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University Physics I · Optional extension · Institutional Extension · Waves & Sound · WS.3

Wave Speed in Strings & Media

Pluck a string harder, or play a higher note, and the pulse still crosses it in the same time. Speed belongs to the string — to its tension and its mass per unit length — and the wavelength adjusts to whatever frequency arrives.

Open the lab ↓Practice & worked examples ↓
01

Build the model

Connect the measurement to the mechanism.

Three numbers describe a travelling wave and they are not equal partners. The source fixes the frequency: drive a string end f times a second and every point along it repeats at f. The medium fixes the speed: a transverse pulse on a stretched string moves at v = √(FT/μ), set by tension and mass per unit length alone, with the source nowhere in the expression.

Only the wavelength is left free, so λ = v/f adjusts to whatever arrives. That is all v = fλ says — an identity between three numbers, not a claim that a higher frequency travels faster. The string result is one case of a pattern common to every mechanical medium: v = √(elastic property / inertial property), giving √(Y/ρ) in a thin rod and √(B/ρ) in a fluid.

It holds while the slopes are small, the tension and density are uniform, and the medium has no bending stiffness — and each condition, once named, tells you what a real string does instead.

Simple definition
Wave speed is the rate at which a disturbance travels through a medium, fixed by the medium's stiffness and inertia — on a stretched string v = √(FT/μ) — and tied to the source frequency by the identity v = fλ.
Example
A guitar's low E string, 0.648 m of speaking length at 71.9 N tension with μ = 6.30 × 10⁻³ kg m⁻¹, carries transverse waves at 107 m s⁻¹. Every note played on it, open or fretted, travels at that one speed.
Speed, frequency, wavelengthv = fλ = ω/k = λ/T

One period advances the pattern by one wavelength. An identity between three numbers, not a causal law.

v in m s⁻¹, f in Hz, λ in m; k = 2π/λ, ω = 2πf

Speed on a stretched stringv = √(FT/μ)

Tension restores, mass per unit length resists. Neither frequency nor amplitude appears.

FT is the tension in N, μ = m/L in kg m⁻¹; v comes out in m s⁻¹

Linear mass densityμ = m/L = ρA = ρπd²/4

Doubling the diameter quadruples μ and so halves the speed at the same tension.

Uniform round wire of density ρ and diameter d; μ in kg m⁻¹

The wave equation it satisfies∂²y/∂t² = v² ∂²y/∂x², with v² = FT/μ

Speed enters as a coefficient in the equation of motion, before any source is chosen.

Any y = f(x − vt) or g(x + vt) is a solution; v is a constant of the string

Elastic over inertialv = √(elastic property / inertial property)

Every mechanical wave speed has this shape — stiffness above, the mass that must be moved below.

String: √(FT/μ). Thin solid rod: √(Y/ρ). Fluid: √(B/ρ).

Small-slope condition(∂y/∂x)² ≪ 1, so 2πA/λ ≪ 1

The derivation holds the tension fixed. Steep slopes stretch the string and speed the crest up.

∂y/∂x is the slope of the snapshot; A/λ = 0.05 already gives a slope of 0.31

01

v = fλ is bookkeeping, not a cause

Three numbers describe a periodic travelling wave, and they come from different places. Drive the end of a string at f oscillations per second and every point along it must repeat at f: the source owns the frequency, and nothing downstream can change it. The string owns the speed. Send a pulse of any shape along it and the pulse advances at one fixed v, which the next section gets from tension and mass per unit length alone. That leaves the wavelength as the only quantity free to adjust, so λ = v/f. Written as v = fλ the relation looks symmetric, but it is an identity between three numbers, not a statement that a higher frequency makes a faster wave. Drive the guitar's low E string at 220 Hz and the wavelength is 106.8/220 = 0.486 m; at 440 Hz it is 0.243 m; the speed is 106.8 m s⁻¹ both times. The graphs say the same thing: a snapshot measures λ along the x-axis, a history graph measures T at one fixed point, and v = λ/T converts between them.

02

Where √(FT/μ) comes from

Take a pulse of any shape moving at speed v along a string under tension FT, and watch it from a frame moving with it. In that frame the shape stands still and the string material streams backwards through it at speed v. Look at a short arc at the top of the pulse, where the string curves with radius R and the arc subtends a small angle Δθ, so its length is RΔθ and its mass is μRΔθ. The tension pulls tangentially at each end of the arc; those two pulls are not parallel, and their resultant points towards the centre of curvature with magnitude 2FT sin(Δθ/2) ≈ FT Δθ. That resultant is the centripetal force on material rounding a bend of radius R at speed v, which is (μRΔθ)v²/R = μv²Δθ. Setting the two equal cancels Δθ and R together: FT = μv², so v = √(FT/μ). Nothing about the pulse survived the cancellation — not its height, not its width, not its shape. The dimensions check: FT/μ is (M L T⁻²)/(M L⁻¹) = L² T⁻², whose root is a speed.

