Fundamental vs Harmonic: What Is the Difference?

The fundamental is a harmonic: it is the first one, the standing wave with the longest wavelength a region allows. Harmonic names the whole ladder of allowed patterns, numbered from that one upward. A region with a node at one end and an antinode at the other has only the odd rungs.

AP Physics: Unit 14 (topics 14.6 Wave Interference and Standing Waves). The whole distinction is AP Physics 2 essential knowledge 14.6.B.2: a standing wave with the longest possible wavelength is called the fundamental or first harmonic, the second-longest wavelength is typically called the second harmonic, the third-longest the third, and so on; however, for a standing wave with a node at one end and an antinode at the other end, only odd harmonics can be established. The allowed set comes from 14.6.B.1.ii, which makes the possible wavelengths depend on the size and boundary conditions of the region, and 14.6.B.1.iii, which names 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 an antinode, so that mapping is assumed standard physics; it also writes closed without saying at how many ends, and the odd-harmonic clause is what fixes the intended reading as closed at one end. 14.6.B.3 directs students to visual representations rather than a formula, and no harmonic formula is printed in the CED or on the equation sheet: the conversions available are the printed lambda equals v over f and the string speed in terms of tension and linear mass density. A sample instructional activity for Topic 14.6 works a 2 metre pipe at 343 metres per second for the first three harmonics open and closed. The word overtone appears in none of the four AP Physics CEDs, and octave appears once in AP Physics 2, in a Unit 14 essential question, undefined. 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.

The distinction, stated once

These two words are not on the same footing, and that is the answer to the question rather than a complication in it.

AP Physics 2 essential knowledge 14.6.B.2 does all of it in three sentences: a standing wave with the longest possible wavelength is called the fundamental or first harmonic. The second-longest wavelength is typically called the second harmonic, the third-longest wavelength is called the third harmonic, and so on. However, for a standing wave with a node at one end and an antinode at the other end, only odd harmonics can be established.

So the CED's own first sentence names the fundamental and the first harmonic as two labels for one thing. Harmonic is the general term for any of the standing wave patterns a confined region allows. Fundamental picks out one particular harmonic, the one at the bottom of the ladder.

The query "fundamental vs harmonic" is therefore not asking you to compare two phenomena. It is asking a question of the same shape as "square vs rectangle", and the useful answer is the relationship rather than a list of differences: every fundamental is a harmonic; most harmonics are not the fundamental.

Two things follow that are worth having straight before any calculation.

Longest wavelength means lowest frequency. Since $\lambda = v/f$ from the equation sheet and the wave speed on a given string or in a given pipe is fixed, the longest wavelength belongs to the lowest frequency. The fundamental is the lowest note the region can make, and every other harmonic is above it.

The ladder is numbered by wavelength, not by what you happen to hear. 14.6.B.2 orders the harmonics by wavelength, longest first, and that ordering is what fixes the numbers. It matters because in some regions the second rung of the ladder does not exist, and calling the next audible tone "the second harmonic" is then wrong. That is the odd-harmonic rule, and it has its own section below.

Node against antinode covers the positions inside a single pattern. This page is about the family of patterns and how it is numbered.

Side by side

Property	Fundamental	Harmonic
What it names	One specific standing wave	Any member of the allowed family
CED wording	The standing wave with the longest possible wavelength (14.6.B.2)	The second-longest is the second harmonic, the third-longest the third, and so on (14.6.B.2)
Also called	First harmonic	The fundamental is one of them
Harmonic number	$n = 1$	$n = 1, 2, 3, \ldots$ , or odd $n$ only in some regions
Wavelength	The longest the region allows	Shorter for every higher $n$
Frequency	The lowest the region allows	A whole-number multiple of the fundamental
How many exist	Exactly one	In principle unlimited
Antinodes on a string fixed at both ends	1	$n$
Nodes on a string fixed at both ends	2	$n + 1$
Is it always present	It is always the lowest available mode	Some rungs are absent when a region has one node end and one antinode end
Printed equation	None	None

The frequency row carries the working rule and it is worth stating as an equation even though the CED does not print one: for any region, $f_n = n f_1$ , because the wavelengths go as $1/n$ and $f = v/\lambda$ at fixed $v$ . That relation is a consequence of 14.6.B.2's numbering, not a separate fact to learn.

