AP Physics 2 · Topic 14.6

Topic 14.6: Wave Interference and Standing Waves

Unit 14: Waves, Sound, and Physical Optics12-15% of the multiple-choice section

Superposition says the displacement where two waves overlap is the sum of their separate displacements. Same direction gives constructive interference, opposite directions destructive. A standing wave is what confined waves leave behind, and the ends of the region decide which wavelengths fit.

AP Physics: Unit 14 (topics 14.6 Wave Interference and Standing Waves). AP Physics 2 Unit 14, Topic 14.6. Two learning objectives. 14.6.A, describe the net disturbance that occurs when two or more wave pulses or waves overlap, carries 14.6.A.1 (wave interference is the interaction of two or more wave pulses or waves), 14.6.A.2 (interacting pulses travel through each other and overlap rather than bouncing off each other), 14.6.A.3 (the resulting displacement can be determined by adding the individual displacements; this is called superposition), 14.6.A.4 (interference may be constructive or destructive) with 14.6.A.4.i, 14.6.A.4.ii and 14.6.A.4.iii, 14.6.A.5 (visual representations are useful in determining the result of two interacting pulses or waves) and 14.6.A.6 (beats arise from the addition of two waves of slightly different frequency) with 14.6.A.6.i, 14.6.A.6.ii (the beat frequency is the difference in the frequencies of the two waves, relevant equation |f_beat| = |f1 - f2|) and 14.6.A.6.iii. 14.6.B, describe the properties of a standing wave, carries 14.6.B.1 (standing waves can result from interference between two waves that are confined to a region and traveling in opposite directions) with 14.6.B.1.i (a node is a point where the amplitude is always zero, an antinode a point where the amplitude is always at maximum), 14.6.B.1.ii (possible wavelengths are determined by the size and boundary conditions of the region) and 14.6.B.1.iii (common regions include pipes with open or closed ends, as well as strings with fixed or loose ends); 14.6.B.2, quoted in full on this page including its exception clause: 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; and 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). The topic prints NO boundary statement; Unit 14's three boundary statements sit under Topics 14.4, 14.5 and 14.9. Suggested skills are 1.B, 1.C, 2.A, 2.D, 3.A and 3.B, six of them, more than any other topic in Unit 14 by the Unit at a Glance count. No harmonic relationship is printed on the AP Physics 2 equation sheet or anywhere in the CED; the word harmonic appears in the CED only at 14.6.B.2, 14.6.B.3 and in the Unit 14 sample instructional activities. Sheet equations used by this topic: |f_beat| = |f1 - f2|, v_string = sqrt(F_T/(m/l)), lambda = v/f and T = 1/f. Path length difference is 14.7 and 14.8 language and does not appear in 14.6. Five of Unit 14's six sample instructional activities are tagged to 14.6. Unit 14 is weighted at 12 to 15 percent of the multiple-choice section across a suggested 14 to 23 class periods.

What Topic 14.6 requires

Topic 14.6 is the only topic in Unit 14 with two learning objectives, and counting the CED's statements it carries 18 of them, nine top-level and nine sub-statements, more than any other topic in the unit. It prints no boundary statement. Unit 14 carries exactly three of those, under Topics 14.4, 14.5 and 14.9, and this is not one of them, so nothing here is fenced off.

14.6.A Describe the net disturbance that occurs when two or more wave pulses or waves overlap.

14.6.A.1 Wave interference is the interaction of two or more wave pulses or waves.
14.6.A.2 When two or more pulses or waves interact with each other, they travel through each other and overlap rather than bouncing off each other.
14.6.A.3 When two or more wave pulses or waves overlap, the resulting displacement can be determined by adding the individual displacements. This is called superposition.
14.6.A.4 Wave interference may be constructive or destructive, with 14.6.A.4.i (same-direction displacements give constructive interference), 14.6.A.4.ii (opposite directions give destructive interference) and 14.6.A.4.iii (interacting waves can produce amplitude variations in the resultant wave).
14.6.A.5 Visual representations of wave pulses or waves are useful in determining the result of two interacting wave pulses or waves.
14.6.A.6 Beats arise from the addition of two waves of slightly different frequency, with 14.6.A.6.i (waves of different frequency are sometimes in phase and sometimes out of phase along the waves, causing periodic amplitude changes), 14.6.A.6.ii (the beat frequency is the difference in the two frequencies, with the relevant equation printed) and 14.6.A.6.iii (tuning forks demonstrate beat frequencies).