03

Read the formula for what is missing

Frequency is absent, so every note on a given string travels at one speed. Amplitude is absent, so a hard pluck and a soft one arrive together. Both surviving dependences are square roots, and that sets the practical scale. Raising a string's pitch one semitone needs the speed up by 2¹⁄¹² = 1.0595, so the tension must rise by 1.0595² = 1.1225 — 12.2%, from 71.9 N to 80.7 N. An octave needs four times the tension, which is why a string near its breaking stress cannot be tuned much higher. The other route is μ. A guitar's high E sounds two octaves above its low E at the same 0.648 m length, so it needs four times the speed and therefore one sixteenth the mass per unit length: 3.94 × 10⁻⁴ against 6.30 × 10⁻³ kg m⁻¹. In plain steel that is a diameter ratio of four, 0.25 mm against 1.01 mm, and 1 mm steel is far too stiff to bend around a nut. Bass strings are wound instead: a thin core takes the tension while a heavy wrap supplies μ.

04

The same shape in every medium

The string result is one instance of a form every mechanical wave speed shares: v = √(elastic property / inertial property). The numerator says how hard the medium pushes back when deformed, the denominator how much mass has to be accelerated. A long thin solid rod carrying a longitudinal wave gives v = √(Y/ρ) with Y the Young modulus; steel at Y = 2.0 × 10¹¹ Pa and ρ = 7850 kg m⁻³ gives 5.0 × 10³ m s⁻¹. A fluid has no shear stiffness, so only bulk compression is available and v = √(B/ρ); water at B = 2.2 × 10⁹ Pa and ρ = 998 kg m⁻³ gives 1.48 × 10³ m s⁻¹, and air near 20 °C gives about 344 m s⁻¹. The ordering looks wrong at first — steel is 6500 times denser than air and still carries sound about 15 times faster — until you compare numerators: steel's stiffness beats air's by a factor near 1.4 × 10⁶, far more than enough to outweigh the density.

05

When the medium is not uniform

Every step so far assumed one tension and one μ along the whole string, and a heavy rope hanging freely from a ceiling has neither. At height y above the free lower end the rope must support the weight below it, so FT = μgy, and μ cancels out of the speed: v(y) = √(gy). A transverse pulse therefore starts almost at rest at the bottom and accelerates as it climbs. Integrating dt = dy/v gives the travel time, t = ∫₀L dy/√(gy) = 2√(L/g), which for a 12.0 m rope with g = 9.81 m s⁻² is 2.21 s, arriving at the top at √(9.81 × 12.0) = 10.9 m s⁻¹. The lesson generalises: v = √(FT/μ) is a local statement, true at each point, and it collapses into a single travel time only when FT and μ really are constant. It also settles what happens where two strings are joined — the speed changes, the driving frequency cannot, so the wavelength must.

06

Stiffness, amplitude, and the end of the ideal string

Three idealisations remain. The derivation held the tension magnitude fixed, which needs small slopes: the neglected term is of order (∂y/∂x)², so a sinusoid with A/λ = 0.05 already carries a maximum slope of 2π(0.05) = 0.31 and about a 5% error. Drive it harder and the crest genuinely stretches the string, raising the local tension and speeding the steep parts up until the shape distorts. Second, a real wire resists bending as well as stretching, which adds a term that makes speed grow with wave number: v(k) = √(FT/μ) √(1 + EIk²/FT), with I = πr⁴/4. For that high E — E = 2.0 × 10¹¹ Pa, r = 0.126 mm, so EI = 4.0 × 10⁻⁵ N m² — the correction moves the fundamental by under one part in 10⁵ but pushes the twentieth partial 0.26% sharp, 4.5 cents. That is dispersion: wavelengths travelling at different speeds. Third, an ideal string never loses energy; a real one damps, so the wave arriving is quieter than the wave sent.

02

Change one variable at a time

Make the relationship visible.

Interactive model
72 N
6.3 g m⁻¹
110 Hz

Drag the driver frequency and watch the wavelength bracket shrink while the 10 ms speed arrow does not move at all — only tension and μ can change that arrow.

Interactive physics modelSnapshot of a string driven at its left end. Tension 72 N and linear density 6.3 grams per metre set the wave speed at 106.9 metres per second, and the driver's 110 hertz then fixes the wavelength at 0.972 metres. Two metres of string are drawn to scale horizontally; the transverse displacement is exaggerated.arrow: 1.07 m — the pulse's run in 10 msv = 106.9 m s⁻¹driver at f = 110 HzFT = 72 N · μ = 6.3 g m⁻¹λ = 0.972 mv = √(FT/μ) — f and A are absentλ = v/f = 0.972 m

WAVE SPEED v106.9 m s⁻¹

WAVELENGTH λ0.972 m

PERIOD T9.09 ms

WAVE NUMBER k6.47 rad m⁻¹

Live interpretationWAVE SPEED v: 106.9 m s⁻¹. WAVELENGTH λ: 0.972 m. PERIOD T: 9.09 ms. WAVE NUMBER k: 6.47 rad m⁻¹

03

Catch the common trap

Explain before calculating.

