60 free, high-quality past-paper-style questions on Wave model for IB Physics HL, each with a full mark scheme. Covers: Electromagnetic and mechanical waves; Wave graphs; The wave equation; and more. For when you have finished the past papers, or want to practise one topic at a time.C.2 at Standard Level →Read the C.2 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.230 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1C-1A-23Electromagnetic and mechanical wavesPaper 1AEasy1 mark
Which statement is correct for both sound waves in air and electromagnetic waves in a vacuum?
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Step 1Sound in air is longitudinal (oscillations of air molecules parallel to the direction of travel); electromagnetic waves are transverse oscillations of electric and magnetic fields.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Sound is a mechanical wave and needs a medium; electromagnetic waves travel through a vacuum, at c = 3.0 × 108 m s−1, far faster than sound (≈ 340 m s−1).
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Step 3What all travelling waves share is that they carry energy from one place to another while the medium (if any) has no resultant displacement.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
ASound in air is longitudinal, not transverse.
BElectromagnetic waves need no medium — light reaches us from the Sun through the vacuum of space.
COnly electromagnetic waves travel at c; sound in air travels at about 340 m s−1.
DCorrect: every travelling wave transfers energy without transferring matter.
2C-1A-37Electromagnetic and mechanical wavesPaper 1AEasy1 mark
A microwave signal of frequency 10 GHz travels through a vacuum. An ultrasonic wave travels through water, in which the speed of sound is 1500 m s−1. The ultrasonic wave has the same wavelength as the microwave signal.
What is the frequency of the ultrasonic wave?
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Step 1For the microwave: λ = c/f = 3.00 × 108/1.0 × 1010 = 0.030 m.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The ultrasound is a mechanical wave and travels at only 1500 m s−1, but has the same wavelength.
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Step 3f = v/λ = 1500/0.030 = 5.0 × 104 Hz.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses the wavelength in millimetres (30) without converting it to metres: 1500/30 = 50 Hz.
BCorrect: λ = 0.030 m, so f = 1500/0.030 = 5.0 × 104 Hz.
CThis assumes the two waves have the same frequency; only their wavelengths are equal, and their speeds differ by a factor of 2 × 105.
DThis inverts the ratio of the speeds: 1.0 × 1010 × (3.00 × 108/1500) = 2.0 × 1015 Hz.
A transverse wave travels in the positive x direction along a rope. The graph shows the displacement y of the rope against position x at time t = 0 (solid line) and at t = 0.010 s (dashed line). In this time the wave moves less than one wavelength.
What is the frequency of the wave?
Displacement y against position x at t = 0 (solid) and t = 0.010 s (dashed), drawn to scale.Show mark scheme
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Step 1From the graph λ = 12 cm, and each crest moves 3.0 cm in the positive x direction in 0.010 s.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2v = 0.030/0.010 = 3.0 m s−1.
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Step 3f = v/λ = 3.0/0.12 = 25 Hz.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: v = 3.0 m s−1 and λ = 0.12 m, so f = 25 Hz.
BThis takes the distance from a crest to the adjacent trough (6 cm) as the wavelength.
CThis reads the shift in the wrong direction (9.0 cm, the distance to the next dashed crest to the left): 9.0 m s−1/0.12 m = 75 Hz.
4IBP2H-021-00-01Electromagnetic and mechanical wavesPaper 1AEasy1 mark
A medical ultrasound probe emits waves of frequency 5.0 MHz into soft tissue. The speed of sound in soft tissue is 1540 m s−1. What is the wavelength of the ultrasound in the tissue?
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λ = v/f = 1540 / (5.0×106) = 3.08×10−4 m ≈ 0.31 mm. The wavelength depends on the speed in tissue, not the speed in air. 0.068 mm comes from using the speed of sound in air (340 m s−1). 3.1 mm comes from a power-of-ten slip. 310 mm comes from treating MHz as kHz.
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Answer B
Answer: B
Common error: Convert the prefix first: 1 MHz = 10⁶ Hz. Use the wave speed in the medium the wave is actually travelling through.
5IBP2H-021-00-02Electromagnetic and mechanical wavesPaper 1AEasy1 mark
Seismic P-waves from an earthquake are detected at a monitoring station on the far side of the Earth, after passing through the liquid outer core. Which statement about P-waves is correct?
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P-waves are longitudinal. The particles oscillate along the direction of travel, so the wave is a series of compressions and rarefactions. Liquids resist compression, so longitudinal waves can pass through them. This is why P-waves cross the liquid outer core. A wave transfers energy, not matter, so the particles do not travel to the station. A perpendicular oscillation would describe S-waves, and S-waves cannot cross liquids.
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Answer A
Answer: A
Exam tip: Link the type of wave to the medium. Longitudinal waves need the medium to resist compression. Transverse mechanical waves need it to resist shear, which liquids and gases cannot do.
6IBP2H-021-01-01Electromagnetic and mechanical wavesPaper 1AEasy1 mark
A medical ultrasound probe emits waves of frequency 2.5 MHz into soft tissue, where the speed of sound is 1540 m s−1. What is the wavelength of the ultrasound in the tissue?
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λ = v/f = 1540 / (2.5 × 106) = 6.2 × 10−4 m (about 0.6 mm). Option B treats MHz as kHz. Option C uses f/v, the inverted ratio. Option D multiplies v by f instead of dividing.
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Answer A
Answer: A
Common error: Convert MHz to Hz (× 10⁶) before using v = fλ. A wavelength below a millimetre is what makes ultrasound useful for imaging small structures.
7IBP2H-021-01-02Electromagnetic and mechanical wavesPaper 1AEasy1 mark
A diver underwater hears a boat engine. The sound passes from air into water, where its speed increases by a factor of about 4.4. Which quantity stays the same as the sound enters the water?
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The frequency is set by the source. Each oscillation arriving at the boundary drives one oscillation in the water, so the frequency is unchanged. The speed increases, so by v = fλ the wavelength also increases by the same factor. That rules out B, C and D.
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Answer A
Answer: A
Exam tip: When any wave crosses into a new medium, frequency is the quantity that stays fixed. Speed and wavelength change together.
8IBP3H-C2-20r34-01Electromagnetic and mechanical wavesPaper 1AEasy1 mark
Which statement about sound waves in air and radio waves in air is correct?
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Radio waves are electromagnetic waves: the electric and magnetic fields oscillate perpendicular to the direction of energy transfer, so they are transverse and can be polarised. Sound in air is a longitudinal pressure wave, with particles oscillating parallel to the direction of travel, so it cannot be polarised. The first option wrongly calls radio waves longitudinal. The second option swaps the two wave types. The last option is wrong because sound is a mechanical wave that needs oscillating particles of a medium.
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Answer C
Answer: C
Exam tip: Polarisation is only possible for transverse waves, so it is a quick test to tell transverse from longitudinal.
9IBP3H-C2-20r34-02The wave equationPaper 1AEasy1 mark
A wave-energy buoy floating at sea rises and falls through 10 complete cycles in 80 s. The distance between neighbouring wave crests is 24 m. What is the speed of the waves?