14.6.B Describe the properties of a standing wave.

14.6.B.1 Standing waves can result from interference between two waves that are confined to a region and traveling in opposite directions, with 14.6.B.1.i (nodes and antinodes), 14.6.B.1.ii (the possible wavelengths are determined by the size and boundary conditions of the region) and 14.6.B.1.iii (common regions include pipes with open or closed ends, as well as strings with fixed or loose ends).
14.6.B.2 Names the harmonics, and carries the odd-harmonic exception quoted in full further down this page.
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.

The suggested skills are 1.B (create quantitative graphs with appropriate scales and units, including plotting data), 1.C (create qualitative sketches of graphs), 2.A (derive a symbolic expression from known quantities), 2.D (predict new values or factors of change using functional dependence), 3.A (create experimental procedures appropriate for a given scientific question) and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim).

Six skills, and no other topic in the unit lists more: 14.8 lists five, 14.1 through 14.5 and 14.9 list four each, 14.7 lists three. Skill 2.A is the tell. It appears here and in 14.7 and 14.8 only, and it is why the harmonic relationships below are things you derive rather than look up.

Unit 14 is weighted at 12 to 15 percent of the multiple-choice section across a suggested 14 to 23 class periods.

Superposition: the one rule the rest of the topic comes from

14.6.A.3 is the engine of the whole topic: when two or more wave pulses or waves overlap, the resulting displacement can be determined by adding the individual displacements. This is called superposition.

Adding. Not averaging, not the larger one winning. If pulse one puts a point of the string at $+4.0 \text{ cm}$ and pulse two puts the same point at $-1.5 \text{ cm}$ , the point sits at $+2.5 \text{ cm}$ .

14.6.A.2 protects that rule from the obvious objection: when two or more pulses or waves interact with each other, they travel through each other and overlap rather than bouncing off each other. A moment after the overlap each pulse is on its way again with its original shape and amplitude.

14.6.A.4 sorts the outcomes into two names, and the sub-statements are about direction of displacement, not about the waves themselves:

14.6.A.4.i When the displacements of the superposed wave pulses or waves are in the same direction, the interaction is called constructive interference.
14.6.A.4.ii When the displacements of the superposed wave pulses or waves are in the opposite directions, the interaction is called destructive interference.

Two consequences students trip on.

Destructive does not mean zero. A $+4.0 \text{ cm}$ displacement meeting a $-1.5 \text{ cm}$ one is destructive and gives $+2.5 \text{ cm}$ . Complete cancellation needs equal magnitudes as well as opposite directions.
The same pair of waves does both at once, in different places. 14.6.A.4.iii says interacting waves can produce amplitude variations in the resultant wave. Constructive here, destructive there, is the normal case, and it is what produces both beats and standing waves.

14.6.A.5 gives the method: visual representations of wave pulses or waves are useful in determining the result of two interacting wave pulses or waves. Draw both on the same axes, pick a position, read both displacements with their signs, and add. That is the whole procedure, and it is skill 1.C.

Beats, and the one equation the sheet prints

14.6.A.6 says beats arise from the addition of two waves of slightly different frequency.

14.6.A.6.i gives the mechanism: waves with different frequencies are sometimes in phase and sometimes out of phase at locations along the waves, causing periodic amplitude changes in the resultant wave. Where the two are in step the displacements add constructively and the sound is loud; where they are out of step it fades.

14.6.A.6.ii gives the number, and this one is printed on the equation sheet:

\left| f_{\text{beat}} \right| = \left| f_1 - f_2 \right|

Absolute value on both sides, so the beat frequency does not tell you which source is higher. Two forks at 512 Hz and 509 Hz beat at 3 Hz, and so do two at 512 Hz and 515 Hz. Resolving that takes a second measurement, which is a standard question shape. 14.6.A.6.iii adds the apparatus: tuning forks are devices commonly used to demonstrate beat frequencies.

The CED's sample exam includes a beats question, its thirteenth multiple-choice item, aligned in the answer key to skill 3.B, learning objective 14.6.A and essential knowledge 14.6.A.6, with the answer A. It shows four graphs of gauge pressure against time, each from a different pair of tuning forks, and asks which pair has the greatest difference in frequencies.

That question is a reading exercise. Such a graph shows a fast oscillation inside a slowly changing envelope, and the envelope is the amplitude variation of 14.6.A.4.iii. The time from one point of quiet to the next is the beat period, and the beat frequency is its reciprocal, so the pair with the shortest time between quiet moments has the largest frequency difference. Reading the fast wiggles instead of the envelope is the mistake the question is built to catch.

