The wave model describes how a local disturbance carries energy through a medium without moving the medium along with it. You need the difference between transverse and longitudinal travelling waves, and wavelength, frequency, period and wave speed linked by v = fλ = λ/T.
Questions often ask you to read displacement–position and displacement–time graphs, to locate compressions and rarefactions in a sound wave, and to compare mechanical and electromagnetic waves, including the orders of magnitude of the electromagnetic spectrum from the data booklet.
18 questions
110 marks
Paper 1A: 9
Paper 1B: 2
Paper 2: 7
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18 practice questions on C.2 Wave model
1C-1A-09
The wave equation·C.2 Wave model
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksCalculate
A sound wave has a frequency of 250 Hz and a wavelength of 1.36 m.
What is its speed?
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Mark
Notes
Step 1v = fλ = 250 × 1.36 = 340 m s⁻¹.
✓ 1
Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThis is λ/f, which is not a speed. Speed is wavelength per period, so the frequency multiplies.
BThis divides the frequency by the wavelength. The wave equation is a product.
CThis adds the two quantities. They have different units and cannot be added.
DCorrect: v = fλ = 250 × 1.36 = 340 m s⁻¹.
Syllabus understandingC.2 — the nature of sound waves, and the wave equation as given by v = fλ = λ/TCommand term: Calculate
2C-1A-10
Reading wave graphs·C.2 Wave model
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
The graph shows the displacement against distance along a transverse wave at one instant. Each point on the wave takes 0.020 s to complete one oscillation.
What is the speed of the wave?
Graph drawn to scaleShow mark scheme
Marking point
Mark
Notes
Step 1From the graph, one complete cycle occupies 0.40 m, so λ = 0.40 m.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2v = λ/T = 0.40/0.020 = 20 m s−1.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis multiplies the wavelength by the period, 0.40 × 0.020. Speed is wavelength divided by period.
BThis reads the wavelength as 0.20 m, the distance between adjacent points of zero displacement, which is only half a wavelength.
CCorrect: λ = 0.40 m and T = 0.020 s, so v = λ/T = 20 m s−1.
DThis is the frequency, 1/0.020 = 50 Hz, quoted as a speed; it has not been multiplied by the wavelength.
Syllabus understandingC.2 — wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/TCommand term: Determine
3C-1A-11
Sound waves·C.2 Wave model
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksState
A loudspeaker produces a sound wave in air.
Which statement about the sound wave is correct?
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Notes
Step 1Sound in air is a longitudinal wave: the molecules oscillate about fixed mean positions along the direction of travel, forming compressions and rarefactions; energy, not air, is carried.
✓ 1
Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AA wave transfers energy, not matter: each molecule only oscillates about its mean position.
BCorrect: sound in air is longitudinal, with the molecules oscillating parallel to the direction of energy transfer.
CThis describes a transverse wave. Sound in air is longitudinal.
DA compression and the nearest rarefaction are half a wavelength apart; one wavelength separates two adjacent compressions.
Syllabus understandingC.2 — the nature of sound waves Command term: State
4C-1A-12
Waves in a new medium·C.2 Wave model
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
Light of frequency f and wavelength λ in a vacuum enters a glass block of refractive index 1.5.
Which row gives the frequency and the wavelength of the light in the glass?
FrequencyWavelength
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Notes
Step 1The frequency is fixed by the source and does not change at the boundary, so it stays f.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The speed falls to c/1.5, so the wavelength λ = v/f falls by the same factor, to λ/1.5 = 2λ/3.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe frequency has been divided by 1.5 as well. Only the speed and the wavelength change when light enters a new medium.
BThe wavelength has been multiplied by 1.5. Light travels more slowly in glass, so at the same frequency its wavelength is shorter.
CCorrect: the frequency is unchanged and the wavelength shrinks in proportion to the speed, to 2λ/3.
DThis keeps fλ equal to the speed in a vacuum by lowering the frequency; in fact the speed falls and the frequency stays the same.
Syllabus understandingC.2 — the nature of electromagnetic waves, and wave speed v applied to wave motion as given by v = fλCommand term: Deduce
5C-1A-13
Transverse & longitudinal·C.2 Wave model
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksState
A wave travels along a stretched horizontal rope. A small ribbon tied to the rope moves up and down as the wave passes.
Which statement about this wave is correct?
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Notes
Step 1The ribbon moves vertically while the wave travels horizontally, so the oscillation is perpendicular to the direction of energy transfer: the wave is transverse.
✓ 1
Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThe direction of energy transfer does not decide the type. What matters is the direction in which the medium oscillates.
BEvery particle in every mechanical wave returns to its starting position; that is true of both types.
CThe ribbon does not travel along the rope. The wave transfers energy, not matter.
DCorrect: in a transverse wave the oscillation of the medium is at right angles to the direction in which the wave travels.
