Physics · Mechanical waves

Wave speed on a stretched string

Q: What does Wave speed on a stretched string describe physically?

$v=\sqrt{\frac{F_T}{\mu}}$ describes the speed of a small transverse disturbance along a taut ideal flexible string. FT is positive tension; μ is moving mass per unit length. A string particle instead moves across the propagation direction and has its own speed depending on amplitude and frequency. The square-root dependence means quadrupling tension doubles v, while quadrupling linear density halves v. Each comparison keeps the other variable unchanged. Raising real tension may stretch the string and change μ. The equation therefore applies to a specified taut state with uniform density and small local slopes, rather than indiscriminately to every cord.

Q: How do I check units and scale for Wave speed on a stretched string?

Use newtons for FT and kg/m for μ. Then $\frac{F_T}{\mu}$ has units $\frac{\mathrm N}{\mathrm{kg\,m^{-1}}}=\mathrm{m^2\,s^{-2}}$, so its square root gives m/s. A value of 3.50 g/m equals 0.00350 kg/m; volume density in kg/m³ cannot be inserted directly. At FT = 35.0 N, $v=\sqrt{10000}\,\mathrm{m/s}=100\,\mathrm{m/s}$. This is propagation speed, not a particle’s transverse speed. Raising it to 150 m/s at the same μ requires tension $0.00350\cdot150^2\,\mathrm N=78.75\,\mathrm N$, a factor of 2.25. That comparison checks the quadratic rearrangement $F_T=\mu v^2$; a merely linear force factor of 1.5 would be incorrect.

Q: When should I combine Wave speed on a stretched string with another relation?

For a source at a specified frequency, combine speed with $\lambda=\frac vf$. At 100 m/s and 50 Hz, $\lambda=2.00\,\mathrm m$; if v rises to 150 m/s while the source stays unchanged, λ becomes 3.00 m. A string fixed at both ends instead adds boundary conditions: $f_n=\frac{nv}{2L}$, with vibrating length L and positive integer mode number n. An arbitrary driving frequency is not automatically natural. A one-way travel time also obeys $t=\frac dv$; an echo travels 2d. Boundaries and measurement path determine the appropriate extra equation. Speed alone gives neither resonance amplitude nor the energy input of a real instrument.

The transverse-wave speed on an ideal flexible string grows with the square root of tension and decreases with the square root of linear mass density.

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Formula

Calculate

v = \frac{F _{T}}{μ}

v in m/s; FT in N; μ in kg/m.

Variables & units

Variables & units
Symbol	Meaning	Unit
$v$	Wave propagation speed	m/s
$FT$	String tension	N
$μ$	Mass per unit length	kg/m

LaTeX

v=\sqrt{\frac{F_T}{\mu}}

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Slightly curved string element μΔx: equal tension magnitudes FT point outward along left and right tangents. Small transverse components add to downward ΔFy. A separate horizontal arrow marks phase speed, not transverse string motion. — Ideal string element with small local slopes: tangential tensions have opposing longitudinal components and a resultant transverse force. Phase speed along the string is v = √(FT/μ); the resultant arrow is schematic.

Wave speed on a string calculator

Find the speed of a transverse wave on a taut string from tension and mass per length, plus linear density and the travel time of a pulse.

Calculation model

Find

String tension F_t

Force with which the string is stretched, e.g. 72 N for a guitar string.

Mass per length μ

Mass per metre of string, e.g. 5 g/m; that is mass divided by length.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for v, substitute, evaluate

1. Rearrange for v
$v=\sqrt{\frac{F_T}{\mu}}$
2. Substitute values (in SI units)
- μ = 5 g/m = 0.005 kg/m
$\sqrt{\frac{72}{0.005}}$
3. Result
v = 120 m/s

Result: Wave speed v

v = 120 m/s

With F_t = 72 N and μ = 5 g/m, v = 120 m/s.

Quick tips

Four times the tension, twice the speed: v grows with the square root of F. That is why a tighter string sounds higher.
Convert mass per length to kg/m: 5 g/m is 0.005 kg/m. Unit check: N divided by kg/m gives m²/s², and its square root gives m/s.

Assumptions and limits

Nearly straight, uniform, ideal flexible taut string with constant positive tension and linear mass density. Small local transverse slopes; losses and bending stiffness neglected. This is propagation speed magnitude, not speed of material particles.

Derivation & background

The linear model requires a taut nearly straight ideal flexible string with constant tension FT and linear density μ. Small local slope $\left|\frac{\partial y}{\partial x}\right|\ll1$ is essential; for a harmonic wave $Ak\ll1$ . $\mu=\frac mL$ applies to a uniform sample in the considered state. v describes propagation along the string, not transverse particle speed.

Exam blueprint

Validity range

Taut nearly straight uniform ideal flexible string, constant tension and linear density, small local slopes and neglected losses. Ideal model is nondispersive; real bending stiffness or large displacement requires checking the model.

Derivation steps

Apply Newton’s law to a short ideal flexible string element with small slopes.

