Answer key · Unit 11

Waves and sound.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 5

Lesson 1 of 5 · About 25 minutes

Wavelength, frequency and speed

What a wave carries and what stays in place, and three numbers that describe it: the frequency the source sets, the speed the medium sets, and the wavelength that fits both.

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The 5 steps and their answers

Step 1 of 5
Open this step
What travels
Practice · Choose
Question
A small leaf floats on the water in the tank. What does it do when the rings reach it?

Correct answer
- Correct Rises and falls and rocks a little, staying in about the same place
Setup, choices and feedback
The Scene at this step
Frequency
8 Hz
Wave speed
25 cm/s
Amplitude
60 %
Show the crests
Off
Choices
Shown in a random order.
- Correct Rises and falls and rocks a little, staying in about the same place
- Wrong Is carried outward with the rings, away from the bead
  Misconception: Thinks the medium travels with the wave: the water flows outward with the ripples, the air streams from the speaker to the ear.
  Feedback from the catalogue
  A wave carries energy, not matter. Each bit of water or each air particle swings about its own place and passes the motion on to its neighbour, and once the wave has gone by it is back almost exactly where it was. What travels is the pattern: the crest, or the compression.
- Wrong Does not move at all: the wave passes underneath it
  Feedback
  If the leaf did not move, the wave would deliver nothing to it. The water beneath it rises and falls as the wave passes, and the leaf with it.
- Wrong Is pulled inward, towards the bead
  Feedback
  Nothing pulls the water towards the bead. Each bit of water oscillates about its own place and does not drift anywhere over time.
Feedback for a right answer
The leaf rises with each crest and sinks with each trough, but does not travel with the wave. The wave delivered energy to it, the motion, without delivering any water.

Hint
Watch one spot of water, not the rings.

Worked solution
1. A wave is a disturbance passed from one bit of water to its neighbour.
2. Each bit of water oscillates about its own place: up at a crest, down at a trough.
3. The leaf floats on one bit of water and does what it does: rises, sinks and stays about where it was.
If you chose that the leaf is carried outward, you thought the water flows with the rings. What travels is the pattern, not the water.
Step 2 of 5
Open this step
The wavelength
Practice · Calculate
Question
The Wave speed is 25 cm/s and the Frequency 8 Hz. What is the wavelength?

Correct answer
3.125 cm ± 0.05 cm
Setup, choices and feedback
The Scene at this step
Frequency
8 Hzfixed
Wave speed
25 cm/sfixed
Amplitude
60 %
Show the crests
On
Accepted within
± 0.05 cm
Computed from the Scene at this step’s setting
Crest to crest, λ = v / f, from a source at rest/nodes/tank/outputs/wavelength
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 200 cm ± 1 cm
  Misconception: Multiplies the speed by the frequency to get the wavelength: λ = v·f.
  Feedback from the catalogue
  The rule is $v = f\lambda$ ⁠, so $\lambda = \dfrac{v}{f}$ ⁠: divide the speed by the frequency. A quick check: a higher frequency should give shorter waves, and $v \times f$ ⁠ does the opposite.
- 0.32 cm ± 0.005 cm
  Misconception: Divides the frequency by the speed: λ = f / v, the formula upside down.
  Feedback from the catalogue
  $\lambda = \dfrac{v}{f}$ ⁠, with the speed on top. Check the units: $\dfrac{\mathrm{m/s}}{\mathrm{1/s}} = \mathrm{m}$ ⁠, a length, while $\dfrac{f}{v}$ ⁠ comes out in $\mathrm{1/m}$ ⁠.
Feedback for a right answer
$\lambda = \dfrac{25}{8} \approx 3.1$ ⁠ centimetres. Each second eight crests leave, and the first of them has already moved on 25 cm, so the gap from crest to crest is an eighth of that.

Hint
$v = f\lambda$ ⁠. Solve for $\lambda$ ⁠.

Worked solution
1. Given: v = 25 cm/s, f = 8 Hz.
2. From $v = f\lambda$ ⁠, $\lambda = \dfrac{v}{f}$ ⁠.
3. $\lambda = \dfrac{25}{8} = 3.125$ ⁠ centimetres.
4. The wavelength is about 3.1 cm.
If you got $200$ ⁠, you multiplied the speed by the frequency. A higher frequency should give shorter waves, so you divide by it.
Step 3 of 5
Open this step
The period
Practice · Calculate
Question
At 8 Hz, how much time passes in reality between two crests passing the same spot on the water?

Correct answer
0.125 s ± 0.002 s
Setup, choices and feedback
The Scene at this step
Frequency
8 Hzfixed
Wave speed
25 cm/sfixed
Amplitude
60 %
Show the crests
On
Accepted within
± 0.002 s
Computed from the Scene at this step’s setting
Time between crests, T = 1 / f/nodes/tank/outputs/period
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 8 s ± 0.01 s
  Misconception: Takes the period to be the frequency in seconds: 8 Hz gives T = 8 s.
  Feedback from the catalogue
  The frequency counts cycles per second, and the period is how many seconds one cycle lasts. Each is the other turned over: $T = \dfrac{1}{f}$ ⁠. Eight crests a second means an eighth of a second from crest to crest, not eight seconds.
Feedback for a right answer
$T = \dfrac{1}{8} = 0.125$ ⁠ seconds. In that time each crest moves on one wavelength: $\dfrac{3.125}{0.125} = 25$ ⁠ centimetres a second, the wave speed.

Hint
Eight crests pass each second. How long for each?

Worked solution
1. Given f = 8 Hz: eight cycles a second.
2. $T = \dfrac{1}{f}$ ⁠.
3. $T = \dfrac{1}{8} = 0.125$ ⁠ seconds.
4. In the Scene time is slowed $8$ ⁠ times, so there a crest passes about once a second; in a real tank it is an eighth of a second.
If you answered 8 s, you took the frequency for the period. Each is the other turned over: many cycles a second means a short cycle.
Step 4 of 5
Open this step
Tapping faster
Practice · Predict, then look
Question
The bead is about to tap twice as fast: 16 Hz instead of 8 Hz, in the same water. What will happen?