The is it always present row is the one that ruins otherwise correct answers. In a region with a node at one end and an antinode at the other, the even rungs are missing entirely, so the second sound the region can make is its third harmonic and sits at three times the fundamental, not two.

The printed equation row is not an oversight in this table. Neither $f_n = n v /(2L)$ nor $\lambda_n = 2L/n$ appears in the AP Physics 2 CED or on its equation sheet, and the next-but-one section explains what to do instead.

The ladder, and how each region builds it

Everything comes out of one requirement: the pattern has to satisfy both ends of the region at once. 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 lists the common regions as pipes with open or closed ends and strings with fixed or loose ends.

A fixed string end and a closed pipe end force a node; a loose string end and an open pipe end force an antinode. The CED does not print that mapping, so it is standard physics being assumed rather than stated, but 14.6.B.2's odd-harmonic clause is phrased in terms of nodes and antinodes precisely so that it applies to any region with those ends.

Region	Ends	$\lambda_1$	Allowed $n$	$f_1$
String, both ends fixed	Node, node	$2L$	every $n$	$v/(2L)$
Pipe, both ends open	Antinode, antinode	$2L$	every $n$	$v/(2L)$
Pipe, one end open one closed	Antinode, node	$4L$	odd $n$ only	$v/(4L)$
String, one end fixed one loose	Node, antinode	$4L$	odd $n$ only	$v/(4L)$

Rows one and two are different apparatus with the same arithmetic, because two nodes and two antinodes constrain a pattern identically: either way, a whole number of half wavelengths has to fit the length.

Rows three and four are the same case in two costumes. A node at one end and an antinode at the other means a quarter wavelength is the shortest pattern that fits, and adding half a wavelength at a time is the only way to keep both ends correct, which is where the odd numbers come from.

A caution about the word closed. The CED writes "pipes with open or closed ends" at 14.6.B.1.iii and a sample instructional activity for Topic 14.6 contrasts a pipe that is open with one that is closed, without saying closed at how many ends. Read it as closed at one end, for two reasons. First, 14.6.B.2's odd-harmonic rule is stated for a region with a node at one end and an antinode at the other, which is exactly a pipe closed at one end and open at the other. Second, a pipe sealed at both ends has nodes at both ends and therefore behaves like row one, which makes the comparison the activity sets up have no content. If a question ever does mean sealed at both ends, it will say so, and you would treat it as a string.

Sketch first, always. 14.6.B.3: visual representations of standing waves are useful in determining the relationships between length of the region, wavelength, frequency, wave speed, and harmonic. Draw the two ends, draw the simplest curve that satisfies both, count how many quarter or half wavelengths fit into $L$ , and only then convert with the printed $\lambda = v/f$ .

The odd harmonics, and why a closed pipe sounds an octave down

14.6.B.2's third sentence is the sentence: however, for a standing wave with a node at one end and an antinode at the other end, only odd harmonics can be established.

The geometry is short. With a node at one end and an antinode at the other, the shortest pattern that satisfies both is a quarter wavelength, so $L = \lambda_1/4$ and $\lambda_1 = 4L$ . To keep a node at one end and an antinode at the other while shortening the wavelength, you have to add half a wavelength at a time, giving $L = 3\lambda/4$ , then $5\lambda/4$ , and so on. The allowed wavelengths are $\lambda_n = 4L/n$ with $n$ odd, and the even values simply do not satisfy both boundary conditions.

So the second harmonic of such a region does not exist. Its second audible tone is the third harmonic, at three times the fundamental frequency. A student who hears two tones and labels them the first and second harmonic will be off by a factor of $3/2$ on everything that follows.

Now the octave. An octave is a factor of two in frequency. The AP Physics 2 CED uses the word once, in the Unit 14 essential question asking why two notes an octave apart sound the same, and never defines it, so state the factor of two yourself if you use the word in an answer.