1Let $y(x,t)$ be transverse displacement and $\mu>0$ mass per length.
2Constant positive tension FT points outward along both end tangents.
3For $\left|\frac{\partial y}{\partial x}\right|\ll1$ , transverse force is approximately $F_T\frac{\partial y}{\partial x}$ .
4The difference over Δx is $F_T\Delta x\frac{\partial^2y}{\partial x^2}$ ; element mass is $\mu\Delta x$ .
5Newton’s law and cancellation give $\frac{\partial^2y}{\partial t^2}=\frac{F_T}\mu\frac{\partial^2y}{\partial x^2}$ .
6For travelling shape $g(x-vt)$ , $v^2=\frac{F_T}\mu$ and speed is $v=\sqrt{\frac{F_T}{\mu}}$ .

Rearrangements

Tension

F_T=\mu v^2

μ is mass per unit length.

Task variant

An ideal flexible string has tension 80 N and linear mass density 0.010 kg/m. The transverse wave has small slopes. Calculate propagation speed.

Approach: $v=\sqrt{\frac{F_T}{\mu}}$ .
Substitution gives $v=\sqrt{\frac{80}{0.010}}\,\mathrm{m/s}=\sqrt{8000}\,\mathrm{m/s}\approx89.4427\,\mathrm{m/s}$ .

Check: $\mu v^2=80\,\mathrm N$ . This is propagation along the string, not a material point’s transverse velocity.

Common mistakes

Ignoring units or assumptions stated for the formula: v in m/s; FT in N; μ in kg/m.

Convert units before substitution and verify the validity range; round only at the end.

Exam context

Relate string instruments, standing waves and wavelengths to frequency.

Formula cluster

Mechanical waves

Related relations: stehende-welle-saite, wellengleichung, harmonische-schwingung-federpendel.

λ_n = 2L/n; f_n = nv/(2L)

Standing wave on a string fixed at both ends

c = λ·f

Wave Equation (c = λ·f)

T = 2π·√(m/k)

Spring Pendulum (Oscillation Period)

Worked example

With $F_T=80\,\mathrm N$ and $\mu=0.010\,\mathrm{kg/m}$ , the small linear transverse-wave model gives $v=\sqrt{\frac{80}{0.010}}\,\mathrm{m/s}=\sqrt{8000}\,\mathrm{m/s}\approx89.4\,\mathrm{m/s}$ . Reversing $\mu v^2=80\,\mathrm N$ checks the unrounded value.

Applications

Relate string instruments, standing waves and wavelengths to frequency.

How do you calculate with Wave speed on a stretched string?

Here is how to work through a typical Wave speed on a stretched string (v = √(FT/μ)) task step by step:

1
Task
An ideal flexible string has tension 80 N and linear mass density 0.010 kg/m. The transverse wave has small slopes. Calculate propagation speed.
Solution path
1. Approach: $v=\sqrt{\frac{F_T}{\mu}}$ .
2. Substitution gives $v=\sqrt{\frac{80}{0.010}}\,\mathrm{m/s}=\sqrt{8000}\,\mathrm{m/s}\approx89.4427\,\mathrm{m/s}$ .
Check: $\mu v^2=80\,\mathrm N$ . This is propagation along the string, not a material point’s transverse velocity.

Understand and apply the equation

A taut string carries a pulse faster under greater tension and more slowly with greater mass per length. The square-root relation describes small transverse waves on an ideal string. Travel-time measurements show how to calculate its quantities and check the model.

A pulse travels while the string stays in place

Briefly move a taut string sideways. The disturbance travels along it while each small material segment is displaced transversely. The speed of the travelling shape depends on how strongly neighbouring segments pull on each other and how much mass must be accelerated.[1]

The model requires Newton’s force law, mass per length, and small slopes. Tension is a force in newtons. Linear mass density tells you how many kilograms occupy one metre of the taut string; it is neither total mass nor density per volume.[1]

Specify tension and mass per unit length

\begin{aligned}v&=\sqrt{\frac{F_T}{\mu}}\\\mu&=\frac mL\end{aligned}

$\mu=\frac{m}{L}$ applies to a uniform string; L is its length in the considered taut state.[1]

$F_T>0$ denotes longitudinal tension magnitude in newtons. $\mu>0$ is linear mass density in $\mathrm{kg/m}$ , not density per volume in $\mathrm{kg/m^3}$ . Total mass alone does not fix it: a shorter string with the same mass contains more mass per length. Varying thickness or composition can change local linear density, so a single constant speed is not automatic.

The formula describes a transverse disturbance propagating along the string. An element moves mainly across the string; its speed additionally depends on frequency and amplitude. Tension supplies restoring action while string mass resists transverse acceleration.

Small slope is the geometric approximation

Let $y(x,t)$ describe small transverse motion about a nearly straight string. Local slope $\frac{\partial y}{\partial x}=\tan\theta$ must have magnitude much smaller than 1. Then $\sin\theta\approx\tan\theta\approx\frac{\partial y}{\partial x}$ and length changes are higher-order effects. The linear model treats tension and mass per projected length as constant.[1]

\begin{aligned}\left|\frac{\partial y}{\partial x}\right|&\ll1\\y&=A\sin(kx-\omega t)\\\max\left|\frac{\partial y}{\partial x}\right|&=Ak\end{aligned}

Small amplitude alone is insufficient at very short wavelength; the relevant product is Ak.