Correct answer
- Correct The wavelength will halve, and the speed will not change
Setup, choices and feedback
The Scene at this step
Frequency
8 Hz
Wave speed
25 cm/sfixed
Amplitude
60 %
Show the crests
On
Choices
- Correct The wavelength will halve, and the speed will not change
- Wrong The waves will travel twice as fast, and the wavelength will not change
  Misconception: Thinks a higher frequency makes the wave travel faster.
  Feedback from the catalogue
  The medium sets the speed: the depth of the water, or the material the sound passes through. The source sets only the frequency. Tap faster and the crests leave more often and travel at the same speed, so they end up closer together: $\lambda = \dfrac{v}{f}$ ⁠ shrinks, and $v$ ⁠ stays.
- Wrong The wavelength will double
  Misconception: Multiplies the speed by the frequency to get the wavelength: λ = v·f.
  Feedback from the catalogue
  The rule is $v = f\lambda$ ⁠, so $\lambda = \dfrac{v}{f}$ ⁠: divide the speed by the frequency. A quick check: a higher frequency should give shorter waves, and $v \times f$ ⁠ does the opposite.
- Wrong The waves will travel faster, and get shorter too
  Feedback
  If the speed went up as well, the wavelength would not halve. But the water has not changed, so the speed stays as it was.
Once the prediction is locked in, the Scene changes
Frequency
16 Hz
The Scene then shows: 0.01563 m · Crest to crest, λ = v / f, from a source at rest/nodes/tank/outputs/wavelength
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The crests leave twice as often and travel at the same speed, 25 cm/s, so the gap between them halves: $\dfrac{25}{16} \approx 1.6$ ⁠ centimetres, 0.016 m.

Worked solution
1. The water has not changed, so the speed stays 25 cm/s.
2. $\lambda = \dfrac{v}{f} = \dfrac{25}{16} \approx 1.56$ ⁠ centimetres.
3. Before, it was $\dfrac{25}{8} \approx 3.13$ ⁠: the wavelength has halved.
If you expected the waves to speed up, you thought the frequency sets the speed. The medium sets the speed; the source sets only how many crests leave each second.
Step 5 of 5
Open this step
Deeper water
Practice · Reach the target
Question
The Frequency is fixed at 8 Hz. Change the Wave speed, as if adding water to the tank or taking some out, until the wavelength is 4 cm, that is 0.04 m, and press Check.

Goal
- 0.04 m ± 0.0005 m
  At the start of the step: 0.025 m
Setup, choices and feedback
The Scene at this step
Frequency
8 Hzfixed
Wave speed
20 cm/s
Amplitude
60 %
Show the crests
On
Read from the Scene at
Crest to crest, λ = v / f, from a source at rest/nodes/tank/outputs/wavelength
One setting that reaches it
Wave speed
32 cm/s
Amplitude
0 %
Feedback for a right answer
At 32 cm/s each crest moves on 32 cm a second, and in that second eight crests leave: $\dfrac{32}{8} = 4$ ⁠ centimetres apart.

Feedback for a wrong answer
Not yet. A higher speed moves the crests further apart, and a lower one brings them closer.

Hint
$v = f\lambda$ ⁠: what speed gives a 4 cm wavelength at 8 Hz?

Worked solution
1. You want λ = 4 cm at f = 8 Hz.
2. $v = f\lambda = 8 \times 4 = 32$ ⁠ centimetres a second.
3. Set Wave speed to 32 cm/s.
If you lowered the speed, the crests came closer together: at the same frequency, slower water gives shorter waves.

Lesson 2 of 5 · About 25 minutes

Sound

Sound is a longitudinal wave of compressions and rarefactions: its wavelength in air, what loudness changes and what it doesn't, what happens when sound passes into water, and echoes.

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The 5 steps and their answers

Step 1 of 5
Open this step
Along the tube
Practice · Choose
Question
How does each particle of air in the tube move as the sound passes through?

Correct answer
- Correct Back and forth, along the direction the sound travels, about the same point
Setup, choices and feedback
The Scene at this step
Pitch
500 Hz
Loudness
50 %
Travelling through
Air
Show the pressure
Off
Choices
Shown in a random order.
- Correct Back and forth, along the direction the sound travels, about the same point
- Wrong Up and down, across the direction the sound travels
  Misconception: Thinks air particles in a sound wave move up and down, across the direction the sound travels, as in a ripple.
  Feedback from the catalogue
  Sound is a longitudinal wave: each particle swings back and forth along the direction the sound travels, which makes the compressions and rarefactions. In a transverse wave, as on a rope, the motion is across the direction of travel. Air does not pull itself back sideways, so sound in air can only be longitudinal.
- Wrong From the loudspeaker to the far end of the tube, along with the sound
  Misconception: Thinks the medium travels with the wave: the water flows outward with the ripples, the air streams from the speaker to the ear.
  Feedback from the catalogue
  A wave carries energy, not matter. Each bit of water or each air particle swings about its own place and passes the motion on to its neighbour, and once the wave has gone by it is back almost exactly where it was. What travels is the pattern: the crest, or the compression.
- Wrong It does not move: only the pressure passes along
  Feedback
  The pressure changes precisely because the particles move: where they come closer together the pressure rises, and where they move apart it falls.
Feedback for a right answer
Each particle swings back and forth along the tube and returns to the same place. That makes compressions and rarefactions that travel on: a longitudinal wave.

Hint
Follow one particle in the Scene for a few cycles.

Worked solution
1. The loudspeaker’s membrane moves back and forth, along the tube.
2. It pushes the particles next to it the same way, and they push the next ones.
3. Each particle swings along the direction the sound travels and stays about its own place: a longitudinal wave.
If you chose up and down, you pictured a wave like a ripple or a rope, a transverse wave. Air does not pull itself back sideways, so sound in air is always longitudinal.
Step 2 of 5
Open this step
The wavelength in air
Practice · Calculate
Question
The loudspeaker plays a 500 Hz note in air, where the speed of sound is 343 m/s. What is the wavelength?