Syllabus understandingC.2 — transverse and longitudinal travelling waves Command term: State
6C-1A-14
Reading wave graphs·C.2 Wave model
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksCalculate
The displacement–time graph shows one point on a wave.
What is the frequency of the wave?
Graph drawn to scaleShow mark scheme
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Notes
Step 1One complete cycle takes 4.0 ms, so T = 4.0 × 10⁻³ s and f = 1/T = 250 Hz.
✓ 1
Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AThis is 1/(0.4 × 10⁻³ s) — the period has been read one order of magnitude too small.
BCorrect: the graph gives T = 4.0 ms = 4.0 × 10⁻³ s, so f = 1/T = 250 Hz.
CThis is the period read straight off the graph, not the frequency.
DThis uses the period in milliseconds without converting. 1 ms = 10⁻³ s.
Syllabus understandingC.2 — the wave equation as given by v = fλ = λ/TCommand term: Calculate
7C-1A-15
The electromagnetic spectrum·C.2 Wave model
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksState
Which statement is correct for all electromagnetic waves?
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Notes
Step 1Electromagnetic waves are transverse oscillations of electric and magnetic fields, which need no medium and all travel at c in a vacuum.
✓ 1
Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThe fields oscillate at right angles to the direction of travel, so electromagnetic waves are transverse, not longitudinal.
BAll electromagnetic waves travel through a vacuum at the same speed, c = 3.00 × 108 m s−1, whatever their frequency.
CElectromagnetic waves need no medium: light reaches the Earth from the Sun through the vacuum of space.
DCorrect: every electromagnetic wave consists of electric and magnetic fields oscillating at right angles to the direction of travel.
Syllabus understandingC.2 — the nature of electromagnetic waves Command term: State
8C-1A-16
Mechanical & electromagnetic waves·C.2 Wave model
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A sound wave and a light wave each pass from air into water.
Which row describes the change in the wavelength of each wave?
Wavelength of the sound waveWavelength of the light wave
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Notes
Step 1For both waves the frequency is set by the source and does not change on entering the water, so λ = v/f changes in proportion to the speed.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Sound is a mechanical wave and travels faster in water (about 1500 m s−1) than in air (about 340 m s−1).
—
Step 3Light travels more slowly in water (n = 1.33) than in air, so the sound wavelength increases while the light wavelength decreases.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: at constant frequency λ ∝ v; sound speeds up in water and light slows down.
BThis treats sound like light. A mechanical wave travels faster in a liquid than in a gas, so the sound wavelength increases.
CThe sound wave is right, but light slows down in water, which has the greater refractive index, so its wavelength decreases.
DBoth have been reversed: sound speeds up in water and light slows down, not the other way round.
Syllabus understandingC.2 — the differences between mechanical waves and electromagnetic waves Command term: Deduce
9C-1A-17
Reading wave graphs·C.2 Wave model
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
A transverse wave travels to the right along a rope. The graph shows the displacement of the rope against distance at one instant. P and Q are two points on the rope.
Which row describes the velocity of P and the velocity of Q at this instant?
Graph drawn to scale
Velocity of PVelocity of Q
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Notes
Step 1P is at maximum displacement, so it is momentarily at rest.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The wave moves to the right, so Q next takes the displacement that the rope just to its left has now, which is negative: Q is moving downwards.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: P, at a crest, is momentarily at rest, and Q is about to take the negative displacement of the rope to its left, so it moves downwards.
BThe points of the rope are not carried along with the wave; in a transverse wave they move only at right angles to the direction of travel.
CThis swaps displacement and velocity: a point at maximum displacement is momentarily at rest, and a point at zero displacement moves fastest.
DP is right, but this is the direction Q would move if the wave travelled to the left.
Data-based question10 steps to full marksDetermine
Two microphones, A and B, are placed in a straight line with a small metal plate that is struck with a hammer. Microphone A is close to the plate and microphone B is a distance d further away. A data-logger starts timing when the signal from A exceeds a set level and stops when the signal from B does, giving the time interval Δt. Each value of Δt is the mean of five strikes and has an uncertainty of ±0.15 ms. The air temperature is 20 °C.
d / m ± 0.01 m
Δt / ms ± 0.15 ms
0.50
1.73
1.00
3.18
1.50
4.71
2.00
6.06
2.50
7.57
3.00
8.99
Graph drawn to scale
(a)
Describe the motion of an air molecule between the microphones as the sound passes.
(1)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the speed of sound v from the graph.
(3)
(d)
The gradients of the lines of maximum and minimum gradient that pass through all the error bars are 3.02 ms m−1 and 2.78 ms m−1. Determine the absolute uncertainty in v.
(2)
(e)
The accepted speed of sound in air at 20 °C is 343 m s−1. Comment on the student's result.
(1)
(f)
The line of best fit does not pass through the origin. State the value of the intercept on the Δt axis and suggest a cause.