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The frequency is 10/80 = 0.125 Hz, and the wavelength is 24 m, so v = fλ = 0.125 × 24 = 3.0 m s−1. Dividing the wavelength by the frequency gives 192 m s−1, which uses λ/f instead of fλ. Multiplying the wavelength by the number of cycles gives 240, which ignores the time. The value 0.33 comes from dividing the period by the wavelength.
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Answer B
Answer: B
Common error: Convert a count of cycles and a time into a frequency (cycles per second) before using v = fλ.
10IBP3H-C2-20r34-04Electromagnetic and mechanical wavesPaper 1AEasy1 mark
An ultraviolet lamp emits waves of wavelength 254 nm and an infrared thermometer detects waves of wavelength 12.7 μm. Which comparison of these two waves in a vacuum is correct?
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All electromagnetic waves travel at the same speed in a vacuum. Since c = fλ, frequency is inversely proportional to wavelength, so f(UV)/f(IR) = λ(IR)/λ(UV) = 12.7 × 10−6 / 254 × 10−9 = 50. The claim that higher frequency means higher speed is false in a vacuum. The claim of equal frequency would require equal wavelengths. Saying the longer wavelength has the greater frequency reverses the inverse relationship.
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Answer D
Answer: D
Exam tip: In a vacuum every electromagnetic wave has the same speed, so compare frequencies through c = fλ.
11C-1A-12Intensity and amplitudePaper 1AMedium1 mark
A small loudspeaker radiates sound uniformly in all directions. At a distance of 2.0 m the amplitude of the pressure variation is A. The intensity of a wave is proportional to the square of its amplitude.
What is the amplitude at a distance of 6.0 m?
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Step 1For a point source the intensity obeys an inverse-square law: I ∝ 1/r², so tripling the distance divides the intensity by 9.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Intensity is proportional to amplitude squared: I ∝ A².
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Step 3So the amplitude falls by √9 = 3: new amplitude = A/3.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis applies the inverse-square law to the amplitude. It is the intensity that falls by 9; the amplitude falls by √9.
BCorrect: the power spreads over a sphere, so the intensity falls by (6.0/2.0)² = 9, and the amplitude by √9 = 3.
CThis takes the intensity to fall in proportion to 1/r (by 3), as if the sound spread over a circle rather than a sphere.
DThis assumes the amplitude is conserved as the wave travels. The energy is conserved, but it spreads over a larger area, so the intensity and the amplitude both fall.
A transverse wave travels along a rope. The distance between a crest and the adjacent trough is d. A point on the rope takes a time t to move from its maximum displacement to its equilibrium position.
What is the speed of the wave?
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Step 1A crest and the adjacent trough are half a wavelength apart: λ = 2d.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Maximum displacement to equilibrium is a quarter of a cycle: T = 4t.
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Step 3v = λ/T = 2d/(4t) = d/(2t).
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis takes d as the whole wavelength (λ = d) with T = 4t.
BCorrect: λ = 2d and T = 4t, so v = d/(2t).
CThis takes the time from maximum displacement to equilibrium as half a period (T = 2t), with λ = 2d.
Graph 1 shows the displacement of the particles of a medium against position at one instant, and graph 2 shows the displacement of one particle against time, for the same wave.
What is the speed of the wave?
Two graphs describing the same transverse wave.Show mark scheme
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Step 1From graph 1 (against position) the wavelength is the repeat distance: λ = 4.0 cm = 0.040 m.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2From graph 2 (against time) the period is the repeat time: T = 0.20 s, so f = 5.0 Hz.
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Step 3v = fλ = 5.0 × 0.040 = 0.20 m s−1.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis is T/λ with the wavelength left in cm (0.20/4.0) — the ratio is inverted and the units are inconsistent.
BCorrect: v = λ/T = 0.040/0.20 = 0.20 m s−1.
CThis uses 8.0 cm, the full length of the position axis (two wavelengths), as the wavelength: 0.080/0.20 = 0.40 m s−1.
A longitudinal wave travels along a long spring. The graph shows, at one instant, the displacement s of the coils against their equilibrium position x. A positive displacement is to the right.
Which point is at the centre of a rarefaction?
Displacement of the particles against their equilibrium position at one instant (positive = displaced to the right). Drawn to scale.Show mark scheme
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Step 1At a rarefaction the coils are furthest apart: the coils on its left are displaced to the left (s < 0) and those on its right are displaced to the right (s > 0).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2This happens where s = 0 and the graph has a positive gradient — point Q (x = 20 cm).
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Step 3At S the coils on either side are displaced towards S (s > 0 on the left, s < 0 on the right): S is the centre of a compression.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AP has the largest displacement to the left, but its neighbours are displaced by almost the same amount, so the spacing there is normal.
BCorrect: the coils on either side of Q are displaced away from it, so the coils are spread out.
CR has the largest displacement to the right; a maximum displacement does not mean a compression or rarefaction.
DAt S the neighbouring coils move towards S: this is a compression, not a rarefaction.
A transverse wave travels to the right along a rope. The graph shows the displacement of the rope against position at one instant. X and Y are points on the rope.
Which row describes the motion of X and of Y at this instant?
A transverse wave on a rope at one instant, travelling to the right (drawn to scale).
XY
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Step 1A short time later the whole waveform has shifted slightly to the right. The part of the rope just to the left of X is below the axis, so X will move down.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2X is at zero displacement, where the speed of an oscillating point is greatest.
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Step 3Y is at a crest (maximum displacement), where it is momentarily at rest before moving down.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: X moves down at maximum speed; Y is instantaneously at rest at maximum displacement.
BY is at maximum displacement, where the speed is zero; zero displacement (not maximum displacement) is where the speed is greatest.
CThis is the direction for a wave travelling to the left; for a wave travelling to the right X takes on the displacement of the rope to its left, which is below it.
DIn a transverse wave the rope moves perpendicular to the direction of travel; the rope itself does not travel with the wave.
A longitudinal wave travels in the positive x direction along a long spring. The graph shows, at one instant, the displacement s of the coils against their equilibrium position x. A positive displacement is in the positive x direction. P is a coil at x = 10 cm.
Which statements about P at this instant are correct?
I. P is momentarily at rest.
II. P is at the centre of a compression.
III. The acceleration of P has its maximum magnitude and is in the negative x direction.
Displacement s of the coils against their equilibrium position x at one instant (drawn to scale). The wave travels in the positive x direction.Show mark scheme
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Step 1P has its maximum displacement, so it is at an extreme of its oscillation: its velocity is zero (I correct).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2In SHM a = −ω²s, so at the maximum positive displacement the acceleration has its maximum magnitude and points in the negative x direction (III correct).
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Step 3The coils on either side of P (at 8 cm and 12 cm, say) have almost the same displacement as P, so the spacing around P is the normal spacing. The centre of a compression is where neighbouring coils are displaced towards each other, at x = 20 cm, where s = 0 and the graph has its steepest negative gradient (II incorrect).