Standing waves: nodes, antinodes, and what the ends decide

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.

Three conditions, all of which matter: two waves, confined to a region, going opposite ways. In practice the second wave is the first one reflected off the far end, which is Topic 14.3 territory.

14.6.B.1.i defines the two features by amplitude, absolutely:

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.

"Always" is what makes the wave standing rather than traveling. In a traveling wave every point takes a turn at maximum displacement; here the nodes never move and the antinodes never stop, and neither goes anywhere along the medium.

Two geometric facts follow, and you need both for every calculation on this page:

Adjacent nodes are half a wavelength apart. So are adjacent antinodes.
A node and the nearest antinode are a quarter of a wavelength apart.

14.6.B.1.ii turns all this into an examinable relationship: the possible wavelengths of a standing wave are determined by the size and boundary conditions of the region to which it is confined. Two inputs, and the second is what separates the cases.

14.6.B.1.iii lists the regions: common regions where standing waves can form include pipes with open or closed ends, as well as strings with fixed or loose ends. Four end types, each forcing one of the two features:

End of the region	What the medium can do there	Forced feature
String fixed to a support	cannot move	node
String with a loose (free) end	free to move fully	antinode
Closed end of a pipe	air cannot move through the wall	node
Open end of a pipe	air free to move	antinode

For a pipe the quantity described is the displacement of the air, since 14.1.A.5 defines a longitudinal wave as one where the disturbance is parallel to the direction of propagation. Air at a sealed end has nowhere to go, so that end is a node.

Everything in the next section is this table applied twice, once to each end.

The harmonic series, case by case

14.6.B.2, quoted in full because the last sentence is the one that gets dropped:

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.

The exception is stated by boundary condition, not by apparatus. "A node at one end and an antinode at the other" covers a pipe closed at one end and also a string fixed at one end with the other loose. The CED never says "closed pipe" in that sentence, and reading it as a pipe rule alone will cost you the string version.

Here is every case 14.6.B.1.iii names, with $L$ the length of the region, $v$ the wave speed in it, and $n$ the harmonic number.

Region	Ends	$\lambda_n$	$f_n$	Harmonics present
String fixed at both ends	node, node	$\dfrac{2L}{n}$	$\dfrac{nv}{2L}$	$n = 1, 2, 3, 4, \ldots$
Pipe open at both ends	antinode, antinode	$\dfrac{2L}{n}$	$\dfrac{nv}{2L}$	$n = 1, 2, 3, 4, \ldots$
Pipe closed at one end	node, antinode	$\dfrac{4L}{n}$	$\dfrac{nv}{4L}$	$n = 1, 3, 5, 7, \ldots$
String fixed at one end, loose at the other	node, antinode	$\dfrac{4L}{n}$	$\dfrac{nv}{4L}$	$n = 1, 3, 5, 7, \ldots$

The top two rows are identical, which surprises people. A string with a node at each end and a pipe with an antinode at each end look nothing alike and share a harmonic series, because what fixes the wavelengths is that the two ends match each other, not which feature they are.

The derivation is two lines each, and skill 2.A means producing it rather than recalling the table.

Matched ends. Consecutive nodes are $\lambda/2$ apart, and so are consecutive antinodes, so the region must hold a whole number of half-wavelengths:

L = n \frac{\lambda_n}{2} \quad \Rightarrow \quad \lambda_n = \frac{2L}{n} \quad \Rightarrow \quad f_n = \frac{v}{\lambda_n} = \frac{nv}{2L}

Mismatched ends. A node and the nearest antinode are $\lambda/4$ apart, so the region must hold an odd number of quarter-wavelengths:

L = n \frac{\lambda_n}{4} \quad (n \text{ odd}) \quad \Rightarrow \quad \lambda_n = \frac{4L}{n} \quad \Rightarrow \quad f_n = \frac{nv}{4L}

Both use $f = v/\lambda$ , the printed relationship $\lambda = v/f$ rearranged; the wave speed guide covers that rearrangement itself.

Two patterns worth carrying into a question. For matched ends, every harmonic is a whole-number multiple of the fundamental and consecutive frequencies are spaced by $v/2L$ . For mismatched ends the fundamental is half that of a matched region of the same length, $v/4L$ against $v/2L$ , and yet consecutive modes are still $v/2L$ apart, because the odd numbers go up in twos.

