Node vs Antinode: What Is the Difference?

On a standing wave a node is a point whose displacement is zero at every instant, so it never moves. An antinode is a point that swings through the full amplitude. They alternate along the pattern a quarter of a wavelength apart, and the boundaries fix which one sits at each end.

AP Physics: Unit 14 (topics 14.6 Wave Interference and Standing Waves). Both definitions come from one sentence, AP Physics 2 essential knowledge 14.6.B.1.i: a node is a point on the standing wave where the amplitude is always zero, and an antinode is a point where the amplitude is always at maximum. They sit under 14.6.B.1, that standing waves can result from interference between two waves confined to a region and traveling in opposite directions, with 14.6.B.1.ii making the allowed wavelengths depend on the size and boundary conditions of the region and 14.6.B.1.iii naming the common regions as pipes with open or closed ends and strings with fixed or loose ends. The CED does not tabulate which end forces a node and which forces an antinode, so that mapping is standard physics the framework assumes; what it does state, at 14.6.B.2, is that a standing wave with a node at one end and an antinode at the other supports only odd harmonics. 14.6.B.3 directs students to visual representations rather than a formula, and no equation for node positions or harmonic frequencies is printed in the CED or on the equation sheet. A sample instructional activity for Topic 14.6 has students find the maximum speed attained at an antinode from amplitude and period data, calculate the wave speed from wavelength and period data, and see that the two are different. Topic 14.6 prints no boundary statement. Unit 14 carries 12 to 15 percent of the multiple-choice section across about 14 to 23 class periods, and standing waves appear in no other AP Physics course: the words standing wave and antinode occur nowhere in the AP Physics 1, C: Mechanics or C: E&M frameworks.

The distinction, stated once

AP Physics 2 puts both definitions in one sentence, essential knowledge 14.6.B.1.i: standing waves have nodes and antinodes. A node is a point on the standing wave where the amplitude is always zero. An antinode is a point on the standing wave where the amplitude is always at maximum.

Read the word amplitude carefully, because the sentence is symmetric and its two halves are not.

For a node, zero amplitude means zero displacement at every instant. Amplitude is the largest displacement a point ever reaches, so a point whose amplitude is zero never reaches any displacement at all. A node is motionless. Not motionless on average, not passing through zero twice a cycle like every other point on any wave: permanently at rest.

For an antinode, maximum amplitude does not mean maximum displacement. The antinode's amplitude is the largest in the pattern, and its displacement swings the whole way from $+A$ to $-A$ and back every cycle. So an antinode is at zero displacement twice per cycle, exactly like a point on a traveling wave, and at those instants it is moving faster than anything else on the string.

That asymmetry is the whole comparison. A node is defined by something it never does. An antinode is defined by something it does more than any other point. One is a statement about all time, the other is a statement about the extremes.

The consequence students miss follows immediately: there is an instant, twice every cycle, when the entire standing wave is flat and every point including every antinode has zero displacement. At that instant a photograph cannot distinguish a node from an antinode. Only the motion can, and worked example two makes the difference numerical.

Everything else on this page is bookkeeping around those two definitions: where the boundaries force each one to sit, how far apart they are, and what a question is really asking when it says "amplitude".

Side by side

Property	Node	Antinode
CED definition	Amplitude is always zero (14.6.B.1.i)	Amplitude is always at maximum (14.6.B.1.i)
Displacement at any instant	Zero, always	Anywhere between $+A$ and $-A$
Displacement when the pattern is flat	Zero	Zero, momentarily
Displacement at maximum extension	Zero	$\pm A$ , the largest on the wave
Speed of the medium there	Zero, always	Zero at the extremes, largest of any point as it passes through zero
What the two component waves are doing there	Permanently opposed	Permanently reinforcing
Spacing to the next one of the same kind	$\lambda/2$	$\lambda/2$
Spacing to the nearest one of the other kind	$\lambda/4$	$\lambda/4$
At a fixed string end or a closed pipe end	This one	Not this one
At a loose string end or an open pipe end	Not this one	This one
Number on a string fixed at both ends in harmonic $n$	$n + 1$	$n$
What it looks like in a long-exposure photograph	A pinch point	A blur the full width of the pattern

Two rows are worth extra attention.

