Unit 14 · Topic 14.6 · about 30 minutes · 7 parts · 12 scored checks, 2 of them multiple choice, plus 2 from earlier lessons

AP Physics 2: Algebra-Based 14.6: Wave Interference and Standing Waves

Q: How do you find the beat frequency?

Subtract the two frequencies: |fbeat| = |f1 - f2|. Two notes at 440 Hz and 436 Hz give 4 beats each second.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 14: 12 to 15% of the multiple-choice section

What you'll be able to do

Describe the net disturbance where two or more pulses or waves overlap: displacements add, constructively or destructively, and two close frequencies produce beats.
Describe a standing wave: its nodes and antinodes, and how the length and the ends of a string or pipe set the wavelengths and harmonics it allows.

Key vocabulary

Term	What it means	The exam trap
Superposition	The rule that where waves overlap, the medium's displacement at each point is the sum of the displacements each wave would cause alone.	Superposition adds displacements, at one point and one instant. It does not add the waves' speeds.
Node	A point on a standing wave where the amplitude is always zero.	A node is not just a point the string happens to pass through at one instant; it never moves at all.
Antinode	A point on a standing wave where the amplitude is always the largest on the wave.	An antinode is not always displaced; it swings through zero like any other point, with the biggest swing.
Fundamental	The standing wave with the longest wavelength a region allows, also called the first harmonic.	The longest wavelength gives the lowest frequency, not the highest.
Beats	The regular swelling and fading of loudness heard when two waves of slightly different frequencies overlap.	Beats are a change in loudness, heard over one note, not a change in pitch.

A straight pipe drawn side on. Its left end is closed by a solid cap and its right end is open. A line below the pipe marks its full length, labeled L. A note reads: sound travels at 340 m/s in the air inside, and the lowest frequency at which the pipe resonates is 125 Hz. — The pipe in Watch it done.

Part 1 Think first Two pulses on one rope

Commit to a guess first. Being wrong now makes the right idea stick better.

Predict first

Two friends hold the ends of a long rope. At the same moment, one flicks an upward pulse along it and the other flicks a downward pulse of the same size and shape. The pulses meet in the middle. What happens after that?

In short. Wave interference is what happens where two or more waves overlap: the medium's displacement there is the sum of the displacements each wave would cause on its own.

Displacements in the same direction add to a bigger one (constructive interference); in opposite directions they partly or fully cancel (destructive interference).
Two waves of slightly different frequencies drift in and out of step, so the loudness swells and fades: beats, at a rate equal to the difference of the frequencies.
Waves reflecting back and forth in a string or pipe can form a standing wave, whose possible wavelengths are set by the region's length and by what its ends allow.

Part 2 Learn it When waves overlap

6a. Adding displacements

Wave interference is what two or more pulses or waves do when they meet. They do not collide like balls. Each one travels through the other and comes out the far side as it went in.

While they overlap, the medium cannot move two ways at once. Its displacement at each point is the sum of the displacements the pulses would give it on their own. This adding is called superposition, and each displacement keeps its sign: up is positive, down is negative.

When the displacements are in the same direction, they add to a larger one: constructive interference.
When they are in opposite directions, they cancel, partly or completely: destructive interference.

In the picture, a pulse 3 cm tall meets a pulse 2 cm deep. Where their peaks coincide, the rope stands $3 + (- 2) = 1$ cm up. Two pulses of the same size and shape, one up and one down, would add to zero there for that instant.

Drawing each pulse on its own, dashed, and then adding their heights point by point is the surest way to find the shape of the rope while they overlap.

One rope drawn at three moments, one above the other. Before: an upward pulse 3 cm tall on the left moves right, and a downward pulse 2 cm deep on the right moves left. While overlapping: the two pulses, drawn dashed, sit at the same place, and the rope itself, drawn solid, stands 1 cm up there. After: the upward pulse, 3 cm tall, is on the right moving right, and the downward pulse, 2 cm deep, is on the left moving left. Heights are drawn larger than true.

Quick check · Calculate

Two pulses, identical in width and shape, travel toward each other along a stretched rope. Pulse 1 is an upward pulse 4.0 cm tall, and pulse 2 is a downward pulse 1.5 cm deep. At the instant their peaks are at the same point of the rope, what is the rope's displacement at that point? Take up as positive.

Your answer cm

6b. Beats

Strike two tuning forks together, one at 440 Hz and one at 437 Hz. You do not hear two separate notes. You hear one note whose loudness swells and fades, again and again. These are beats, and tuning forks are the usual way to show them.

Beats come from superposition. The two sound waves have slightly different frequencies, so at your ear they drift in and out of step. When they are in phase, crest meeting crest, they add constructively and the sound is loud. A little later one wave has gained half a cycle on the other; they are out of phase, crest meeting trough, and the sound is faint. Then they come back into step. The amplitude of the combined wave rises and falls periodically.

