Physics · Waves and resonance

Standing waves on a string

Q: What determines wave speed v?

The ideal flexible-string model gives $v=\sqrt{\frac{F_T}{\mu}}$. FT is positive tension in newtons and μ linear mass density in kg/m. For a uniform sample in its considered taut state, $\mu=\frac mL$. Quadrupling tension doubles speed at fixed density; quadrupling density halves speed at fixed tension. The model needs a taut nearly straight string and small local slopes. Transverse speed of an individual string element is a different quantity depending additionally on amplitude and frequency. Real bending stiffness or varying tension can produce dispersion, so the ideal evenly spaced harmonic frequency sequence must then be checked.

For a uniform ideal flexible string with two fixed ends and constant v, boundary conditions give discrete wavelengths and natural frequencies.

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Formula

Calculate

λ_{n} = \frac{2 L}{n}, f_{n} = \frac{n v}{2 L}

L and λ in metres, v in m/s, f in hertz and n a positive integer.

Variables & units

Variables & units
Symbol	Meaning	Unit
$L$	Vibrating string length	m
$n$	Mode number or harmonic	dimensionless
$v$	Wave speed on the string	m/s
$f_{\mathrm{n}}$	Natural frequency of mode n	Hz
$λ_{\mathrm{n}}$	Wavelength of mode n	m

LaTeX

\lambda_n=\frac{2L}{n}, \quad f_n=\frac{n v}{2L}

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A standing string mode with three half-waves and nodes at both fixed ends and between them; the length L and node spacing λ₃/2 are marked. — Snapshot of the third ideal mode with two fixed ends: L = 3λ₃/2. Nodes stay fixed while displacements change in time. The dashed profile shows the opposite phase half a period later.

Standing wave on a string calculator

Find the natural frequency, wavelength, string length or wave speed of a string fixed at both ends for the n-th mode.

Find

Vibrating length between fixed ends L

Vibrating length between the fixed ends, e.g. 65 cm on a guitar.

Wave speed v on the string

How fast a wave travels along the string, e.g. 143 m/s; not the speed of sound in air.

Mode number n

Number of the mode, always a whole number: 1 for the fundamental, 2 for the first overtone.

1

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

Result: Natural frequency f_n

f_n = 110 Hz

With L = 65 cm, v = 143 m/s and n = 1, f_n = 110 Hz.

Working · rearrange for fₙ, substitute, evaluate

1. Rearrange for f_n
$f_n=\frac{nv}{2L}$
2. Substitute values (in SI units)
- L = 65 cm = 0.65 m
$f_n=\frac{1\cdot143}{2\cdot 0.65}$
3. Result
f_n = 110 Hz

Mode 1: 1 antinode, 0 internal nodes and two endpoint nodes (2 nodes in total). Adjacent nodes are λ_n/2 apart; the nearest antinode is λ_n/4 away.

Quick tips

A whole number of half-wavelengths always fits on the string: L = n · λ/2. For the fundamental the wavelength is twice the string length.
The natural frequencies are multiples of the fundamental: f₂ = 2 · f₁, f₃ = 3 · f₁. Half the string length gives twice the frequency, one octave up.

Assumptions and limits

Uniform ideal flexible string fixed at both ends, with positive vibrating length L and constant positive propagation speed v. Small slopes and a nondispersive model. n is a positive integer mode number. v describes the opposing travelling components, not a string element’s transverse velocity. This is not a pipe or free-end model.

Derivation & background

Both fixed ends are nodes at every time. A mode contains n half-wavelengths: $L=\frac{n\lambda_n}{2}$ . The ideal nondispersive string has $v=\sqrt{\frac{F_T}{\mu}}$ and $f_n=\frac v{\lambda_n}=\frac{nv}{2L}$ . Nodes stay fixed in space; v is speed of the opposing travelling components. Other boundaries or air columns require their own mode calculation.

Exam blueprint

Validity range

Uniform ideal flexible linearly vibrating string with two ideal fixed ends, constant propagation speed and small slopes. n is a positive integer. Other boundaries, bending stiffness or losses can change modes and resonance peaks.

Derivation steps

Apply fixed endpoint conditions at every time to a standing mode.