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AII is incorrect: at the maximum displacement the coils near P are displaced by almost the same amount, so the spacing there is normal, not compressed.
BCorrect: P is at rest at an extreme of its oscillation and its acceleration −ω²s is greatest in magnitude and in the negative x direction.
CI is correct: a coil at maximum displacement is momentarily at rest. II is incorrect.
DThis treats the maximum of the displacement graph as a compression, as if it were a picture of a transverse wave. II is incorrect.
A transverse wave travels to the right along a taut climbing rope. At one instant, particle X on the rope is at its equilibrium position. The nearest crest is a quarter of a wavelength to the left of X. What is the motion of X at this instant?
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The waveform moves to the right. After a quarter of a period, the crest now a quarter-wavelength to the left of X will reach X. X must therefore be rising towards a crest, so it is moving upwards. It is at equilibrium, so its speed is at a maximum, not zero. The particles of a transverse wave move perpendicular to the direction of travel, not to the right.
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Answer A
Answer: A
Exam tip: Imagine shifting the waveform slightly in the direction of travel. Whatever displacement moves onto the particle tells you which way it is heading.
18IBP2H-021-00-04Longitudinal wavesPaper 1AMedium1 mark
A longitudinal wave travels along a Slinky spring. Displacement to the right is taken as positive. At one instant, a coil at position P has zero displacement. Coils just to the left of P are displaced to the right. Coils just to the right of P are displaced to the left. What is located at P?
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Coils on both sides of P are displaced towards P, so the coils crowd together there. P is therefore the centre of a compression, where the spacing is smallest. A rarefaction would need coils displaced away from P. The displacement at P is zero, not a maximum. The spacing is undisturbed at points of maximum displacement, not at P.
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Answer B
Answer: B
Common error: On a displacement–distance graph for a longitudinal wave, compressions and rarefactions occur at zero displacement, not at the peaks. Work out which way the neighbouring particles are displaced.
19IBP2H-021-00-05Electromagnetic and mechanical wavesPaper 1AMedium1 mark
A radio signal and a sound wave both pass from air into seawater. Which row correctly describes the changes to each wave?
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Frequency is set by the source and does not change at a boundary, so λ = v/f changes in the same way as v. EM waves travel fastest in a vacuum and more slowly in water. Sound, a mechanical wave, travels faster in water (about 1500 m s−1) than in air (about 340 m s−1). So the radio wavelength decreases and the sound wavelength increases. Assuming a denser medium slows every wave gives the second option. Assuming frequency changes gives the last option.
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Answer A
Answer: A
Common error: Do not assume all waves slow down in denser media. Mechanical waves usually travel faster in liquids and solids, while EM waves travel more slowly.
A buoy rides on water waves in a harbour. Its displacement–time graph shows 5 complete oscillations in 8.0 s. A displacement–distance graph of the water surface at one instant shows that the distance from the first crest to the fifth crest is 4.8 m. What is the speed of the waves?
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From the first crest to the fifth crest there are 4 wavelengths, so λ = 4.8/4 = 1.2 m. The frequency is f = 5/8.0 = 0.625 Hz. So v = fλ = 0.625 × 1.2 = 0.75 m s−1. 0.60 m s−1 comes from dividing 4.8 m by 5 instead of 4. 1.9 m s−1 uses the period (1.6 s) in place of the frequency. 3.0 m s−1 uses 4.8 m as the wavelength.
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Answer B
Answer: B
Exam tip: Count the gaps between crests, not the crests themselves. Five crests span four wavelengths.
21IBP2H-021-01-03Intensity and amplitudePaper 1AMedium1 mark
Active noise-cancelling headphones reduce the amplitude of a low-frequency hum reaching the eardrum to 40% of its original value. The frequency is unchanged. What is the new intensity of the hum, as a percentage of the original intensity?
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Intensity is proportional to amplitude squared: I ∝ A2. The new intensity is (0.40)2 = 0.16 of the original, which is 16%. Option B assumes I ∝ A. Option C takes a square root instead of squaring. Option D doubles the fraction.
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Answer A
Answer: A
Common error: Square the amplitude ratio to get the intensity ratio. Square-root the intensity ratio to get the amplitude ratio. Do not mix the two up.
22IBP2H-021-01-04Longitudinal wavesPaper 1AMedium1 mark
A sound wave travels through air in the +x direction. Displacement of air molecules is taken as positive in the +x direction. At one instant, molecule P has zero displacement. Molecules just to the left of P are displaced in the +x direction, and molecules just to the right of P are displaced in the −x direction. Which statement describes the position of P at this instant?
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The molecules on both sides of P are displaced towards P, so the air is crowded together there. That makes P the centre of a compression. B would need the neighbours displaced away from P. C and D describe points of maximum displacement, but P has zero displacement.
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Answer A
Answer: A
Exam tip: On a displacement–distance graph for a longitudinal wave, compressions and rarefactions sit at zero-displacement points. Check which way the neighbouring molecules point to tell them apart.
An engineer sends a transverse wave along a taut steel suspension cable. The wave speed is 12 m s−1 and the frequency is 4.0 Hz. Two clamps on the cable are 1.5 m apart along the direction of travel. What is the phase difference between the oscillations of the two clamps?
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λ = v/f = 12/4.0 = 3.0 m. The separation is 1.5/3.0 = 0.5 of a wavelength. Phase difference = 2π(Δx/λ) = 2π × 0.5 = π rad, so the clamps oscillate in antiphase. Option A comes from taking λ = 12 m, which ignores the frequency: 2π × 1.5/12 = π/4. Option B comes from multiplying the fraction 0.5 by π instead of 2π. Option D comes from treating the 1.5 m separation as one whole wavelength.
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Answer C
Answer: C
Exam tip: Find λ first, then express the separation as a fraction of λ. Multiply that fraction by 2π to get the phase difference.
24IBP2H-021-01-06The wave equationPaper 1AMedium1 mark
In a 100 m sprint, a timekeeper stands at the finish line. The timekeeper wrongly starts the stopwatch on hearing the starting pistol rather than on seeing its flash. The speed of sound in air is 340 m s−1. How does the recorded time compare with the true time?
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The sound takes 100/340 ≈ 0.29 s to reach the finish. Light arrives almost instantly. The watch therefore starts 0.29 s late and the recorded time is 0.29 s too short. B gets the direction wrong. C doubles the path as if it were an echo. D uses v/d instead of d/v.
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Answer A
Answer: A
Common error: Starting the clock late makes the recorded interval shorter, not longer.
25IBP3H-C2-20r34-05Electromagnetic and mechanical wavesPaper 1AMedium1 mark
A speaker above a swimming pool emits a tone of frequency 740 Hz. The speed of sound is 340 m s−1 in air and 1480 m s−1 in water. What is the wavelength of the sound after it enters the water?
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The frequency is set by the source and does not change at the boundary, so λ = v/f = 1480/740 = 2.0 m. The wavelength in air is 340/740 = 0.46 m, so quoting that value ignores the change of speed. Scaling the air wavelength by the inverse speed ratio gives 0.11 m, which wrongly makes the wavelength shorter in the faster medium. The value 4.4 is the ratio of the speeds, not a wavelength.