Compare two pipes of the same length $L$ . The one open at both ends has $\lambda_1 = 2L$ and $f_1 = v/(2L)$ . The one closed at one end has $\lambda_1 = 4L$ and $f_1 = v/(4L)$ . Dividing, the closed pipe's fundamental is exactly half the open pipe's, which is one octave lower, and it is lower for the same length of pipe. Worked example one puts the numbers on a two-metre pipe and gets $85.75 \ \mathrm{Hz}$ against $42.875 \ \mathrm{Hz}$ .

That is a real design fact and not a curiosity: closing one end of a pipe buys an octave of pitch without adding any length. It is why a stopped organ pipe can be half the height of an open one making the same note.

The two effects are separate and both get tested. Closing one end halves the fundamental frequency, and it also deletes every even harmonic. A question can ask about either without mentioning the other, and worked example three exploits the second one to identify a pipe from two of its frequencies.

What is not printed, and what the CED wants instead

No harmonic formula exists anywhere in AP Physics 2. The Waves, Sound, and Optics group of the equation sheet prints fifteen lines and none of them is $f_n = nv/(2L)$ or $\lambda_n = 2L/n$ . The framework does not state either relation either. Everything in the table two sections above was derived from geometry rather than looked up.

Essential knowledge 14.6.B.3 says what to do instead, and it reads like an instruction: visual representations of standing waves are useful in determining the relationships between length of the region, wavelength, frequency, wave speed, and harmonic. The intended route is sketch, read $\lambda_n$ off the sketch, then convert.

The two printed relations that do the converting are both short:

\lambda = \frac{v}{f} \qquad v_{\text{string}} = \sqrt{\frac{F_T}{m/\ell}}

The second is the one that lets a string question start from tension and mass rather than from a given speed, and worked example two uses it in reverse.

The word "overtone" appears in none of the four AP Physics CEDs. Not in AP Physics 2, and not in AP Physics 1, C: Mechanics or C: E&M. Many textbooks number the same ladder a second way, calling the fundamental's next neighbour the first overtone, so that the third harmonic of a closed pipe is its first overtone. That numbering is legitimate outside AP and it is a reliable way to get an AP answer wrong, because the two schemes disagree about every rung above the first. Use harmonic numbers on this exam and say "harmonic" in your written answers.

Nor does any of this appear in the other three courses. The words standing wave, antinode, interference and diffraction occur nowhere in the AP Physics 1, C: Mechanics or C: E&M frameworks. Waves and their harmonics are AP Physics 2 material only, filed under Unit 14, which carries 12 to 15 percent of that exam across about 14 to 23 class periods.

One thing the CED does supply is a full worked scenario. A sample instructional activity for Topic 14.6 gives a two-metre pipe in a room where the speed of sound is $343 \ \mathrm{m/s}$ and asks students to draw the first three harmonics for the pipe open and for the pipe closed, calculate the frequencies for both, plot the two sets on a single number line with different symbols, and describe the pattern. That activity is the topic's own summary of everything above, and it is worked out in full below.

When it costs a mark

Treating the fundamental and the first harmonic as different things. 14.6.B.2 names them as one: a standing wave with the longest possible wavelength is called the fundamental or first harmonic. A question that gives you one has given you the other.
Calling the second sound a closed pipe makes its second harmonic. It is the third harmonic, three times the fundamental, because 14.6.B.2 allows only odd harmonics when a region has a node at one end and an antinode at the other.
Using overtone numbering. The first overtone of a closed pipe is its third harmonic, so the two schemes differ by more than a name. The AP CEDs never use the word overtone, so neither should your answer.
Assuming a longer pipe or string means a higher frequency. It is the reverse. The fundamental wavelength scales with the length, so $f_1$ scales as $1/L$ : doubling the length halves the fundamental.
Applying $\lambda_1 = 2L$ to a pipe closed at one end. That region has $\lambda_1 = 4L$ . Sketching the ends first prevents this, because a quarter-wave curve looks nothing like a half-wave one.
Confusing which quantity is fixed. On one string at one tension, the wave speed is the same for every harmonic. What changes from harmonic to harmonic is the wavelength and therefore the frequency. Changing the tension changes the speed and moves every harmonic together.
Reaching for a harmonic formula. None is printed. Sketch the pattern, read the wavelength, use $\lambda = v/f$ , as 14.6.B.3 directs.
Reading "closed pipe" as sealed at both ends. In this context it means closed at one end and open at the other. A pipe sealed at both ends has a node at each end and behaves exactly like a string fixed at both ends.
Saying "an octave" without saying what it means. The CED uses the word once, in the Unit 14 essential question, and never defines it. Write "a factor of two in frequency" alongside it.