"Speed of the medium there" is the row that resolves most confusion, because it is the only property where the two are opposite in a way a photograph cannot show. A node is at rest at every instant. An antinode is at rest only at the two turning points, and at the moment the pattern is flat it is the fastest-moving point on the wave. A CED sample instructional activity for Topic 14.6 is built on exactly this: it asks students to find the maximum speed attained at an antinode from amplitude and period data, calculate the wave speed separately from wavelength and period data, and see that the two are different.

The count row is a bookkeeping fact you can read off a sketch rather than remember. A string fixed at both ends in its third harmonic has three antinodes and four nodes, because the two forced ends are both nodes and the pattern alternates. Worked example one draws it out.

The case that separates them: one cycle, two points

Pick one node and one antinode on the same standing wave and follow both of them through a single period. Use the sheet's printed $x(t) = A\cos(\omega t) = A\cos(2\pi f t)$ for the antinode, starting the clock at maximum displacement.

Time	Node displacement	Antinode displacement	What the pattern looks like
$0$	$0$	$+A$	Fully extended one way
$T/8$	$0$	$+0.71A$	Partly extended
$T/4$	$0$	$0$	Completely flat
$3T/8$	$0$	$-0.71A$	Partly extended the other way
$T/2$	$0$	$-A$	Fully extended the other way
$3T/4$	$0$	$0$	Completely flat again
$T$	$0$	$+A$	Back to the start

The node column has one value in it. That is what "the amplitude is always zero" means, and it is why the pair is not a symmetric contrast: one of these points has no story.

The antinode column contains zero twice, at $T/4$ and $3T/4$ . Those are the two instants when the whole standing wave is flat, and they are the reason a single snapshot is not enough to identify anything. At $T/4$ the node and the antinode have exactly the same displacement. They are still completely different points, because the node is stationary and the antinode is moving at its top speed.

That top speed is the second half of the separation, and it is worth putting a number on. Worked example two takes an antinode swinging $4.0 \ \mathrm{mm}$ either side at $240 \ \mathrm{Hz}$ and finds it moving at a few metres per second at the flat instant, while the wave on the same string travels at $240 \ \mathrm{m/s}$ . Two speeds, differing by a factor of about forty, in one experiment. The medium is barely moving; the pattern's defining speed belongs to the traveling waves that built it.

Energy is the third way to see it. At $t = T/4$ nothing is displaced, so no part of the string is stretched away from its rest shape and the wave's energy is entirely kinetic. All of that kinetic energy is in the fast-moving parts of the string, which are the regions around the antinodes. At $t = 0$ nothing is moving and the energy is entirely in the deformation, largest where the string has been pulled furthest, which is again around the antinodes. The node contributes nothing at either instant.

Which one the boundary forces, and where the CED stops

Essential knowledge 14.6.B.1.ii says the possible wavelengths of a standing wave are determined by the size and boundary conditions of the region to which it is confined, and 14.6.B.1.iii names the common regions: pipes with open or closed ends, as well as strings with fixed or loose ends.

Four kinds of end, and the CED does not tabulate which gets which. 14.6.B.1.iii lists the regions and stops. So the mapping below is standard physics that the framework assumes rather than prints, and it is worth knowing that it is an assumption you are importing.

End of the region	What sits there	Why
String, fixed end	Node	The end is clamped, so it cannot move
String, loose end	Antinode	The end is free, so it moves most
Pipe, closed end	Node	Air cannot move through the seal
Pipe, open end	Antinode	Air there is freest to move

The two string rows are self-evident: a clamped point has zero displacement by construction, which is exactly the definition of a node. The two pipe rows need one clarification, because sound is a longitudinal wave and the word displacement means something different there.