1Supports lie at $x=0$ and x = L; y must always be zero there.
2Equal opposite travelling waves add to $y=2A\sin(kx)\cos(\omega t)$ .
3The left node is satisfied; the right requires $\sin(kL)=0$ .
4Thus $kL=n\pi$ with $n=1,2,\ldots$ ; n counts half-wavelengths.
5Using $k=\frac{2\pi}{\lambda}$ gives $\lambda_n=\frac{2L}n$ .
6In the nondispersive model $\omega=vk$ , hence $f_n=\frac{nv}{2L}$ . $v=\sqrt{\frac{F_T}{\mu}}$ follows from string force balance.

Rearrangements

Calculate a string’s natural frequency

f_n=nf_1

Ideal string fixed at both ends; wave speed is assumed constant.

Task variant

An ideal uniform string is fixed at both ends. Its length is 0.75 m and travelling-wave speed 120 m/s. Find wavelength and frequency of the second mode.

Approach: $\lambda_n=\frac{2L}n$ , $f_n=\frac v{\lambda_n}$ .
For $n=2$ , $\lambda_2=\frac{2\cdot0.75}{2}\,\mathrm m=0.75\,\mathrm m$ , and $f_2=\frac{120}{0.75}\,\mathrm{Hz}=160\,\mathrm{Hz}$ .
Cross-check: $\frac{nv}{2L}=\frac{2\cdot120}{1.50}\,\mathrm{Hz}=160\,\mathrm{Hz}$ .

Common mistakes

Using a quarter wavelength instead of a half wavelength between fixed ends.

Each interval between adjacent nodes is $\frac\lambda2$ ; the string must contain n such segments.

Exam context

String harmonics and resonance of standing waves.

Formula cluster

Boundary conditions and resonance

Standing waves quantize the possible wavelengths through boundary conditions.

d·sin(α_k) = k·λ

Double-Slit Interference

c = λ·f

Wave Equation (c = λ·f)

T = 2π·√(m/k)

Spring Pendulum (Oscillation Period)

Worked example

An ideal string fixed at both ends has $L=0.75\,\mathrm m$ , $v=120\,\mathrm{m/s}$ . For $n=2$ , $\lambda_2=\frac{2L}2=0.75\,\mathrm m$ and $f_2=\frac v{\lambda_2}=160\,\mathrm{Hz}$ . Check: $\frac{nv}{2L}=\frac{2\cdot120}{1.50}\,\mathrm{Hz}=160\,\mathrm{Hz}$ .

Applications

String instruments, resonance measurements, mode identification and the relation between wavelength, frequency and boundary conditions.

How do you calculate with Standing wave on a string fixed at both ends?

Here is how to work through a typical Standing wave on a string fixed at both ends (λₙ = 2L/n; fₙ = nv/(2L)) task step by step:

1
Task
An ideal uniform string is fixed at both ends. Its length is 0.75 m and travelling-wave speed 120 m/s. Find wavelength and frequency of the second mode.
Solution path
1. Approach: $\lambda_n=\frac{2L}n$ , $f_n=\frac v{\lambda_n}$ .
2. For $n=2$ , $\lambda_2=\frac{2\cdot0.75}{2}\,\mathrm m=0.75\,\mathrm m$ , and $f_2=\frac{120}{0.75}\,\mathrm{Hz}=160\,\mathrm{Hz}$ .
3. Cross-check: $\frac{nv}{2L}=\frac{2\cdot120}{1.50}\,\mathrm{Hz}=160\,\mathrm{Hz}$ .

Understand and apply the equation

In a standing wave, nodes remain at fixed positions while the string between them oscillates. End conditions select the possible wavelengths and frequencies. Counted antinodes, measured frequencies, and a local amplitude show how to read the pattern.