Correct answer
0.686 m ± 0.005 m
Setup, choices and feedback
The Scene at this step
Pitch
500 Hzfixed
Loudness
50 %
Travelling through
Airfixed
Show the pressure
On
Accepted within
± 0.005 m
Computed from the Scene at this step’s setting
Compression to compression, λ = v / f/nodes/wave/outputs/wavelength
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 171500 m ± 100 m
  Misconception: Multiplies the speed by the frequency to get the wavelength: λ = v·f.
  Feedback from the catalogue
  The rule is $v = f\lambda$ ⁠, so $\lambda = \dfrac{v}{f}$ ⁠: divide the speed by the frequency. A quick check: a higher frequency should give shorter waves, and $v \times f$ ⁠ does the opposite.
- 1.458 m ± 0.005 m
  Misconception: Divides the frequency by the speed: λ = f / v, the formula upside down.
  Feedback from the catalogue
  $\lambda = \dfrac{v}{f}$ ⁠, with the speed on top. Check the units: $\dfrac{\mathrm{m/s}}{\mathrm{1/s}} = \mathrm{m}$ ⁠, a length, while $\dfrac{f}{v}$ ⁠ comes out in $\mathrm{1/m}$ ⁠.
Feedback for a right answer
$\lambda = \dfrac{343}{500} \approx 0.69$ ⁠ metres: about 69 cm from compression to compression.

Hint
$\lambda = \dfrac{v}{f}$ ⁠.

Worked solution
1. Given: v = 343 m/s, f = 500 Hz.
2. $\lambda = \dfrac{v}{f}$ ⁠.
3. $\lambda = \dfrac{343}{500} = 0.686$ ⁠ metres.
4. The wavelength is about 0.69 m.
If you got a huge number like $171\,500$ ⁠, you multiplied instead of dividing. Check the units: $\dfrac{\mathrm{m/s}}{\mathrm{1/s}} = \mathrm{m}$ ⁠.
Step 3 of 5
Open this step
Louder
Practice · Predict, then look
Question
Loudness is turned up from 30% to 100%. What will happen to the speed at which the sound travels along the tube?

Correct answer
- Correct It will not change
Setup, choices and feedback
The Scene at this step
Pitch
500 Hzfixed
Loudness
30 %
Travelling through
Airfixed
Show the pressure
On
Choices
- Correct It will not change
- Wrong It will increase: a loud sound travels faster
  Misconception: Thinks a louder sound (a larger amplitude) travels faster.
  Feedback from the catalogue
  Loudness is the amplitude: how far each particle swings. The speed of sound is set by the medium alone, so a whisper and a shout reach you at the same moment. A louder sound carries more energy; it does not arrive sooner.
- Wrong It will decrease: the particles have further to go
  Feedback
  The particles do swing further, but also faster, so the compression is passed on at the same rate. The medium sets the speed.
Once the prediction is locked in, the Scene changes
Loudness
100 %
The Scene then shows: 343 m/s · Speed of sound in the medium/nodes/wave/outputs/waveSpeed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The particles swing further and the compressions are denser, but the gaps between them, the wavelength, have not moved, and the speed is still 343 m/s. A shout and a whisper reach you at the same time.

Worked solution
1. The speed of sound is set by the medium: how stiff and how dense it is. The air has not changed.
2. Loudness is the amplitude, how far each particle swings. It does not appear in $v = f\lambda$ ⁠.
3. So the speed stays 343 m/s, and the wavelength stays $\dfrac{343}{500} \approx 0.69$ ⁠ metres.
If you expected the sound to speed up, think of thunder: the loud crack and the quiet rumble after it travel at the same speed, so they arrive from the same direction and in the same order.
Step 4 of 5
Open this step
From air into water
Practice · Predict, then look
Question
The 500 Hz note passes from air into water, where the speed of sound is 1480 m/s. What will happen?

Correct answer
- Correct The wavelength will grow, and the frequency will not change
Setup, choices and feedback
The Scene at this step
Pitch
500 Hzfixed
Loudness
50 %
Travelling through
Air
Show the pressure
On
Choices
- Correct The wavelength will grow, and the frequency will not change
- Wrong The frequency will rise, and the wavelength will not change
  Misconception: Thinks a sound's frequency changes when it passes into another medium.
  Feedback from the catalogue
  The source sets the frequency: every compression that reaches the boundary between the two media carries on across it, one by one, so the same number of compressions pass each second on both sides. What changes is the speed, and with it the wavelength: $\lambda = \dfrac{v}{f}$ ⁠.
- Wrong The wavelength will shrink, because sound is slower in water
  Misconception: Thinks sound travels slower in water or steel than in air, because they are denser and harder to push through.
  Feedback from the catalogue
  Usually the opposite: sound travels about four times faster in water than in air, and about seventeen times faster in steel. What decides it is how stiff the material is, how hard it pushes back when squeezed. Water and steel are far stiffer than air, and that outweighs their density.
- Wrong Nothing will change
  Feedback
  Something has to change: the speed in water is different, and $v = f\lambda$ ⁠. If $f$ ⁠ stays, $\lambda$ ⁠ changes.
Once the prediction is locked in, the Scene changes
Travelling through
Water
The Scene then shows: 2.96 m · Compression to compression, λ = v / f/nodes/wave/outputs/wavelength
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The loudspeaker still sends 500 compressions a second, but in water each one moves on faster, so they are further apart: $\dfrac{1480}{500} \approx 3$ ⁠ metres instead of $0.69$ ⁠.

Worked solution
1. The source sets the frequency: 500 Hz in water too.
2. The speed in water is 1480 m/s, about $4.3$ ⁠ times that in air.
3. $\lambda = \dfrac{1480}{500} = 2.96$ ⁠ metres, against $0.686$ ⁠ in air.
4. The wavelength grows $4.3$ ⁠ times, as the speed does.
If you expected sound to slow down in water, that is understandable: water is denser. But it is also far stiffer, pushing back hard when squeezed, and that is what decides it.
Step 5 of 5
Open this step
Echo
Practice · Calculate
Question
You stand facing a cliff, clap, and hear the echo 2 s later. The speed of sound in air is 343 m/s. How far away is the cliff?

Correct answer
343 m ± 3 m
Setup, choices and feedback
The Scene at this step
Pitch
500 Hz
Loudness
50 %
Travelling through
Air
Show the pressure
Off
Accepted within
± 3 m
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 686 m ± 3 m
  Misconception: Forgets that an echo goes to the wall and back, so takes the whole distance travelled as the distance to the wall.
  Feedback from the catalogue
  An echo covers the distance twice: to the wall and back. So $v \times t$ ⁠ is twice the distance to the wall, and you halve it.
Feedback for a right answer
The sound travelled 343 × 2 = 686 m, but that is the way to the cliff and back. The cliff is half that away, 343 m.