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Answer C
Answer: C
Common error: When a wave changes medium, frequency stays the same; speed and wavelength change in proportion.
26IBP3H-C2-20r34-06Electromagnetic and mechanical wavesPaper 1AMedium1 mark
A pulse of ultrasound is sent into soft tissue, in which its speed is 1500 m s−1. The pulse reflects from a boundary and the echo returns to the probe after 40 μs. How deep is the boundary?
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The pulse travels to the boundary and back, so the total path is 1500 × 40 × 10−6 = 0.060 m and the depth is half of this, 0.030 m = 3.0 cm. Forgetting to halve gives 6.0 cm. Doubling the path instead gives 12 cm. Using the speed of sound in air, 340 m s−1, gives 0.68 cm.
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Answer A
Answer: A
Exam tip: In pulse-echo problems the time recorded is for a round trip, so halve the distance travelled.
27IBP3H-C2-20r34-07Sound wavesPaper 1AMedium1 mark
A sound wave of frequency 170 Hz travels through air at 340 m s−1. Two air particles lie along the direction of travel, 0.25 m apart. What is the phase difference between their oscillations?
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The wavelength is 340/170 = 2.0 m. The separation is 0.25/2.0 = 1/8 of a wavelength, which corresponds to a phase difference of (1/8) × 2π = π/4 rad. The value 0.125 is the fraction of a cycle, not an angle in radians. A phase difference of π/2 would need a separation of a quarter of a wavelength, 0.50 m. A phase difference of π would need half a wavelength, 1.0 m.
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Answer B
Answer: B
Common error: A fraction of a wavelength must be multiplied by 2π to give the phase difference in radians.
A sound wave travels through air. The graph shows how the displacement s of one air molecule M varies with time. A positive displacement is in the direction in which the sound travels, and T is the period.
At which time is M at the centre of a compression?
Displacement of one air molecule against time; time is shown as a multiple of the period T. Positive displacement is in the direction in which the sound travels.Show mark scheme
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Step 1At the centre of a compression (or rarefaction) the molecule itself has zero displacement, so only T/2 and T are possible.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2As the wave moves forward, M takes on the displacement of the molecules just behind it. At t = T M is moving forwards (displacement increasing), so the molecules just behind it are displaced forwards, towards M; the molecules just ahead of M were displaced backwards, also towards M.
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Step 3Molecules on both sides are displaced towards M: M is at the centre of a compression at t = T. (At T/2 M moves backwards and is at the centre of a rarefaction.)
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AAt T/4 M has its maximum forward displacement; its neighbours are displaced by nearly the same amount, so the air there is at normal density.
BAt T/2 M has zero displacement but is moving backwards, against the direction of travel: this is the centre of a rarefaction.
CAt 3T/4 M has its maximum backward displacement; this is not a compression or rarefaction.
DCorrect: zero displacement while moving in the direction of travel of the wave marks the centre of a compression.
29IBP2S-020-00-09Longitudinal wavesPaper 1AHard1 mark
A longitudinal wave travels along a long spring in the +x direction. At one instant, the displacement s of the coils is given by s = A sin(2πx/λ). A positive value of s means a coil is displaced in the +x direction. At which position is the centre of a compression?
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Just to the left of x = λ/2, s is positive, so those coils are displaced towards λ/2. Just to the right, s is negative, so those coils are also displaced towards λ/2. The coils crowd together there, so x = λ/2 is the centre of a compression. At x = 0 the neighbouring coils move apart, giving a rarefaction. At λ/4 and 3λ/4 the displacement is a maximum, but neighbouring coils move in the same direction, so there is no change in density.
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Answer B
Answer: B
Common error: Maximum displacement is not a compression. Compressions and rarefactions occur where the displacement is zero; look at which way the neighbouring particles are displaced.
A transverse wave of wavelength 0.80 m travels in the +x direction along a cable. At t = 0 a crest is at x = 0.20 m. At t = 0.050 s a crest is at x = 0.30 m. Which of the following is a possible frequency of the wave?
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The crest at x = 0.30 m need not be the original crest. In 0.050 s the waveform moved d = 0.10 + 0.80n metres, where n = 0, 1, 2, … So v = d/0.050 = 2.0 + 16n m s−1 and f = v/λ = 2.5 + 20n Hz. With n = 1, f = 22.5 Hz. 17.5 Hz would require the wave to move 0.70 m in the −x direction. 5.0 Hz and 10 Hz would shift the crests by 0.20 m and 0.40 m, which does not place a crest at 0.30 m.
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Answer D
Answer: D
Exam tip: Two snapshots of a periodic wave only fix the shift up to a whole number of wavelengths. Write the general shift as Δx + nλ.
31IBP2H-021-00-08Longitudinal wavesPaper 1AHard1 mark
A sound wave travels in the +x direction through air. At one instant, the centre of a compression is at position Q. Which describes the air particle at Q at this instant?
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Notes
At the centre of a compression, particles on either side are displaced towards Q, so Q has zero displacement. Zero displacement in simple harmonic oscillation means maximum speed. As the pattern moves in +x, the particle at Q next takes the displacement of the particle just behind it, which is displaced forwards (+x). So the particle at Q is moving in the direction of wave travel. In a rarefaction the particle moves the opposite way.
✓ 1
Answer B
Answer: B
Common error: A compression is not where particles are stationary. The particles there move fastest, forwards with the wave.
32IBP2H-021-01-07Intensity and amplitudePaper 1AHard1 mark
A small source emits sound uniformly in all directions, with no absorption. Microphone M1 is 2.0 m from the source and microphone M2 is 6.0 m from the source. Which row gives the ratio of amplitudes at M1 and M2, and the ratio of intensities at M1 and M2?
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Notes
For a point source, I ∝ 1/r2, so I1/I2 = (6.0/2.0)2 = 9. Since I ∝ A2, A ∝ 1/r, giving A1/A2 = 3. B swaps the two relationships. C assumes the intensity ratio equals the amplitude ratio. D applies the inverse-square law to amplitude and then squares again.
✓ 1
Answer A
Answer: A
Exam tip: For a point source, amplitude falls as 1/r and intensity falls as 1/r².
33IBP2H-021-01-08Longitudinal wavesPaper 1AHard1 mark
A microphone records the displacement–time graph of an air molecule as a sound wave passes. The trace is sinusoidal and completes exactly 5 cycles in 2.0 ms. A simultaneous measurement shows that a compression and the nearest rarefaction are 0.068 m apart. What is the speed of the sound?
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Notes
f = 5 / (2.0 × 10−3) = 2500 Hz. A compression and the nearest rarefaction are half a wavelength apart, so λ = 2 × 0.068 = 0.136 m. Then v = fλ = 2500 × 0.136 = 340 m s−1. Option A uses f = 1/(2.0 ms), ignoring the 5 cycles. Option B takes 0.068 m as the full wavelength. Option D takes it as a quarter-wavelength.