Why a pattern can stay still while the string moves

A plucked string sends disturbances towards its ends, where they reflect. Suitable waves travelling in opposite directions can combine so that certain positions always remain still while others oscillate. “Standing” refers to this stationary pattern, not to every string segment being still or the component waves having stopped travelling.[1]

You should distinguish wavelength as a spatial distance from period as the time for a complete cycle. For an ideal uniform string fixed at both ends, we will then count half-wavelengths fitting between its ends. Mode numbering starts at one. Internal nodes exclude the two endpoints, which are already fixed.[1]

Specify vibrating length and both endpoints

L is vibrating length between two ideal fixed supports, not the full length of a coiled string. Transverse displacement obeys $y(0,t)=y(L,t)=0$ at every time. A node stays at zero displacement; an antinode has the mode’s greatest local oscillation amplitude.[1]

\begin{aligned}\lambda_n&=\frac{2L}{n}\\f_n&=\frac{nv}{2L},\quad n=1,2,3,\ldots\end{aligned}

This sequence applies to a uniform ideal flexible string fixed at both ends with constant v.

n counts half-wavelengths along L. Fundamental mode $n=1$ has one antinode between endpoint nodes. Mode two has two antinodes and one internal node. In general, mode n has n antinodes and $n-1$ internal nodes; endpoint nodes are counted separately.

Counterpropagating components create a standing shape

\begin{aligned}y_1&=A\sin(kx-\omega t)\\y_2&=A\sin(kx+\omega t)\\y&=2A\sin(kx)\cos(\omega t)\end{aligned}

Equal amplitudes and frequencies of the two opposite travelling components produce this pure standing form.[1]

The sum follows from a trigonometric identity. Spatial factor $\sin(kx)$ stays fixed while $\cos(\omega t)$ supplies time evolution. This satisfies $y(0,t)=0$ . To satisfy $y(L,t)=0$ for every time, require $\sin(kL)=0$ and hence $kL=n\pi$ . A snapshot with endpoints coincidentally at zero is insufficient as a boundary condition.

\begin{aligned}k_n&=\frac{n\pi}{L}\\\lambda_n&=\frac{2\pi}{k_n}=\frac{2L}{n}\\\omega_n&=vk_n\end{aligned}

The ideal nondispersive string has $\omega=vk$ , giving $f_n=\frac{\omega_n}{2\pi}$ .

Spatial nodes do not travel along the string, but its components still propagate oppositely at v. “Speed of the standing wave” is therefore ambiguous: v in fn is the component propagation speed rather than speed of fixed nodes.

The wave-speed relation connects wavelength and frequency of the travelling components through c = λ·f. For the string this propagation speed is called v; the fixed nodes themselves do not travel at v.[1]

Locate nodes, antinodes and relative signs

\begin{aligned}x_{j,\mathrm{node}}&=\frac{jL}{n},\quad j=0,\ldots,n\\x_{j,\mathrm{anti}}&=\frac{(2j+1)L}{2n},\quad j=0,\ldots,n-1\end{aligned}

Adjacent nodes are λn/2 apart; a neighbouring antinode is λn/4 from a node.

All points within one node interval oscillate in phase. In the next interval the spatial factor $\sin\left(\frac{n\pi x}{L}\right)$ changes sign, so those points move in the opposite direction. When $\cos(\omega_nt)=0$ , the whole string crosses equilibrium simultaneously. Permanent nodes are still only the positions whose spatial factor vanishes at every time.

The largest antinode amplitude of this superposition is $2A$ . At an arbitrary position it is $2A\left|\sin\left(\frac{n\pi x}{L}\right)\right|$ . A sensor at a node detects no transverse motion in the ideal model even though the same mode oscillates strongly elsewhere. A weak signal can therefore result from an unsuitable measurement position. The excitation position similarly influences which modes are excited strongly.

Third mode at three instants

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The third example uses maximum standing-wave amplitude

6\,\mathrm{mm}

. The zero profile at

t=\frac T4

does not make all points permanent nodes. Transverse and longitudinal scales differ for readability.[1]

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Connect natural frequencies with tension and density

\begin{aligned}v&=\sqrt{\frac{F_T}{\mu}}\\f_n&=\frac{n}{2L}\sqrt{\frac{F_T}{\mu}}\end{aligned}

Taut uniform ideal flexible string, constant tension and small local slopes.[2]

At fixed n with unchanged FT and μ, halving vibrating length doubles frequency. Quadrupling tension doubles frequency at fixed L and μ. Increasing displacement amplitude alone does not shift natural frequencies in the ideal linear model; real tension may nevertheless increase at large displacement.