(2)
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Notes
Part (a)
It oscillates about a fixed (equilibrium) position, parallel to the direction in which the sound travels
✓ 1
Must refer to oscillation AND the parallel direction. Do not accept 'moves along with the wave'.
Part (b)
Single straight line through all error bars, not forced through the origin
✓ 1
Part (c)
Gradient from a large triangle = 2.90 ms m−1
✓ 1
Accept 2.85 to 3.00 ms m−1.
v = 1/gradient = 1/(2.90 × 10−3 s m−1)
✓ 1
Must convert ms to s, or show the power of ten.
v = 344 m s−1
✓ 1
Accept 333 to 351 m s−1. Allow ECF from the gradient. Award [3] for CNA with gradient seen.
Part (d)
vmax = 1000/2.78 = 360 m s−1 AND vmin = 1000/3.02 = 331 m s−1
✓ 1
Note that the least steep line gives the largest speed.
Δv = (360 − 331)/2 ≈ ±14 m s−1
✓ 1
Accept ±12 to ±16 m s−1. Accept the fractional-uncertainty method using the difference from the best gradient.
Part (e)
343 m s−1 lies within the range v ± Δv (about 330 to 359 m s−1), so the result agrees with the accepted value
✓ 1
Allow ECF from (c) and (d). Must refer to the uncertainty.
Part (f)
Intercept ≈ 0.29 ms
✓ 1
Accept 0.10 to 0.40 ms.
A constant (systematic) delay in the timing, e.g. the signal from B takes longer to reach the trigger level than the signal from A, or a delay in the logger or cables
✓ 1
Accept 'the sound must travel a little further to microphone B than d' only if it refers to a constant extra distance. Do not accept 'reaction time'.
Answers: (c) v ≈ 344 m s−1 · (d) Δv ≈ ±14 m s−1 · (f) ≈ 0.29 ms (the remaining parts are explanations — see the table above)
Syllabus understandingC.2 — the nature of sound waves; wavelength λ, frequency f, time period T, and wave speed v applied to wave motion as given by v = fλ = λ/T; Tool 3 — uncertainty in a gradient Command term: Determine
11C-2-04
Transverse & longitudinal·C.2 Wave model
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksExplain
A loudspeaker at the left-hand end of a long tube produces a sound wave of frequency 850 Hz that travels to the right through the air in the tube. Figure 1 shows, at one instant, the displacement s of layers of air from their undisturbed positions against the undisturbed position x of each layer. A displacement to the right is positive.
Figure 1 — drawn to scale
(a)
(i)
State the wavelength of the sound.
(1)
(ii)
Hence show that the speed of sound in the air in the tube is about 340 m s−1.
(1)
(b)
On Figure 1, label with the letter C the position of the centre of a compression and with the letter R the position of the centre of a rarefaction.
(2)
(c)
Explain, with reference to Figure 1, why the air at the position you labelled C is compressed.
(2)
(d)
Outline how the total distance moved by one layer of air during one period of the wave compares with the distance travelled by the wave in the same time, and what this shows about a wave.
(3)
(e)
A student says that the wave must be transverse because the graph in Figure 1 has the shape of a transverse wave. Explain why the student is wrong.
(2)
(f)
The frequency of the loudspeaker is increased to 1700 Hz. State the new distance between the centre of a compression and the centre of the nearest rarefaction.
(1)
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Notes
Part (a)(i)
λ = 0.40 m
✓ 1
Accept 0.39–0.41 m
Part (a)(ii)
v = fλ = 850 × 0.40 «= 340 m s−1»
✓ 1
Must see full substitution OR answer to 3 s.f. Allow ECF from (a)(i)
Part (b)
C at x = 0.20 m, 0.60 m or 1.00 m «where s = 0 and the graph slopes downwards»
✓ 1
Accept ±0.02 m
R at x = 0, 0.40 m or 0.80 m «where s = 0 and the graph slopes upwards»
✓ 1
Accept ±0.02 m. Award [1] max if C and R are interchanged
Part (c)
layers just to the left of C have positive displacements «are displaced to the right» and layers just to the right of C have negative displacements «are displaced to the left»
✓ 1
Allow ECF from (b)
so layers on both sides are displaced towards C, are closer together than normal, and the density of the air there is greater
✓ 1
OWTTE
Part (d)
a layer moves 4 × amplitude = 4 × 6.0 μm = 24 μm in one period «and returns to its starting position»
✓ 1
Accept 2.4 × 10−5 m
the wave travels one wavelength, 0.40 m, in one period «about 1.7 × 104 times further»
✓ 1
Allow ECF from (a)(i)
the air oscillates about fixed positions with no net movement, while energy is transferred along the tube by the wave
✓ 1
OWTTE. Do not accept 'the air moves with the wave'
Part (e)
the graph shows the displacement of each layer, which is along the tube «parallel to x», plotted on the vertical axis; it is not a picture of the air
✓ 1
OWTTE
the layers oscillate parallel to the direction in which the wave travels, so the wave is longitudinal
✓ 1
Do not award MP2 for 'sound is longitudinal' alone
Part (f)
0.10 m «the speed is unchanged, so λ = 340/1700 = 0.20 m and the distance is λ/2»
✓ 1
Accept 10 cm
Answers: (a)(i) λ = 0.40 m · (a)(ii) 340 m s−1 · (d) 24 μm compared with 0.40 m · (f) 0.10 m (the remaining parts are explanations — see the table above)
Syllabus understandingC.2 — transverse and longitudinal travelling waves; the nature of sound waves; wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T Command term: Explain
12C-2-05
Sound waves·C.2 Wave model
Paper 2Medium10 marks
Short answer & extended response10 steps to full marksCalculate
A blue whale produces a low-pitched call of frequency 17 Hz. A hydrophone (an underwater microphone) on a research ship records the call.