In a pipe, AP Physics 2 works in displacement, so draw displacement. The antinode at an open end is a displacement antinode: the air there swings back and forth furthest. In terms of pressure it is the opposite, because an open end is held at atmospheric pressure and cannot vary much, so a displacement antinode is a pressure node. Both descriptions are correct and they are mirror images. Reading a diagram as pressure when it was drawn as displacement puts every node and antinode in the wrong place.

The one place the CED does commit is the consequence rather than the mapping. 14.6.B.2 says that for a standing wave with a node at one end and an antinode at the other end, only odd harmonics can be established. That sentence describes a region by its ends rather than by naming a pipe, which is the framework being careful: the rule is about the boundary conditions, and it applies to a pipe closed at one end and to a string with one fixed and one loose end in exactly the same way. Fundamental against harmonic works that ladder in full.

Practical order of operations. Sketch the two ends first, using the table. Draw the simplest curve that satisfies both. Read the wavelength off the sketch by counting how many quarter or half wavelengths fit the length. Only then reach for $\lambda = v/f$ . 14.6.B.3 endorses exactly that route, saying visual representations of standing waves are useful in determining the relationships between length of the region, wavelength, frequency, wave speed, and harmonic.

The quarter-wavelength geometry, and how to count

Nodes and antinodes alternate, and the spacings follow from that alone.

Node to next node: $\lambda/2$ . Exactly one antinode sits between them.
Antinode to next antinode: $\lambda/2$ . Exactly one node sits between them.
Node to nearest antinode: $\lambda/4$ . Half of the previous spacing, because the antinode sits midway.

Those three facts turn a measurement into a wavelength without any formula. Measure the distance between two pinch points on a vibrating string, double it, and you have $\lambda$ . Worked example three does this and then converts with the printed $\lambda = v/f$ .

Counting on a string fixed at both ends. Both ends are forced to be nodes. The $n$ th harmonic has $n$ antinodes and therefore $n + 1$ nodes, and its wavelength is $\lambda_n = 2L/n$ , because $n$ half wavelengths have to fit the length exactly.

Harmonic	Antinodes	Nodes	$\lambda$
1	1	2	$2L$
2	2	3	$L$
3	3	4	$2L/3$
4	4	5	$L/2$

Counting in a pipe closed at one end works the same way but from a different starting shape. One end is a node and the other an antinode, so the shortest pattern that fits is a single quarter wavelength and $\lambda_1 = 4L$ . Adding half a wavelength at a time keeps both ends correct, which is why the allowed patterns are $\lambda_n = 4L/n$ for odd $n$ only, and it is 14.6.B.2's odd-harmonic rule arriving from geometry.

A warning about counting from a photograph. A long-exposure photograph of a vibrating string shows the antinodes as wide blurs and the nodes as pinch points, which makes counting easy. A short-exposure photograph catches one instant, and if that instant happens to be one of the two flat moments per cycle, the string looks straight and there is nothing to count at all. If it catches an intermediate instant, the picture is a curve whose zero crossings are the nodes, so counting still works, but the amplitudes shown are smaller than the true amplitudes.

When it costs a mark

Saying an antinode is always at maximum displacement. 14.6.B.1.i says its amplitude is always at maximum, which is a statement about the size of its swing, not about where it is right now. An antinode passes through zero displacement twice per cycle.
Saying a node is at zero displacement "at that moment". A node is at zero displacement at every moment, and that permanence is the entire definition. Any point on any wave is momentarily at zero twice a cycle; only a node stays there.
Treating a node as a place where nothing is happening. The string is present, it is under strain, and it is the region of steepest slope on the wave when the pattern is fully extended. What is true is that the point does not move.
Reading a pipe diagram as pressure when it was drawn as displacement. An open end is a displacement antinode and a pressure node at the same time. AP Physics 2 works in displacement, so draw displacement and say which you have drawn.
Assuming a fixed end could be an antinode. A clamped point cannot move, so its amplitude is zero, so it is a node. This is not a convention, it is the definition applied to a constraint.
Using $\lambda/2$ for the node-to-antinode distance. That is the node-to-node distance. A node and its neighbouring antinode are $\lambda/4$ apart.
Miscounting nodes on a string. The forced ends count. A string fixed at both ends in its second harmonic has three nodes, not one, because the two ends are nodes as well as the point in the middle.
Confusing the medium's speed with the wave's speed. The maximum speed of the string at an antinode and the speed at which a wave travels along that string are different quantities with different values, and a CED sample activity for Topic 14.6 exists specifically to make students calculate both and compare them.
Expecting a printed equation. Nothing in the AP Physics 2 CED or on its equation sheet gives node positions, antinode positions or harmonic frequencies directly. 14.6.B.3 tells you the intended method: draw the pattern, read the wavelength, convert with $\lambda = v/f$ .