For $L=0.75\,\mathrm m$ and $v=120\,\mathrm{m/s}$ , the fundamental is $f_1=80\,\mathrm{Hz}$ . The second mode has $\lambda_2=L=0.75\,\mathrm m$ , $f_2=160\,\mathrm{Hz}$ , and a midpoint internal node. The fundamental instead has $\lambda_1=2L=1.50\,\mathrm m$ . Always identify the mode number when comparing the two shapes.

Wave speed on a stretched string connects tension with linear mass density. Use values for the considered taut state; the natural-mode calculation below assumes this propagation speed is constant.[2]

Two original calculations on mode shape and resonance spacing

Third mode from string data

Given: L = 0.800 m, FT = 64.0 N, μ = 0.00400 kg/m; fixed ends, uniform ideal flexible string; $n=3$ .
Find: Find v, f3, λ3, nodes and antinodes.

Approach: Both ends are fixed: use string speed and the third mode.
$\begin{aligned}v&=\sqrt{\frac{F_T}{\mu}}\\f_n&=\frac{nv}{2L}\\\lambda_n&=\frac{2L}{n}\end{aligned}$
Substitute the given values with matching units.
$v=\sqrt{\frac{64}{0.004}}=\sqrt{16000}\approx126.491106\,\mathrm{m/s}$
Fit three half-wavelengths along L.
$\begin{aligned}\lambda_3&=\frac{1.6}{3}\approx0.533333\,\mathrm m\\{} f_3&=\frac{3v}{1.6}\approx237.170825\,\mathrm{Hz}\end{aligned}$
Use integer and half-integer subdivisions.
$x_{\mathrm{node}}=0,\ 0.266667,\ 0.533333,\ 0.800000\,\mathrm m$
Place antinodes between nodes.
$x_{\mathrm{anti}}=0.133333,\ 0.400000,\ 0.666667\,\mathrm m$

Result: Three antinodes, two internal nodes and two endpoint nodes; f3 ≈ 237.17 Hz and λ3 ≈ 0.53333 m.

Check: Unrounded values give $f_3\lambda_3=v$ . Both endpoints satisfy $\sin\left(\frac{3\pi x}{L}\right)=0$ ; adjacent nodes are spaced by $\frac L3=\frac{\lambda_3}{2}$ . The three antinodes thus match the mode number.

Identify two adjacent resonances

Given: Ideal uniform string fixed at both ends, L = 0.900 m and μ = 0.00600 kg/m. Two adjacent natural frequencies are 135 and 180 Hz.
Find: Find f1, mode numbers, v and FT.

Approach: The explicitly adjacent modes differ by the fundamental frequency.
$\begin{aligned}f_1&=f_{n+1}-f_n\\v&=2Lf_1\\F_T&=\mu v^2\end{aligned}$
Substitute the given values with matching units.
$f_1=180-135=45\,\mathrm{Hz}$
Divide both frequencies by f1.
$\begin{aligned}n_a&=\frac{135}{45}=3\\{} n_b&=\frac{180}{45}=4\end{aligned}$
Obtain propagation speed from the fundamental.
$v=2Lf_1=1.8\cdot45=81\,\mathrm{m/s}$
Find tension from μv².
$F_T=0.00600\cdot81^2=39.366\,\mathrm N$

Result: Fundamental is 45 Hz; measured modes are $n=3$ and 4; v = 81 m/s; arithmetic tension is 39.366 N.

Check: Check: $\frac{3\cdot81}{1.8}=135\,\mathrm{Hz}$ and $\frac{4\cdot81}{1.8}=180\,\mathrm{Hz}$ . “Adjacent” is essential. If a missed mode lies between them, the frequency spacing is a multiple of the fundamental, making this inference ambiguous.

Opposing energy flows and local oscillation

A pure standing wave made from equal opposing components has no time-averaged net energy flux along an ideal string. This does not mean it contains no energy. Elements have kinetic energy and the displaced taut string stores potential energy; these contributions vary over a period.[3]

A real resonator loses energy. Sustained motion therefore needs input replacing those losses. An observed pattern may be dominated by a standing mode without every actual component having equal strength. “Standing” alone proves neither zero losses nor a particular energy or amplitude.

Explore further: driven resonance

Natural frequencies describe free linear motion with specified boundaries. A driven response also depends on damping, excitation position and coupling. Resonance peaks can be broad and damping may shift them slightly from ideal values. fn therefore does not predict an infinite real amplitude.