When the two words are used as if they were rivals

The question shape "fundamental vs harmonic" comes from somewhere real, and it is worth naming the three habits that produce it.

Habit one: hearing "harmonics" as "the extra ones". In everyday musical language, the harmonics of a note often means the tones above it, with the fundamental treated as the note itself and the harmonics as its colouring. That usage is common and it is not the CED's. 14.6.B.2 counts the fundamental as harmonic number one, so a question asking how many harmonics are below $500 \ \mathrm{Hz}$ includes it.

Habit two: numbering from the second one. Overtone numbering starts counting after the fundamental, so the second harmonic is the first overtone. Both schemes are internally consistent and they disagree everywhere above the bottom rung. The AP CEDs contain the word overtone zero times, which settles which scheme to use on this exam.

Habit three: assuming the ladder always has every rung. Because the string fixed at both ends is the first case anyone meets, and it does have every rung, the pattern $f_n = nf_1$ with $n = 1, 2, 3, \ldots$ gets learned as though it were the definition of a harmonic. It is not. 14.6.B.2's "however" clause exists to break that habit, and worked example three is a question that cannot be answered without it.

Where the two words genuinely need care in one sentence. "The fundamental frequency" is a number belonging to a particular region: $v/(2L)$ or $v/(4L)$ depending on the ends. "The harmonics" is a set of numbers, all whole multiples of that one, possibly with the even members missing. So a sentence like "the harmonics are multiples of the fundamental" is correct, while "the harmonics are above the fundamental" is correct only if you have already agreed that the fundamental is not one of them, which the CED has not.

A safe formulation for a written answer. Name the harmonic number, state the wavelength you read off your sketch, and give the frequency. "The third harmonic has $\lambda = 4L/3$ and $f_3 = 3f_1$ " is unambiguous under either numbering scheme, because it commits to a number rather than to a word.

The CED's own two-metre pipe, open and then closed

A sample instructional activity for Topic 14.6 puts a $2 \ \mathrm{m}$ long pipe in a room where the speed of sound in air is $343 \ \mathrm{m/s}$ , and asks for the first three harmonics with the pipe open and with the pipe closed, and for the frequencies of each. Work both cases, then describe the pattern the two sets of frequencies make.

Read the ends first. Open at both ends means an antinode at each end. Closed means closed at one end and open at the other, so a node at the sealed end and an antinode at the open one; that reading is what makes 14.6.B.2's odd-harmonic clause apply and what makes the two cases differ at all.
Open pipe. An antinode at each end means a whole number of half wavelengths fits the length: $\lambda_n = \dfrac{2L}{n} = \dfrac{4.00}{n} \ \mathrm{m}$ , with every $n$ allowed.
$n = 1$ : $\lambda_1 = 4.00 \ \mathrm{m}$ and $f_1 = \dfrac{343}{4.00} = 85.75 \ \mathrm{Hz}$ .
$n = 2$ : $\lambda_2 = 2.00 \ \mathrm{m}$ and $f_2 = \dfrac{343}{2.00} = 171.5 \ \mathrm{Hz}$ .
$n = 3$ : $\lambda_3 = \dfrac{4.00}{3} = 1.333 \ \mathrm{m}$ and $f_3 = \dfrac{343}{4.00/3} = 257.25 \ \mathrm{Hz}$ .
Closed pipe. A node at one end and an antinode at the other means an odd number of quarter wavelengths fits: $\lambda_n = \dfrac{4L}{n} = \dfrac{8.00}{n} \ \mathrm{m}$ , with $n$ odd only.
$n = 1$ : $\lambda_1 = 8.00 \ \mathrm{m}$ and $f_1 = \dfrac{343}{8.00} = 42.875 \ \mathrm{Hz}$ .
$n = 3$ : $\lambda_3 = \dfrac{8.00}{3} = 2.667 \ \mathrm{m}$ and $f_3 = \dfrac{343}{8.00/3} = 128.625 \ \mathrm{Hz}$ .
$n = 5$ : $\lambda_5 = 1.60 \ \mathrm{m}$ and $f_5 = \dfrac{343}{1.60} = 214.375 \ \mathrm{Hz}$ .
Note that the first three harmonics of the closed pipe are numbers $1$ , $3$ and $5$ , not $1$ , $2$ and $3$ . There is no second harmonic to draw, which is the point of drawing them.
The pattern on a number line. The open pipe's frequencies are spaced $85.75 \ \mathrm{Hz}$ apart, at $1$ , $2$ and $3$ times the fundamental. The closed pipe's are spaced $85.75 \ \mathrm{Hz}$ apart as well, but starting from $42.875 \ \mathrm{Hz}$ , so they interleave with the open pipe's rather than coinciding with them: $42.875$ , $85.75$ , $128.625$ , $171.5$ , $214.375$ , $257.25$ alternates closed, open, closed, open, closed, open.
The octave. $\dfrac{42.875}{85.75} = 0.500$ exactly, so the closed pipe's fundamental is one octave below the open pipe's, from the same $2 \ \mathrm{m}$ of pipe. The general result is $\dfrac{v/(4L)}{v/(2L)} = \dfrac{1}{2}$ , independent of both $L$ and $v$ .
A physical check on the closed pipe's fundamental: an $8.00 \ \mathrm{m}$ wavelength in a $2 \ \mathrm{m}$ pipe is not a contradiction, because only a quarter of that wavelength has to fit inside the pipe. The rest of the pattern is a shape the boundary conditions imply, not a thing that has to be housed.