✓ 1
Answer C
Answer: C
Common error: Compression to adjacent rarefaction is λ/2. A quarter-wavelength separates a compression from a point of maximum displacement.
The displacement s of the particles in a medium carrying an ultrasound wave varies with position x (in m) and time t (in s) as s = s0 sin[2π(3.0 × 106t − 2000x)]. What is the speed of the wave?
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Notes
Comparing with s = s0 sin[2π(ft − x/λ)] gives f = 3.0 × 106 Hz and 1/λ = 2000 m−1, so λ = 5.0 × 10−4 m. Then v = fλ = 3.0 × 106 × 5.0 × 10−4 = 1.5 × 103 m s−1. Multiplying f by 2000 instead of dividing gives 6.0 × 109. Dividing 2000 by f gives 6.7 × 10−4. Multiplying the correct speed by 2π, by treating the coefficients as angular quantities, gives 9.4 × 103.
✓ 1
Answer D
Answer: D
Exam tip: When the phase is written as 2π(ft − x/λ), read f and 1/λ directly from the coefficients of t and x.
A ripple tank is lit from above so that the wavefronts form an image on a screen below the tank. The image is 1.5 times larger than the real wave pattern. For each frequency f the student freezes the image with a stroboscope and measures the distance across five wavelengths, 5λ, on the screen with a ruler; each reading has an uncertainty of ±0.2 cm.
A graph of λ against 1/f is a straight line through the origin.
f / Hz
10
14
18
22
26
30
5λ / cm
11.9
8.6
6.7
5.4
4.6
3.9
λ / cm
2.38
1.72
1.34
1.08
0.92
0.78
f−1 / s
0.100
0.0714
0.0556
0.0455
0.0385
(a)
(i)
Complete the table.
(1)
(b)
(i)
Using the full range of the data, determine the speed of the image of the waves on the screen.
(2)
(c)
(i)
Determine the speed of the water waves in the tank.
(1)
(d)
(i)
Calculate the percentage uncertainty in λ at 30 Hz. Suggest how the student could halve this uncertainty using the same ruler.
(2)
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Notes
Part (a)(i)
1/30 = 0.0333 s
✓ 1
Accept 0.033.
Part (b)(i)
Gradient of λ against f−1 is v: (2.38 − 0.78) × 10−2/(0.100 − 0.0333)
✓ 1
The 10 Hz and 30 Hz rows must be used (full range); use of the value from (a): allow ECF. Adjacent rows only: [1 max].
v = 0.24 m s−1
✓ 1
Accept 0.23–0.25 m s−1.
Part (c)(i)
Every wavelength on the screen is 1.5 times the real one while the frequency is unchanged, so vtank = 0.24/1.5 = 0.16 m s−1
✓ 1
Multiplying by 1.5: [0]. Allow ECF from (b).
Part (d)(i)
0.2/3.9 × 100 = 5 %
✓ 1
Accept 5.1 %.
Measure across 10 wavelengths (about 7.8 cm) instead of 5, so the same ±0.2 cm is a fraction half as large: ≈ 2.6 %
✓ 1
A change of instrument or "repeat readings" alone: [0] for this mark.
Answers: (a)(i) 0.0333 s · (b)(i) 0.24 m s−1 · (c)(i) 0.16 m s−1 · (d)(i) 5 %; measure 10λ (the remaining parts are explanations — see the table above)
A loudspeaker emits sound of frequency 2500 Hz. Two small microphones lie on a straight line in front of it. One is fixed and the other is moved away from the loudspeaker so that their separation d increases. Both signals are displayed on an oscilloscope, and the phase difference Δφ between them is measured for each separation. The separation is read from a scale fixed to the bench.
The graph shows Δφ against d with the line of best fit.
d / cm
2.0
4.0
6.0
8.0
10.0
12.0
Δφ / rad
1.16
2.04
3.00
3.89
4.75
5.76
Phase difference Δφ against microphone separation d with the line of best fit (drawn to scale).
(a)
(i)
Determine the gradient of the line in rad m−1 and hence the wavelength of the sound.
(2)
(b)
(i)
Hence determine the speed of sound.
(1)
(c)
(i)
The frequency is known to ±1 % and the wavelength to ±3 %. Determine the absolute uncertainty in the speed of sound.
(2)
(d)
(i)
The model predicts a line through the origin. Use the graph to determine the zero error in the separation scale and state its effect on your answer to (b).
(2)
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Notes
Part (a)(i)
Large triangle, e.g. (0, 0.23) to (13.5, 6.41): gradient = 0.457 rad cm−1 = 45.7 rad m−1
✓ 1
Accept 44.5–47.0 rad m−1.
A separation of one wavelength gives 2π, so λ = 2π/gradient = 0.137 m
✓ 1
Accept 0.134–0.141 m. Allow ECF.
Part (b)(i)
v = fλ = 2500 × 0.137 = 343 m s−1
✓ 1
Accept 334–353 m s−1. Allow ECF from (a).
Part (c)(i)
Percentage uncertainty = 1 % + 3 % = 4 %
✓ 1
Δv = 0.04 × 343 = ±10 m s−1
✓ 1
Absolute uncertainty to 1 s.f. (±14 m s−1 also accepted). Allow ECF from (b).
Part (d)(i)
Intercept 0.23 rad corresponds to 0.23/gradient = 0.5 cm: the true separation is ≈ 0.5 cm more than the scale reading
✓ 1
Accept 0.4–0.6 cm. Allow ECF from (a).
A constant offset changes only the intercept; the gradient, λ and v are unaffected
✓ 1
Answers: (a)(i) 45.7 rad m−1; 0.137 m · (b)(i) 343 m s−1 · (c)(i) ±10 m s−1 · (d)(i) ≈ 0.5 cm; no effect (the remaining parts are explanations — see the table above)
A rope passes over a pulley at one end of a bench, and a load hanging from the end of the rope sets the tension F. A pulse is made at the other end, and a video camera recording 250 frames per second shows the pulse passing two markers 3.00 m apart. The time t taken by the pulse to travel between the markers is found from the frame count to ±0.004 s.
A student suggests that the speed of the pulse is v = KFn, where K and n are constants, and plots lg v against lg F. Theory predicts v = √(F/μ), where μ is the mass per unit length of the rope.
F / N
t / s
lg(F / N)
lg(v / m s−1)
1.5
0.536
0.176
0.748
3.0
0.380
0.477
0.897
5.0
0.296
0.699
1.006
8.0
0.232
0.903
1.112
12.0
0.190
1.079
18.0
0.155
1.255
1.287
lg v against lg F with the line of best fit (drawn to scale).
(a)
(i)
Calculate the missing value of lg(v / m s−1) for F = 12.0 N.
(1)
(b)
(i)
Determine the gradient of the line and deduce whether the data support the theoretical relationship.
(2)
(c)
(i)
The rope breaks if the tension exceeds 40 N. Use the line of best fit to predict the greatest speed at which a pulse can travel along this rope.