A hypothetical 1 mm amplitude at $\lambda=0.20\,\mathrm m$ gives $Ak\approx0.0314$ . The same amplitude at $\lambda=2\,\mathrm{mm}$ gives $Ak\approx3.14$ , outside the small-slope approximation. This comparison separates geometric scale from absolute displacement.

Balance transverse forces on a short element

Isolate an element between x and x + Δx. Tensions point outward along the local tangents. In the linear model their horizontal magnitudes are FT, giving transverse resultant FT[∂xy(x+Δx) − ∂xy(x)]. For a smooth shape this becomes FT∂²xy Δx. Element mass is $\mu\Delta x$ .[1]

\begin{aligned}\mu\Delta x\,\partial_t^2y&=F_T\Delta x\,\partial_x^2y\\\partial_t^2y&=\frac{F_T}{\mu}\partial_x^2y\end{aligned}

Apply Newton’s law to transverse motion; Δx cancels.

\begin{aligned}y(x,t)&=g(x-vt)\\\partial_t^2y&=v^2g^{\prime\prime}\\\partial_x^2y&=g^{\prime\prime}\end{aligned}

A smooth travelling disturbance solves the equation when $v^2=\frac{F_T}\mu$ .

The positive speed magnitude is therefore a square root. Direction is selected by $g(x-vt)$ or g(x+vt). The force-to-mass ratio follows from the model assumptions; it is not a universal formula for any rope, bending vibration or slack string.

Double speed requires four times the tension

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At fixed

\mu=0.004\,\mathrm{kg/m}

v=\sqrt{\frac{F_T}{\mu}}

. The curve starts at positive tension; a slack string is outside the model. Changes to taut length or mass density are excluded here.[1]

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Mass units, rearrangements and scaling

\begin{aligned}F_T&=\mu v^2\\\mu&=\frac{F_T}{v^2}\\\left[\frac{F_T}{\mu}\right]&=\frac{\mathrm N}{\mathrm{kg\,m^{-1}}}=\mathrm{m^2\,s^{-2}}\end{aligned}

Solving for μ needs $v>0$ ; taking the square root gives m/s.

1 g/m equals 0.001 kg/m. If mass is measured in grams and length in centimetres, convert them separately. Quadrupling tension at fixed μ doubles v; quadrupling μ at fixed tension halves v. Doubling speed requires four times the tension when μ is unchanged.

For $F_T=80\,\mathrm N$ and $\mu=0.010\,\mathrm{kg/m}$ , $v=\sqrt{8000}\,\mathrm{m/s}\approx89.4427\,\mathrm{m/s}$ . Increasing real tension can lengthen the string and change mass per length. Keeping $\mu$ fixed is therefore an explicit model condition, not an automatic property of every experiment.

Here, string tension means tensile force $F_T$ in newtons. Mechanical normal stress is instead force per cross-sectional area, $\sigma=F_T/A$ , in pascals. Without the appropriate area, a value in pascals cannot be inserted as tension; linear mass density $\mu$ does not supply that area either.[6][1]

Two original calculations: transit time and retensioning

A uniform string sample

Given: A uniform sample in the considered state has mass 12.0 g over 2.40 m. FT = 72 N; a transverse disturbance travels 1.50 m; flexible linear model.
Find: Find μ, v and one-way transit time.

Approach: The uniform taut sample determines density. The pulse travels a one-way path.
$\begin{aligned}\mu&=\frac mL\\v&=\sqrt{\frac{F_T}{\mu}}\\t&=\frac dv\end{aligned}$
Substitute the given values with matching units.
$\mu=\frac{0.0120}{2.40}=0.00500\,\mathrm{kg/m}$
Take the square root of tension divided by linear density.
$v=\sqrt{\frac{72}{0.00500}}=120\,\mathrm{m/s}$
Divide stationary distance by speed.
$t=\frac{1.50}{120}=0.0125\,\mathrm s$

Result: μ = 5.00 g/m, v = 120 m/s and transit time 12.5 ms.

Check: $\mu v^2=0.005\cdot14400\,\mathrm N=72\,\mathrm N$ reproduces tension. $vt=1.50\,\mathrm m$ checks the one-way path. An echo covering the same outward and return paths would require $25.0\,\mathrm{ms}$ , twice as long.

Increase speed by a factor of 1.5

Given: μ = 0.00350 kg/m remains fixed in the model; initial tension 35.0 N; desired speed 150 m/s. The same source stays at f = 50 Hz.
Find: Find new tension and both wavelengths.

Approach: Keep density fixed but change tension: square the speed factor first.
$\begin{aligned}v_1&=\sqrt{\frac{F_{T,1}}\mu}\\F_{T,2}&=F_{T,1}\left(\frac{v_2}{v_1}\right)^2\\\lambda_i&=\frac{v_i}{f}\end{aligned}$
Substitute the given values with matching units.
$v_1=\sqrt{\frac{35.0}{0.00350}}=100\,\mathrm{m/s}$
Insert desired speed into the rearranged equation.
$F_{T,2}=0.00350\cdot150^2=78.75\,\mathrm N$
Find wavelengths at the same frequency.
$\begin{aligned}\lambda_1&=\frac{100}{50}=2.00\,\mathrm m\\{} \lambda_2&=\frac{150}{50}=3.00\,\mathrm m\end{aligned}$

Result: New tension is arithmetically 78.75 N, a factor of 2.25; wavelength increases from 2 to 3 m.