Open: $85.75$ , $171.5$ and $257.25 \ \mathrm{Hz}$ for $n = 1$ , $2$ , $3$ . Closed: $42.875$ , $128.625$ and $214.375 \ \mathrm{Hz}$ for $n = 1$ , $3$ , $5$ , with no second harmonic. The two sets interleave, and the closed pipe's fundamental is exactly half the open pipe's, one octave lower for the same length.

A guitar string: from the fundamental to the tension and the fourth harmonic

A string of length $\ell = 0.640 \ \mathrm{m}$ and mass $2.00 \ \mathrm{g}$ is fixed at both ends and tuned so that its fundamental is $220 \ \mathrm{Hz}$ . (a) Find the fundamental wavelength and the wave speed. (b) Find the tension, using the relation printed on the AP Physics 2 sheet. (c) Find the frequency and wavelength of the fourth harmonic, and say how many nodes and antinodes it has.

(a) Both ends are fixed, so both are nodes, and the longest pattern that fits is one half wavelength: $\lambda_1 = 2\ell = 2(0.640) = 1.280 \ \mathrm{m}$ .
From the printed $\lambda = v/f$ , $v = \lambda_1 f_1 = (1.280)(220) = 281.6 \ \mathrm{m/s}$ .
(b) The sheet prints $v_{\text{string}} = \sqrt{\dfrac{F_T}{m/\ell}}$ . The linear mass density is $\dfrac{m}{\ell} = \dfrac{2.00 \times 10^{-3} \ \mathrm{kg}}{0.640 \ \mathrm{m}} = 3.125 \times 10^{-3} \ \mathrm{kg/m}$ .
Squaring and rearranging, $F_T = \left(\dfrac{m}{\ell}\right)v^2 = (3.125 \times 10^{-3})(281.6)^2 = (3.125 \times 10^{-3})(79298.56) = 247.8 \ \mathrm{N}$ .
Check it by going back the other way: $\sqrt{247.8 / 3.125 \times 10^{-3}} = \sqrt{79296} = 281.6 \ \mathrm{m/s}$ , which is the speed we started from.
(c) Since both ends are nodes, every harmonic exists, and $\lambda_n = 2\ell/n$ . For $n = 4$ : $\lambda_4 = \dfrac{1.280}{4} = 0.320 \ \mathrm{m}$ .
The wave speed does not depend on the harmonic, because it is set by the tension and the linear mass density and neither has changed. So $f_4 = \dfrac{v}{\lambda_4} = \dfrac{281.6}{0.320} = 880 \ \mathrm{Hz}$ , which is $4f_1 = 4(220)$ as the numbering requires. It is two octaves above the fundamental, since $880/220 = 4 = 2^2$ .
For $n = 4$ there are $4$ antinodes and $5$ nodes, counting both clamped ends. The nodes sit at $0$ , $0.160$ , $0.320$ , $0.480$ and $0.640 \ \mathrm{m}$ , spaced $\lambda_4/2 = 0.160 \ \mathrm{m}$ apart.
One thing not to reach for here: there is no printed $f_n = nv/(2\ell)$ . The chain used was geometry to get $\lambda_n$ , then the printed $\lambda = v/f$ , which is the route 14.6.B.3 describes.