(2)
(d)
(i)
Determine the percentage uncertainty in v for F = 18.0 N. Suggest how the student could reduce this uncertainty using the same camera.
(2)
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Notes
Part (a)(i)
v = 3.00/0.190 = 15.79 m s−1, so lg v = 1.198
✓ 1
Accept 1.20. Answer to 3 d.p. not required.
Part (b)(i)
Gradient from a large triangle = 0.50
✓ 1
Accept 0.48–0.52. Award [0] for a gradient from two adjacent table values only.
The theory gives v = F½/√μ, so lg v = ½ lg F − ½ lg μ: the gradient n ≈ 0.5 supports the theory
✓ 1
Conclusion must be consistent with their gradient. Allow ECF from their gradient.
Part (c)(i)
Equation of the line: lg v = 0.50 lg F + 0.66; with lg 40 = 1.602, lg v = 0.50 × 1.602 + 0.66 = 1.46
✓ 1
Intercept read from the graph: accept 0.65–0.67. Allow ECF from (b). The prediction lies beyond the data, so the equation of the line (not a reading from the graph) is expected.
v = 101.46 = 29 m s−1
✓ 1
Accept 28–30 m s−1. Using 1.46 as the speed scores [0] for this mark.
Part (d)(i)
% uncertainty in v = % uncertainty in t = 0.004/0.155 × 100 = 2.6 %
✓ 1
Accept 3 %. The uncertainty in the 3.00 m distance is negligible.
Place the markers further apart (use a longer rope), or time the pulse over several reflections, so that the measured time is longer while the ±0.004 s stays the same
✓ 1
The suggestion must increase the measured time. "Repeat the readings" alone is not enough here.
Answers: (a)(i) 1.198 · (b)(i) 0.50; supports · (c)(i) 29 m s−1 · (d)(i) 2.6 % (the remaining parts are explanations — see the table above)
38C-1B-06Intensity and amplitudePaper 1BHard7 marks
A small filament lamp is placed in a darkened room. A radiation sensor, which responds to all wavelengths, measures the intensity I of the radiation at distance r from the filament. With the lamp switched off, the sensor reads 0.40 W m−2 because of radiation from the warm surroundings; this reading does not change during the experiment.
The student wants to test whether the radiation from the filament obeys an inverse-square law.
r / m
0.200
0.300
0.400
0.500
0.600
Sensor reading / W m−2
24.47
10.91
6.42
4.25
3.03
(a)
(i)
Determine the intensity due to the lamp alone at r = 0.600 m.
(1)
(b)
(i)
Deduce whether the data support the inverse-square law.
(2)
(c)
(i)
Show that the total power radiated by the filament is about 12 W.
(1)
(d)
(i)
The filament has a surface area of 6.0 × 10−6 m² and an emissivity of 0.40. Determine the temperature of the filament.
(2)
(ii)
The power is uncertain by ±8 %. Determine the absolute uncertainty in the temperature.
(1)
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Notes
Part (a)(i)
3.03 − 0.40 = 2.63 W m−2
✓ 1
Subtraction of the background must be seen.
Part (b)(i)
Ir² calculated, after subtracting the background, for at least three distances, e.g. 0.963, 0.963, 0.947 W
✓ 1
Allow ECF from (a). Using uncorrected readings gives 0.979, 1.027, 1.091 (rising): [1 max] for a correct conclusion from these.
The values are constant within ≈ 2 %, so I ∝ 1/r²: the law is supported
✓ 1
The conclusion must refer to the constancy of their product (or ratio).
Part (c)(i)
P = 4πr²I = 4π × 0.958 = 12.0 W
✓ 1
Must use a background-corrected value; allow ECF from (b). A 2-s.f. value or the full substitution is required.
Percentage uncertainty in T = ¼ × 8 % = 2 %; ΔT = 0.02 × 3060 ≈ ±60 K
✓ 1
Absolute uncertainty to 1 s.f. Using 4 × 8 % = 32 % is a named error: [0]. Allow ECF from (d)(i).
Answers: (a)(i) 2.63 W m−2 · (b)(i) Ir² ≈ 0.96 W (constant) · (d)(i) 3.06 × 103 K · (d)(ii) ±60 K (the remaining parts are explanations — see the table above)
During a storm an observer sees a flash of lightning and hears the thunder 6.0 s later. The air temperature is 30 °C. The speed of sound in air at kelvin temperature T is v = √(γRT/M), where γ = 1.40 and M = 0.0290 kg mol−1 is the molar mass of air.
(a)
(i)
Show that the speed of sound in the air is about 350 m s−1.
(2)
(ii)
Calculate the distance to the lightning.
(1)
(iii)
Explain why the travel time of the light can be ignored.
(1)
(b)
(i)
State two differences between the light and the sound, as waves.
(2)
(ii)
The thunder contains frequencies from 50 Hz to 250 Hz. Determine the range of wavelengths and explain why the thunder can be heard behind a large building that blocks the view of the lightning.
(2)
(iii)
The lightning also emits electromagnetic waves of frequency 30 kHz. State the region of the electromagnetic spectrum to which they belong.
(1)
(c)
(i)
The lightning flash transfers a charge of 5.0 C in 30 μs through a potential difference of 1.0 × 108 V. Determine the mean current and the energy transferred.
(2)
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Notes
Part (a)(i)
T = 30 + 273 = 303 K
✓ 1
Use of 30 K: [0] for the question part.
v = √(1.40 × 8.31 × 303/0.0290) = 349 m s−1
✓ 1
Answer to at least 3 s.f. required.
Part (a)(ii)
d = 349 × 6.0 = 2.1 × 103 m
✓ 1
Allow ECF from (a)(i).
Part (a)(iii)
The light takes 2.1 × 103/(3.00 × 108) ≈ 7 μs, which is negligible compared with 6.0 s
✓ 1
A numerical comparison or "about a million times faster" required. Allow ECF from (a)(ii).
Part (b)(i)
Light is an electromagnetic wave (oscillating electric and magnetic fields) and can travel through a vacuum; sound is a mechanical wave and needs a medium
✓ 1
Any two distinct differences, one mark each.
Light is transverse; sound is longitudinal (or: light travels about 106 times faster)
✓ 1
Part (b)(ii)
λ = v/f: from 1.4 m (250 Hz) to 7.0 m (50 Hz)
✓ 1
Allow ECF from (a)(i).
These wavelengths are comparable with the size of buildings and gaps, so the sound diffracts around the building
✓ 1
Reference to diffraction and comparable size needed.
Part (b)(iii)
Radio waves (λ = 1.0 × 104 m)
✓ 1
Part (c)(i)
I = Δq/Δt = 5.0/30 × 10−6 = 1.7 × 105 A
✓ 1
Energy = qV = 5.0 × 1.0 × 108 = 5.0 × 108 J
✓ 1
Answers: (a)(i) 349 m s−1 · (a)(ii) 2.1 × 103 m · (b)(ii) 1.4 m to 7.0 m · (b)(iii) radio · (c)(i) 1.7 × 105 A; 5.0 × 108 J (the remaining parts are explanations — see the table above)
A transverse wave travels to the right along a long horizontal rope. The graph shows the displacement y of the rope against the position x along it at one instant, t = 0. P and Q are two points on the rope.