Speed of sound in seawater = 1.50 × 103 m s−1; speed of sound in air = 340 m s−1.
(a)
Part of the sound of the call passes from the sea into the air. State why the frequency of the sound is the same in the air as in the water.
(1)
(b)
Calculate the wavelength of the call in seawater and in air.
(2)
(c)
The whale is 36 km from the ship.
(i)
Calculate the time taken for the call to travel through the water to the hydrophone.
(1)
(ii)
The hydrophone records the call twice: first after it has travelled through the rock of the sea bed, in which the speed of sound is 4.50 × 103 m s−1, and then directly through the water. Assume that both paths are 36 km long. Determine the time between the two arrivals.
(2)
(d)
(i)
People talk on the deck of the ship. Show that sound from their voices that meets the sea surface at an angle of incidence greater than about 13° cannot enter the water.
(1)
(ii)
Sound from the whale reaches the sea surface from below at an angle of incidence of 8.0°. Calculate the angle of refraction of the sound in the air.
(2)
(e)
Suggest why a diver under the sea near the ship hears very little of the conversation on the deck.
(1)
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Notes
Part (a)
the frequency is set by the source «the whale» / every oscillation that reaches the surface continues into the air, so the number of oscillations per second does not change
✓ 1
OWTTE
Part (b)
water: λ = 1500/17 = 88 m
✓ 1
Accept 88.2 m
air: λ = 340/17 = 20 m
✓ 1
Award [1] max if both are correct but interchanged
Part (c)(i)
t = 36 × 103/1.50 × 103 = 24 s
✓ 1
Part (c)(ii)
time through the rock = 36 × 103/4.50 × 103 = 8.0 s
✓ 1
interval = 24 − 8.0 = 16 s
✓ 1
Allow ECF from (c)(i). Award [2] for CNA
Part (d)(i)
sin c = 340/1500 «= 0.227», so c = 13.1° «beyond this angle the sound is totally reflected»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (d)(ii)
sin θ = (340/1500) × sin 8.0°
✓ 1
Do not accept use of 1500/340
θ = 1.8°
✓ 1
Accept 1.8°. Award [2] for CNA
Part (e)
most of the sound meets the surface at angles of incidence greater than the critical angle «13°», so it is totally reflected and does not enter the water
✓ 1
Accept 'very little of the sound is transmitted into the water' with a reason based on reflection at the surface
Answers: (b) 88 m in seawater; 20 m in air · (c)(i) 24 s · (c)(ii) 16 s · (d)(i) c = 13.1° · (d)(ii) 1.8° (the remaining parts are explanations — see the table above)
Syllabus understandingC.2 — the nature of sound waves; wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T (with C.3 — Snell's law, critical angle and total internal reflection) Command term: Calculate
13C-2-06
Mechanical & electromagnetic waves·C.2 Wave model
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksDescribe
During a thunderstorm an observer sees a flash of lightning and hears the thunder 4.2 s later. The speed of sound in air is 340 m s−1.
(a)
Describe the nature of the light from the flash as an electromagnetic wave.
(2)
(b)
(i)
Calculate the distance between the observer and the lightning.
(1)
(ii)
Show that the time taken by the light to reach the observer can be neglected in this calculation.
(1)
(c)
The lightning channel is almost vertical and about 3 km long. Explain why the thunder is heard as a rumble lasting several seconds rather than as a single short bang.
(2)
(d)
Lightning also emits radio waves of frequency 10 kHz. Calculate the wavelength of these radio waves.
(1)
(e)
Lightning also occurs in the atmosphere of the planet Jupiter. Radio telescopes on the Earth detect radio waves from this lightning.
(i)
Explain why the radio waves from lightning on Jupiter can reach the Earth but the sound of the thunder cannot.
