Preview
Normal Modes
Standing waves and harmonics on a fixed-fixed string
Pin a string at both ends and you have done something quietly violent to it: you have forbidden almost every motion it could ever make. Pluck it, strike it, bow it — whatever you do, the two pinned ends refuse to move, and only the shapes that respect that refusal can survive. Those shapes are the normal modes, and they form a ladder with no missing rungs: one arch fitting half a wavelength between the walls, two arches fitting two halves, three fitting three. Each rung sings its own note, and the notes are not arbitrary — the second rung sings exactly twice the fundamental, the third exactly three times, an integer ladder climbing by the string's own master note. Marin Mersenne measured this in 1636; every guitar, violin, and piano built since is an application of it. In this lab the string is yours to interrogate. Three levers wait: the wave speed, which stretches or shrinks every note on the ladder at once while the shape refuses to notice; the string length, which retunes the whole instrument the way a guitarist's finger does — shorter string, higher note, always the inverse; and the harmonic number itself, which walks you rung by rung up the ladder while red node markers hold perfectly still and blue antinode markers ride the crests. Watch the nodes. They are the places the string has agreed never to move — and counting them wrong is the most common way to fail this unit. Your prediction before you touch anything: what does doubling the string's length do to its master note?
What you'll be able to do
- Run the harmonic accounts with the panel's own rows: at the factory state (n=1, L=1.0 m, v=20 m/s) f₁ = nv/2L = 10.00 Hz and λ₁ = 2L/n = 2.000 m (Eq. 16.15/16.16, verified); walk the integer ladder n 1→5 and read fₙ = nf₁ climbing 10.00 → 50.00 Hz in exact integer multiples while λₙ falls 2.000 → 0.400 m; price all three presets — Fundamental (10.00 Hz, 2.000 m), 2nd Harmonic (20.00 Hz, 1.000 m), 5th Harmonic (37.50 Hz at L=2.0 m, v=30 m/s, λ = 0.800 m) — against the same two formulas, the OpenStax Example 16.7 lineage (a 2.00 m lab string at 57.15 m/s sings 14.29/28.58/42.87 Hz)
- Explain WHY the string's frequencies are discrete: 'The symmetrical boundary conditions (a node at each end) dictate the possible frequencies that can excite standing waves' (§16.6, verified) — only shapes that pin nodes at both fixed ends survive, so the string fits exactly n half-wavelengths, nodes sit at x = mλ/2 (n+1 of them including the endpoints) and antinodes at odd multiples of λ/4 (n of them); count them live on the marker spheres — 2 nodes/1 antinode at n=1, 6 nodes/5 antinodes at n=5 — and connect off-resonance driving to the verified resonance sentence: driven at a non-mode frequency, 'the amplitude of the vibrations will be much smaller than the amplitude at resonance'
- Keep the three levers' jobs separate: wave speed sets the frequency scale only (sweep v 10→50 m/s at fixed L and f₁ climbs linearly 5.00 → 25.00 Hz while λ₁ stays pinned at 2.000 m — λₙ = 2L/n has no v in it); string length sets the geometry inversely (sweep L 0.5→3.0 m and f₁ falls 20.00 → 3.33 Hz along v/2L — 'the lengths of the strings are changed by pressing down on the strings', §16.3 verified); the harmonic number walks the integer ladder; and amplitude is nowhere in the mode structure — the engine holds A = 1.2 across all modes, the exact convention OpenStax Fig 16.29 declares ('kept constant for visualization'), while a lab string's amplitude would DECREASE with frequency
Formulas
Make a prediction
Default state: n = 1, L = 1.0 m, v = 20 m/s — the panel reads f₁ = 10.00 Hz and λ₁ = 2.000 m. You drag String Length from 1.0 m to 2.0 m (Harmonic and Wave Speed untouched). What happens to the fundamental?
No grading here — commit to a guess, then scroll down and test it yourself.
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Your prediction
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Answer: f₁ halves to 5.00 Hz and λ₁ doubles to 4.000 m — f₁ = v/(2L) has L in the denominator, so twice the string means half the note and twice the wavelength
f₁ = v/(2L) = 20/(2×2.0) = 5.00 Hz, and λ₁ = 2L = 4.000 m — the string still fits exactly one half-wavelength between its fixed ends, but the half-wavelength is now twice as long, and a wave traveling at unchanged speed takes twice as long to cross it: half the frequency (OpenStax Eq. 16.15/16.16, verified). Option B is the direct-proportion instinct this gate targets — the HTML's own quiz Q2 banks on it with its 'Doubles' distractor — and every guitarist carries the counter in their left hand: 'To play notes other than the fundamental notes, the lengths of the strings are changed by pressing down on the strings' (OpenStax §16.3, verified) — fretting SHORTENS the speaking length to RAISE the note, the inverse law in daily use. The full sweep makes the same point 26 times: L 0.5 → 3.0 m runs f₁ 20.00 → 3.33 Hz, falling the whole way. Option C half-remembers the right physics: wave speed does set the note — but as the numerator of v/(2L), not the whole fraction; hold v fixed and double L, and the denominator wins 2-to-1. Run the drag live and watch the Frequency row fall 10.00 → 5.00 while the Wavelength row climbs 2.000 → 4.000 — the shape keeps its single arch the entire time, because the boundary conditions (a node at each end) never stopped ruling which shapes may exist.
Quiz (0/3)
At L = 2.0 m and v = 20 m/s, what frequency does the 3rd harmonic sing — and how many nodes does it pin?
Double the string length with everything else fixed. What happens to the fundamental, and why isn't it 'more room, higher note'?
How do normal modes relate to a guitar string's harmonics?
You can now
- Run the harmonic accounts with the panel's own rows: at the factory state (n=1, L=1.0 m, v=20 m/s) f₁ = nv/2L = 10.00 Hz and λ₁ = 2L/n = 2.000 m (Eq. 16.15/16.16, verified); walk the integer ladder n 1→5 and read fₙ = nf₁ climbing 10.00 → 50.00 Hz in exact integer multiples while λₙ falls 2.000 → 0.400 m; price all three presets — Fundamental (10.00 Hz, 2.000 m), 2nd Harmonic (20.00 Hz, 1.000 m), 5th Harmonic (37.50 Hz at L=2.0 m, v=30 m/s, λ = 0.800 m) — against the same two formulas, the OpenStax Example 16.7 lineage (a 2.00 m lab string at 57.15 m/s sings 14.29/28.58/42.87 Hz)
- Explain WHY the string's frequencies are discrete: 'The symmetrical boundary conditions (a node at each end) dictate the possible frequencies that can excite standing waves' (§16.6, verified) — only shapes that pin nodes at both fixed ends survive, so the string fits exactly n half-wavelengths, nodes sit at x = mλ/2 (n+1 of them including the endpoints) and antinodes at odd multiples of λ/4 (n of them); count them live on the marker spheres — 2 nodes/1 antinode at n=1, 6 nodes/5 antinodes at n=5 — and connect off-resonance driving to the verified resonance sentence: driven at a non-mode frequency, 'the amplitude of the vibrations will be much smaller than the amplitude at resonance'
- Keep the three levers' jobs separate: wave speed sets the frequency scale only (sweep v 10→50 m/s at fixed L and f₁ climbs linearly 5.00 → 25.00 Hz while λ₁ stays pinned at 2.000 m — λₙ = 2L/n has no v in it); string length sets the geometry inversely (sweep L 0.5→3.0 m and f₁ falls 20.00 → 3.33 Hz along v/2L — 'the lengths of the strings are changed by pressing down on the strings', §16.3 verified); the harmonic number walks the integer ladder; and amplitude is nowhere in the mode structure — the engine holds A = 1.2 across all modes, the exact convention OpenStax Fig 16.29 declares ('kept constant for visualization'), while a lab string's amplitude would DECREASE with frequency