Hint
In those two seconds the sound went to the cliff and came back.

Worked solution
1. Given: v = 343 m/s, t = 2 s.
2. The distance the sound covered: $343 \times 2 = 686$ ⁠ metres.
3. That is there and back, so the distance to the cliff is $\dfrac{686}{2} = 343$ ⁠ metres.
If you answered 686 m, you forgot that the echo covers the distance twice.

Lesson 3 of 5 · About 25 minutes

Interference

Two waves that meet add up: constructive and destructive interference, the path difference that decides where the water is calm, and how many calm lines fit between two sources.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Two waves meet
Practice · Choose
Question
A crest from one bead and an equal trough from the other arrive together at the same spot on the water. What happens?

Correct answer
- Correct At that moment the water there is flat, and afterwards each wave carries on as before
Setup, choices and feedback
The Scene at this step
Source separation
8 cm
Frequency
10 Hz
Show the calm lines
Off
Choices
Shown in a random order.
- Correct At that moment the water there is flat, and afterwards each wave carries on as before
- Wrong The two waves cancel each other and vanish, and no wave goes on past that spot
  Misconception: Thinks two waves that meet bounce off each other, or cancel each other for good.
  Feedback from the catalogue
  Waves pass through each other. Where they meet, their displacements add (the principle of superposition): an equal crest and trough give flat water for a moment. After they meet, each wave goes on as if the other had not been there, with the same shape and in the same direction.
- Wrong The waves collide, and each goes back the way it came
  Misconception: Thinks two waves that meet bounce off each other, or cancel each other for good.
  Feedback from the catalogue
  Waves pass through each other. Where they meet, their displacements add (the principle of superposition): an equal crest and trough give flat water for a moment. After they meet, each wave goes on as if the other had not been there, with the same shape and in the same direction.
- Wrong A crest twice as tall forms there
  Feedback
  A double crest forms where crest meets crest. A trough lowers the water by exactly as much as an equal crest raises it, and the two add up to zero.
Feedback for a right answer
The heights add: $(+1) + (-1) = 0$ ⁠, so for a moment the water is flat. But the waves pass through each other, and each goes on unchanged. This is destructive interference.

Hint
By the principle of superposition, the heights add. What is $(+1) + (-1)$ ⁠?

Worked solution
1. The principle of superposition: where two waves meet, the height is the sum of the heights.
2. A crest of height $+1$ ⁠ and a trough of depth $-1$ ⁠: $1 + (-1) = 0$ ⁠.
3. The water at that spot is flat, but only at that spot and at that moment. Each wave goes on as if the other had not been there.
If you chose that the waves vanish or turn back, you thought of them as balls that collide. Waves pass through each other.
Step 2 of 5
Open this step
The path difference
Practice · Choose
Question
The wavelength in the tank is 2.5 cm. At which point does the water hardly move?

Correct answer
- Correct 20 cm from one bead and 23.75 cm from the other
Setup, choices and feedback
The Scene at this step
Source separation
8 cmfixed
Frequency
10 Hzfixed
Show the calm lines
On
Choices
Shown in a random order.
- Correct 20 cm from one bead and 23.75 cm from the other
- Wrong 20 cm from one bead and 25 cm from the other
  Misconception: Swaps the conditions: thinks the water is calm where the path difference is a whole number of wavelengths.
  Feedback from the catalogue
  When the path difference is a whole number of wavelengths, the two waves arrive in step, crest with crest, and build each other up: constructive interference. The water is calm where the difference is half a wavelength, one and a half wavelengths and so on, so that a crest meets a trough.
- Wrong 15 cm from both beads
  Feedback
  A point equally far from both beads is on the centre line. The crests from both arrive there together, so the water there swings the most.
- Wrong 18 cm from one bead and 20 cm from the other
  Feedback
  The difference is $2$ ⁠ centimetres, $0.8$ ⁠ of a wavelength: neither crest on crest nor crest on trough. The water there moves, less than on the centre line.
Feedback for a right answer
The path difference is 23.75 − 20 = 3.75 cm, that is $\dfrac{3.75}{2.5} = 1.5$ ⁠ wavelengths. A crest from one bead always arrives with a trough from the other: a point on a calm line.

Hint
Work out the path difference at each point, and divide it by the wavelength.

Worked solution
1. For each point, work out $\Delta$ ⁠ and divide by λ = 2.5 cm.
2. $23.75 - 20 = 3.75$ ⁠, that is $1.5\lambda$ ⁠: a half wavelength plus a whole one. Calm.
3. $25 - 20 = 5$ ⁠, that is $2\lambda$ ⁠: crest with crest. Strong.
4. $15 - 15 = 0$ ⁠: the centre line, strong. $20 - 18 = 2$ ⁠, that is $0.8\lambda$ ⁠: in between.
If you chose the difference of $2\lambda$ ⁠, you swapped the conditions: a whole number of wavelengths gives constructive interference, and halves give destructive.
Step 3 of 5
Open this step
How many calm lines
Practice · Calculate
Question
The Source separation is 8 cm and the wavelength 2.5 cm. How many calm lines are there in all?

Correct answer
6 ± 0.1
Setup, choices and feedback
The Scene at this step
Source separation
8 cmfixed
Frequency
10 Hzfixed
Show the calm lines
On
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Calm (nodal) lines between two point sources with no barrier: path differences where the waves arrive half a cycle apart, inside ±separation; 0 otherwise/nodes/tank/outputs/nodalLines
Wrong answers it recognises
- 3 ± 0.1
  Misconception: Counts the calm lines on one side of the centre line only.
  Feedback from the catalogue
  Each path difference comes with two signs: a point nearer the right bead, and a point just as much nearer the left one. So the calm lines come in pairs, one on each side of the centre line.
Feedback for a right answer
Three path differences fit, 1.25, 3.75 and 6.25 cm, and each appears twice, once on each side of the centre line: six lines.