Where the two coincide, and why that lulls you

They coincide once per half cycle, exactly, and that is the trap.

The flat instant. Twice every period the whole standing wave passes through its rest shape and every point on it, node and antinode alike, has zero displacement. A question that shows one snapshot and asks which points are nodes has no answer if the snapshot was taken then. A question that shows one snapshot at any other instant does have an answer, and it is the zero crossings.

This is why the safe way to identify a node is by what it does over time rather than by a reading. A node is a point that is at zero in every snapshot. An antinode is a point that reaches the largest value in some snapshot. Both statements are quantified over time; neither can be checked from one picture.

The second lull is the word amplitude. It is used in two ways in wave problems and the CED's definition uses one of them. In 14.6.B.1.i, "the amplitude is always zero" and "the amplitude is always at maximum" describe a property of a location on the standing wave: how big a swing that location makes. In a sentence like "the amplitude of the wave is $4.0 \ \mathrm{mm}$ ", it means the biggest swing anywhere in the pattern, which is the antinode's. If a question gives you an amplitude without saying where, it means the antinode's.

The third lull is that both are consequences of the same thing. Nodes and antinodes are not two mechanisms. 14.6.B.1 says standing waves can result from interference between two waves that are confined to a region and traveling in opposite directions, and both features come out of that single interference: the node is where the two are permanently opposed, the antinode where they permanently reinforce. Constructive against destructive interference is the same distinction one topic earlier, before the region and the boundaries pin the pattern in place.

Where they genuinely never coincide is the speed. A node's speed is zero at every instant of every cycle. An antinode's speed is zero only at the two turning points, and at the flat instant it is the largest on the wave. There is no moment at which a node and an antinode are doing the same thing, and if a question seems to say otherwise, it is describing displacement and not motion.

Locating every node and antinode on a string

A string of length $L = 1.50 \ \mathrm{m}$ is fixed at both ends. Waves travel along it at $v = 240 \ \mathrm{m/s}$ . It is driven in its third harmonic. (a) Find the wavelength. (b) Find the frequency. (c) Give the position of every node and every antinode, measured from the left-hand end. (d) State the node-to-node and node-to-antinode spacings and check them against the wavelength.

(a) Both ends are fixed, so both ends are nodes. The third harmonic fits three half wavelengths into the length, so $3\left(\dfrac{\lambda}{2}\right) = L$ and $\lambda = \dfrac{2L}{3} = \dfrac{2(1.50)}{3} = 1.00 \ \mathrm{m}$ .
(b) The sheet prints $\lambda = v/f$ , so $f = \dfrac{v}{\lambda} = \dfrac{240 \ \mathrm{m/s}}{1.00 \ \mathrm{m}} = 240 \ \mathrm{Hz}$ . For reference the fundamental would have $\lambda_1 = 2L = 3.00 \ \mathrm{m}$ and $f_1 = 240/3.00 = 80 \ \mathrm{Hz}$ , and $240$ is three times $80$ as the numbering requires.
(c) Nodes sit every half wavelength starting at the fixed left end, so at $0$ , $0.50$ , $1.00$ and $1.50 \ \mathrm{m}$ . That is four nodes, and the last one is the fixed right-hand end, as it must be.
Antinodes sit midway between consecutive nodes, so at $0.25$ , $0.75$ and $1.25 \ \mathrm{m}$ . That is three antinodes, one per half wavelength, matching the harmonic number.
(d) Node to node is $0.50 \ \mathrm{m}$ , which is $\lambda/2 = 1.00/2$ . Node to nearest antinode is $0.25 \ \mathrm{m}$ , which is $\lambda/4 = 1.00/4$ . Both check.
Count check: four nodes and three antinodes for $n = 3$ , which is the general $n + 1$ and $n$ . The pattern alternates node, antinode, node, antinode, node, antinode, node from left to right, seven features in all.
One property that is easy to lose in the arithmetic: every one of those four nodes has zero displacement at every instant, including the two that are not clamped. The points at $0.50$ and $1.00 \ \mathrm{m}$ are free to move as far as the string is concerned, and they do not, because 14.6.B.1's two counter-propagating waves are permanently opposed there.