Check: $\sqrt{2.25}=1.5$ confirms the speed factor. Both wavelengths reproduce their speeds at $f=50\,\mathrm{Hz}$ . Source frequency is fixed here; natural frequencies of a string fixed at both ends would instead rise with speed.

Driving and natural frequencies play different roles

A travelling harmonic wave with supplied frequency has $\lambda=\frac vf$ . The material model sets speed, while the source sets frequency. A length fixed at both ends instead has natural frequencies $f_n=\frac{nv}{2L}$ . Greater tension shifts these upwards if length and mass density stay unchanged. Source frequency and natural frequency are therefore different specifications.[2]

f_n=\frac{n}{2L}\sqrt{\frac{F_T}{\mu}}

For an ideal flexible string fixed at both ends, $n=1,2,\ldots$

Not every driving frequency is a natural frequency. A real driven system also responds away from resonance, generally with different amplitude and phase. Boundaries, excitation position and losses determine that response; speed alone does not give resonance amplitude.

A standing wave on a string fixed at both ends adds boundary conditions to this propagation speed. Its discrete natural frequencies belong to the vibrating length; they replace neither a freely specified driving frequency nor the small-slope check.[2][1]

Explore further: measurement uncertainty and inference

A simple one-way measurement gives $v=\frac d{\Delta t}$ . An out-and-back echo instead covers $2d$ . Timing zero must match launch of the disturbance; instrument delay is not an extra string path. Also check whether the reading identifies one clean pulse or several overlapping reflections. The third example shows the effect of a known delay explicitly.

Weigh a representative length of the actual string. A wound string cannot be described by a guessed core density; μ includes all moving mass per length. A heavy sensor or attached mass may change boundaries and reflect waves. Such additions cannot silently be absorbed into one uniform μ.

The wave relation for speed, wavelength and frequency gives $v=\lambda f$ for a travelling harmonic wave. A transit measurement instead tracks an identifiable pulse over its actual path. Phase and group speed agree in the ideal nondispersive string model; frequency alone cannot reconstruct unknown reflection paths.[5][1]

Flexibility, losses and dispersion limit the model

Real strings have bending stiffness. At short wavelengths its contribution can alter the tension-only model and cause dispersion. Large slopes, spatially varying tension, nonuniform linear density, and nonlinear extension also exceed $v=\sqrt{\frac{F_T}{\mu}}$ alone. The simple result requires a nearly straight, taut, ideal flexible string undergoing small transverse motion.[1][3]

The ideal model is nondispersive: $\omega=vk$ and phase and group speeds agree. Losses or real stiffness can alter pulse shape and mode spacing. With zero tension the taut-string assumption is lost; a formal zero speed does not replace a model of a slack cord.

A round-trip time determines tension

Inferring tension squares the speed. A modest timing error can therefore appreciably alter the inferred tension. This echo example separates instrument delay from actual travel time.[1][4]

A round-trip time determines tension

Given: A uniform taut string has length 2.00 m and mass 8.00 g. A small transverse pulse travels to the reflecting end and back. The instrument reads 45.0 ms, including an independently known 5.0 ms delay. Other paths and reflection delays are excluded in this ideal model.
Find: Calculate propagation speed and tension; assess the error if the delay is not removed.

Approach: first obtain actual travel time, then use the doubled path and taut-string mass density.
$\begin{aligned}t_{\mathrm{travel}}&=t_{\mathrm{read}}-t_{\mathrm{delay}}\\v&=\frac{2L}{t_{\mathrm{travel}}}\\\mu&=\frac mL\\F_T&=\mu v^2\end{aligned}$
Substitute consistent SI units.
$\begin{aligned}t_{\mathrm{travel}}&=0.0400\,\mathrm s\\\mu&=\frac{0.00800}{2.00}\,\mathrm{kg/m}=0.00400\,\mathrm{kg/m}\\v&=\frac{4.00}{0.0400}\,\mathrm{m/s}=100\,\mathrm{m/s}\\F_T&=0.00400\cdot100^2\,\mathrm N=40.0\,\mathrm N\end{aligned}$
Without correction, a smaller speed and its square would be used.
$\begin{aligned}v_{\mathrm{wrong}}&=\frac{4.00}{0.0450}\,\mathrm{m/s}\approx88.8889\,\mathrm{m/s}\\F_{\mathrm{wrong}}&\approx31.6049\,\mathrm N\\\frac{F_{\mathrm{wrong}}}{F_T}&=\left(\frac{0.0400}{0.0450}\right)^2=\frac{64}{81}\end{aligned}$

Result: The model gives $v=100\,\mathrm{m/s}$ and $F_T=40.0\,\mathrm N$ . Omitting the correction would underestimate tension by approximately 20.99%.

Check: Check: $\sqrt{\frac{40}{0.004}}=100\,\mathrm{m/s}$ and $\frac4{100}\,\mathrm s+0.005\,\mathrm s=0.045\,\mathrm s$ . The delay must be known independently; one travel-time reading cannot determine both tension and an unknown delay.

Five solved exercises

Calculate · Foundation

FT = 50 N and μ = 0.00200 kg/m. Find v.