Hint
A calm line is $\Delta = \tfrac12\lambda, \tfrac32\lambda, \dots$ ⁠, and $\Delta$ ⁠ must be smaller than the source separation. Don't forget both sides.

Worked solution
1. Calm lines: $\Delta = 0.5\lambda, 1.5\lambda, 2.5\lambda, \dots$ ⁠, that is $1.25, 3.75, 6.25, 8.75, \dots$ ⁠ centimetres.
2. $\Delta$ ⁠ must be smaller than the source separation, 8 cm: that leaves $1.25$ ⁠, $3.75$ ⁠ and $6.25$ ⁠.
3. Each difference appears twice, nearer the right bead or nearer the left one.
4. $3 \times 2 = 6$ ⁠ calm lines.
If you answered $3$ ⁠, you counted only one side of the centre line.
Step 4 of 5
Open this step
Further apart
Practice · Predict, then look
Question
The beads are about to move apart, from 4 cm to 12 cm, at the same frequency. What will happen to the calm lines?

Correct answer
- Correct There will be more lines, closer together
Setup, choices and feedback
The Scene at this step
Source separation
4 cm
Frequency
10 Hzfixed
Show the calm lines
On
Choices
- Correct There will be more lines, closer together
- Wrong There will be fewer lines, wider apart
  Feedback
  That is what happens when the beads come closer. The further apart they are, the larger the path difference can get, and the more half-wavelengths fit in it.
- Wrong The same lines, just somewhere else
  Feedback
  The number of lines depends on how many half-wavelengths fit in the source separation, so it changes when the separation changes.
Once the prediction is locked in, the Scene changes
Source separation
12 cm
The Scene then shows: 10 · Calm (nodal) lines between two point sources with no barrier: path differences where the waves arrive half a cycle apart, inside ±separation; 0 otherwise/nodes/tank/outputs/nodalLines
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At $4$ ⁠ centimetres only the differences $1.25$ ⁠ and $3.75$ ⁠ fit: four lines. At $12$ ⁠ centimetres five differences fit, up to $11.25$ ⁠: ten lines, closer together.

Worked solution
1. The wavelength is 2.5 cm; the calm lines are at $\Delta = 1.25, 3.75, 6.25, 8.75, 11.25, \dots$ ⁠ centimetres.
2. At 4 cm: $1.25$ ⁠ and $3.75$ ⁠ are smaller than $4$ ⁠. Two differences, four lines.
3. At 12 cm: five differences, up to $11.25$ ⁠. Ten lines.
4. More lines in the same tank, so they are closer together.
Step 5 of 5
Open this step
Only two lines
Practice · Reach the target
Question
The Frequency is fixed at 10 Hz. Change the Source separation until only two calm lines are left, one on each side, and press Check.

Goal
- 2 ± 0.1
  At the start of the step: 6
Setup, choices and feedback
The Scene at this step
Source separation
8 cm
Frequency
10 Hzfixed
Show the calm lines
On
Read from the Scene at
Calm (nodal) lines between two point sources with no barrier: path differences where the waves arrive half a cycle apart, inside ±separation; 0 otherwise/nodes/tank/outputs/nodalLines
One setting that reaches it
Source separation
2 cm
Feedback for a right answer
Only one path difference, half a wavelength, is smaller than the source separation, so one pair of lines is left. Around the centre line a wide beam of rough water spreads out.

Feedback for a wrong answer
Not yet. Too many lines? Bring the beads closer. No lines at all? Move them a little apart.

Hint
Only Δ = 1.25 cm should fit inside the source separation, and 3.75 cm should not.

Worked solution
1. The wavelength is 2.5 cm: the calm lines are at $\Delta = 1.25, 3.75, \dots$ ⁠ centimetres.
2. For one pair of lines you need $1.25 < d$ ⁠ and $3.75 \ge d$ ⁠.
3. Any separation above 1.25 cm and up to 3.75 cm works; in the Scene, 3 cm for example.
If you moved the beads apart, the lines only multiplied: a larger separation allows more path differences.

Lesson 4 of 5 · About 25 minutes

Standing waves

A wave reflected back and forth between two ends settles into a pattern that does not travel: nodes and antinodes, the wavelength of each harmonic, the frequencies of the harmonics, and how a string's length sets its pitch.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Nodes and antinodes
Practice · Calculate
Question
The string vibrates in its third harmonic. How many nodes are on it, counting both ends?

Correct answer
4 ± 0.1
Setup, choices and feedback
The Scene at this step
Length
65.5 cm
Harmonic
3 nfixed
Break down the sound
Nodes and antinodes
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Nodes on the string, its two ends included: n + 1/nodes/music/outputs/nodes
Wrong answers it recognises
- 3 ± 0.1
  Misconception: Thinks the n-th harmonic has n nodes on the string, forgetting that both ends are nodes.
  Feedback from the catalogue
  In the $n$ ⁠-th harmonic, $n$ ⁠ half-waves fit on the string, and $n$ ⁠ pieces need $n + 1$ ⁠ boundary points: the two fixed ends and $n - 1$ ⁠ nodes between them. So there are $n + 1$ ⁠ nodes and $n$ ⁠ antinodes.
Feedback for a right answer
Four: the two bridges and two more nodes between them, which split the string into three half-waves. And in the middle of each half there is an antinode, three in all.

Hint
Count the orange dots on the string, or work it out: how many boundary points do three pieces need?

Worked solution
1. In the $n$ ⁠-th harmonic, $n$ ⁠ half-waves fit on the string; here $n = 3$ ⁠.
2. Three pieces need four boundary points: $n + 1$ ⁠.
3. Two of them are the ends of the string, and two lie between. There are $4$ ⁠ nodes.
If you answered $3$ ⁠, you counted the half-waves, or the antinodes, not the nodes. The fixed ends are nodes too.
Step 2 of 5
Open this step
The wavelength
Practice · Calculate
Question
The string is 65.5 cm long and vibrates in its third harmonic. What is the wavelength?