(a) $\lambda_1 = 1.280 \ \mathrm{m}$ and $v = 281.6 \ \mathrm{m/s}$ . (b) $F_T = 248 \ \mathrm{N}$ . (c) $\lambda_4 = 0.320 \ \mathrm{m}$ and $f_4 = 880 \ \mathrm{Hz}$ , with $4$ antinodes and $5$ nodes.

Identifying a pipe from two of its resonances

A pipe resonates at $300 \ \mathrm{Hz}$ and again at $500 \ \mathrm{Hz}$ , with no resonance anywhere between them. The speed of sound is $343 \ \mathrm{m/s}$ . (a) Decide whether the pipe is open at both ends or closed at one. (b) Find the fundamental frequency. (c) Find the length of the pipe. (d) Predict the next resonance above $500 \ \mathrm{Hz}$ .

(a) Test the open-pipe hypothesis first. A pipe open at both ends has every harmonic, so consecutive resonances differ by exactly $f_1$ . That would make $f_1 = 500 - 300 = 200 \ \mathrm{Hz}$ . But $300$ is not a whole-number multiple of $200$ , so the hypothesis fails.
Now the closed-pipe hypothesis. A pipe with a node at one end and an antinode at the other has only odd harmonics by 14.6.B.2, so consecutive resonances are two rungs apart and differ by $2f_1$ . That gives $2f_1 = 200 \ \mathrm{Hz}$ and $f_1 = 100 \ \mathrm{Hz}$ .
Check both given frequencies against it: $300 = 3 \times 100$ and $500 = 5 \times 100$ , and both multipliers are odd, as required. The pipe is closed at one end, and the two resonances are its third and fifth harmonics.
(b) $f_1 = 100 \ \mathrm{Hz}$ . Note that this frequency was not observed and does not have to be: the question gave two harmonics and the fundamental follows from the spacing.
(c) A node at one end and an antinode at the other gives $\lambda_1 = 4L$ . From the printed $\lambda = v/f$ , $\lambda_1 = \dfrac{343}{100} = 3.43 \ \mathrm{m}$ , so $L = \dfrac{3.43}{4} = 0.8575 \ \mathrm{m}$ , which is $0.86 \ \mathrm{m}$ to two significant figures.
Sanity check the length against the harmonics directly: $\lambda_3 = 4L/3 = 1.1433 \ \mathrm{m}$ gives $f_3 = 343/1.1433 = 300 \ \mathrm{Hz}$ , and $\lambda_5 = 4L/5 = 0.686 \ \mathrm{m}$ gives $f_5 = 343/0.686 = 500 \ \mathrm{Hz}$ . Both match.
(d) The next odd harmonic is $n = 7$ , so $f_7 = 7(100) = 700 \ \mathrm{Hz}$ . Note what is not there: $600 \ \mathrm{Hz}$ would be the sixth harmonic, and even harmonics cannot be established in this region at all.
The whole question turns on the odd-harmonic clause. A student who assumes every harmonic exists gets $f_1 = 200 \ \mathrm{Hz}$ , then finds $300 \ \mathrm{Hz}$ unaccounted for, and has no route to a length.

(a) Closed at one end, because consecutive resonances differ by $2f_1$ rather than $f_1$ . (b) $f_1 = 100 \ \mathrm{Hz}$ , and $300$ and $500 \ \mathrm{Hz}$ are its third and fifth harmonics. (c) $L = 0.8575 \ \mathrm{m}$ , about $0.86 \ \mathrm{m}$ . (d) $700 \ \mathrm{Hz}$ , the seventh harmonic; there is no resonance at $600 \ \mathrm{Hz}$ .