(2)
(ii)
On one occasion Jupiter is 6.3 × 1011 m from the Earth. Calculate the time taken for the radio waves to travel from Jupiter to the Earth.
(1)
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Notes
Part (a)
oscillating electric and magnetic fields
✓ 1
at right angles to each other and to the direction of travel «transverse», travelling at c without needing a medium
✓ 1
Accept either 'transverse' or 'needs no medium / travels through a vacuum' for MP2
Part (b)(i)
d = 340 × 4.2 = 1.4 × 103 m
✓ 1
Accept 1428 m
Part (b)(ii)
t = 1.4 × 103/3.00 × 108 = 4.8 × 10−6 s, which is negligible compared with 4.2 s
✓ 1
Must see the substitution AND a comparison with 4.2 s
Part (c)
sound from different parts of the channel travels different distances to the observer
✓ 1
sound is slow, so path differences of hundreds of metres give arrival times that differ by seconds «1 km ≈ 3 s», whereas the light from all parts arrives almost at the same time
✓ 1
OWTTE
Part (d)
λ = 3.00 × 108/1.0 × 104 = 3.0 × 104 m
✓ 1
Part (e)(i)
sound is a mechanical wave: it needs a medium «particles that pass on the oscillation», and the space between Jupiter and the Earth is «almost» a vacuum
✓ 1
radio waves are electromagnetic waves «oscillating electric and magnetic fields», which need no medium and travel through a vacuum
✓ 1
MP2 needs a reference to travel through a vacuum / no medium. OWTTE
Part (e)(ii)
t = 6.3 × 1011/3.00 × 108 = 2.1 × 103 s «35 minutes»
✓ 1
Accept 35 min
Answers: (b)(i) 1.4 × 103 m · (b)(ii) 4.8 × 10−6 s · (d) 3.0 × 104 m · (e)(ii) 2.1 × 103 s (the remaining parts are explanations — see the table above)
Syllabus understandingC.2 — the nature of electromagnetic waves, the nature of sound waves, and the differences between mechanical waves and electromagnetic waves Command term: Describe
14C-1B-14
The wave equation·C.2 Wave model
Paper 1BHard12 marks
Data-based question12 steps to full marksDetermine
A university hydraulics laboratory publishes data from a long wave flume. A paddle at one end produces long water waves of period 6.0 s. Two wave-height gauges, (4.00 ± 0.01) m apart, record the time t for a wave crest to travel from the first gauge to the second, for different still-water depths h. The published uncertainty in each time is ±0.02 s.
A student uses the published data to test the model v = √(gh) for the speed v of water waves whose wavelength is much greater than the depth of the water.
h / m ± 0.001 m
t / s ± 0.02 s
v / m s−1
log( h / m )
log( v / m s−1 )
0.040
6.38
0.627
−1.398
−0.203
0.080
4.52
0.885
−1.097
−0.053
0.120
3.68
1.09
−0.921
0.036
0.160
3.19
0.240
2.61
1.53
−0.620
0.185
0.320
2.25
1.78
−0.495
0.250
Graph drawn to scale
(a)
Show that the absolute uncertainty in v for h = 0.040 m is about 0.004 m s−1.
(1)
(b)
Complete the table for h = 0.160 m.
(2)
(c)
Explain why, if the model is correct, the graph of log v against log h is a straight line of gradient 0.5.
(2)
(d)
Draw the line of best fit on the graph.
(1)
(e)
Determine the gradient of your line and state whether the data support the model.
(2)
(f)
Determine g using your line of best fit.
(2)
(g)
Calculate the wavelength of the waves when h = 0.320 m, and hence comment on whether the condition for the model is met.
(2)
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Marking point
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Notes
Part (a)
v = 4.00/6.38 = 0.627 m s−1; Δv/v = 0.02/6.38 + 0.01/4.00 = 0.0056, so Δv = 0.0035 m s−1
✓ 1
Must see both fractional uncertainties added AND an answer to at least 2 s.f.
Part (b)
v = 4.00/3.19 = 1.25 m s−1
✓ 1
log( h / m ) = −0.796 AND log( v / m s−1 ) = 0.098
✓ 1
Accept 2 or 4 decimal places. Accept 0.097 for log v if the rounded speed 1.25 m s−1 is used. Allow ECF for log v.
Part (c)
Taking logs: log v = ½ log h + ½ log g
✓ 1
Accept log v = 0.5 log h + log √g.
This has the form y = mx + c with g constant, so the graph is a straight line whose gradient is 0.5
✓ 1
OWTTE.
Part (d)
Single straight line with points evenly distributed on both sides
✓ 1
Part (e)
gradient = 0.50
✓ 1
Accept 0.48 to 0.52.
The gradient equals 0.5 within the precision of the graph and all points lie close to a straight line, so the data support the model
✓ 1
Allow ECF from the gradient.