Correct answer
43.67 cm ± 0.1 cm
Setup, choices and feedback
The Scene at this step
Length
65.5 cmfixed
Harmonic
3 nfixed
Break down the sound
Nodes and antinodes
Accepted within
± 0.1 cm
Computed from the Scene at this step’s setting
The sounding harmonic's wavelength, λ = 2L / n/nodes/music/outputs/wavelength
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 21.83 cm ± 0.1 cm
  Misconception: Takes the distance between two neighbouring nodes to be a whole wavelength, so λ = L / n.
  Feedback from the catalogue
  Between two neighbouring nodes there is half a wavelength: the string bulges up in one half and down in the next. A whole wave is one loop up and one loop down. So $\dfrac{\lambda}{2} = \dfrac{L}{n}$ ⁠, that is $\lambda = \dfrac{2L}{n}$ ⁠.
Feedback for a right answer
$\lambda = \dfrac{2 \times 65.5}{3} \approx 43.7$ ⁠ centimetres. The board shows half of that, 21.8 cm, the distance from node to node.

Hint
Between two neighbouring nodes there is half a wavelength. $\lambda = \dfrac{2L}{n}$ ⁠.

Worked solution
1. Given: L = 65.5 cm, $n = 3$ ⁠.
2. Three half-waves fit the length: $3 \cdot \dfrac{\lambda}{2} = L$ ⁠.
3. $\lambda = \dfrac{2L}{3} = \dfrac{131}{3} \approx 43.7$ ⁠ centimetres.
If you answered 21.8 cm, you took the distance between two nodes to be a whole wavelength. It is only half: one loop up, without the loop down that completes it.
Step 3 of 5
Open this step
The harmonics
Practice · Calculate
Question
The tuner shows the string’s fundamental frequency. What will the frequency of the third harmonic be, at the same length?

Correct answer
785.5 Hz ± 2 Hz
Setup, choices and feedback
The Scene at this step
Length
65.5 cmfixed
Harmonic
1 nfixed
Break down the sound
Vibration
Accepted within
± 2 Hz
Computed from the Scene at this step’s setting
The sounding harmonic, fₙ = n·v / 2L (the tuner's frequency)/nodes/music/outputs/frequency
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 87.28 Hz ± 1 Hz
  Misconception: Divides the fundamental by n for the n-th harmonic, because its wavelength is n times shorter.
  Feedback from the catalogue
  In the $n$ ⁠-th harmonic the wavelength is $n$ ⁠ times shorter, and the speed on the string does not change. By $f = \dfrac{v}{\lambda}$ ⁠, a wave $n$ ⁠ times shorter means a frequency $n$ ⁠ times higher: $f_n = n f_1$ ⁠.
Feedback for a right answer
$3 \times 262 \approx 785$ ⁠ hertz: the note G, an octave and a fifth above the C of the fundamental.

Hint
$f_n = n f_1$ ⁠: the wavelength is $n$ ⁠ times shorter, and the speed does not change.

Worked solution
1. Given: f₁ ≈ 262 Hz, and the speed of waves on the string does not change.
2. In the third harmonic the wavelength is $3$ ⁠ times shorter, so by $f = \dfrac{v}{\lambda}$ ⁠ the frequency is $3$ ⁠ times higher.
3. $f_3 = 3 \times 261.8 \approx 785$ ⁠ hertz.
If you got about 87 Hz, you divided by $3$ ⁠. A shorter wave at the same speed means a higher frequency.
Step 4 of 5
Open this step
A shorter string
Practice · Predict, then look
Question
The string will be shortened from 80 cm to 40 cm, in its fundamental. What will happen to the frequency?

Correct answer
- Correct It will double: an octave higher
Setup, choices and feedback
The Scene at this step
Length
80 cm
Harmonic
1 nfixed
Break down the sound
Vibration
Choices
- Correct It will double: an octave higher
- Wrong It will halve: an octave lower
  Misconception: Thinks a shorter string sounds lower.
  Feedback from the catalogue
  A shorter string sounds higher. In the fundamental half a wave fits on it, so the wavelength is $2L$ ⁠. Half the string, half the wavelength, and at the same speed twice the frequency: $f_1 = \dfrac{v}{2L}$ ⁠. That is why a guitarist presses a string against a fret, shortening it, to play a higher note.
- Wrong It will grow four times
  Feedback
  The fundamental frequency is $\dfrac{v}{2L}$ ⁠: it depends on $L$ ⁠ itself, not on its square. Half the length gives twice the frequency.
- Wrong It will not change
  Feedback
  A shorter length fits a shorter half-wave, so the wavelength changes, and the frequency with it.
Once the prediction is locked in, the Scene changes
Length
40 cm
The Scene then shows: 428.8 Hz · The sounding harmonic, fₙ = n·v / 2L (the tuner's frequency)/nodes/music/outputs/frequency
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$f_1 = \dfrac{v}{2L}$ ⁠, so half the length gives twice the frequency: from about 214 Hz to about 429 Hz, an octave up.

Worked solution
1. In the fundamental half a wave fits on the string: $\lambda = 2L$ ⁠.
2. At 80 cm: $\lambda = 1.6$ ⁠ metres, and $f_1 = \dfrac{343}{1.6} \approx 214$ ⁠ hertz.
3. At 40 cm: $\lambda = 0.8$ ⁠ metres, and $f_1 = \dfrac{343}{0.8} \approx 429$ ⁠ hertz.
4. Twice the frequency: an octave.
If you expected the note to drop, think of a guitar: the player shortens the string to play higher.
Step 5 of 5
Open this step
Tuning to A
Practice · Reach the target
Question
The Harmonic is fixed at $2$ ⁠. Change Length until the tuner shows 440 Hz, and press Check.

Goal
- 440 Hz ± 2 Hz
  At the start of the step: 523.7 Hz
Setup, choices and feedback
The Scene at this step
Length
65.5 cm
Harmonic
2 nfixed
Break down the sound
Vibration
Read from the Scene at
The sounding harmonic, fₙ = n·v / 2L (the tuner's frequency)/nodes/music/outputs/frequency
One setting that reaches it
Length
78 cm
Feedback for a right answer
At 78 cm two half-waves fit on the string, and the wavelength is $78$ ⁠ centimetres: $\dfrac{343}{0.78} \approx 440$ ⁠ hertz.

Feedback for a wrong answer
Not yet. Frequency too high? Lengthen the string. Too low? Shorten it.