Show hint

$\frac{F_T}{\mu}$ has units m²/s².

Learning goal: Calculate speed from tension and line density.

Show solution and calculation

Approach: the positive square root gives speed magnitude.
$v=\sqrt{25000}\,\mathrm{m/s}\approx158.114\,\mathrm{m/s}$ .

Check: $\mu v^2=50\,\mathrm N$ reproduces tension.

\begin{aligned}v&=\sqrt{\frac{F_T}\mu}\\&=\sqrt{\frac{50}{0.00200}}\,\mathrm{m/s}\\&\approx158.114\,\mathrm{m/s}\end{aligned}

Rearrange · Application

A uniform 3.00 m sample has mass 9.00 g; measured speed is 80.0 m/s in the ideal model. Find FT.

Show hint

Calculate μ in kg/m first.

Learning goal: Infer tension using specimen mass.

Show solution and calculation

Approach: divide mass by taut length first, then infer tension from speed squared.
$\mu=0.00300\,\mathrm{kg/m}$ , $F_T=19.2\,\mathrm N$ .

Check: $\sqrt{\frac{19.2}{0.00300}}=80\,\mathrm{m/s}$ . Length refers to the measured taut state.

\begin{aligned}\mu&=\frac mL=\frac{0.00900}{3.00}\,\mathrm{kg/m}=0.00300\,\mathrm{kg/m}\\F_T&=\mu v^2=0.00300\cdot80.0^2\,\mathrm N=19.2\,\mathrm N\end{aligned}

Transfer · Application

FT stays fixed while μ rises from 4 to 9 g/m. How does v change?

Show hint

Compare using the square root of a ratio.

Learning goal: Apply square-root scaling for changed density.

Show solution and calculation

Approach: constant tension cancels in the ratio.
$\frac{v_2}{v_1}=\sqrt{\frac49}=\frac23$ .
Speed falls by one third; $\frac49$ would incorrectly assume linear rather than square-root dependence.

\begin{aligned}\frac{v_2}{v_1}&=\sqrt{\frac{\mu_1}{\mu_2}}\\&=\sqrt{\frac49}=\frac23\end{aligned}

Interpret · Application

A small transverse wave on an ideal string has FT=80 N and μ=0.005 kg/m. Marked material points move up and down. Does every point move along the string at propagation speed? Calculate that speed and explain the distinction.

Show hint

Wave speed tracks a phase feature rather than a material particle.

Learning goal: Distinguish transverse particle motion from propagation speed.

Show solution and calculation

Approach: The string model gives approximately 126.491 m/s for phase propagation.
Material points oscillate transversely with their own time-dependent velocities set by amplitude and frequency.
Tension and line density alone do not give that speed.
Rapid pattern propagation does not imply comparable longitudinal flow of string material.

v_p=\sqrt{\frac{F_T}{\mu}}=\sqrt{\frac{80}{0.005}}\,\mathrm{m\,s^{-1}}\approx126.491\,\mathrm{m\,s^{-1}}

[1]

Error analysis · Application

A sinusoidal wave has amplitude A=2 cm and wavelength 10 cm. Someone treats the linear string speed as exact because 2 cm looks small. Check maximum slope and assess the small-slope assumption.

Show hint

Compare amplitude to wavelength through kA, not to an arbitrary length.

Learning goal: Check small slope before applying linear wave speed.

Show solution and calculation

Approach: Maximum slope of a sine profile is wave number times amplitude.
Here it is about 1.25664, nowhere near much less than one.
The linear small-slope derivation is not justified; tension and geometry can change dynamically.
An apparently small centimetre value alone cannot justify exactness.

\max|\partial_xy|=kA=\frac{2\pi A}{\lambda}=\frac{2\pi(0.02)}{0.10}\approx1.25664

[1]

In brief

Determine mass per taut length and use tension in newtons. Linear-model speed increases with the square root of tension. Travel-time measurements require the correct path and known delays. Small absolute displacement alone does not guarantee small slope.[1][4]

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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Change history

October 2, 2026Distinguished tensile force from mechanical stress and added relevant wave connections; preserved all existing calculations, exercises, figures and model limits.
October 1, 2026Distinguished propagation from material motion; added echo inference with a known instrument delay and a tension–speed curve. Preserved previous calculations, model limits, and all five exercises.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Wave speed on a stretched string. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/wellengeschwindigkeit-saite

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Frequently asked questions about Wave speed on a stretched string

What does Wave speed on a stretched string describe physically?+

$v=\sqrt{\frac{F_T}{\mu}}$ describes the speed of a small transverse disturbance along a taut ideal flexible string. FT is positive tension; μ is moving mass per unit length. A string particle instead moves across the propagation direction and has its own speed depending on amplitude and frequency. The square-root dependence means quadrupling tension doubles v, while quadrupling linear density halves v. Each comparison keeps the other variable unchanged. Raising real tension may stretch the string and change μ. The equation therefore applies to a specified taut state with uniform density and small local slopes, rather than indiscriminately to every cord.