Part (f)
Uses a point on the line in log v = ½ log h + ½ log g, e.g. at log( h / m ) = −1.0, log( v / m s−1 ) = −0.004
✓ 1
Or reads the intercept of a line extended to log h = 0.
log g = 2 × (−0.004 + 0.500) = 0.992, so g = 9.82 m s−2
✓ 1
Accept 9.6 to 10.1 m s−2. Allow ECF. Award [2] for CNA.
Part (g)
λ = vT = 1.78 × 6.0 = 10.7 m
✓ 1
Allow the use of v = √(gh) = 1.77 m s−1.
λ/h ≈ 33: the wavelength is much greater than the depth, so the condition is met (even for the deepest water)
✓ 1
Must compare λ with h.
Answers: (a) Δv ≈ 0.0035 m s−1 · (b) 1.25 m s−1; −0.796; 0.098 · (e) 0.50 · (f) g ≈ 9.82 m s−2 · (g) λ ≈ 10.7 m (the remaining parts are explanations — see the table above)
Syllabus understandingC.2 — wavelength λ, frequency f, time period T, and wave speed v applied to wave motion as given by v = fλ = λ/T; the differences between mechanical waves and electromagnetic waves; Tool 3 — linearizing a power law with a log–log graph Command term: Determine
15C-2-20
Intensity & amplitude·C.2 Wave model
Paper 2Medium14 marks
Short answer & extended response14 steps to full marksDetermine
An infrared heat lamp hangs above a pen of young chicks to keep them warm. The tungsten filament of the lamp has a temperature of 2400 K and radiates a total power of 250 W. Treat the filament as a black body.
(a)
(i)
Calculate the wavelength at which the power radiated by the filament is a maximum.
(1)
(ii)
State the region of the electromagnetic spectrum in which this wavelength lies.
(1)
(iii)
Describe the nature of the electromagnetic waves emitted by the lamp.
(2)
(b)
Assume that the lamp radiates uniformly in all directions and that none of the radiation is absorbed by the air.
(i)
Show that the intensity of the radiation 0.50 m from the lamp is about 80 W m−2.
(1)
(ii)
The chicks need an intensity of at least 35 W m−2 at floor level. The farmer hangs the lamp 0.80 m above the floor. Determine whether the chicks are kept warm enough.
(3)
(iii)
The upper half of the glass bulb has a reflecting coating, so the lamp does not radiate uniformly in all directions. Suggest how this affects your conclusion in (b)(ii).
(2)
(c)
(i)
Determine the surface area of the filament.
(2)
(ii)
Explain why most of the energy radiated by the lamp is not visible light.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
λmax = 2.9 × 10−3/2400 = 1.2 × 10−6 m
✓ 1
Accept 1.21 × 10−6 m or 1.2 μm.
Part (a)(ii)
Infrared
✓ 1
Allow ECF from (a)(i).
Part (a)(iii)
Oscillating electric and magnetic fields, at right angles to each other
✓ 1
The oscillations are perpendicular to the direction of travel «transverse waves» OR they travel through a vacuum at 3.00 × 108 m s−1
✓ 1
Either statement for MP2.
Part (b)(i)
I = P/4πd² = 250/(4π × 0.50²) = 79.6 «W m−2»
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
I = 250/(4π × 0.80²)
✓ 1
ALTERNATIVE: inverse-square scaling, I = 79.6 × (0.50/0.80)².
I = 31 W m−2
✓ 1
Accept 31.0–31.1 W m−2. Allow ECF from (b)(i) in the alternative.
31 W m−2 is less than 35 W m−2, so the chicks are not kept warm enough «the lamp must be lowered»
✓ 1
MP3 only scores if an intensity at 0.80 m has been calculated. Allow ECF for the conclusion.
Part (b)(iii)
The coating reflects radiation downwards, so the intensity below the lamp is greater than the value calculated for uniform emission «up to about twice as large»
✓ 1
So the intensity at floor level may be above 35 W m−2: the conclusion of (b)(ii) is not reliable «the chicks may be warm enough»
✓ 1
OWTTE. MP2 must refer to the conclusion.
Part (c)(i)
L = σAT⁴, so A = 250/(5.67 × 10−8 × 2400⁴)
✓ 1
A = 1.3 × 10−4 m²
✓ 1
Accept 1.33 × 10−4 m². Award [2] for CNA.
Part (c)(ii)
The emission peaks at about 1.2 μm, a longer wavelength than the longest visible wavelength «about 700 nm»
✓ 1
Allow ECF from (a)(i).
Only a small fraction of the black-body spectrum «of the area under the curve» lies between about 400 nm and 700 nm, so most of the power is emitted as infrared
✓ 1
OWTTE.