The closed pipe, and why the second harmonic is missing

This is the trap in Topic 14.6, and it is a naming trap on top of a physics trap.

The physics. A pipe closed at one end has a node there and an antinode at the open end. Try to build a mode with two quarter-wavelengths in it, meaning $L = 2(\lambda/4) = \lambda/2$ . Half a wavelength puts the same feature at both ends, so if one end is a node the other is a node too. But the open end has to be an antinode, so the mode cannot exist. The same argument kills every even $n$ , which is what 14.6.B.2 states.

The naming. 14.6.B.2's first two sentences set up a rule by ordering: longest wavelength is the first harmonic, second-longest is typically the second harmonic, and so on. Then the "however" clause overrides it for node-antinode regions. So for a closed pipe:

Mode	Wavelength	Frequency	Called
longest	$4L$	$\dfrac{v}{4L}$	fundamental, first harmonic
second-longest	$\dfrac{4L}{3}$	$\dfrac{3v}{4L}$	third harmonic
third-longest	$\dfrac{4L}{5}$	$\dfrac{5v}{4L}$	fifth harmonic

The second-longest wavelength in a closed pipe is the third harmonic, not the second. The word "typically" in the CED's second sentence is doing the work: the ordering rule is the usual naming, and the odd-harmonic clause is the exception that governs here. The harmonic number is the count of quarter-wavelengths in the region, and for this geometry that count is always odd.

This matters for reading questions as much as answering them. The CED's own sample activity asks students to draw the first three harmonics of a closed pipe and calculate the frequencies. "The first three harmonics" there means the first three that can be established, so $n = 1$ , $3$ and $5$ . Anyone who writes $n = 1, 2, 3$ has answered a different question.

Two checks catch the error fast. Count quarter-wavelengths, not loops: sketch the mode, count how many fit between the ends, and that count is $n$ , which comes out odd on its own for a node-antinode region. Test the ends: if your drawing puts a node where there must be an antinode, the mode is not allowed, whatever the arithmetic said.

What the equation sheet prints, and what you derive

The AP Physics 2 equation sheet carries 129 entries across seven groups. Counting the Waves, Sound, and Optics group gives 15, and four of those belong to Topic 14.6 or are needed by it:

Printed on the sheet	Role in Topic 14.6
$\lvert f_{\text{beat}} \rvert = \lvert f_1 - f_2 \rvert$	printed as the relevant equation at 14.6.A.6.ii
$v_{\text{string}} = \sqrt{\dfrac{F_T}{m / \ell}}$	the wave speed on a string, cited at 14.1.A.3.ii
$\lambda = \dfrac{v}{f}$	cited at 14.2.A.3, and the bridge from wavelength to frequency
$T = \dfrac{1}{f}$	cited at 14.2.A.1.ii

Two more in the same group, $x(t) = A\cos(\omega t)$ and $y(x) = A\cos\left(\frac{2\pi x}{\lambda}\right)$ , are the sinusoidal descriptions from 14.2.A.2.

Not printed anywhere on the sheet, and this is the important part:

$\lambda_n = 2L/n$ and $f_n = nv/(2L)$ .
$\lambda_n = 4L/n$ and $f_n = nv/(4L)$ .
Any statement of which harmonics a given boundary condition allows.
Any value for the speed of sound in air. The constants block prints the speed of light and nothing for sound, so a question needing the speed of sound has to supply it.

The CED gives the same answer. The word "harmonic" appears in only two places in the whole AP Physics 2 CED: essential knowledge 14.6.B.2 and 14.6.B.3, and the Unit 14 sample instructional activities. No harmonic relationship is printed as a relevant or derived equation anywhere.

That is deliberate, and 14.6.B.3 gives the intended route: visual representations of standing waves are useful in determining the relationships between length of the region, wavelength, frequency, wave speed, and harmonic. Sketch the mode, count the quarter-wavelengths between the ends, read off $\lambda$ in terms of $L$ , and convert with $f = v/\lambda$ . That is skill 2.A. A topic with no printed formula and a derivation skill attached is telling you what it will ask.

The labs this topic is built around

Unit 14 lists six sample instructional activities in the CED, and five of them are tagged to Topic 14.6. Reading them tells you what the free-response version of this topic looks like.