Ideal point transverse excitation at a mode’s node does not couple to its displacement. A sensor at that node may miss the mode. The two frequencies assumed adjacent in the example must therefore truly be adjacent, not merely the two strongest visible peaks.

Other end conditions create other sequences

At an ideal free end, transverse force is zero, giving zero slope in the simple flexible model. A fixed-free taut string then has odd quarter-wave modes: $\lambda_j=\frac{4L}{2j-1}$ and $f_j=\frac{(2j-1)v}{4L}$ . An actual free termination must physically maintain tension; it is a different boundary problem from two fixed ends.

Air-column boundaries concern pressure or particle motion depending on representation; a pressure node is not a displacement node. Do not transfer a string diagram unchecked to a pipe. String stiffness, large slopes, varying μ, compliant supports and losses can also modify the ideal harmonic sequence.[1]

One sensor lies at an antinode, the other at a node

Mode frequency describes the whole string; a sensor observes only its own position. The following spatial factor determines where a mode is visible and where it has no transverse motion.[1][2]

One sensor lies at an antinode, the other at a node

Given: An ideal string fixed at both ends has L=0.90 m and v=90 m/s and oscillates only in mode $n=3$ . Displacement is maximal at t=0: y(x,0)=(6.0 mm)sin(3πx/L). Sensors are at x=0.15 m and x=0.30 m.
Find: Find frequency, local amplitudes, and velocities after a quarter period.

Approach: the mode sets frequency and spatial factor. Evaluate displacement and velocity separately.
$\begin{aligned}f_3&=\frac{3v}{2L}\\y(x,t)&=Y\sin\left(\frac{3\pi x}{L}\right)\cos\omega t\\v_y(x,t)&=-Y\omega\sin\left(\frac{3\pi x}{L}\right)\sin\omega t\end{aligned}$
The sensors correspond to spatial phases π/2 and π.
$\begin{aligned}f_3&=150\,\mathrm{Hz},&T&=\frac1{150}\,\mathrm s\\Y_1&=6.0\,\mathrm{mm},&Y_2&=0\\v_{y,1}\left(\frac T4\right)&=-(0.006)(300\pi)\,\mathrm{m/s}\\&\approx-5.65487\,\mathrm{m/s}\\v_{y,2}\left(\frac T4\right)&=0\end{aligned}$

Result: At 0.15 m there is an antinode with 6.0 mm amplitude; at 0.30 m there is a permanent node. After a quarter period displacement is zero everywhere, yet the first sensor moves downward.

Check: Slope remains small: $Yk=0.006\cdot\frac{3\pi}{0.90}\approx0.0628$ . Material speed $5.65\,\mathrm{m/s}$ is not propagation speed $90\,\mathrm{m/s}$ . Y is the maximum standing-wave amplitude, not the amplitude of one travelling component.

Five exercises on mode, length and position

Calculate · Foundation

L = 1.20 m and v = 96 m/s. Find f1, f4 and λ4 for fixed ends.

Show hint

Mode number is a positive integer.

Learning goal: Find eigenfrequencies and modal wavelength.

Show solution and calculation

Approach: fixed-end conditions give the fundamental, and four times that value gives mode four.
Thus $f_1=40\,\mathrm{Hz}$ , $f_4=160\,\mathrm{Hz}$ , and $\lambda_4=0.600\,\mathrm m$ .

Check: $160\cdot0.600=96\,\mathrm{m/s}$ ; four half-wavelengths fill the string.

\begin{aligned}f_1&=\frac v{2L}=\frac{96}{2.40}\,\mathrm{Hz}=40\,\mathrm{Hz}\\f_4&=4f_1=160\,\mathrm{Hz}\\\lambda_4&=\frac{2L}4=0.600\,\mathrm m\end{aligned}

Interpret · Application

A string fixed at both ends has three internal nodes. Identify the mode and count antinodes.

Show hint

There are $n-1$ internal nodes; count endpoints separately.

Learning goal: Map internal nodes to mode number.

Show solution and calculation

Approach: three internal nodes make four intervals between five total nodes.
Thus $n=4$ , with four antinodes.
Internal nodes are at $\frac L4$ , $\frac L2$ , and $\frac{3L}4$ ; $\lambda_4=\frac L2$ .