Answers: (a)(i) λmax = 1.2 × 10−6 m · (b)(i) 79.6 W m−2 · (b)(ii) I = 31 W m−2 — not warm enough · (c)(i) A = 1.3 × 10−4 m² (the remaining parts are explanations — see the table above)
Syllabus understandingC.2 — the nature of electromagnetic waves; B.1 — the emission spectrum of a black body and Wien's displacement law as given by λmaxT = 2.9 × 10−3 m K; the Stefan–Boltzmann law as given by L = σAT⁴; the concept of apparent brightness as given by b = L/4πd² Command term: Determine
16C-2-21
Transverse & longitudinal·C.2 Wave model
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDetermine
An earthquake produces two kinds of seismic wave that travel through the Earth's crust: P waves, which are longitudinal, and S waves, which are transverse. In the crust P waves travel at 6.0 km s−1 and S waves travel at 3.5 km s−1.
(a)
Describe the motion of the particles of the rock as a P wave passes and as an S wave passes.
(2)
(b)
A seismometer at a monitoring station records the first P wave from an earthquake 24.0 s before the first S wave. Determine the distance from the station to the earthquake.
(3)
(c)
In an early-warning system, a station 30 km from an earthquake detects the first P wave. A computer takes 3.0 s to analyse the signal and then sends an alert by radio to a city 150 km from the earthquake. Radio waves travel at 3.00 × 108 m s−1. Determine the time between the arrival of the alert at the city and the arrival of the S waves there.
(3)
(d)
A P wave has a frequency of 2.0 Hz. Calculate its wavelength.
(1)
(e)
Suggest why the distances from at least three stations are needed to locate the earthquake.
(1)
(f)
The speed of P waves in the rock between the earthquake and the station in (b) may be 5.8 km s−1 rather than 6.0 km s−1. Determine the distance using this speed and comment on the reliability of your answer to (b).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
P wave: the rock particles oscillate parallel to the direction in which the wave travels «forming compressions and rarefactions»
✓ 1
S wave: the rock particles oscillate perpendicular to the direction in which the wave travels
✓ 1
Both MPs need 'oscillate' or 'vibrate' about a fixed position. OWTTE.
Part (b)
d/3.5 − d/6.0 = 24.0 «with d in km»
✓ 1
Award [0] for 24.0 × 3.5 or 24.0 × 6.0.
d × (0.286 − 0.167) = 24.0, so d = 24.0/0.119
✓ 1
d = 202 km «2.0 × 105 m»
✓ 1
Accept 200–202 km. Award [3] for CNA.
Part (c)
The P wave reaches the station after 30/6.0 = 5.0 s, so the alert is sent at 8.0 s «the radio travel time of about 10−3 s is negligible»
✓ 1
The S waves reach the city after 150/3.5 = 42.9 s
✓ 1
Warning time = 42.9 − 8.0 = 35 s
✓ 1
Accept 34–35 s. Award [3] for CNA.
Part (d)
λ = v/f = 6.0 × 103/2.0 = 3.0 × 103 m
✓ 1
Accept 3.0 km.
Part (e)
One distance only places the earthquake somewhere on a circle around the station; the circles from three stations intersect at a single point
✓ 1
OWTTE.
Part (f)
d = 24.0/(1/3.5 − 1/5.8) = 212 km
✓ 1
Accept 211–212 km.
A 3 % change in the P-wave speed changes the distance by about 5 %, so the answer to (b) is only reliable to within about 10 km «the method is sensitive to the assumed speeds»
✓ 1
Allow ECF from (b). Accept any sensible comment that compares the two distances.
Answers: (b) 202 km · (c) 35 s · (d) 3.0 × 103 m · (f) 212 km (the remaining parts are explanations — see the table above)
Syllabus understandingC.2 — transverse and longitudinal travelling waves; the differences between mechanical waves and electromagnetic waves; wavelength λ, frequency f, time period T, and wave speed v applied to wave motion as given by v = fλ = λ/T Command term: Determine
17C-2-22
Reading wave graphs·C.2 Wave model
Paper 2Hard11 marks
Short answer & extended response11 steps to full marksDeduce
A transverse wave travels along a long horizontal rope. P and Q are two points on the rope 0.90 m apart. Graph 1 shows how the vertical displacement y of P (solid line) and of Q (dashed line) varies with time t. Upward displacements are positive. It is known that the distance PQ is less than half a wavelength.
Graph 1 — drawn to scaleGraph 2 — for part (e)
(a)
State the amplitude and the frequency of the wave.
(2)
(b)
Assume first that the wave travels from P towards Q.
(i)
Show that the speed of the wave is 4.5 m s−1.
(1)
(ii)
Calculate the wavelength.
(1)
(c)
Deduce that the wave cannot be travelling from Q towards P.
(3)
(d)
Describe the motion of P between t = 0 and t = 0.20 s.
(2)
(e)
On Graph 2, sketch the shape of the rope between P and Q at t = 0.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Amplitude = 4.0 cm
✓ 1
Unit required.
f = 1/0.80 = 1.25 Hz
✓ 1
Accept 1.2–1.3 Hz.