How can I check a calculation involving Wave speed on a stretched string?+

First check μ from a representative mass and length in the considered state. For 12.0 g over 2.40 m, $\mu=\frac{0.0120}{2.40}\,\mathrm{kg/m}=0.00500\,\mathrm{kg/m}$ . At FT = 72 N, $v=\sqrt{\frac{72}{0.00500}}\,\mathrm{m/s}=120\,\mathrm{m/s}$ . Back-substitution $\mu v^2=72\,\mathrm N$ independently checks the square-root calculation. Over a known one-way distance of 1.50 m, travel time is 12.5 ms; $vt$ returns 1.50 m. An outbound-and-return echo instead needs 25.0 ms. Compare the correct distance and account for instrument delays. Successful substitution alone does not additionally prove that bending stiffness and losses are negligible.

What assumptions and limits apply to Wave speed on a stretched string?+

The model requires a nearly straight taut uniform ideal flexible string with constant tension. Small local slope is essential: $\left|\frac{\partial y}{\partial x}\right|\ll1$ . For a harmonic wave this becomes $Ak\ll1$ with $k=\frac{2\pi}{\lambda}$ . A 1 mm amplitude at $\lambda=0.20\,\mathrm m$ gives $Ak\approx0.0314$ ; at $\lambda=2\,\mathrm{mm}$ it would give $Ak\approx3.14$ , which is not small. Real bending stiffness can make short waves dispersive. Large slopes, nonuniform density, changing tension and losses need additional model terms. At FT = 0 the taut-string assumption fails; a formally calculated zero speed does not describe an arbitrary slack cord.

How do I check units and scale for Wave speed on a stretched string?+

Use newtons for FT and kg/m for μ. Then $\frac{F_T}{\mu}$ has units $\frac{\mathrm N}{\mathrm{kg\,m^{-1}}}=\mathrm{m^2\,s^{-2}}$ , so its square root gives m/s. A value of 3.50 g/m equals 0.00350 kg/m; volume density in kg/m³ cannot be inserted directly. At FT = 35.0 N, $v=\sqrt{10000}\,\mathrm{m/s}=100\,\mathrm{m/s}$ . This is propagation speed, not a particle’s transverse speed. Raising it to 150 m/s at the same μ requires tension $0.00350\cdot150^2\,\mathrm N=78.75\,\mathrm N$ , a factor of 2.25. That comparison checks the quadratic rearrangement $F_T=\mu v^2$ ; a merely linear force factor of 1.5 would be incorrect.

When should I combine Wave speed on a stretched string with another relation?+

For a source at a specified frequency, combine speed with $\lambda=\frac vf$ . At 100 m/s and 50 Hz, $\lambda=2.00\,\mathrm m$ ; if v rises to 150 m/s while the source stays unchanged, λ becomes 3.00 m. A string fixed at both ends instead adds boundary conditions: $f_n=\frac{nv}{2L}$ , with vibrating length L and positive integer mode number n. An arbitrary driving frequency is not automatically natural. A one-way travel time also obeys $t=\frac dv$ ; an echo travels 2d. Boundaries and measurement path determine the appropriate extra equation. Speed alone gives neither resonance amplitude nor the energy input of a real instrument.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 1, §16.3: Wave Speed on a Stretched String

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Retain Wave speed on a stretched string for exams

Create a curated FSRS exam set for v = √(FT/μ): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Wave speed on a stretched string

The transverse-wave speed on an ideal flexible string grows with the square root of tension and decreases with the square root of linear mass density.

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Symbol

Meaning

Unit

v

Wave propagation speed

m/s

FT

String tension

μ

Mass per unit length

kg/m

Derivation & background

Exam blueprint

Validity range

Derivation steps

Apply Newton’s law to a short ideal flexible string element with small slopes.

1Let $y(x,t)$ be transverse displacement and $\mu>0$ mass per length.
2Constant positive tension FT points outward along both end tangents.
3For $\left|\frac{\partial y}{\partial x}\right|\ll1$ , transverse force is approximately $F_T\frac{\partial y}{\partial x}$ .
4The difference over Δx is $F_T\Delta x\frac{\partial^2y}{\partial x^2}$ ; element mass is $\mu\Delta x$ .
5Newton’s law and cancellation give $\frac{\partial^2y}{\partial t^2}=\frac{F_T}\mu\frac{\partial^2y}{\partial x^2}$ .
6For travelling shape $g(x-vt)$ , $v^2=\frac{F_T}\mu$ and speed is $v=\sqrt{\frac{F_T}{\mu}}$ .

How do you calculate with Wave speed on a stretched string?

Here is how to work through a typical Wave speed on a stretched string (v = √(FT/μ)) task step by step:

Task

An ideal flexible string has tension 80 N and linear mass density 0.010 kg/m. The transverse wave has small slopes. Calculate propagation speed.

Solution path

Approach: $v=\sqrt{\frac{F_T}{\mu}}$ .
Substitution gives $v=\sqrt{\frac{80}{0.010}}\,\mathrm{m/s}=\sqrt{8000}\,\mathrm{m/s}\approx89.4427\,\mathrm{m/s}$ .

Check: $\mu v^2=80\,\mathrm N$ . This is propagation along the string, not a material point’s transverse velocity.