(a) $\lambda = 1.00 \ \mathrm{m}$ . (b) $f = 240 \ \mathrm{Hz}$ . (c) Nodes at $0$ , $0.50$ , $1.00$ and $1.50 \ \mathrm{m}$ ; antinodes at $0.25$ , $0.75$ and $1.25 \ \mathrm{m}$ . (d) Node to node $0.50 \ \mathrm{m} = \lambda/2$ , node to antinode $0.25 \ \mathrm{m} = \lambda/4$ .

How fast the string moves at an antinode, against how fast the wave moves

On the string above, driven at $f = 240 \ \mathrm{Hz}$ , an antinode swings $4.0 \ \mathrm{mm}$ either side of the rest line. (a) Find the period, and the displacement of the antinode at $t = 0$ , $T/8$ , $T/4$ and $T/2$ , taking $t = 0$ at maximum displacement. (b) Find the average speed of the antinode over the quarter cycle from full extension to the flat instant. (c) Compare that with the wave speed of $240 \ \mathrm{m/s}$ , and say what the node is doing throughout.

(a) The sheet prints $T = 1/f$ , so $T = \dfrac{1}{240} = 4.167 \times 10^{-3} \ \mathrm{s}$ , about $4.17 \ \mathrm{ms}$ .
The sheet also prints $x(t) = A\cos(\omega t) = A\cos(2\pi f t)$ . With $A = 4.0 \ \mathrm{mm}$ and the clock started at maximum:
$t = 0$ : $x = 4.0\cos(0) = +4.0 \ \mathrm{mm}$ . $t = T/8$ : $x = 4.0\cos(45^\circ) = +2.8 \ \mathrm{mm}$ . $t = T/4$ : $x = 4.0\cos(90^\circ) = 0$ . $t = T/2$ : $x = 4.0\cos(180^\circ) = -4.0 \ \mathrm{mm}$ .
(b) Over the quarter cycle the antinode travels from $+4.0 \ \mathrm{mm}$ to $0$ , a distance of $4.0 \times 10^{-3} \ \mathrm{m}$ , in a time $T/4 = 1.042 \times 10^{-3} \ \mathrm{s}$ .
Average speed $= \dfrac{4.0 \times 10^{-3}}{1.042 \times 10^{-3}} = 3.84 \ \mathrm{m/s}$ . Equivalently $4Af = 4(0.0040)(240)$ , which is a useful form because it needs only the amplitude and the frequency, the two quantities the CED's sample activity says to measure.
The peak speed, reached at the flat instant, is larger than the average because the motion is sinusoidal: it equals $2\pi f A = 2\pi(240)(0.0040) = 6.0 \ \mathrm{m/s}$ . That relation is not printed on the AP Physics 2 sheet and follows from the slope of the cosine, so quote the average if you need a value the sheet supports and the peak only if the question asks for a maximum.
(c) The wave speed is $240 \ \mathrm{m/s}$ , so the string's own motion is slower by a factor of about $62$ using the average and about $40$ using the peak. This is the point of the Topic 14.6 sample activity, which has students find the maximum speed attained at an antinode from amplitude and period data, calculate the wave speed from wavelength and period data, and see that they are different.
Throughout all of this the node does nothing. Its displacement is zero at $t = 0$ , $T/8$ , $T/4$ , $T/2$ and every other instant, so its speed is zero too, average and peak alike. That is the difference the table in the section above records in one column of identical entries.
Worth noticing why the two speeds are unrelated in size. The wave speed on a string is set by $v_{\text{string}} = \sqrt{F_T/(m/\ell)}$ , printed on the sheet, so it depends on tension and linear mass density and not at all on how hard the string is being driven. The antinode's speed is proportional to the amplitude, which the driver sets. Turn the driver up and the antinode moves faster while the wave speed does not change.