Long springs. Students create a standing wave, find the wave speed from wavelength and period data, separately find the maximum speed reached at an antinode from amplitude and period data, and see that the two differ. Wave speed and particle speed are not the same quantity, which connects to simple harmonic motion: each point oscillates while the pattern stands still.
Straws. Students blow across a straw while a tone-detecting app registers the fundamental frequency of the standing wave inside, use that with the straw length to get the speed of sound, then cut the straw shorter and repeat to linearize the data. Skills 3.A and 1.B in one activity: you cannot measure the speed of sound directly, so you measure a length and a frequency and let the standing-wave relationship do the rest.
Four-square problem solving. A 2 m long pipe where the speed of sound is 343 m/s: draw the first three harmonics and calculate the frequencies with the pipe open, then closed, then plot both sets on a number line and describe the pattern. Worked out below.

Two more, glass bottles tuned with water to play a song and a write-and-switch drill on which of frequency, wavelength and wave speed changes when another is held fixed, round out the five. The write-and-switch one is skill 2.D in miniature.

The Instructional Approaches section adds one more, under Construct an Argument: a long string runs from an oscillator over a pulley to a hanging object of mass $m$ , the oscillator runs at frequency $f$ , and two full standing waves form between oscillator and pulley. Students argue whether $m$ should be raised or lowered to fit more standing waves at the same frequency. That is the third worked example below.

How Topic 14.6 is tested

The AP Physics 2 exam runs 3 hours: 42 multiple-choice questions in 85 minutes for half the score, and 4 free-response questions in 95 minutes for the other half. A calculator is allowed on both sections, and Unit 14's 12 to 15 percent weighting applies to the multiple-choice section.

The six suggested skills line up with the free-response question types unusually well. The Experimental Design and Analysis question is scored on 1.B, 2.B, 2.D and 3.A, and Topic 14.6 lists three of those four. The Mathematical Routines question uses 1.A, 1.C, 2.A, 2.B, 3.B and 3.C, and Topic 14.6 lists three of those six.

What a question is likely to ask: add two pulses at an instant and name the interference (14.6.A.3, 14.6.A.4); get a beat frequency or work backwards from one to an unknown source (14.6.A.6.ii); identify the harmonic from a drawing and the ends of the region (14.6.B.2, 14.6.B.3); derive $f_n$ symbolically for a stated boundary condition (skill 2.A); predict a factor of change when the tension doubles or the pipe is closed (skill 2.D); design a measurement of wave speed or of the speed of sound (skill 3.A).

Two things commonly bundled with interference belong to other topics.

Path length difference is 14.7 and 14.8, not 14.6. The symbol $\Delta D$ and every equation built on it appear under Topic 14.7 at 14.7.A.4.ii and 14.7.A.4.iii and under Topic 14.8 at 14.8.A.1.iii and 14.8.A.1.iv. Topic 14.6 never uses the term: it gives you superposition, and 14.7 and 14.8 apply it to light through openings.
Thin-film interference is 14.9, whose boundary statement limits the quantitative work to waves normal to the incident surface.

One reason to work from the current CED, effective Fall 2024: the odd-harmonic clause in 14.6.B.2 is a single sentence, and it decides a large share of the standing-wave questions you will meet.

Adding two pulses, then finding an unknown frequency from beats

(a) Two pulses travel toward each other on the same string. Pulse P has a peak displacement of $+4.0 \text{ cm}$ and pulse Q has a peak displacement of $1.5 \text{ cm}$ . Find the net displacement when their peaks coincide, for Q pointing upward and for Q pointing downward, and name each interaction. (b) A tuning fork of known frequency 512 Hz is sounded with a second fork of unknown frequency and 3.0 beats per second are heard. A small piece of putty is stuck to the second fork, which lowers its frequency, and the beat rate rises to 6.0 beats per second. Find the original frequency of the second fork. (c) Give the beat period in the first case.