Check: four half-wavelengths equal L.

\begin{aligned}N_{\mathrm{internal}}&=n-1\\n&=3+1=4\\N_{\mathrm{antinodes}}&=4,&N_{\mathrm{nodes}}&=5\\\lambda_4&=\frac{2L}4=\frac L2\end{aligned}

Transfer · Application

A fixed string’s vibrating length changes from 0.80 m to 0.50 m while v stays fixed. How does the same mode frequency change?

Show hint

Compare $\frac1L$ at fixed n and v.

Learning goal: Transfer length changes at fixed wave speed.

Show solution and calculation

Approach: at fixed mode number and speed, frequency is inversely proportional to length.
$\frac{f_{\mathrm{new}}}{f_{\mathrm{old}}}=\frac{0.80}{0.50}=1.60$ , a 60% increase.
This requires unchanged wave speed; keeping tension and linear density fixed is one way to meet that condition.

\begin{aligned}f_n&=\frac{nv}{2L}\\\frac{f_{\mathrm{new}}}{f_{\mathrm{old}}}&=\frac{L_{\mathrm{old}}}{L_{\mathrm{new}}}=\frac{0.80}{0.50}=1.60\end{aligned}

Interpret · Application

An ideal string fixed at both ends oscillates in a single $n=3$ mode. At one instant its displacement is zero everywhere, so a photo shows no antinodes. Are all positions now permanent nodes? Describe internal nodes, antinodes and velocities at that instant.

Show hint

A node stays unmoving for all time, not only in one photograph.

Learning goal: Distinguish mode shape from an instantaneous zero profile.

Show solution and calculation

Approach: Separate spatial sine and time factors.
Permanent internal nodes lie at one and two thirds of length.
Three antinodes lie at one sixth, one half and five sixths.
At a zero-profile crossing their speed magnitudes are maximal; only true nodes remain at zero velocity as well.

\begin{aligned}y(x,t)&=A\sin\left(\frac{3\pi x}{L}\right)\cos\omega t\\{}x_{\mathrm{nodes}}&\in\left\{0,\frac L3,\frac{2L}3,L\right\}\\{}\partial_ty&=-A\omega\sin\left(\frac{3\pi x}{L}\right)\sin\omega t\end{aligned}

[1]

Error analysis · Challenge

For an ideal string with one fixed end and one transverse-force-free end someone uses the both-fixed fundamental $\frac{v}{2L}$ . Check whether the associated half sine wave fits the free end. Give the correct fundamental wavelength and frequency. An appropriate ideal end support maintains axial tension.

Show hint

The fixed end is a node; the free end an antinode with zero slope.

Learning goal: Check boundary conditions before using integer mode formulas.

Show solution and calculation

Approach: A half sine wave would put a node at the other end and a nonzero slope there.
The ideal free end instead requires a quarter wave.
Fundamental wavelength is therefore four L and frequency is half the incorrectly used value.
Different boundary conditions produce a different set of modes.

\begin{aligned}y(0)&=0,&\partial_xy(L)&=0\\{}k_1L&=\frac\pi2\\{}\lambda_1&=4L,&f_1&=\frac v{4L}\end{aligned}

[1]

In brief

Check end conditions first. With two fixed ends, n half-wavelengths fit along the string: n antinodes and $n-1$ internal nodes. Spatial and temporal factors describe different properties. A flat snapshot does not erase the pattern of nodes and antinodes.[1][2]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

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.

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.

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Change history

October 2, 2026Linked travelling components and string propagation speed in context; preserved boundaries, nodes, sensor example and diagram.
October 1, 2026Clarified nodes, antinodes, and snapshots; added a sensor example and quantitative third-mode graph. Preserved previous mode calculations and all five exercises.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Standing wave on a string fixed at both ends. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/stehende-welle-saite

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Frequently asked questions about Standing wave on a string fixed at both ends

Why do only certain wavelengths fit on the string?+

Both fixed endpoints must be nodes at every time. For standing shape $y=2A\sin(kx)\cos(\omega t)$ , $x=0$ satisfies this condition; x = L additionally requires $\sin(kL)=0$ . Therefore $kL=n\pi$ and $\lambda_n=\frac{2L}n$ with positive integer mode number n. A snapshot where the whole string happens to cross equilibrium is insufficient by itself. Mode n contains n half-wavelengths, n antinodes and $n-1$ internal nodes. Other boundaries, such as an ideal free endpoint, permit a different sequence. Check the mode shape against actual supports before calculating its frequency.