A pulse travels while the string stays in place

Specify tension and mass per unit length

\begin{aligned}v&=\sqrt{\frac{F_T}{\mu}}\\\mu&=\frac mL\end{aligned}

$\mu=\frac{m}{L}$ applies to a uniform string; L is its length in the considered taut state.[1]

Small slope is the geometric approximation

\begin{aligned}\left|\frac{\partial y}{\partial x}\right|&\ll1\\y&=A\sin(kx-\omega t)\\\max\left|\frac{\partial y}{\partial x}\right|&=Ak\end{aligned}

Small amplitude alone is insufficient at very short wavelength; the relevant product is Ak.

Balance transverse forces on a short element

\begin{aligned}\mu\Delta x\,\partial_t^2y&=F_T\Delta x\,\partial_x^2y\\\partial_t^2y&=\frac{F_T}{\mu}\partial_x^2y\end{aligned}

Apply Newton’s law to transverse motion; Δx cancels.

\begin{aligned}y(x,t)&=g(x-vt)\\\partial_t^2y&=v^2g^{\prime\prime}\\\partial_x^2y&=g^{\prime\prime}\end{aligned}

A smooth travelling disturbance solves the equation when $v^2=\frac{F_T}\mu$ .

Double speed requires four times the tension

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At fixed

\mu=0.004\,\mathrm{kg/m}

v=\sqrt{\frac{F_T}{\mu}}

. The curve starts at positive tension; a slack string is outside the model. Changes to taut length or mass density are excluded here.[1]

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Mass units, rearrangements and scaling

\begin{aligned}F_T&=\mu v^2\\\mu&=\frac{F_T}{v^2}\\\left[\frac{F_T}{\mu}\right]&=\frac{\mathrm N}{\mathrm{kg\,m^{-1}}}=\mathrm{m^2\,s^{-2}}\end{aligned}

Solving for μ needs $v>0$ ; taking the square root gives m/s.

Two original calculations: transit time and retensioning

A uniform string sample

Given: A uniform sample in the considered state has mass 12.0 g over 2.40 m. FT = 72 N; a transverse disturbance travels 1.50 m; flexible linear model.
Find: Find μ, v and one-way transit time.

Approach: The uniform taut sample determines density. The pulse travels a one-way path.
$\begin{aligned}\mu&=\frac mL\\v&=\sqrt{\frac{F_T}{\mu}}\\t&=\frac dv\end{aligned}$
Substitute the given values with matching units.
$\mu=\frac{0.0120}{2.40}=0.00500\,\mathrm{kg/m}$
Take the square root of tension divided by linear density.
$v=\sqrt{\frac{72}{0.00500}}=120\,\mathrm{m/s}$
Divide stationary distance by speed.
$t=\frac{1.50}{120}=0.0125\,\mathrm s$

Result: μ = 5.00 g/m, v = 120 m/s and transit time 12.5 ms.

Increase speed by a factor of 1.5

Given: μ = 0.00350 kg/m remains fixed in the model; initial tension 35.0 N; desired speed 150 m/s. The same source stays at f = 50 Hz.
Find: Find new tension and both wavelengths.

Approach: Keep density fixed but change tension: square the speed factor first.
$\begin{aligned}v_1&=\sqrt{\frac{F_{T,1}}\mu}\\F_{T,2}&=F_{T,1}\left(\frac{v_2}{v_1}\right)^2\\\lambda_i&=\frac{v_i}{f}\end{aligned}$
Substitute the given values with matching units.
$v_1=\sqrt{\frac{35.0}{0.00350}}=100\,\mathrm{m/s}$
Insert desired speed into the rearranged equation.
$F_{T,2}=0.00350\cdot150^2=78.75\,\mathrm N$
Find wavelengths at the same frequency.
$\begin{aligned}\lambda_1&=\frac{100}{50}=2.00\,\mathrm m\\{} \lambda_2&=\frac{150}{50}=3.00\,\mathrm m\end{aligned}$

Result: New tension is arithmetically 78.75 N, a factor of 2.25; wavelength increases from 2 to 3 m.

Driving and natural frequencies play different roles

f_n=\frac{n}{2L}\sqrt{\frac{F_T}{\mu}}

For an ideal flexible string fixed at both ends, $n=1,2,\ldots$

Explore further: measurement uncertainty and inference

Flexibility, losses and dispersion limit the model

A round-trip time determines tension

Inferring tension squares the speed. A modest timing error can therefore appreciably alter the inferred tension. This echo example separates instrument delay from actual travel time.[1][4]

A round-trip time determines tension

Given: A uniform taut string has length 2.00 m and mass 8.00 g. A small transverse pulse travels to the reflecting end and back. The instrument reads 45.0 ms, including an independently known 5.0 ms delay. Other paths and reflection delays are excluded in this ideal model.
Find: Calculate propagation speed and tension; assess the error if the delay is not removed.