Declare the sign convention and hold it: upward displacement is positive, downward is negative. Pulse P is $+4.0 \text{ cm}$ throughout.
(a) Apply 14.6.A.3, which says the resulting displacement is found by adding the individual displacements. With Q upward, $+4.0 + 1.5 = +5.5 \text{ cm}$ . The displacements are in the same direction, so by 14.6.A.4.i this is constructive interference.
With Q downward, $+4.0 + (-1.5) = +2.5 \text{ cm}$ . The displacements are in opposite directions, so by 14.6.A.4.ii this is destructive interference. Note that the result is not zero: destructive means opposing, and full cancellation would need equal magnitudes.
In both cases 14.6.A.2 applies afterwards: the pulses travel through each other and continue with their original shapes, so a moment later there is again a $4.0 \text{ cm}$ pulse and a $1.5 \text{ cm}$ pulse, moving apart.
(b) Start from the printed relationship $\left| f_{\text{beat}} \right| = \left| f_1 - f_2 \right|$ at 14.6.A.6.ii. With $f_1 = 512 \text{ Hz}$ and $f_{\text{beat}} = 3.0 \text{ Hz}$ , the absolute value leaves two candidates: $509 \text{ Hz}$ or $515 \text{ Hz}$ .
Use the second measurement to choose. If $f_2$ had been 515 Hz, lowering it moves it toward 512 Hz and the beat rate would fall. If $f_2$ was 509 Hz, lowering it moves it further away and the beat rate rises.
The rate rose from 3.0 Hz to 6.0 Hz, so the original frequency was $f_2 = 509 \text{ Hz}$ . Check it: after loading, $|512 - f_2'| = 6.0 \text{ Hz}$ with $f_2'$ below 509 gives $f_2' = 506 \text{ Hz}$ , which is 3 Hz lower than 509 Hz. Consistent.
(c) The beat period is the time between successive quiet moments, the reciprocal of the beat frequency: $T_{\text{beat}} = \dfrac{1}{3.0 \text{ Hz}} = 0.33 \text{ s}$ . That is the envelope spacing on the gauge-pressure graph the CED's sample question 13 asks students to read.

(a) Q upward: $+5.5 \text{ cm}$ , constructive interference (14.6.A.4.i). Q downward: $+2.5 \text{ cm}$ , destructive interference (14.6.A.4.ii), and not zero because the magnitudes are unequal. (b) The second fork was originally at 509 Hz: loading it lowered the frequency further from 512 Hz, which is why the beat rate rose rather than fell. (c) $T_{\text{beat}} = 0.33 \text{ s}$ .

The CED's own pipe problem: open at both ends versus closed at one

A pipe 2.0 m long sits in a room where the speed of sound in air is 343 m/s. Find the wavelength and frequency of the first three harmonics (a) with the pipe open at both ends and (b) with one end closed. (c) Plot both sets of frequencies on a number line and describe the pattern. This is the CED's Four-Square Problem Solving activity for Topic 14.6.

(a) Fix the boundary conditions first. Both ends open means an antinode at each end, so matched ends, so the region holds a whole number of half-wavelengths: $L = n\lambda_n / 2$ , giving $\lambda_n = 2L/n$ and $f_n = nv/(2L)$ with $n = 1, 2, 3, \ldots$
With $L = 2.0 \text{ m}$ and $v = 343 \text{ m/s}$ , the fundamental is $\lambda_1 = 2(2.0) = 4.0 \text{ m}$ and $f_1 = \dfrac{343}{4.0} = 85.75 \text{ Hz}$ , which is $85.8 \text{ Hz}$ to three significant figures.
Second harmonic: $\lambda_2 = \dfrac{2(2.0)}{2} = 2.0 \text{ m}$ , $f_2 = \dfrac{343}{2.0} = 171.5 \text{ Hz}$ , or $172 \text{ Hz}$ . Third: $\lambda_3 = \dfrac{2(2.0)}{3} = 1.33 \text{ m}$ , $f_3 = \dfrac{3(343)}{2(2.0)} = 257.25 \text{ Hz}$ , or $257 \text{ Hz}$ .
(b) One end closed means a node at the closed end and an antinode at the open end. Mismatched ends, so the region holds an odd number of quarter-wavelengths: $\lambda_n = 4L/n$ and $f_n = nv/(4L)$ with $n = 1, 3, 5$ only, by the odd-harmonic clause of 14.6.B.2. The first three that can be established are therefore $n = 1$ , $3$ and $5$ , not $n = 1, 2, 3$ .
Fundamental: $\lambda_1 = 4(2.0) = 8.0 \text{ m}$ and $f_1 = \dfrac{343}{8.0} = 42.875 \text{ Hz}$ , which is $42.9 \text{ Hz}$ .
Third harmonic: $\lambda_3 = \dfrac{4(2.0)}{3} = 2.67 \text{ m}$ and $f_3 = \dfrac{3(343)}{4(2.0)} = 128.625 \text{ Hz}$ , which is $129 \text{ Hz}$ . Fifth harmonic: $\lambda_5 = \dfrac{4(2.0)}{5} = 1.60 \text{ m}$ and $f_5 = \dfrac{5(343)}{4(2.0)} = 214.375 \text{ Hz}$ , which is $214 \text{ Hz}$ .
Check every frequency against $f = v/\lambda$ independently rather than trusting the $n$ formula. Open pipe: $343 / 4.0 = 85.75$ , $343 / 2.0 = 171.5$ , $343 / 1.3333 = 257.25$ . Closed pipe: $343 / 8.0 = 42.875$ , $343 / 2.6667 = 128.625$ , $343 / 1.60 = 214.375$ . All six agree.
(c) Open set: 85.75, 171.5, 257.25 Hz. Closed set: 42.875, 128.625, 214.375 Hz. Three patterns come off the number line. The closed fundamental is exactly half the open one, since $v/4L$ is half of $v/2L$ . The spacing between consecutive modes is 85.75 Hz in both sets, because the closed pipe's odd numbers step up by two lots of 42.875 Hz. And every closed frequency sits halfway between two open ones, at 0.5, 1.5 and 2.5 times the open fundamental.
Sanity check the physics: closing one end halves the fundamental, so the note drops. That is why a pipe stopped at one end sounds an octave below an open pipe of the same length, and it answers the unit's essential question about notes an octave apart.

(a) Open at both ends: $\lambda = 4.0 \text{ m}$ at $85.8 \text{ Hz}$ , $2.0 \text{ m}$ at $172 \text{ Hz}$ , $1.33 \text{ m}$ at $257 \text{ Hz}$ (harmonics $n = 1, 2, 3$ ). (b) Closed at one end: $\lambda = 8.0 \text{ m}$ at $42.9 \text{ Hz}$ , $2.67 \text{ m}$ at $129 \text{ Hz}$ , $1.60 \text{ m}$ at $214 \text{ Hz}$ (harmonics $n = 1, 3, 5$ , since even harmonics cannot be established). (c) The closed fundamental is half the open one, both ladders have the same 85.75 Hz spacing between consecutive modes, and each closed frequency falls exactly halfway between two open frequencies.

More loops on a string at the same frequency (skills 2.A and 2.D)

A string of linear density $2.0 \times 10^{-3} \text{ kg/m}$ runs 1.50 m from an oscillator, over a pulley, to a hanging object of mass 0.50 kg. The oscillator vibrates at a fixed frequency and exactly two full standing-wave loops form between the oscillator and the pulley. Take $g = 9.8 \text{ m/s}^2$ . (a) Find the oscillator's frequency. (b) Should the hanging mass be increased or decreased to fit three loops at the same frequency, and to what value? (c) Give the general relationship between the number of loops and the hanging mass.

Set up the boundary conditions. The string is held at the oscillator and at the pulley, so there is a node at each end. Matched ends, so $\lambda_n = 2L/n$ and $f_n = nv/(2L)$ , with $n$ equal to the number of loops.
(a) The hanging object is in equilibrium, so the tension equals its weight: $F_T = mg = (0.50)(9.8) = 4.9 \text{ N}$ . Then take the wave speed from the sheet equation cited at 14.1.A.3.ii: $v = \sqrt{\dfrac{F_T}{m/\ell}} = \sqrt{\dfrac{4.9}{2.0 \times 10^{-3}}} = \sqrt{2450} = 49.5 \text{ m/s}$ . Units check: newtons over kilograms per metre is $\text{m}^2/\text{s}^2$ , whose square root is m/s.
Two loops means $n = 2$ , so $\lambda_2 = \dfrac{2(1.50)}{2} = 1.50 \text{ m}$ and $f = \dfrac{v}{\lambda_2} = \dfrac{49.497}{1.50} = 33.0 \text{ Hz}$ .
(b) Answer the direction qualitatively first, which is what the CED's Construct an Argument activity asks for. At fixed $f$ , three loops instead of two means a shorter wavelength, and $\lambda = v/f$ then requires a smaller wave speed, so a smaller tension, so a smaller hanging mass.
Now the value, by functional dependence rather than re-substituting. Setting $f_n = nv/(2L)$ equal for the two cases with $f$ and $L$ fixed makes $nv$ constant, so $v_3 / v_2 = 2/3$ . Since $v = \sqrt{F_T / \mu}$ , tension scales as $v^2$ , so $F_{T,3}/F_{T,2} = (2/3)^2 = 4/9$ , and the mass scales the same way because $F_T = mg$ . So $m_3 = \dfrac{4}{9}(0.50) = 0.22 \text{ kg}$ .
Verify from scratch rather than trusting the ratio. $F_{T,3} = (0.2222)(9.8) = 2.178 \text{ N}$ , so $v_3 = \sqrt{2.178 / 0.0020} = 33.0 \text{ m/s}$ . With $n = 3$ , $\lambda_3 = 2(1.50)/3 = 1.00 \text{ m}$ and $f = 33.0 / 1.00 = 33.0 \text{ Hz}$ , matching part (a).
(c) Combine the two relationships symbolically, which is skill 2.A. From $f = \dfrac{n}{2L}\sqrt{\dfrac{mg}{\mu}}$ with $f$ , $L$ and $\mu$ fixed, $n\sqrt{m}$ is constant, so $m \propto 1/n^2$ . Doubling the number of loops needs a quarter of the mass.