What is the fundamental frequency?+

In the ideal string fixed at both ends, the fundamental is the lowest positive natural frequency: $f_1=\frac v{2L}$ . Its mode contains half a wavelength along the vibrating length, giving $\lambda_1=2L$ , one central antinode and only the endpoint nodes. Higher natural frequencies are $f_n=nf_1$ for a uniform ideal flexible nondispersive string. For example, L = 0.75 m and v = 120 m/s give a fundamental of 80 Hz and second mode at 160 Hz. An arbitrary driving frequency need not be natural. Real resonance amplitude and peak width additionally depend on damping and excitation.

What determines wave speed v?+

The ideal flexible-string model gives $v=\sqrt{\frac{F_T}{\mu}}$ . FT is positive tension in newtons and μ linear mass density in kg/m. For a uniform sample in its considered taut state, $\mu=\frac mL$ . Quadrupling tension doubles speed at fixed density; quadrupling density halves speed at fixed tension. The model needs a taut nearly straight string and small local slopes. Transverse speed of an individual string element is a different quantity depending additionally on amplitude and frequency. Real bending stiffness or varying tension can produce dispersion, so the ideal evenly spaced harmonic frequency sequence must then be checked.

Does the formula also apply to a pipe?+

Do not transfer the string formula to an air column without checking. Fixed string ends are nodes of transverse displacement. A pipe instead involves acoustic pressure and particle motion; in the simple sound model, a pressure node corresponds to a displacement antinode. Two ideal open ends have different physical boundary conditions even when a similar frequency sequence occurs. One ideal open and one closed end produces odd quarter-wave modes rather than the fixed-string sequence. Real pipes also need end corrections and the appropriate sound speed. Identify the measured quantity, both boundary conditions and effective length before choosing a mode formula.

How does frequency change if the string length is halved?+

For the same mode number and unchanged propagation speed, halving vibrating length doubles natural frequency through $f_n=\frac{nv}{2L}$ . At the same time $\lambda_n=\frac{2L}n$ halves. This concerns the distance between the two effective fixed supports. Tension and linear density must remain unchanged, or v must otherwise stay constant. If shortening a real string also changes tension, the length rule alone is insufficient. Comparing different modes additionally requires their n factors. Driving frequency is a separate input: changing a natural frequency does not automatically change the rhythm of a source that continues driving at the same frequency.

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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 Volume 1, §16.6: Standing Waves and Resonance

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Standing waves on a string

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Standing wave on a string calculator

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Boundary conditions and resonance

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How do you calculate with Standing wave on a string fixed at both ends?

Why a pattern can stay still while the string moves

Specify vibrating length and both endpoints

Counterpropagating components create a standing shape

Locate nodes, antinodes and relative signs

Third mode at three instants

Connect natural frequencies with tension and density

Two original calculations on mode shape and resonance spacing

Third mode from string data

Identify two adjacent resonances

Opposing energy flows and local oscillation

Explore further: driven resonance

Other end conditions create other sequences

One sensor lies at an antinode, the other at a node

One sensor lies at an antinode, the other at a node

Five exercises on mode, length and position

L = 1.20 m and v = 96 m/s. Find f1, f4 and λ4 for fixed ends.

A string fixed at both ends has three internal nodes. Identify the mode and count antinodes.

A fixed string’s vibrating length changes from 0.80 m to 0.50 m while v stays fixed. How does the same mode frequency change?

An ideal string fixed at both ends oscillates in a single n=3n=3n=3 mode. At one instant its displacement is zero everywhere, so a photo shows no antinodes. Are all positions now permanent nodes? Describe internal nodes, antinodes and velocities at that instant.

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Related formulas

More formulas: Oscillations and waves

Retain Standing wave on a string fixed at both ends for exams

An ideal string fixed at both ends oscillates in a single $n=3$ mode. At one instant its displacement is zero everywhere, so a photo shows no antinodes. Are all positions now permanent nodes? Describe internal nodes, antinodes and velocities at that instant.