assuming a thicker string always sounds higher — the opposite holds at fixed $T$ and $l$ : larger mass per unit length $m$ lowers $f_0$ , since $f_0\propto1/\sqrt{m}$ , which is why bass strings are thick.
Treating a change in string material or gauge as automatically a change in $T$ or $l$ — swapping to a heavier-gauge wire changes $m$ while $l$ and $T$ can remain exactly as before, so only the mass-per-unit-length law applies.

Ask the tutor: Want a walkthrough of why $\lambda_0=2l$ with an actual sine-wave sketch, or another proportional-reasoning tension problem?

Why is the fundamental wavelength 2l and not l?

Because a full wavelength contains one complete node-antinode-node-antinode-node cycle, but the simplest pattern that fits between two fixed nodes only needs one antinode in the middle — that is just half of a full wavelength. So the string's physical length l corresponds to half a wavelength, meaning the wavelength itself is 2l. Drawing the actual sine-shaped loop on the string usually makes this click faster than memorising the formula.

Why do bass guitar strings look thicker than treble strings?

Because frequency depends on mass per unit length as $f_0\propto1/\sqrt{m}$ , at the same length and tension a thicker (heavier per metre) string vibrates slower, giving a lower note. Rather than making bass strings absurdly long or slack, instrument makers simply increase the mass per unit length by winding extra wire around a thicker core, which keeps the string playable while dropping the pitch.

Does the sonometer experiment prove the formula, or just check it?

Both, in a school-lab sense. The sonometer lets you vary l, T and m one at a time (holding the other two fixed) and match the wire's note to a tuning fork of known frequency by sliding the bridge until resonance occurs — confirmed by a paper rider fluttering off the wire. This directly verifies the three separate proportionalities $f_0\propto1/l$ , $f_0\propto\sqrt{T}$ and $f_0\propto1/\sqrt{m}$ , which together are equivalent to the full formula $f_0=\frac{1}{2l}\sqrt{T/m}$ .