Hint
$f_2 = \dfrac{2v}{2L} = \dfrac{v}{L}$ ⁠. What $L$ ⁠ gives $440$ ⁠?

Worked solution
1. In the second harmonic $\lambda = \dfrac{2L}{2} = L$ ⁠, so $f_2 = \dfrac{v}{L}$ ⁠.
2. $L = \dfrac{v}{f} = \dfrac{343}{440} \approx 0.78$ ⁠ metres.
3. Set Length to 78 cm.
If you chose 39 cm, that is the length that gives 440 Hz in the fundamental. In the second harmonic it gives the octave above, about 880 Hz.

Lesson 5 of 5 · About 25 minutes

The Doppler effect

A moving source squeezes the waves in front of it and stretches them behind: the wavelength and frequency ahead of the source and behind it, why a passing siren drops in pitch, and what happens at the wave speed.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
A moving source
Practice · Predict, then look
Question
The bead is about to start moving at half the wave speed, still tapping 6 times a second. What will happen to the crests?

Correct answer
- Correct They will crowd together in front of the bead and spread apart behind it
Setup, choices and feedback
The Scene at this step
Source speed
0 × wave speed
Frequency
6 Hzfixed
Show the crests
On
Choices
- Correct They will crowd together in front of the bead and spread apart behind it
- Wrong They will spread apart in front of the bead and crowd together behind it
  Misconception: Swaps ahead and behind: thinks the waves stretch in front of a moving source and bunch up behind it.
  Feedback from the catalogue
  The source chases the waves it sent forward, so in front of it each new crest starts closer to the one before: there the crests are crowded and the frequency is higher. Behind it the source moves away from its waves, and they stretch out: a lower frequency.
- Wrong The rings will stay evenly spaced, and move along with the bead
  Feedback
  Each ring spreads from the spot where the bead was when it made it, and is not dragged along. The bead moves on, so each new ring starts somewhere else.
Once the prediction is locked in, the Scene changes
Source speed
0.5 × wave speed
The Scene then shows: 0.02083 m · Crest spacing straight ahead of a moving source, λ(1 − u/v); 0 at and above the wave speed/nodes/tank/outputs/wavelengthAhead
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
In front of the bead the gap between crests halves, about 2.1 cm instead of 4.2 cm. Behind it, it grows one and a half times, to about 6.3 cm.

Worked solution
1. With the bead still, $\lambda = \dfrac{25}{6} \approx 4.17$ ⁠ centimetres in every direction.
2. In each cycle, $\dfrac{1}{6}$ ⁠ of a second, the crest before moves on $4.17$ ⁠ centimetres, and the bead moves half that the same way.
3. In front of the bead: $4.17 - 2.08 \approx 2.08$ ⁠ centimetres. Behind it: $4.17 + 2.08 = 6.25$ ⁠ centimetres.
If you chose the opposite, think about who is chasing whom: the bead chases the crests it sent forward, so they crowd together there.
Step 2 of 5
Open this step
What you hear ahead of the source
Practice · Calculate
Question
The bead taps at 6 Hz and moves at half the wave speed. At what frequency do the crests reach a spot that stands in front of it?

Correct answer
12 Hz ± 0.1 Hz
Setup, choices and feedback
The Scene at this step
Source speed
0.5 × wave speedfixed
Frequency
6 Hzfixed
Show the crests
On
Accepted within
± 0.1 Hz
Computed from the Scene at this step’s setting
Frequency a still listener straight ahead of the source receives, f / (1 − u/v); NaN at and above the wave speed, when no wave gets ahead/nodes/tank/outputs/frequencyAhead
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 9 Hz ± 0.1 Hz
  Misconception: Multiplies by (1 + u/v) for an approaching source instead of dividing by (1 − u/v).
  Feedback from the catalogue
  The crests in front of a moving source are closer together: in each cycle the source moves on by $uT$ ⁠, so the wavelength ahead of it is $\lambda(1 - \tfrac{u}{v})$ ⁠. The wave travels at the same speed $v$ ⁠, so the frequency is $f' = \dfrac{v}{\lambda(1 - u/v)} = \dfrac{f}{1 - u/v}$ ⁠. For a source much slower than its waves this is close to $f(1 + \tfrac{u}{v})$ ⁠, but at half the wave speed the difference is large: twice, not one and a half times.
- 4 Hz ± 0.1 Hz
  Misconception: Swaps ahead and behind: thinks the waves stretch in front of a moving source and bunch up behind it.
  Feedback from the catalogue
  The source chases the waves it sent forward, so in front of it each new crest starts closer to the one before: there the crests are crowded and the frequency is higher. Behind it the source moves away from its waves, and they stretch out: a lower frequency.
Feedback for a right answer
$f' = \dfrac{6}{1 - 0.5} = 12$ ⁠ hertz: twice as much, an octave higher if this were sound. Behind the bead only $\dfrac{6}{1.5} = 4$ ⁠ hertz arrive.

Hint
$f' = \dfrac{f}{1 - u/v}$ ⁠, and $\dfrac{u}{v} = 0.5$ ⁠.

Worked solution
1. Given: f = 6 Hz, $\dfrac{u}{v} = 0.5$ ⁠.
2. In front of the source the wavelength is $\lambda(1 - 0.5)$ ⁠, half, and the wave travels at the same speed.
3. $f' = \dfrac{f}{1 - u/v} = \dfrac{6}{0.5} = 12$ ⁠ hertz.
If you got $9$ ⁠, you multiplied by $1 + 0.5$ ⁠. When the wavelength halves, the frequency doubles, not one and a half times.
Step 3 of 5
Open this step
Ten hertz ahead
Practice · Reach the target
Question
The Frequency is fixed at 6 Hz. Change Source speed until a spot in front of the bead receives the crests at 10 Hz, and press Check.

Goal
- 10 Hz ± 0.1 Hz
  At the start of the step: 6 Hz
Setup, choices and feedback
The Scene at this step
Source speed
0 × wave speed
Frequency
6 Hzfixed
Show the crests
On
Read from the Scene at
Frequency a still listener straight ahead of the source receives, f / (1 − u/v); NaN at and above the wave speed, when no wave gets ahead/nodes/tank/outputs/frequencyAhead
One setting that reaches it
Source speed
0.4 × wave speed
Feedback for a right answer
At $0.4$ ⁠ of the wave speed the wavelength in front of the bead is $0.6$ ⁠ of what it was, so the frequency there is $\dfrac{1}{0.6}$ ⁠ times higher: $\dfrac{6}{0.6} = 10$ ⁠ hertz.