The number of loud moments each second, the beat frequency, is the difference of the two frequencies:

| f_{beat} | = | f_{1} - f_{2} |

For 440 Hz and 437 Hz that is 3 Hz: three swells every second. The beat frequency tells you how far apart the two frequencies are, but not which one is higher. Musicians tune by listening to beats: as two notes get closer, the beats slow down, and when the notes match, the beats stop.

Quick check · Fill the statement

Complete the explanation of beats. Tap a blank, then an answer.

Two tuning forks with slightly different frequencies sound together. At some moments their waves reach your ear , and the sound is loud; at others they arrive , and the sound is faint. The number of loud moments each second equals the of the two frequencies.

6c. Standing waves

Shake one end of a string tied to a post. Waves run to the post, reflect, and run back, so two waves of the same frequency travel through each other in opposite directions. For the right frequencies they superpose into a standing wave: a pattern that oscillates in place instead of traveling along the string.

A standing wave has two kinds of special point.

Nodes, where the two waves always cancel. The amplitude there is always zero, so the string never moves. A fixed end is always a node.
Antinodes, where the two waves always add constructively. The amplitude there is always the largest anywhere on the string.

Neighboring nodes are half a wavelength apart, and each antinode sits midway between two nodes. Between them, each point swings up and down with an amplitude somewhere between zero and the antinode's. In the picture the string is one wavelength long, so it has two loops, three nodes and two antinodes.

Every point between two neighboring nodes moves in step: they all reach their highest point together. Nothing travels along the string, even though energy is stored in its motion.

Tap to open the picture full size

Quick check · Rank it

A string 1.20 m long is fixed at both ends and vibrates in a standing wave with a wavelength of 0.80 m. Point P is 0.20 m from the left end, Q is 0.30 m from it, R is 0.40 m from it and S is 0.60 m from it. Rank the points by the amplitude of their oscillation.

Largest amplitude first

Give each one a place, 1 for first. Give two the same place if they are equal.

Point P, 0.20 m from the left end
Point Q, 0.30 m from the left end
Point R, 0.40 m from the left end
Point S, 0.60 m from the left end

Your ranking: none yet

6d. Harmonics on strings and in pipes

A string or pipe holds only the standing waves whose pattern fits its length and its ends. A fixed string end or a closed pipe end is a node; a loose string end or an open pipe end is an antinode.

The standing wave with the longest wavelength that fits is the fundamental, or first harmonic. The next longest is the second harmonic, then the third, and so on. With $v$ fixed by the medium, $f = \frac{v}{λ}$ , so a shorter wavelength means a higher frequency.

Node at both ends (a string fixed at both ends), or antinode at both ends (a pipe open at both ends): the fundamental fits half a wavelength, $L = \frac{λ_{1}}{2}$ . Every harmonic is possible, $λ_{n} = \frac{2 L}{n}$ and $f_{n} = n f_{1}$ .
Node at one end, antinode at the other (a pipe closed at one end): the fundamental fits a quarter wavelength, $L = \frac{λ_{1}}{4}$ . Each extra loop adds half a wavelength, so only odd harmonics can form: $f_{1}$ , $3 f_{1}$ , $5 f_{1}$ .

Drawing the pattern, the right point at each end and loops between, is the quickest way to connect length, wavelength, frequency, speed and harmonic.

What each region of length L allows
Region	Its ends	Fundamental wavelength	Harmonics possible
String fixed at both ends	node and node	2L	every one: 1, 2, 3, ...
Pipe open at both ends	antinode and antinode	2L	every one: 1, 2, 3, ...
Pipe closed at one end	node and antinode	4L	odd only: 1, 3, 5, ...

Quick check · Select all

In which two of these can only odd harmonics be established? Pick two.

Select 2 answers, then check.

Common mistakes

The mistake	Why it is wrong	Do this instead
Thinking two pulses that meet bounce off each other, or destroy each other.	Pulses pass through each other; while they overlap the displacements add, and afterward each carries on unchanged.	Add the displacements point by point while they overlap, then let each pulse continue as it was.
Reading the beat frequency as one of the two frequencies, or as their sum.	Beats count how often the waves come back into step, which is the difference of the frequencies.	Use $\| f_{1} - f_{2} \|$ , and remember it does not say which frequency is higher.
Giving a pipe closed at one end every harmonic, or the same fundamental as an open pipe of the same length.	A closed end is a node and an open end an antinode, so the fundamental fits a quarter wavelength and only odd harmonics form.	Draw the pattern with the right kind of point at each end before writing any wavelength.

Part 3 Watch it done How long is a closed pipe?