Displacement y of the rope against position x at t = 0 (drawn to scale). The wave travels in the positive x direction.
(a)
(i)
State the amplitude and the wavelength of the wave.
(1)
(ii)
A point on the rope completes 5 oscillations in 2.0 s. Calculate the speed of the wave.
(1)
(b)
(i)
State and explain the direction of motion of Q at t = 0.
(2)
(ii)
Determine the phase difference between the oscillations of P and Q. Give your answer in radians.
(2)
(c)
(i)
Determine the displacement of Q at t = 0.10 s.
(3)
(ii)
Calculate the maximum speed of a point on the rope, and explain why it is different from the speed of the wave.
(2)
(d)
(i)
Outline how the wave transfers energy along the rope even though no part of the rope moves along it.
(1)
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Notes
Part (a)(i)
Amplitude 4.0 cm and wavelength 1.60 m
✓ 1
Both needed.
Part (a)(ii)
f = 5/2.0 = 2.5 Hz, so v = fλ = 2.5 × 1.60 = 4.0 m s−1
✓ 1
Allow ECF from (a)(i).
Part (b)(i)
Q is moving upwards (in the positive y direction)
✓ 1
The wave profile moves to the right, so Q next takes the displacement of the rope just to its left, which is higher (less negative) than Q
✓ 1
Accept a sketch of the profile shifted slightly to the right. Award [0] for "upwards" with no valid reason.
In 0.10 s the wave moves vt = 4.0 × 0.10 = 0.40 m (a quarter of a wavelength)
✓ 1
Allow ECF from (a)(ii).
Q then has the displacement that the point at x = 1.00 − 0.40 = 0.60 m had at t = 0
✓ 1
ALT: 0.10 s = T/4, and Q moves up from −2.8 cm through equilibrium.
y = +2.8 cm
✓ 1
Accept +2.7 to +2.9 cm from the graph. Sign required.
Part (c)(ii)
vmax = 2πfA = 2π × 2.5 × 0.040 = 0.63 m s−1
✓ 1
Allow ECF from (a).
The wave speed is the speed at which the profile (and energy) moves along the rope; the point only oscillates transversely about a fixed position, with a speed that depends on the amplitude and frequency
✓ 1
Some reference to the different meanings is needed.
Part (d)(i)
Each part of the rope pulls on (does work on) the next, so oscillation energy is passed along while each part only oscillates about its equilibrium position
✓ 1
Accept "the disturbance, not the material, travels".
Answers: (a)(i) 4.0 cm; 1.60 m · (a)(ii) 4.0 m s−1 · (b)(ii) 3π/4 rad (2.36 rad) · (c)(i) +2.8 cm · (c)(ii) 0.63 m s−1(the remaining parts are explanations — see the table above)
41IBP2H-021-00-09Electromagnetic and mechanical wavesPaper 2Easy5 marks
Long-distance internet links use infrared signals of wavelength 1550 nm (in a vacuum) sent along glass optical fibres. The speed of light in the fibre core is 2.04×108 m s−1.
(a)
State one difference between electromagnetic waves and mechanical waves.
(1)
(b)
Calculate the frequency of the infrared signal.
(2)
(c)
Calculate the wavelength of the signal inside the fibre core.
(2)
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Notes
Part (a)
Electromagnetic waves can travel through a vacuum, whereas mechanical waves need a material medium. Also accept: EM waves are oscillating electric and magnetic fields, whereas mechanical waves are oscillations of the particles of a medium.
✓ 1
Part (b)
f = c/λ = 3.00×108 / 1550×10−9
✓ 1
f = 1.94×1014 Hz
✓ 1
Part (c)
The frequency is unchanged, so λ = v/f = 2.04×108 / 1.94×1014. Also accept λ = λ0v/c = 1550×10−9 × 2.04×108 / 3.00×108.
✓ 1
λ = 1.05×10−6 m (about 1050 nm). Allow error carried forward from (b).
✓ 1
Common error: When a wave enters a new medium, its frequency stays the same. Only its speed and wavelength change.
43C-2-22Transverse and longitudinal wavesPaper 2Medium11 marks
An earthquake produces longitudinal P waves and transverse S waves that travel through the Earth's crust. The graph shows the travel times of the two waves to six seismic stations at different distances from the source, with lines of best fit.
Travel time against distance for P waves and S waves (drawn to scale).
(a)
(i)
Use the graph to determine the speeds of the P waves and of the S waves.
(2)
(ii)
At another station the S wave arrives 30.0 s after the P wave. Determine the distance of this station from the source.
(3)
(iii)
The P wave reaches this station at 14:02:50. Determine the time at which the earthquake occurred.
(1)
(b)
(i)
Describe the motion of a small region of rock at the station as (1) the P wave and (2) the S wave passes.
(2)
(ii)
The S wave has a frequency of 0.80 Hz. Calculate its wavelength.
(1)
(iii)
The S wave makes the ground at the station move in simple harmonic motion with amplitude 2.0 cm. Determine the maximum horizontal force needed to make a building of mass 2.0 × 106 kg move with the ground.
(2)
Show mark scheme
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Notes
Part (a)(i)
P waves: speed = 1/gradient = 400 km/64.5 s = 6.2 km s−1
✓ 1
Accept 6.0–6.4 km s−1. Gradient itself (0.16 s km−1) quoted as speed: [0].
S waves: 400 km/111 s = 3.6 km s−1
✓ 1
Accept 3.5–3.7 km s−1.
Part (a)(ii)
d/vS − d/vP = 30.0 s
✓ 1
d = 30.0/(1/3.6 − 1/6.2)
✓ 1
Allow ECF from (a)(i).
d = 258 km
✓ 1
Accept 240–275 km. Reading the graph where the lines are 30 s apart is an acceptable alternative.
Part (a)(iii)
Travel time of the P wave = 258/6.2 = 42 s, so the earthquake occurred at about 14:02:08
✓ 1
Allow ECF from (a)(ii).
Part (b)(i)
(1) It oscillates about its equilibrium position parallel to the direction in which the wave travels
✓ 1
(2) It oscillates about its equilibrium position perpendicular to the direction in which the wave travels
✓ 1
Reference to oscillation about a fixed position is needed once.
Part (b)(ii)
λ = v/f = 3.6 × 103/0.80 = 4.5 × 103 m
✓ 1
Allow ECF from (a)(i).
Part (b)(iii)
amax = ω²x0 = (2π × 0.80)² × 0.020 = 0.51 m s−2
✓ 1
F = mamax = 2.0 × 106 × 0.51 = 1.0 × 106 N
✓ 1
Answers: (a)(i) 6.2 km s−1; 3.6 km s−1 · (a)(ii) 258 km · (a)(iii) 14:02:08 · (b)(ii) 4.5 × 103 m · (b)(iii) 1.0 × 106 N (the remaining parts are explanations — see the table above)
A ship's echo sounder sends pulses of 50.0 kHz sound straight down to the seabed. It records the time taken for each echo to return. The table gives the speed of sound in seawater at different temperatures.