Feedback for a wrong answer
Not yet. A faster source squeezes the crests in front of it more, and raises the frequency there.

Hint
$10 = \dfrac{6}{1 - u/v}$ ⁠. What must $1 - \dfrac{u}{v}$ ⁠ be?

Worked solution
1. You want $f' = 10$ ⁠ hertz with $f = 6$ ⁠ hertz.
2. $1 - \dfrac{u}{v} = \dfrac{f}{f'} = \dfrac{6}{10} = 0.6$ ⁠.
3. $\dfrac{u}{v} = 0.4$ ⁠: set Source speed to $0.4$ ⁠.
If you tried about $0.65$ ⁠, you assumed $f' = f(1 + \tfrac{u}{v})$ ⁠. There the frequency in front of the bead is already about 17 Hz.
Step 4 of 5
Open this step
A passing siren
Practice · Choose
Question
An ambulance with its siren on drives at a steady speed along the road, passes you and drives away. What do you hear?

Correct answer
- Correct A steady high note while it approaches, a drop as it passes, and a steady low note as it drives away
Setup, choices and feedback
The Scene at this step
Source speed
0.5 × wave speed
Frequency
6 Hz
Show the crests
On
Choices
Shown in a random order.
- Correct A steady high note while it approaches, a drop as it passes, and a steady low note as it drives away
- Wrong A note that rises and rises as the ambulance comes closer, and falls as it drives away
  Misconception: Thinks the pitch keeps rising as a source approaches at a steady speed, and falls steadily as it leaves.
  Feedback from the catalogue
  The frequency you hear depends on how fast the source is coming towards you, not on how far away it is. A source approaching at a steady speed sounds a steady high note; it only gets louder. The note drops as the source passes you, then stays low and steady while it moves away.
- Wrong The siren itself plays a higher note when it goes fast, and everyone by the road hears that note
  Misconception: Thinks the source itself sounds a different note when it moves (the siren works harder at speed).
  Feedback from the catalogue
  The siren plays the same note all the time, and the driver, moving with it, hears no change. The change happens on the way: the source’s motion squeezes the waves in front of it and stretches them behind, so a listener standing still receives a different frequency from the one the source sends.
- Wrong The same note all the time; it is only louder when the ambulance is close
  Feedback
  The loudness does change, but so does the pitch: the crests in front of the ambulance are closer together, so they reach the ear at a higher frequency.
Feedback for a right answer
While the ambulance approaches at a steady speed, the crests in front of it are equally crowded all the time, so the note is high and steady. As it passes you go from in front of it to behind it, and the note drops to a steady low value.

Hint
The frequency in front of the source is $\dfrac{f}{1 - u/v}$ ⁠. Does the distance appear in it?

Worked solution
1. In front of the source: $f' = \dfrac{f}{1 - u/v}$ ⁠. Behind it: $f' = \dfrac{f}{1 + u/v}$ ⁠.
2. Both formulas hold only the source’s speed, not its distance. At a steady speed, the frequency is steady.
3. So: a steady high note, a drop at the moment it passes, a steady low note.
If you expected the note to keep rising, you mixed up pitch with loudness: the note gets louder as the ambulance comes closer, not higher.
Step 5 of 5
Open this step
A real siren
Practice · Calculate
Question
A siren plays 700 Hz, and the ambulance drives straight towards you at 30 m/s. The speed of sound is 343 m/s. What frequency do you hear?

Correct answer
767.1 Hz ± 1.5 Hz
Setup, choices and feedback
The Scene at this step
Source speed
0.5 × wave speed
Frequency
6 Hz
Show the crests
On
Accepted within
± 1.5 Hz
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 761.2 Hz ± 1 Hz
  Misconception: Multiplies by (1 + u/v) for an approaching source instead of dividing by (1 − u/v).
  Feedback from the catalogue
  The crests in front of a moving source are closer together: in each cycle the source moves on by $uT$ ⁠, so the wavelength ahead of it is $\lambda(1 - \tfrac{u}{v})$ ⁠. The wave travels at the same speed $v$ ⁠, so the frequency is $f' = \dfrac{v}{\lambda(1 - u/v)} = \dfrac{f}{1 - u/v}$ ⁠. For a source much slower than its waves this is close to $f(1 + \tfrac{u}{v})$ ⁠, but at half the wave speed the difference is large: twice, not one and a half times.
- 643.7 Hz ± 1 Hz
  Misconception: Swaps ahead and behind: thinks the waves stretch in front of a moving source and bunch up behind it.
  Feedback from the catalogue
  The source chases the waves it sent forward, so in front of it each new crest starts closer to the one before: there the crests are crowded and the frequency is higher. Behind it the source moves away from its waves, and they stretch out: a lower frequency.
Feedback for a right answer
$f' = \dfrac{700}{1 - 30/343} \approx 767$ ⁠ hertz. Once the ambulance has passed, you will hear $\dfrac{700}{1 + 30/343} \approx 644$ ⁠ hertz: a drop of more than 120 Hz.

Hint
$f' = \dfrac{f}{1 - u/v}$ ⁠, with $u = 30$ ⁠ and $v = 343$ ⁠.

Worked solution
1. Given: f = 700 Hz, u = 30 m/s, v = 343 m/s.
2. $\dfrac{u}{v} = \dfrac{30}{343} \approx 0.0875$ ⁠.
3. $f' = \dfrac{700}{1 - 0.0875} = \dfrac{700}{0.9125} \approx 767$ ⁠ hertz.
If you got about 761 Hz, you multiplied by $1 + \dfrac{u}{v}$ ⁠ instead of dividing by $1 - \dfrac{u}{v}$ ⁠. At a low speed the difference is small, but it is there.

Across the unit · 21 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Wavelength, frequency and speed

What travels

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

The wavelength

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The period

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Tapping faster

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Deeper water

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Sound

Along the tube

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

The wavelength in air

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Louder

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

From air into water

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Echo

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Interference

Two waves meet

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution