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SCIENCE · INVESTIGATE AND EXPLAIN

Build the harmonic series from fixed-end modes

Count loops and nodes, relate allowed wavelengths to string length and compare harmonic frequencies.

Learning outcome: Apply wavelength = 2L/n and frequency = n times the fundamental, while distinguishing mode shape from a driven response or a complete instrument sound.

Starting recap

A fixed end is a displacement node. Each loop between adjacent nodes contains half a wavelength.

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Try

Can a fixed-end string have a mode whose shape has an antinode at one endpoint?

Open the experiment

Predict

At 16 N, predict wavelengths, frequencies and node counts for modes 1, 2 and 3.

Experiment

Keep tension at 16 N and capture modes 1, 2 and 3. Count loops and endpoints, compare the equal frequency spacing, and inspect mode 5. Then change tension without changing mode and explain which features change.

Select an allowed mode of a fixed-end string

Worked teaching model · synthetic values · Uniform flexible string, fixed length 2.000 m, linear mass density 0.010 kg/m, both ends fixed, small transverse displacement and constant tension. Wave speed is sqrt(tension / linear density). No stiffness, stretching-induced density change, damping or drive response. Each shape is a separate single mode shown at a displacement extreme with assigned peak amplitude 0.010 m, not predicted resonance amplitude. The graph uses 161 spatial samples and does not animate time. Changing tension holds length and density fixed; the five displayed modes are examples, not all possible ideal modes.

Reset keeps your prediction, captures and practice history.

Mode 2 · y displacement (m) · one displacement extreme-0.01200.01200.511.52Position x (m) · endpoints fixed · vertical scale exaggerated
Allowed frequencies · equal spacing 10.000 Hz020406080n = 1Selected n = 2n = 3n = 4n = 5Frequency (Hz) · no amplitude or loudness prediction

The shape shows one selected mode at a displacement extreme. Adjacent loops point in opposite directions. Changing tension shifts the frequency scale but leaves the selected mode's geometry unchanged because length and boundary conditions stay fixed. Actual plucked strings can contain several modes at once.

String properties and selected mode
QuantityValueRelationship
Tension16.000 NConstant force magnitude
Vibrating length2.000 mDistance between fixed ends
Linear mass density0.010 kg/mMass per unit length, not volume density
Vibrating mass0.020 kgLinear density × length
Wave speed40.000000 m/ssqrt(tension / linear density)
Selected wavelength2.000000 m2L / n
Selected frequency20.000000 Hzn × fundamental
Selected period0.050000 s1 / frequency
Nodes including endpoints0.000, 1.000, 2.000 mSpacing L/n
Antinodes0.500, 1.500 mCenters of the loops
First five ideal string modes
Mode nWavelength (m)Frequency (Hz)
14.00000010.000000
2 (selected)2.00000020.000000
31.33333330.000000
41.00000040.000000
50.80000050.000000

Wave speed 40.000 m/s; mode 2 frequency 20.000 Hz; wavelength 2.000 m.

Fundamental 10.000 Hz; selected period 0.050000 s. 2 half-wavelength loops fit between fixed endpoints. Shape amplitude is assigned for comparison and is not a driven response.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Fit the pattern between two fixed ends

A fixed string endpoint cannot have transverse displacement. Both ends must therefore be nodes. Between neighboring nodes there is one half-wavelength loop. If a string of length L contains n such loops, L = nλ/2, so the allowed wavelength is λ = 2L/n. The mode number n is a positive integer.

The first mode has one loop and no interior node. The second has two loops and one interior node. Mode n has n loops, n antinodes and n + 1 nodes when both endpoints are included. Count the open markers rather than mistaking every change of curve direction for a node.

The browser’s 2 m string is drawn at one displacement extreme. Its vertical scale is exaggerated. Adjacent loops point in opposite directions at this instant; after half a period all their signs would reverse. The endpoints remain fixed throughout the motion, even though this view does not animate time.

Build frequencies from geometry and speed

Keep tension at 16 N. The model’s wave speed is 40 m/s. The first mode has wavelength 4 m, giving frequency 40/4 = 10 Hz. This is the fundamental frequency f1 = c/(2L).

Mode 2 has wavelength 2 m and frequency 20 Hz. Mode 3 has wavelength 4/3 m, approximately 1.333333 m, and frequency 30 Hz. In general, fn = n f1 for this ideal uniform flexible string with fixed ends. The first five modes at this tension are 10, 20, 30, 40 and 50 Hz.

The bars use a fixed 0–80 Hz scale, so changing tension changes their lengths honestly. Selecting a mode changes which bar is labeled and which shape appears; it does not remove the other allowed natural frequencies. The five displayed modes are a bounded teaching selection, not the entire ideal series.

Keep mode number, harmonic and overtone distinct

In this ideal harmonic series, mode 1 is the first harmonic or fundamental. Mode 2 is the second harmonic and the first overtone. Mode 3 is the third harmonic and second overtone. The word “third” therefore means different frequencies in “third harmonic” and “third overtone.” State the convention.

At the default 16 N, mode 2 has nodes at x = 0, 1 and 2 m, antinodes at 0.5 and 1.5 m, wavelength 2 m, frequency 20 Hz and period 0.050 s. Moving to mode 3 adds one loop and one interior node, increases frequency to 30 Hz and reduces wavelength. Wave speed remains 40 m/s because tension and linear density did not change.

Optional: connect a shape with its time dependence

A single ideal mode can be written y(x,t) = A sin(nπx/L) cos(2πfn t). Here A is the peak displacement assigned to that mode, x is distance along the string, L is vibrating length, n is a positive integer, fn is its frequency and t is time. Both endpoint values are zero because sin(0) and sin(nπ) vanish. The displayed shape is the t = 0 slice with A = 0.010 m. That amplitude is chosen for visualization, not calculated from a driver.

A frequency list is not an instrument’s complete sound

A pluck generally excites several modes, with amplitudes depending on how and where the string is disturbed. The bridge, body and surrounding air influence the emitted sound. This model displays one mode at a time and computes its ideal natural frequency; it does not predict loudness, timbre, damping or acoustic radiation.

A driver near a natural frequency can transfer energy efficiently under suitable conditions, but the amplitude of the response depends on driving and losses. The assigned 0.010 m peak in every shape is not a resonance curve. Real string stiffness, nonuniformity and support motion can also shift frequencies away from exact integer multiples.

Transfer to a shorter vibrating length

A fixed-end string has a 10 Hz fundamental. Its vibrating length is halved while tension and linear density remain unchanged. Predict the new fundamental and third harmonic. Then explain why halving the amount of vibrating string does not halve its linear density.

Check the harmonic transfer

Wave speed is unchanged, while f1 = c/(2L) doubles to 20 Hz. The new third harmonic is 60 Hz. For a uniform string, halving length also halves vibrating mass, so their ratio, mass per length, stays unchanged. This ideal comparison does not model a finger's compliance or changes in tension when a real string is pressed.

OpenStax standing waves and resonance explains fixed-end modes. The lesson’s controls hold the stated material parameters constant so that geometry and tension effects can be compared separately.

Apply

An ideal fixed-end string has a fundamental frequency of 10 Hz. Its third harmonic is:

Public practice checks a specific claim. It does not establish independent mastery.

Use the endpoint condition to explain why mode number must be a positive integer. Distinguish harmonic number from overtone number and explain why the assigned shape amplitude does not predict instrument loudness.

Continue

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