Frequently asked questions

What is the difference between the fundamental and a harmonic?

The fundamental is a harmonic, specifically the first one. AP Physics 2 essential knowledge 14.6.B.2 says a standing wave with the longest possible wavelength is called the fundamental or first harmonic, the second-longest wavelength is typically called the second harmonic, the third-longest the third harmonic, and so on. So harmonic is the general name for any of the standing wave patterns a confined region allows, and fundamental picks out the bottom rung of that ladder. Longest wavelength means lowest frequency, so the fundamental is also the lowest note the region can make, and every other harmonic sits at a whole-number multiple of it.

Is the fundamental frequency the same as the first harmonic?

Yes. AP Physics 2 essential knowledge 14.6.B.2 gives both names in the same clause, calling the standing wave with the longest possible wavelength the fundamental or first harmonic. They are two labels for one pattern, so a question that gives you the first harmonic has given you the fundamental. Be careful with a different numbering scheme found in some textbooks, which calls the second harmonic the first overtone and counts upward from there. The word overtone does not appear in any of the four AP Physics course descriptions, so use harmonic numbers on this exam.

Why does a pipe closed at one end only have odd harmonics?

Because of the boundary conditions at its two ends. AP Physics 2 essential knowledge 14.6.B.2 states the rule directly: for a standing wave with a node at one end and an antinode at the other end, only odd harmonics can be established. A closed end forces a node, since air cannot move through the seal, and an open end forces a displacement antinode. The shortest pattern that satisfies both is a single quarter wavelength, so the fundamental wavelength is four times the pipe length, and the only way to shorten the wavelength while keeping both ends correct is to add half a wavelength at a time. That gives allowed wavelengths of 4L over n for odd n only. The practical consequence is that such a pipe has no second harmonic, so its second audible tone is the third harmonic at three times the fundamental.

Why is a closed pipe an octave lower than an open pipe of the same length?

Because its fundamental wavelength is twice as long. A pipe open at both ends has an antinode at each end, so a half wavelength fits the length and the fundamental wavelength is 2L, giving a fundamental frequency of v over 2L. A pipe closed at one end has a node at one end and an antinode at the other, so only a quarter wavelength fits and the fundamental wavelength is 4L, giving v over 4L. The ratio is exactly one half, independent of the length and of the speed of sound, and a factor of two in frequency is what an octave means. For a 2 metre pipe with sound at 343 metres per second the two fundamentals are 85.75 hertz open and 42.875 hertz closed. Note that the AP Physics 2 CED uses the word octave only once, in a Unit 14 essential question, and never defines it.

What is the formula for the nth harmonic?

AP Physics 2 does not print one, and that is deliberate. Neither f-n equals n times v over 2L nor lambda-n equals 2L over n appears in the CED or on the equation sheet. Essential knowledge 14.6.B.3 gives the intended method instead: visual representations of standing waves are useful in determining the relationships between length of the region, wavelength, frequency, wave speed, and harmonic. So sketch the two ends, draw the simplest curve that satisfies both, count how many quarter or half wavelengths fit the length to get the wavelength, and convert with the printed lambda equals v over f. Working this way also protects you from applying the wrong formula to a region with one node end and one antinode end, where the even harmonics do not exist.

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 along the pattern. The fundamental therefore has a single antinode in the middle and two nodes at the ends. For a pipe closed at one end the counting starts from a different shape, with a node at the sealed end and an antinode at the open one, so the fundamental holds only a quarter of a wavelength: one node and one antinode. Counting from a sketch is safer than memorising either result, since the answer depends entirely on which end forces which feature.

Do the harmonics of a string change if you tighten it?

They all move together, and they stay in the same whole-number ratios. The wave speed on a string is set by the printed relation v equals the square root of the tension divided by the linear mass density, so tightening the string raises the speed. The allowed wavelengths depend only on the length and the boundary conditions, so they do not change at all. With lambda fixed and v raised, every frequency rises by the same factor, which is why tuning a guitar string shifts the whole harmonic series rather than rearranging it. Shortening the string by fretting it changes the wavelengths instead, and again moves every harmonic together.