(a) $T = 4.17 \ \mathrm{ms}$ ; displacements $+4.0$ , $+2.8$ , $0$ and $-4.0 \ \mathrm{mm}$ . (b) Average speed over the quarter cycle is $3.84 \ \mathrm{m/s}$ , with a peak of $6.0 \ \mathrm{m/s}$ at the flat instant. (c) Both are tiny beside the $240 \ \mathrm{m/s}$ wave speed, and the node's speed is zero at every instant.

Working backwards from the spacing of the pinch points

A long-exposure photograph of a vibrating string shows a row of pinch points $0.42 \ \mathrm{m}$ apart. The string is being driven at $500 \ \mathrm{Hz}$ and is fixed at both ends, with a total length of $2.10 \ \mathrm{m}$ . (a) Find the wavelength and the wave speed. (b) Find the distance from a pinch point to the nearest blur. (c) Find the harmonic number and count the nodes and antinodes. (d) Find the fundamental frequency.

(a) The pinch points are the nodes, because a node has zero amplitude at all times and so leaves a sharp mark on a long exposure while every other point smears. Adjacent nodes are half a wavelength apart, so $\lambda = 2(0.42) = 0.84 \ \mathrm{m}$ .
The sheet prints $\lambda = v/f$ , so $v = \lambda f = (0.84)(500) = 420 \ \mathrm{m/s}$ .
(b) The blurs are the antinodes, and a node and its nearest antinode are a quarter wavelength apart: $\dfrac{0.84}{4} = 0.21 \ \mathrm{m}$ . It is also half of the $0.42 \ \mathrm{m}$ node spacing, which is the quicker route from the measurement.
(c) The harmonic number is how many half wavelengths fit the string: $n = \dfrac{L}{\lambda/2} = \dfrac{2.10}{0.42} = 5$ , so this is the fifth harmonic.
For $n = 5$ there are $5$ antinodes and $6$ nodes, and the two end nodes are the clamped ends. Check against the length: $6$ nodes spaced $0.42 \ \mathrm{m}$ apart span $5(0.42) = 2.10 \ \mathrm{m}$ , which is the whole string. The count is consistent.
(d) The fundamental has $\lambda_1 = 2L = 4.20 \ \mathrm{m}$ , so $f_1 = \dfrac{v}{\lambda_1} = \dfrac{420}{4.20} = 100 \ \mathrm{Hz}$ .
Cross-check with the harmonic number: the fifth harmonic of a string fixed at both ends is $5f_1 = 5(100) = 500 \ \mathrm{Hz}$ , which is the driving frequency given. The two routes agree.
Notice that nothing here needed a formula for node positions, because there is none in the CED or on the sheet. Everything came from 14.6.B.3's route: read the geometry off the picture, then convert with $\lambda = v/f$ .

(a) $\lambda = 0.84 \ \mathrm{m}$ and $v = 420 \ \mathrm{m/s}$ . (b) $0.21 \ \mathrm{m}$ . (c) The fifth harmonic, with $6$ nodes and $5$ antinodes. (d) $f_1 = 100 \ \mathrm{Hz}$ , and $5f_1 = 500 \ \mathrm{Hz}$ as required.

Frequently asked questions

What is the difference between a node and an antinode?