(a) $f = 33.0 \text{ Hz}$ , from $F_T = 4.9 \text{ N}$ , $v = 49.5 \text{ m/s}$ and $\lambda_2 = 1.50 \text{ m}$ . (b) Decrease it, to $m = 0.22 \text{ kg}$ : more loops at a fixed frequency needs a shorter wavelength, so a slower wave, so less tension. (c) $m \propto 1/n^2$ at fixed frequency, length and linear density, so going from two loops to three scales the mass by $(2/3)^2 = 4/9$ .

Frequently asked questions

What is superposition in AP Physics 2?

Essential knowledge 14.6.A.3 in the AP Physics 2 CED says that when two or more wave pulses or waves overlap, the resulting displacement can be determined by adding the individual displacements, and that this is called superposition. Add the displacements with their signs. Statement 14.6.A.2 adds that the pulses travel through each other rather than bouncing off each other, so after the overlap each continues with its original shape and amplitude.

What is the difference between constructive and destructive interference?

It is the direction of the displacements, not the size of the result. Essential knowledge 14.6.A.4.i in the AP Physics 2 CED says that when the displacements of the superposed waves are in the same direction, the interaction is constructive interference, and 14.6.A.4.ii says that when they are in opposite directions it is destructive interference. Destructive does not mean zero: a +4.0 cm displacement meeting a -1.5 cm one is destructive and gives +2.5 cm. Full cancellation also needs equal magnitudes.

What is the standing wave formula for a string fixed at both ends?

The allowed wavelengths are lambda_n = 2L/n and the allowed frequencies are f_n = nv/(2L), with n = 1, 2, 3 and so on. A fixed end forces a node, so with a node at each end the string holds a whole number of half-wavelengths. Neither relationship is printed on the AP Physics 2 equation sheet or anywhere in the CED: essential knowledge 14.6.B.3 expects you to get them from a drawing of the mode, and skill 2.A for this topic is deriving a symbolic expression.

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

Because a closed end must be a node and an open end must be an antinode, and only an odd number of quarter-wavelengths can fit between a node and an antinode. Any even harmonic would put the same feature at both ends, which the boundary conditions forbid. Essential knowledge 14.6.B.2 in the AP Physics 2 CED states it as a general rule about geometry, not about pipes: for a standing wave with a node at one end and an antinode at the other end, only odd harmonics can be established. That covers a string fixed at one end with a loose other end too.

Is a closed pipe's second-longest wavelength called the second or third harmonic?

The third. Essential knowledge 14.6.B.2 in the AP Physics 2 CED first gives the general naming rule, that the second-longest wavelength is typically called the second harmonic, and then overrides it: 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 pipe of length L has wavelengths 4L, 4L/3 and 4L/5, called the first, third and fifth harmonics. So the first three harmonics of a closed pipe are n = 1, 3 and 5.

How do you find the beat frequency?

Subtract the two frequencies and take the magnitude. The AP Physics 2 equation sheet prints |f_beat| = |f1 - f2|, and the CED gives it as the relevant equation under essential knowledge 14.6.A.6.ii, which says the beat frequency is the difference in the frequencies of the two waves. The absolute value means a 3 Hz beat against a 512 Hz fork could come from either 509 Hz or 515 Hz, so identifying an unknown source needs a second measurement, such as loading one fork and seeing whether the beat rate rises or falls.

What are nodes and antinodes?

Essential knowledge 14.6.B.1.i in the AP Physics 2 CED defines them by amplitude: 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. The word always is what distinguishes a standing wave from a traveling one, where every point takes a turn at maximum displacement. Adjacent nodes are half a wavelength apart, and a node and the nearest antinode are a quarter of a wavelength apart.