Waves & Optics

Standing Waves and Harmonics

Definition

Standing waves are fixed patterns that appear when a wave is confined between two boundaries, because a length accepts only wavelengths making a whole number of half waves. That condition, L = n lam / 2, lets a string or air column sound at a whole-number ladder of frequencies, f = n v / (2L), and nothing between them.

Press a finger onto a guitar string and the note jumps. Lift it and the note drops back. Nothing about the string itself has changed — only how much of it is free to vibrate between two fixed points.

A trapped wave is choosy. Between its two ends a string will not sound at any pitch you ask it for: it accepts a short list, set by its length, and refuses everything else. Working out that list is what this page is about.

What Is a Standing Wave?

A standing wave is a vibration pattern that stays in one place: it is built from a wave reflecting back along its own path, and the result has points that never move at all. Nothing travels along the length, and no net energy is carried from one end to the other.

Send a pulse down a rope tied to a post and it comes back at you. Keep sending, and the outgoing wave and its own reflection add together all along the rope. At certain frequencies they add into a shape that simply sits there, swinging in place.

The points that never move are called nodes. Halfway between each pair of them the medium moves furthest, and those places are the antinodes. Away from the nodes every point of the medium traces simple harmonic motion of its own, swinging about the position it rests at; how far it swings depends on where along the length it sits, and that arrangement of large and small swings never shifts.

Some syllabuses call it a stationary wave. Same object, same mathematics — the word is simply making the point that the thing does not go anywhere.

Hold on to that, because everything below follows from it. A travelling wave hands a crest from one place to the next; a standing wave hands nothing along its length. The two ends are what make the difference, and they are what the formula is built out of.

The Standing Wave Formula: L = n Lambda / 2

The standing wave condition is L = n lam / 2: the length between the two ends must hold a whole number of half wavelengths. Rearrange it and the wavelength of mode n is lam = 2L / n, which fixes the frequency of that mode at f = n v / (2L).

L = n lam / 2
lam_n = 2L / n — f_n = n v / (2L) — f_n = n f_1 — f_1 = v / (2L)
Symbol What it is SI unit A 0.800 m string at 400 m/s, mode 1
LLength of the vibrating string or air column, measured between the two endsmetre (m)0.800 m
vWave speed in the medium — an input here, and a property of the medium rather than of the modemetre per second (m/s)400 m/s
nMode number, also called the harmonic number. A whole number, never a fractiondimensionless1
lam_nWavelength of mode n, fixed by the length and the mode number alonemetre (m)1.60000 m
f_nFrequency of mode n — how many times a second the whole pattern completes a cyclehertz (Hz)250.000 Hz
f_1Fundamental frequency: mode 1 of the same length and speed, whichever mode you happen to be looking athertz (Hz)250.000 Hz
node spacingHalf the wavelength of the mode, whatever the two ends are; where two or more nodes exist, that is the distance between adjacent onesmetre (m)0.800000 m

Put real numbers in and the condition does the rest. A 0.800 m string along which waves run at 400 m/s has a fundamental wavelength of 2 x 0.800 m = 1.600 m, or 1.60000 m at the six significant figures the simulator below prints its readings to, and a fundamental frequency of 400 / 1.600 = 250.000 Hz. Type the same two boxes into the standing waves calculator and it will run the relation backwards too, returning the length a wanted pitch needs or the wave speed a measured mode implies.

One slip accounts for most wrong answers here, and it is the 2. Half a wavelength fits a fundamental, not a whole one, so the fundamental wavelength is twice the length and never equal to it. Get that factor wrong and every frequency on the page is out by two.

A diagram of the first three standing wave modes on one 0.800 m string along which waves run at 400 m/s, fixed at both ends. The same length is drawn three times to one shared horizontal scale, with a hatched wall at each end and a dashed rest line through the middle of each row. The top row carries one half wave, with node dots at both walls, one antinode marked at the centre, and labels reading n = 1, one half wave, lam_1 = 1.60000 m and f_1 = 250.000 Hz. The middle row carries two half waves, with three node dots and two antinodes marked, labelled n = 2, two half waves, lam_2 = 0.800000 m and f_2 = 500.000 Hz. The bottom row carries three half waves, with four node dots and three antinodes marked, labelled n = 3, three half waves, lam_3 = 0.533333 m and f_3 = 750.000 Hz. A dimension line beneath all three reads L = 0.800 m, the same in all three rows, and a legend identifies the dots as nodes that never move and the ticks as antinodes that swing furthest, with a footnote reading L = n lam / 2, so lam_n = 2L / n and f_n = n f_1.
Figure 1: the same 0.800 m string in its first three modes, drawn to one shared horizontal scale. The length never changes; the number of half waves squeezed into it does, and the frequency climbs 250.000, 500.000, 750.000 Hz with it.

Here is the same idea as something you can push around. Set the length, climb the mode number, change the medium, then switch what the two ends are and watch which numbers move and which refuse to.

Standing Waves Lab

Nodes and Antinodes: How Many and Where

A node is a point of the medium that never moves; an antinode sits midway between two nodes, where the movement is greatest. On a string fixed at both ends in mode n there are n + 1 nodes and n antinodes, and neighbouring nodes always sit half a wavelength apart.

The counts depend on what the two ends are, and on nothing else. Fixed at both ends puts a node at each end. Free at both ends puts an antinode at each end instead, so the counts swap over: n nodes and n + 1 antinodes. One end fixed and the other free gives n of each.

Positions are easier than they look. Take a 1.200 m string along which waves run at 240 m/s, in its fourth mode. The wavelength is 2 x 1.200 m / 4 = 2.400 m / 4 = 0.600000 m, so the node spacing is half of that: 0.300000 m.

Measure from one end and the five nodes fall at 0, 0.300, 0.600, 0.900 and 1.200 m. The four antinodes sit exactly between consecutive pairs, at 0.150, 0.450, 0.750 and 1.050 m. Count the nodes including both ends — in practice, forgetting the two ends is the commonest way to come out one short.

That half-wavelength spacing earns its keep in the laboratory. Because it holds whatever the ends are doing, the distance between two still points on a real vibrating object hands you the wavelength on the spot, and the wavelength then gives you the frequency as soon as the wave speed is known.

Harmonics: the Whole-Number Ladder

The modes of a string fixed at both ends are whole-number multiples of its fundamental: f_n = n f_1. A 0.800 m string at 400 m/s has a fundamental of 250.000 Hz, so its first four modes are 250.000, 500.000, 750.000 and 1000.00 Hz.

Mode number n (= half waves) Name Wavelength lam_n Frequency f_n Nodes and antinodes
1Fundamental, or first harmonic1.60000 m250.000 Hz2 nodes, 1 antinode
2Second harmonic, or first overtone0.800000 m500.000 Hz3 nodes, 2 antinodes
3Third harmonic, or second overtone0.533333 m750.000 Hz4 nodes, 3 antinodes
4Fourth harmonic, or third overtone0.400000 m1000.00 Hz5 nodes, 4 antinodes

The whole table is written in hertz on purpose, rather than switching the last row to kilohertz, because the point of it is the 1 : 2 : 3 : 4 running down the frequency column. Two things in it are worth a second look.

The wavelength column falls as 1, then a half, then a third, then a quarter of 1.60000 m, while the frequency column climbs in equal steps of 250.000 Hz. That reciprocal pairing is exactly what f = v / lam demands of a fixed wave speed. The node count, meanwhile, just shadows the mode number, always one more than it.

The ladder itself does not shift when you climb it. Take a 1.200 m string along which waves run at 240 m/s: its fundamental is 240 / (2 x 1.200) = 240 / 2.400 = 100.000 Hz, and that figure is the same whether you are looking at its fourth mode (4 x 100.000 = 400.000 Hz) or its fifth (5 x 100.000 = 500.000 Hz).

Where does the wave speed itself come from? It is a property of the medium rather than of the mode, and this page takes it as given — what sets the wave speed in different media is a subject of its own, and for a string it involves quantities that appear nowhere here. The relation tying a speed to a frequency and a wavelength belongs with the travelling waves that make a standing one possible.

Change what the two ends are, though, and the ladder moves bodily. Free both ends instead of fixing them and not one number changes — same wavelength, same frequency, same fundamental — while the picture turns inside out and the counts swap over. Fix one end and free the other and something stranger happens, which is the next section.

A diagram comparing a 0.500 m air column carrying sound at 343 m/s open at both ends with the same column stopped at one end, both drawn to one shared horizontal scale of 200 pixels to 1.00000 m, where a closed end is a displacement node and an open end a displacement antinode. The top row shows the open column with an open mouth at each end, an antinode tick at each end, one node dot at the centre, and the remaining half of the wave continued beyond the column as a faint dashed curve; brackets beneath read the column, 0.500 m, half a wavelength, and one whole wavelength, lam = 1.00000 m, with labels to the right reading open at both ends, lam = 1.00000 m twice the column, f_1 = 343.000 Hz, and 1 node, 2 antinodes. The bottom row shows the same column with a hatched wall at the left end, a node dot there, an antinode tick at the open right end, and the remaining three quarters of the wave continued as a faint dashed curve; brackets beneath read the column, 0.500 m, a QUARTER of a wavelength, and one whole wavelength, lam = 2.00000 m, with labels to the right reading closed at one end, lam = 2.00000 m four times the column, f_1 = 171.500 Hz, and 1 node, 1 antinode. The footnote reads 171.500 Hz x 2 = 343.000 Hz exactly, stopping one end drops the pitch a whole octave, and the closed column sounds 171.500, 514.500 and 857.500 Hz, the odd multiples only, so 343.000 Hz is not one of its modes.
Figure 2: the same 0.500 m air column at 343 m/s, open at both ends and then stopped at one end, drawn to one shared horizontal scale. Half a wavelength fits the open column and only a quarter fits the stopped one, so the stopped column sounds 171.500 Hz against 343.000 Hz.

Why a Pipe Closed at One End Sounds an Octave Lower

A pipe closed at one end sounds an octave below the same pipe open at both ends, because a closed end must be a node and an open end an antinode, so a quarter of a wavelength fits where half of one did before. A doubled wavelength halves the frequency.

Run the numbers on a 0.500 m air column carrying sound at 343 m/s. Open at both ends, the fundamental wavelength is 2 x 0.500 m = 1.000 m, giving 343 / 1.000 = 343.000 Hz. Stop one end and it becomes 4 x 0.500 m = 2.000 m, giving 343 / 2.000 = 171.500 Hz — exactly half the pitch, from the same pipe.

That 343 m/s is the speed of sound in air at 20 C, the figure this site publishes for that medium.

The second surprise is what the stopped pipe leaves out. Its modes are the odd multiples only — 171.500, 514.500 and 857.500 Hz for that column — so the 343.000 Hz the open pipe gives as its fundamental is not a mode of the stopped one at all. The even-numbered rungs are not faint on a stopped pipe. The boundary never lets them in.

What the two ends are, and what it covers Mode condition Wavelength of mode 1 Fundamental Nodes and antinodes, mode 1 First four modes
Fixed at both ends — A string clamped at both endsL = n lam / 21.60000 m250.000 Hz2 nodes, 1 antinode250.000, 500.000, 750.000, 1000.00 Hz
Free at both ends — A pipe open at both ends, or a bar free at both ends vibrating lengthwise rather than in bendingL = n lam / 21.60000 m250.000 Hz1 node, 2 antinodes250.000, 500.000, 750.000, 1000.00 Hz
One end fixed, one free — A pipe closed at one endL = (2n-1) lam / 43.20000 m125.000 Hz1 node, 1 antinode125.000, 375.000, 625.000, 875.000 Hz

Those three rows are one 0.800 m length at 400 m/s, with only the ends changed. Read the first two side by side: fixing both ends and freeing both ends give exactly the same wavelength, the same frequency and the same fundamental, and only the node and antinode counts swap, because one pattern is the mirror image of the other.

The third row is the odd one out in two ways. Its fundamental is exactly half the other two — 125.000 Hz against 250.000 Hz — and its ladder climbs in odd multiples, 125.000, 375.000, 625.000, 875.000 Hz, with every even rung missing. That is the same octave drop and the same missing harmonics as the 0.500 m column above, stated for a different length.

A rank of organ pipes of different lengths, the standing waves inside each one setting its pitch
Pipes of different lengths in one rank. Each length admits its own set of wavelengths, and that is all a pitch is.

Why the Shape Does Not Care How Fast the Wave Travels

Changing the medium changes every frequency and leaves the shape untouched, because lam = 2L / n has no v in it at all. The length and the mode number set the picture; the medium only sets how fast it runs.

A one metre column, free at both ends, makes the point cleanly. Fill it with water, where sound runs at 1482 m/s, and the fundamental wavelength is 2 x 1.000 m = 2.000 m with a pitch of 1482 / 2.000 = 741.000 Hz. Take the same one metre length in steel, where sound runs at 5960 m/s, and the wavelength is still 2.000 m while the pitch climbs to 5960 / 2.000 = 2980 Hz, which is 2.98000 kHz.

Same wavelength, same node positions, the same 1.00000 m node spacing, the same single node with an antinode at each end. The only thing the medium touched was the pitch.

Sound travels far further through water than radar or light does, which is the whole premise of sonar and the reason the sea floor is surveyed with sound rather than light. That is a fact about how far each kind of wave penetrates, and lam = 2L / n has no bearing on it either way. What the 1482 m/s above does account for is why a water-filled column sounds so much higher than an air-filled one of the same length.

Where Standing Waves Break Down

The mode condition is exact arithmetic, so what fails is never the sum. What fails is a boundary that is not really there, a medium that is not uniform, or a geometry being asked a question it cannot answer.

  • It needs a reflection. A wave running along an endless medium never settles into a standing pattern; it is the return from a boundary that does. One boundary is enough for the pattern but not for the ladder: with only one end there is no L, so L = n lam / 2 has no length to fix and no shortlist of frequencies follows from it.
  • It is a geometry, not an excitation. L = n lam / 2 says which modes fit. It says nothing at all about how a mode is set going, how strongly it answers, or how long it keeps sounding after you stop — those are separate questions with separate physics, and no figure on this page addresses any of them.
  • The wave speed must be the same all along the length. A string of varying thickness, or an air column whose temperature changes from one end to the other, has no single wavelength and so no clean ladder. Use a length over which the medium really is uniform.
  • The ideal ladder assumes a perfectly flexible medium. A real string resists being bent as well as being stretched, which pushes its short-wavelength modes slightly above exact whole-number multiples. That follows from what the model leaves out rather than from a measurement, and no figure for the size of it on any particular instrument is published on this site, so none is given here — verify before use.
  • A real open end is not exactly an antinode. The air just outside an open mouth is dragged into the motion, which puts the true displacement antinode a short way past the opening. A real pipe therefore behaves as though it were a little longer than the L you measured, and sounds slightly flat against v / (2L), or against v / (4L) when one end is stopped. Same rule: the mechanism is sound, the size of the correction is not published here, so no number for it appears — verify before use.
  • The formula is arithmetic, so it reaches states nothing can hear. Push the length down and the speed and mode number up, and a mode walks straight past the top of human hearing. Pull the other way and it drops below the bottom: a five-metre column stopped at one end, with the wave speed wound down to 10 m/s — the floor of the simulator’s speed slider rather than any real medium — sits at 0.500000 Hz.

Those last two limits have numbers attached, and this site publishes them: roughly 20 Hz to 20,000 Hz for a healthy young ear. Below 20 Hz a mode is infrasound. Above 20,000 Hz it is ultrasound, the band medical imaging probes work in — above the threshold of human hearing, and in practice far above it.

Common Misconceptions About Standing Waves

“A standing wave travels”

It does not, and the name is the clue. The pattern sits still and carries no net energy from one end to the other, which is precisely what separates it from a travelling wave, where a crest really does move along.

“The nodes drift along the medium”

Nodes are fixed points. On the 1.200 m string in its fourth mode they sit at 0, 0.300, 0.600, 0.900 and 1.200 m and stay there for as long as that mode is sounding. Watch a node rather than a crest and the stillness is obvious.

“The second harmonic has twice the wavelength of the first”

It has half. Climbing the ladder means squeezing more half waves into the same length, so lam_2 = L while lam_1 = 2L. On the 0.800 m string those are 0.800000 m against 1.60000 m, and the frequency doubles rather than halves.

“Any frequency will set up a standing wave”

Only the frequencies that fit will. The 0.800 m string at 400 m/s has modes at 250.000, 500.000, 750.000 and 1000.00 Hz, and nothing between 250.000 Hz and 500.000 Hz is a mode of it at all. There is no partial mode and no n = 2.7.

“A closed pipe sounds higher because less of it is open”

It sounds lower, by an octave. Stopping one end turns that end into a node facing the antinode still at the mouth, and a node-to-antinode run is a quarter of a wavelength rather than a half. The same length therefore holds a quarter wave instead of a half, the wavelength doubles and the pitch halves: 171.500 Hz where the open pipe gave 343.000 Hz.

Worked Problems

Problem 1
A string 0.800 m long is clamped at both ends, and waves run along it at 400 m/s. Find the wavelength and the frequency of its fundamental.
Show Solution
Solution: Step 1: the fundamental is mode 1, so the condition L = n lam / 2 becomes L = lam / 2. One half wave fills the length, and the wavelength is twice it. Step 2: lam_1 = 2L = 2 x 0.800 m = 1.600 m, which is 1.60000 m to six figures. Step 3: the frequency follows from f = v / lam, so f_1 = 400 m/s / 1.600 m = 250.000 Hz. Answer: lam_1 = 1.60000 m and f_1 = 250.000 Hz. That pair is the whole ladder’s first rung: every other mode of this string is a whole-number multiple of 250.000 Hz.
Problem 2
The same 0.800 m string at 400 m/s is now sounding its third harmonic. Find the wavelength and the frequency of that mode, and say how many nodes and antinodes it has.
Show Solution
Solution: Step 1: for mode n the wavelength is lam_n = 2L / n, and the frequency is a whole-number multiple of the fundamental, f_n = n f_1. Step 2: lam_3 = 1.600 m / 3 = 0.533333 m to six figures. Keep the exact value for the next step. Step 3: f_3 = 3 f_1 = 3 x 250.000 Hz = 750.000 Hz. Do NOT divide 400 by the rounded 0.533333: that gives 750.000469 Hz, and because it still prints as 750.000 Hz at six figures the error hides itself completely. Work from the exact fundamental instead. Step 4: fixed at both ends, mode 3 has n + 1 = 4 nodes and n = 3 antinodes, with the nodes 0.266667 m apart. Answer: lam_3 = 0.533333 m, f_3 = 750.000 Hz, with 4 nodes and 3 antinodes. The fundamental has not moved: it is still 250.000 Hz.
Problem 3
A string has to sound 440 Hz in its fundamental, in a medium along which waves run at 320 m/s. How long must the vibrating part be?
Show Solution
Solution: Step 1: for the fundamental, f_1 = v / (2L). Rearranged for the length, L = v / (2 f_1). Step 2: substitute with units. 2 x 440 Hz = 880 Hz, so L = 320 m/s / 880 Hz. Step 3: L = 0.363636 m to six figures. Carry the unrounded value if you need to check it: feeding the printed 0.363636 m back through f = v / (2L) returns 440.000 Hz at six figures, but the exact length is what reproduces 440 Hz exactly. Answer: L = 0.363636 m. A longer string than this sounds flat rather than sharp, because the frequency falls as the length grows.
Problem 4
A 1.200 m string is clamped at both ends, with waves running along it at 240 m/s. In its fourth mode, where exactly are the nodes and antinodes, and what frequency does it sound?
Show Solution
Solution: Step 1: the wavelength first. lam_4 = 2L / n = 2 x 1.200 m / 4 = 2.400 m / 4 = 0.600000 m. Step 2: the node spacing is half a wavelength, 0.300000 m, which is also L / n = 1.200 m / 4. With both ends fixed there are n + 1 = 5 nodes, measured from one end at 0, 0.300, 0.600, 0.900 and 1.200 m. Step 3: the n = 4 antinodes sit midway between consecutive nodes, at 0.150, 0.450, 0.750 and 1.050 m. Step 4: the frequency. f_4 = n v / (2L) = 4 x 240 m/s / (2 x 1.200 m) = 960 / 2.400 = 400.000 Hz. The fundamental of the same string is 240 / 2.400 = 100.000 Hz, and 4 x 100.000 = 400.000 Hz agrees. Answer: 5 nodes at 0, 0.300, 0.600, 0.900 and 1.200 m, 4 antinodes at 0.150, 0.450, 0.750 and 1.050 m, and f_4 = 400.000 Hz. Both ends count as nodes, which is why there are five and not four.
Problem 5
A 0.500 m air column carries sound at 343 m/s. Find its fundamental open at both ends, then closed at one end, and list the first three modes of the closed column.
Show Solution
Solution: Step 1: open at both ends there is an antinode at each end and the condition is L = n lam / 2, so lam_1 = 2L = 2 x 0.500 m = 1.000 m. Then f_1 = 343 m/s / 1.000 m = 343.000 Hz. Step 2: closed at one end there is a node at the closed end and an antinode at the open one, so only a quarter wavelength fits: lam_1 = 4L = 4 x 0.500 m = 2.000 m. Then f_1 = 343 m/s / 2.000 m = 171.500 Hz. Step 3: compare them. 343.000 Hz divided by 171.500 Hz is exactly 2, so stopping one end drops the column by precisely one octave. Step 4: the closed column’s modes are the odd multiples of its own fundamental: 171.500 Hz, then 3 x 171.500 = 514.500 Hz, then 5 x 171.500 = 857.500 Hz. The open column’s 343.000 Hz is not among them. Answer: 343.000 Hz open, 171.500 Hz closed, exactly an octave lower, with the closed column’s first three modes at 171.500, 514.500 and 857.500 Hz. Its even harmonics are absent altogether rather than merely weak.
Problem 6
A 0.900 m string clamped at both ends is sounding its second harmonic at 300 Hz. How fast do waves run along it, and what is its fundamental?
Show Solution
Solution: Step 1: the second harmonic has n = 2, so lam_2 = 2L / n = 2 x 0.900 m / 2 = 1.800 m / 2 = 0.900000 m. In mode 2 the wavelength equals the length. Step 2: rearrange f = n v / (2L) for the speed: v = 2 L f / n = 2 x 0.900 m x 300 Hz / 2 = 540 / 2 = 270.000 m/s. Step 3: check it. The fundamental is f_1 = v / (2L) = 270 / 1.800 = 150.000 Hz, and 2 x 150.000 = 300.000 Hz, which is the measurement you started from. Answer: v = 270.000 m/s, with a fundamental of 150.000 Hz. Take that 300 Hz for a fundamental by mistake and the speed comes out twice as large, because the mode number divides the whole answer.
Problem 7
The simulator above opens on a 0.800 m length with waves running at 400 m/s, in mode 1, fixed at both ends. Without changing a control, state its wavelength, its frequency, its fundamental and its node and antinode counts, and say which of those the mode number would move.
Show Solution
Solution: Step 1: identify the state. Mode 1 of a length fixed at both ends IS the fundamental, so the mode’s own frequency and the fundamental must be the same number here and nowhere else on the ladder. Step 2: read the four quantities off. The wavelength is 1.60000 m, the frequency is 250.000 Hz, the fundamental is 250.000 Hz, and there are two nodes with one antinode between them. The nodes are 0.800000 m apart, which is half the wavelength and, in mode 1, the whole length. Step 3: now predict. Raising the mode number alone moves the wavelength, the frequency, the node spacing and the counts; it leaves the fundamental exactly where it is, because the fundamental depends on the length and the speed and not on which mode you are looking at. Answer: 1.60000 m, 250.000 Hz, 250.000 Hz, and two nodes with one antinode. Nothing in that had to be derived — it is the opening state read straight off, and the fundamental is the one figure the mode slider cannot touch.

Frequently Asked Questions

What is the formula for a standing wave on a string?
The condition is L = n lam / 2, where L is the length between the two fixed ends, n is a whole mode number and lam is the wavelength. Rearranged it gives lam = 2L / n and f = n v / (2L), with v the wave speed in the medium. A 0.800 m string at 400 m/s therefore sounds 250.000 Hz in its fundamental.
What is the difference between a node and an antinode?
A node is a point of the medium that never moves, and an antinode is a point halfway between two nodes, where the movement is greatest. They alternate along the length, half a wavelength apart node to node. A string fixed at both ends has a node at each end; a pipe open at both ends has an antinode at each end instead.
What is the fundamental frequency?
The fundamental frequency is the lowest frequency at which a length will stand a wave: mode 1, with f_1 = v / (2L) for a string fixed at both ends. Every other mode of that length is a whole-number multiple of it. It depends on the length and the wave speed only, so climbing to a higher mode leaves it exactly where it was.
Why do only certain frequencies form standing waves?
Because the two ends only accept a whole number of half wavelengths between them. Any other wavelength arrives back out of step with itself after reflecting and cancels rather than building, so it never establishes a pattern. That is the whole content of L = n lam / 2, and it is why a 0.800 m string at 400 m/s has nothing between 250.000 Hz and 500.000 Hz.
How many nodes does the third harmonic have?
Four, on a string fixed at both ends, with three antinodes between them. The rule is n + 1 nodes and n antinodes in mode n, counting both fixed ends as nodes. For a 0.800 m string at 400 m/s the third harmonic has a wavelength of 0.533333 m, a frequency of 750.000 Hz and nodes spaced 0.266667 m apart.
Why does a pipe closed at one end sound an octave lower?
Because a closed end has to be a displacement node and an open end an antinode, so only a quarter of a wavelength fits the pipe instead of a half. The wavelength doubles, which halves the frequency. A 0.500 m column at 343 m/s gives 343.000 Hz open and 171.500 Hz stopped, and the stopped pipe keeps only the odd harmonics.
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