Two wires of the same material, one twice the diameter of the other, are joined end to end and stretched under a single tension. A continuous sinusoidal wave is sent from the thin wire into the thick one. In the thick wire, compared with the thin one, the wave has:

Choose an answer to test the model.

04

Practice & worked examples

Reason from the model, then test the result.

01 · EasyA steel wire of speaking length 1.80 m has mass 12.6 g and is pulled to a tension of 88.0 N. Find its linear mass density, the transverse wave speed on it, and the wavelength of a 150 Hz wave sent along it.
  1. Linear mass density first, because the speed formula wants mass per unit length, not mass: μ = m/L = 0.0126 kg ÷ 1.80 m = 7.00 × 10⁻³ kg m⁻¹.
  2. Speed from the medium alone: v = √(FT/μ) = √(88.0 ÷ (7.00 × 10⁻³)) = √(1.2571 × 10⁴ m² s⁻²) = 112.12 m s⁻¹, which is 112 m s⁻¹ to three significant figures.
  3. The 150 Hz belongs to whatever is shaking the end, not to the wire, so it cannot touch v. Use v = fλ in the only direction left: λ = v/f.
  4. λ = 112.12 m s⁻¹ ÷ 150 s⁻¹ = 0.747 m.
  5. Check what did not happen: 150 Hz never entered step 2. Drive the same wire at 300 Hz and v is still 112 m s⁻¹ while λ halves to 0.374 m.

Answerμ = 7.00 × 10⁻³ kg m⁻¹, v = 112 m s⁻¹, λ = 0.747 m.

02 · MediumA 0.650 m string with μ = 3.90 × 10⁻⁴ kg m⁻¹ is tuned so that its fundamental — half a wavelength on the string — sounds at 330 Hz. Find the tension. The player then wants the same string to sound 350 Hz. Find the new tension and the percentage change.
  1. The fundamental fits half a wavelength between the ends, so λ₁ = 2L = 1.300 m, and the string must carry waves at v = f₁λ₁ = 330 × 1.300 = 429.0 m s⁻¹.
  2. Invert v = √(FT/μ) by squaring: FT = μv² = (3.90 × 10⁻⁴)(429.0²) = (3.90 × 10⁻⁴)(1.84041 × 10⁵) = 71.8 N.
  3. Retuning changes neither the length nor the wire, so λ₁ = 1.300 m still and v′ = 350 × 1.300 = 455.0 m s⁻¹.
  4. F′ = μv′² = (3.90 × 10⁻⁴)(2.07025 × 10⁵) = 80.7 N.
  5. Ratio route, no recomputation needed: at fixed L and μ, FT ∝ v² ∝ f², so F′/F = (350/330)² = 1.1249.
  6. So a 6.1% rise in pitch costs a 12.5% rise in tension. The square root that makes speed only weakly sensitive to tension runs backwards as a square law when you tune.

AnswerFT = 71.8 N at 330 Hz and F′ = 80.7 N at 350 Hz — 12.5% more tension for 6.1% more frequency.

03 · HardA uniform rope of mass 3.60 kg and length 8.00 m hangs vertically from a ceiling with a 12.0 kg crate on its lower end. Take g = 9.81 m s⁻². Find the transverse wave speed at the bottom of the rope and at the top, then the time a pulse takes to climb from the crate to the ceiling.
  1. Linear mass density of the rope: μ = m/L = 3.60 ÷ 8.00 = 0.450 kg m⁻¹, uniform along its length.
  2. Tension is not uniform. At height y above the lower end the rope carries the crate plus the rope hanging below that point: FT(y) = (M + μy)g = 9.81(12.0 + 0.450y) N.
  3. Bottom, y = 0: FT = 9.81 × 12.0 = 117.7 N, so v = √(117.7 ÷ 0.450) = √261.6 = 16.2 m s⁻¹.
  4. Top, y = 8.00 m: FT = 9.81 × 15.6 = 153.0 N, so v = √(153.0 ÷ 0.450) = √340.0 = 18.4 m s⁻¹. The pulse speeds up as it climbs.
  5. v = √(FT/μ) is a local statement, so no single speed gives the travel time. Integrate dt = dy/v with v(y) = √(g(a + y)) and a = M/μ = 12.0 ÷ 0.450 = 26.67 m: t = ∫₀L dy/√(g(a+y)) = (2/√g)(√(a+L) − √a).
  6. t = (2 ÷ 3.132)(√34.67 − √26.67) = (0.6386)(5.8878 − 5.1640) = 0.462 s. Strip the crate off and a → 0, and the result collapses to the free-rope 2√(L/g) = 1.81 s — 3.91 times longer, because the bottom end is then slack and the pulse starts from rest.

Answerv = 16.2 m s⁻¹ at the bottom and 18.4 m s⁻¹ at the top; the pulse takes 0.462 s to climb.