Predict first, then watch each step, then finish the problem yourself.

In the picture at the top of this lesson, a pipe is closed at its left end and open at its right. Sound travels at 340 m/s in the air inside, and the lowest frequency at which a standing wave forms in the pipe is 125 Hz. How long is the pipe?

Predict first

Before you open the first step, predict: how long is the pipe?

Step 1 · Mark the ends
The closed end is a node, since the air there cannot move. The open end is an antinode. So every pattern in this pipe has a node at one end and an antinode at the other.
Step 2 · Pick the pattern
The lowest frequency is the fundamental, the longest wavelength that fits. With a node at one end and an antinode at the other, the simplest pattern is a quarter of a wavelength: $L = \frac{λ_{1}}{4}$ .
Step 3 · Find the wavelength
$λ_{1} = \frac{v}{f_{1}} = \frac{340 m/s}{125 Hz} = 2.72$ m.
Why this step?
Why is the fundamental's pattern a quarter of a wavelength, when a string fixed at both ends starts with a half?
Neighboring nodes are half a wavelength apart, and an antinode sits midway between them, a quarter wavelength from each. A region with a node at one end and an antinode at the other spans the distance from a node to its nearest antinode.
Your step · Calculate

Finish it. A pipe closed at one end and open at the other has a fundamental frequency of 125 Hz, and sound travels at 340 m/s in the air inside, so the fundamental's wavelength is 2.72 m. How long is the pipe?

Your answer m

Part 4 Try it Wavelengths, a capped tube and a fork

Predict first: from one harmonic to the next on a string fixed at both ends, does the wavelength fall by the same amount each time, by less each time, or by more each time? Then sketch it, predict what capping a tube does, and judge a claim about beats.

Try it · Sketch the graph

A string 1.50 m long is fixed at both ends. Sketch the wavelength of its standing waves against the harmonic number, from the first harmonic to the fourth. The first point is fixed at 3.00 m, the wavelength of the fundamental.

Drag a point up or down, or tap it and use the arrow keys. The shape is checked, and so are the values the question lets you work out.

Try it · Multiple choice

A tube open at both ends has a fundamental frequency $f$ . A cap is then fitted over one end, and nothing else changes. What is the tube's new fundamental frequency?

Try it · Decide and justify

A tuning fork marked 512 Hz and a second fork with no marking are struck together, and the sound swells and fades 6 times each second. A student says the second fork must vibrate at 518 Hz. Is she right?

Which reason supports your answer?

Part 5 Explain it The harmonics of a guitar string

On the exam this is handwritten, so answer on paper first. Nothing you type here leaves this page. Then check your answer against the model and each scoring point.

Free response is 50% of the exam score. This one is shaped like Question 2, Translation Between Representations.

Free-response practice · Question 2 style

A guitar string of length $L$ is fixed at both ends, and waves travel along it at speed $v$ .

(a) On paper, draw the string in its third harmonic at its two extreme shapes. Label every node N and every antinode A. Do this by hand before you open the model.

(b) Derive an expression for the frequency $f_{n}$ of the string's $n$ th harmonic in terms of $n$ , $v$ and $L$ .

(c) For one string, $L = 0.65$ m and $v = 260$ m/s. On paper, graph $f_{n}$ against $n$ for $n = 1$ to $4$ , with a scale and units on each axis.

(d) Say whether the shape of your graph in (c) agrees with your expression in (b), and what the slope of the graph tells you about the string.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) Compare your drawing with the model below. The string forms three equal loops: four nodes, the two fixed ends and two points between, and three antinodes, one in the middle of each loop.

(b) The ends are both nodes, and neighboring nodes are half a wavelength apart, so the $n$ th harmonic fits $n$ half-wavelengths: $L = \frac{n λ_{n}}{2}$ , so $λ_{n} = \frac{2 L}{n}$ . Then $f_{n} = \frac{v}{λ_{n}} = \frac{n v}{2 L}$ .

(c) $f_{n} = \frac{n (260 m/s)}{2 (0.65 m)} = n (200 Hz)$ : 200, 400, 600 and 800 Hz. See the graph below: four points on a straight line through the origin.

(d) They agree. The expression says $f_{n}$ is proportional to $n$ , and the points lie on a straight line through the origin. The slope, 200 Hz per harmonic, is $\frac{v}{2 L}$ , the fundamental frequency of the string.

A model drawing. A string fixed at both ends, drawn at its two extreme shapes, forming three equal loops. The four points where the shapes cross, the two ends and two points between, are labeled N. The middle of each loop, where the string swings farthest, is labeled A, three in all. — One correct drawing. Yours needs three loops, with N at the four crossing points, ends included, and A at the middle of each loop.