Part (b)(i)
Q repeats the motion of P 0.20 s later, so the wave travels 0.90 m in 0.20 s: v = 0.90/0.20 = 4.5 «m s−1»
✓ 1
Must see full substitution OR answer to at least 3 s.f. The time of 0.20 s must be read from the graph.
Part (b)(ii)
λ = vT = 4.5 × 0.80 = 3.6 m
✓ 1
Accept λ = 4 × 0.90 m «PQ is a quarter of a wavelength». Allow ECF from (b)(i).
Part (c)
If the wave travelled from Q to P, P would repeat the motion of Q 0.60 s later
✓ 1
v = 0.90/0.60 = 1.5 m s−1, so λ = 1.5 × 0.80 = 1.2 m
✓ 1
Then PQ = 0.90 m would be more than half a wavelength «0.60 m», which contradicts the information, so the wave travels from P to Q
✓ 1
MP3 only scores if a wavelength «or speed» for travel from Q to P has been calculated. OWTTE.
Part (d)
P moves upwards, perpendicular to the rope «and to the direction of travel of the wave», from its equilibrium position to +4.0 cm
✓ 1
Its speed decreases from a maximum at t = 0 to zero at t = 0.20 s
✓ 1
OWTTE.
Part (e)
Curve starting at y = 0 at P and reaching y = −4.0 cm at Q, with no other turning point
✓ 1
Smooth quarter of a sine wave: steepest at P and with zero gradient «horizontal» at Q
✓ 1
Do not accept a straight line. Allow ECF from (b)(ii) only for a wavelength greater than 1.8 m.
Answers: (a) 4.0 cm; 1.25 Hz · (b)(i) 4.5 m s−1 · (b)(ii) 3.6 m · (c) Q → P would need λ = 1.2 m < 2 × 0.90 m (the remaining parts are explanations — see the table above)
Syllabus understandingC.2 — transverse and longitudinal travelling waves; wavelength λ, frequency f, time period T, and wave speed v applied to wave motion as given by v = fλ = λ/T Command term: Deduce
18C-2-23
Sound waves·C.2 Wave model
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksCalculate
African elephants communicate with low-frequency rumbles of frequency 14 Hz. Some bats hunt insects by emitting pulses of ultrasound of frequency 45 kHz and detecting the echoes. The speed of sound in air is 340 m s−1. Humans can hear sounds with frequencies between about 20 Hz and 20 kHz.
(a)
Calculate the period of the elephant's rumble and the time it takes to travel 5.0 km through the air.
(2)
(b)
State why humans cannot hear either the elephant's rumble or the bat's ultrasound.
(1)
(c)
Calculate the wavelength in air of the elephant's rumble and of the bat's ultrasound.
(2)
(d)
Suggest, with reference to diffraction, why elephants can hear each other's rumbles through dense forest several kilometres away.
(2)
(e)
Assume that a bat can detect an object only if the object is at least as large as the wavelength of the ultrasound. Determine whether a bat using 45 kHz ultrasound can detect a mosquito with a wingspan of 5.0 mm, and calculate the minimum frequency it must use to detect the mosquito.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
T = 1/14 = 0.071 s
✓ 1
Accept 0.07 s.
t = 5.0 × 103/340 = 15 s
✓ 1
Accept 14.7 s.
Part (b)
14 Hz is below 20 Hz «infrasound» and 45 kHz is above 20 kHz «ultrasound»
✓ 1
Both needed.
Part (c)
Elephant: λ = 340/14 = 24 m
✓ 1
Accept 24.3 m.
Bat: λ = 340/45 × 103 = 7.6 × 10−3 m
✓ 1
Accept 7.56 mm.
Part (d)
The wavelength «24 m» is comparable to or larger than the trees and other obstacles
✓ 1
Allow ECF from (c).
so the sound diffracts significantly around them and spreads into the regions behind them instead of being blocked
✓ 1
Do not award MP2 for 'low frequencies are absorbed less' alone.
Part (e)
The wavelength «7.6 mm» is larger than the 5.0 mm wingspan
✓ 1
Allow ECF from (c).
so the bat cannot detect the mosquito with 45 kHz ultrasound
✓ 1
MP2 only scores if a comparison is made.
Minimum frequency f = 340/5.0 × 10−3 = 6.8 × 104 Hz «68 kHz»
✓ 1
Answers: (a) 0.071 s and 15 s · (c) 24 m and 7.6 × 10−3 m · (e) cannot detect; minimum 6.8 × 104 Hz (the remaining parts are explanations — see the table above)
Syllabus understandingC.2 — the nature of sound waves; wavelength λ, frequency f, time period T, and wave speed v applied to wave motion as given by v = fλ = λ/T; C.3 — wave diffraction around a body and through an aperture Command term: Calculate
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