A node never moves and an antinode moves the most. AP Physics 2 essential knowledge 14.6.B.1.i states both in one sentence: a node is a point on the standing wave where the amplitude is always zero, and an antinode is a point on the standing wave where the amplitude is always at maximum. Because amplitude is the largest displacement a point ever reaches, a node with zero amplitude has zero displacement at every instant, so it is permanently at rest. An antinode has the largest amplitude in the pattern and swings all the way from plus the amplitude to minus it every cycle, which means it passes through zero displacement twice per cycle. Nodes and antinodes alternate along the pattern a quarter of a wavelength apart.

Is an antinode always at maximum displacement?

No, and this is the most common misreading of the definition. AP Physics 2 essential knowledge 14.6.B.1.i says the amplitude at an antinode is always at maximum, which describes the size of that point's swing, not where the point is at any given moment. An antinode oscillates continuously between plus and minus the amplitude, so it reaches maximum displacement only twice per cycle and is at zero displacement twice per cycle. At those zero moments the whole standing wave is momentarily flat, and the antinode is then moving faster than any other point on the wave. A node, by contrast, is at zero displacement at every instant without exception.

How far apart are nodes and antinodes?

Adjacent nodes are half a wavelength apart, adjacent antinodes are half a wavelength apart, and a node and its nearest antinode are a quarter of a wavelength apart. The pattern alternates, so exactly one antinode sits between any two neighbouring nodes and it sits midway between them. This makes the spacings a fast measurement tool: measure the distance between two pinch points on a vibrating string, double it, and you have the wavelength, then convert with the printed lambda equals v over f. The AP Physics 2 CED prints no equation for node positions, and essential knowledge 14.6.B.3 says instead that visual representations of standing waves are useful in determining the relationships between length of the region, wavelength, frequency, wave speed, and harmonic.

Is a fixed end of a string a node or an antinode?

A node. A clamped point cannot move, so its amplitude is zero, which is the definition of a node in essential knowledge 14.6.B.1.i. The same reasoning gives the other three cases: a loose string end is free to move and is an antinode, the closed end of a pipe is a node because the air cannot move through the seal, and the open end of a pipe is a displacement antinode because the air there is freest to move. Essential knowledge 14.6.B.1.iii names those four kinds of region, pipes with open or closed ends and strings with fixed or loose ends, but the CED does not print the table of which end gets which, so this mapping is standard physics the framework assumes.

Why does a standing wave look flat at some instants?

Because every point on it, antinodes included, passes through zero displacement twice per cycle. The nodes are at zero permanently, and at two moments in each period the antinodes and everything between them happen to be at zero as well, so the medium is momentarily in its rest shape. Nothing has stopped, though. At that instant the wave's energy is entirely kinetic and the fastest-moving parts of the medium are the regions around the antinodes, while the nodes are still at rest as always. This is why identifying nodes from a single photograph can fail: if the shutter catches a flat instant there is nothing to see, and you have to watch the motion over time rather than read one picture.

How many nodes and antinodes does the nth harmonic have?

On a string fixed at both ends, the nth harmonic has n antinodes and n plus 1 nodes, because both ends are forced to be nodes and the features alternate. The third harmonic, for example, has three antinodes and four nodes, two of which are the clamped ends. The pattern fits n half wavelengths into the length, so the wavelength is 2L over n. A pipe closed at one end starts from a different shape, with a node at the closed end and an antinode at the open one, so its shortest pattern is a single quarter wavelength and its fundamental wavelength is 4L. Counting from a sketch is more reliable than memorising the counts, which is the method essential knowledge 14.6.B.3 recommends.

Is the speed at an antinode the same as the wave speed?

No, and the two are usually very different. The wave speed is how fast the disturbance travels along the medium, set for a string by the printed relation v equals the square root of the tension divided by the linear mass density, and it does not depend on how hard the string is driven. The maximum speed at an antinode is how fast that piece of medium moves back and forth, and it is proportional to the amplitude, which the driver controls. A string carrying waves at 240 metres per second can have antinodes moving at only a few metres per second. A CED sample instructional activity for Topic 14.6 asks students to measure both, calculating the wave speed from wavelength and period data and the maximum antinode speed from amplitude and period data, and to see that they are different.