Frequency of each harmonic against harmonic number

The points lie on a straight line through the origin with a slope of 200 Hz per harmonic.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Draws three equal loops with nodes at both fixed ends and two between, four in all, and an antinode in the middle of each loop.
Uses nodes half a wavelength apart to write $L = \frac{n λ_{n}}{2}$ and derives $f_{n} = \frac{n v}{2 L}$ .
Calculates 200, 400, 600 and 800 Hz and plots them with a labeled scale and units on each axis.
States that the straight line through the origin agrees with $f_{n}$ being proportional to $n$ .
Identifies the slope, 200 Hz per harmonic, as $\frac{v}{2 L}$ , the fundamental frequency.

Part 6 Check yourself A string driven at 150 Hz

Both case questions use the picture.

A small vibrator shakes one end of a stretched string at 150 Hz, and the far end is tied to a post. The vibrator moves its end so little that both ends act as fixed points. The picture shows the string at its two extreme shapes.

A string 1.8 m long runs from a small vibrator on the left, labeled 150 Hz, to a post on the right. It is drawn at its two extreme shapes. They cross at both ends and at two points between, dividing the string into three equal loops. A line below the string marks its length, 1.8 m.

Case question 1 · Calculate

Using the picture: how fast do waves travel along this string?

Your answer m/s

Case question 2 · Multiple choice

Using the picture: what produces the pattern on the string?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

A student describes two upward pulses, 3 cm and 2 cm tall, that meet on a long rope. Tap the first line that goes wrong.

Check yourself · Put it in order

A student wants to measure the speed of sound in air with a 512 Hz tuning fork and a long tube standing in deep water, so that the water closes its bottom end. She can raise the tube to lengthen the air column. Tap her steps in order.

Part 7 Keep it Bring back earlier ideas

The Doppler effect from the last lesson and reflection at a boundary from 14.3 are still part of the course. Retrieving them now, between new ideas, is what makes them easy to find later.

From 14.5 The Doppler Effect · Select all

In which two situations is the observed frequency greater than the rest frequency? Pick two.

Select 2 answers, then check.

From 14.3 Boundary Behavior of Waves and Polarization · Sort it

Each card describes a pulse on a stretched string or rope reaching a boundary. Tap a card, then how the reflected pulse comes back.

Upright

Inverted

In three lines

Overlapping waves pass through each other, and the medium's displacement is the sum of theirs: constructive in the same direction, destructive in opposite ones.
Two close frequencies drift in and out of phase, giving beats at $| f_{1} - f_{2} |$ .
A standing wave has nodes half a wavelength apart; its ends decide the harmonics: every one with alike ends, odd only with a node at one end and an antinode at the other.

Exam strategy for this topic

Section I, Multiple Choice: 42 questions in 85 minutes.
Section II, Free Response: 4 questions in 95 minutes.
This lesson's Explain it is shaped like free-response question 2, Translation Between Representations.
A four-function, scientific, or graphing calculator is allowed on both sections of the exam.

For any standing-wave question, draw the pattern first: the right kind of point at each end, then loops between. Write the length as a number of half or quarter wavelengths from the drawing, and only then use $v = f λ$ .

Questions students ask

What is the difference between constructive and destructive interference?

Where overlapping waves displace the medium in the same direction, their displacements add to a larger one: constructive. In opposite directions they cancel, partly or completely: destructive.

Do two waves bounce off each other when they meet?

No. They pass through each other. While they overlap the displacements add, and afterward each wave carries on unchanged.

How do you find the beat frequency?

Subtract the two frequencies: $| f_{beat} | = | f_{1} - f_{2} |$ . Two notes at 440 Hz and 436 Hz give 4 beats each second.

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

Its closed end is a node and its open end an antinode, so its length must be an odd number of quarter wavelengths: one, three, five. Even harmonics would need the same kind of point at both ends.

How far apart are the nodes of a standing wave?

Neighboring nodes are half a wavelength apart, and each antinode is midway between two nodes, a quarter wavelength from each.

Course alignment, for teachers

AP Physics 2: Algebra-Based topic 14.6, Unit 14: Waves, Sound, and Physical Optics.

Skill 1.B: Create quantitative graphs with appropriate scales and units, including plotting data.
Skill 1.C: Create qualitative sketches of graphs that represent features of a model or the behavior of a physical system.
Skill 2.A: Derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway.
Skill 2.D: Predict new values or factors of change of physical quantities using functional dependence between variables.
Skill 3.A: Create experimental procedures that are appropriate for a given scientific question.
Skill 3.B: Apply an appropriate law, definition, theoretical relationship, or model to make a claim.

This lesson maps all 18 essential knowledge statements of the topic to the part that teaches each one: 18 are checked by a scored question. The map is in the lesson file.