Approach: first obtain actual travel time, then use the doubled path and taut-string mass density.
$\begin{aligned}t_{\mathrm{travel}}&=t_{\mathrm{read}}-t_{\mathrm{delay}}\\v&=\frac{2L}{t_{\mathrm{travel}}}\\\mu&=\frac mL\\F_T&=\mu v^2\end{aligned}$
Substitute consistent SI units.
$\begin{aligned}t_{\mathrm{travel}}&=0.0400\,\mathrm s\\\mu&=\frac{0.00800}{2.00}\,\mathrm{kg/m}=0.00400\,\mathrm{kg/m}\\v&=\frac{4.00}{0.0400}\,\mathrm{m/s}=100\,\mathrm{m/s}\\F_T&=0.00400\cdot100^2\,\mathrm N=40.0\,\mathrm N\end{aligned}$
Without correction, a smaller speed and its square would be used.
$\begin{aligned}v_{\mathrm{wrong}}&=\frac{4.00}{0.0450}\,\mathrm{m/s}\approx88.8889\,\mathrm{m/s}\\F_{\mathrm{wrong}}&\approx31.6049\,\mathrm N\\\frac{F_{\mathrm{wrong}}}{F_T}&=\left(\frac{0.0400}{0.0450}\right)^2=\frac{64}{81}\end{aligned}$

Result: The model gives $v=100\,\mathrm{m/s}$ and $F_T=40.0\,\mathrm N$ . Omitting the correction would underestimate tension by approximately 20.99%.

Five solved exercises

Calculate · Foundation

FT = 50 N and μ = 0.00200 kg/m. Find v.

Show hint

$\frac{F_T}{\mu}$ has units m²/s².

Learning goal: Calculate speed from tension and line density.

Show solution and calculation

Approach: the positive square root gives speed magnitude.
$v=\sqrt{25000}\,\mathrm{m/s}\approx158.114\,\mathrm{m/s}$ .

Check: $\mu v^2=50\,\mathrm N$ reproduces tension.

\begin{aligned}v&=\sqrt{\frac{F_T}\mu}\\&=\sqrt{\frac{50}{0.00200}}\,\mathrm{m/s}\\&\approx158.114\,\mathrm{m/s}\end{aligned}

Rearrange · Application

A uniform 3.00 m sample has mass 9.00 g; measured speed is 80.0 m/s in the ideal model. Find FT.

Show hint

Calculate μ in kg/m first.

Learning goal: Infer tension using specimen mass.

Show solution and calculation

Approach: divide mass by taut length first, then infer tension from speed squared.
$\mu=0.00300\,\mathrm{kg/m}$ , $F_T=19.2\,\mathrm N$ .

Check: $\sqrt{\frac{19.2}{0.00300}}=80\,\mathrm{m/s}$ . Length refers to the measured taut state.

\begin{aligned}\mu&=\frac mL=\frac{0.00900}{3.00}\,\mathrm{kg/m}=0.00300\,\mathrm{kg/m}\\F_T&=\mu v^2=0.00300\cdot80.0^2\,\mathrm N=19.2\,\mathrm N\end{aligned}

Transfer · Application

FT stays fixed while μ rises from 4 to 9 g/m. How does v change?

Show hint

Compare using the square root of a ratio.

Learning goal: Apply square-root scaling for changed density.

Show solution and calculation

Approach: constant tension cancels in the ratio.
$\frac{v_2}{v_1}=\sqrt{\frac49}=\frac23$ .
Speed falls by one third; $\frac49$ would incorrectly assume linear rather than square-root dependence.

\begin{aligned}\frac{v_2}{v_1}&=\sqrt{\frac{\mu_1}{\mu_2}}\\&=\sqrt{\frac49}=\frac23\end{aligned}

Interpret · Application

A small transverse wave on an ideal string has FT=80 N and μ=0.005 kg/m. Marked material points move up and down. Does every point move along the string at propagation speed? Calculate that speed and explain the distinction.

Show hint

Wave speed tracks a phase feature rather than a material particle.

Learning goal: Distinguish transverse particle motion from propagation speed.

Show solution and calculation

Approach: The string model gives approximately 126.491 m/s for phase propagation.
Material points oscillate transversely with their own time-dependent velocities set by amplitude and frequency.
Tension and line density alone do not give that speed.
Rapid pattern propagation does not imply comparable longitudinal flow of string material.

v_p=\sqrt{\frac{F_T}{\mu}}=\sqrt{\frac{80}{0.005}}\,\mathrm{m\,s^{-1}}\approx126.491\,\mathrm{m\,s^{-1}}

[1]

Error analysis · Application

A sinusoidal wave has amplitude A=2 cm and wavelength 10 cm. Someone treats the linear string speed as exact because 2 cm looks small. Check maximum slope and assess the small-slope assumption.

Show hint

Compare amplitude to wavelength through kA, not to an arbitrary length.

Learning goal: Check small slope before applying linear wave speed.

Show solution and calculation

Approach: Maximum slope of a sine profile is wave number times amplitude.
Here it is about 1.25664, nowhere near much less than one.
The linear small-slope derivation is not justified; tension and geometry can change dynamically.
An apparently small centimetre value alone cannot justify exactness.

\max|\partial_xy|=kA=\frac{2\pi A}{\lambda}=\frac{2\pi(0.02)}{0.10}\approx1.25664

[1]

In brief

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Distinguished tensile force from mechanical stress and added relevant wave connections; preserved all existing calculations, exercises, figures and model limits.
October 1, 2026Distinguished propagation from material motion; added echo inference with a known instrument delay and a tension–speed curve. Preserved previous calculations, model limits, and all five exercises.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

Cite this reference page

Quanta. Wave speed on a stretched string. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/wellengeschwindigkeit-saite

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