Temperature / °C: 0, 10, 20, 30 Speed of sound / m s−1: 1449, 1490, 1522, 1546
The echo sounder is programmed to assume a speed of 1490 m s−1.
(a)
Outline the nature of the sound wave as it travels through the seawater.
(2)
(b)
Calculate the wavelength of the pulse in water at 20 °C.
(2)
(c)
The echo returns 0.400 s after the pulse is emitted. The water is actually at 20 °C. Determine the difference between the true depth and the depth shown by the echo sounder.
(3)
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Notes
Part (a)
It is a longitudinal wave: water molecules oscillate parallel to the direction of energy transfer.
✓ 1
This produces alternating compressions and rarefactions (regions of higher and lower pressure).
✓ 1
Part (b)
λ = v/f = 1522 / 50.0×103
✓ 1
λ = 3.04×10−2 m (3.04 cm)
✓ 1
Part (c)
The pulse travels down and back, so depth = v × t/2
✓ 1
Displayed depth = 1490 × 0.200 = 298 m; true depth = 1522 × 0.200 = 304.4 m
✓ 1
Difference = 6.4 m, with the true depth greater than the displayed depth
✓ 1
Common error: Echo times cover the journey there and back. Halve the time before finding a distance.
A sinusoidal transverse wave travels along a long zip-line cable from point A towards point B. B is 0.75 m further along the cable than A. The displacement–time graph for point A has amplitude 4.0 cm and period 0.25 s. At t = 0, A has zero displacement and its displacement is increasing. The wave speed is 12 m s−1.
Displacement–time graph for point A
(a)
Calculate the wavelength of the wave.
(2)
(b)
Determine the phase difference between the oscillations of A and B.
(2)
(c)
Describe the displacement–time graph for point B from t = 0 to t = 0.25 s.
(3)
(d)
Calculate the average speed of point A over one complete period. Compare it with the wave speed.
(2)
Show mark scheme
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Notes
Part (a)
f = 1/T = 1/0.25 = 4.0 Hz
✓ 1
λ = v/f = 12/4.0 = 3.0 m
✓ 1
Part (b)
Separation as a fraction of a wavelength = 0.75/3.0 = 1/4
✓ 1
Phase difference = π/2 rad (90°), with B lagging A. ECF from (a).
✓ 1
Part (c)
Sinusoidal, with the same amplitude (4.0 cm) and the same period (0.25 s) as A
✓ 1
Shifted later by T/4 = 0.0625 s compared with A
✓ 1
At t = 0, B is at −4.0 cm (a trough). It passes through zero, moving upward, at t = 0.0625 s, and reaches +4.0 cm at t = 0.125 s. ECF from (b).
✓ 1
Part (d)
Distance moved in one period = 4A = 0.16 m, so average speed = 0.16/0.25 = 0.64 m s−1
✓ 1
This is much less than 12 m s−1: the particle speed and the wave speed are different quantities
✓ 1
Exam tip: In one full oscillation a particle travels four amplitudes. Never confuse particle speed with wave speed.
46IBP2H-021-00-13Electromagnetic and mechanical wavesPaper 2Medium6 marks
A rover on Mars communicates with Earth using radio waves of frequency 8.4 GHz. It also carries a microphone that records the sound made when its laser strikes a nearby rock.
(a)
Explain why radio waves, but not sound waves, can travel from Mars to Earth.
(2)
(b)
Calculate the wavelength of the radio waves.
(1)
(c)
On one day, Mars is 2.25×1011 m from Earth. Calculate the time for a signal to travel from the rover to Earth.
(1)
(d)
The laser hits a rock 3.0 m from the microphone. The sound is recorded 12.4 ms after the laser pulse is fired. Determine the speed of sound in the Martian atmosphere, stating and justifying any assumption you make.
(2)
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Part (a)
Sound is a mechanical wave that needs particles of a medium to oscillate
✓ 1
Interplanetary space is effectively a vacuum, but EM waves (oscillating electric and magnetic fields) need no medium
✓ 1
Part (b)
λ = 3.00×108 / 8.4×109 = 3.6×10−2 m
✓ 1
Part (c)
t = 2.25×1011 / 3.00×108 = 750 s
✓ 1
Part (d)
Assumption: the light travel time (about 10−8 s) is negligible compared with 12.4 ms, so the whole delay is due to the sound
✓ 1
v = 3.0 / 12.4×10−3 = 240 m s−1 (accept 242 m s−1)
✓ 1
Exam tip: When light and sound share a delay, show briefly that the light travel time is negligible before ignoring it.
Outline why the wavelength decreases as the frequency increases.
(1)
(b)
Determine the speed of sound in the steel, using the data.
(2)
(c)
A worker strikes the rail 1.2 km from an engineer. The sound reaches the engineer both through the rail and through the air. The speed of sound in air is 340 m s−1. Calculate the time interval between the two arrivals.
(3)
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Part (a)
The wave speed depends only on the medium, so it is constant; since λ = v/f, λ is inversely proportional to f
✓ 1
Part (b)
Calculates fλ for the rows: 5900, 5960, 5880, 5920 m s−1
✓ 1
Mean v ≈ 5.9 × 103 m s−1
✓ 1
Part (c)
tair = 1200/340 = 3.53 s
✓ 1
trail = 1200/5.9 × 103 = 0.203 s (ECF from (b))
✓ 1
Δt = 3.3 s
✓ 1
Exam tip: When 'using the data' is asked, use more than one row and average, rather than relying on a single reading.
48IBP2H-021-01-11Intensity and amplitudePaper 2Medium7 marks
At an outdoor concert, a loudspeaker tower can be modelled as a point source. It radiates 12 W of sound power uniformly in all directions.
(a)
Calculate the sound intensity at a distance of 25 m from the tower.
(2)
(b)
Safety rules limit the intensity at the front barrier to 1.0 × 10−2 W m−2. Determine the minimum distance between the tower and the barrier.
(2)
(c)
The barrier is placed at this minimum distance. Determine the ratio of the sound amplitude at the barrier to the sound amplitude at 25 m from the tower.
(2)
(d)
Outline one reason why the measured intensity at 25 m could be less than the value calculated for that distance.
(1)
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Part (a)
I = P/(4πr2) = 12/(4π × 252)
✓ 1
I = 1.5 × 10−3 W m−2
✓ 1
Part (b)
r = √(P/(4πI)) = √(12/(4π × 0.010))
✓ 1
r = 9.8 m
✓ 1
Part (c)
A ∝ √I, so A ∝ 1/r; equivalently, ratio = √(Ibarrier/I25 m)
✓ 1
Ratio = 25/9.8 ≈ 2.6. Allow error carried forward from the earlier answers.
✓ 1
Part (d)
Any one of the following :
✓ 1
• Some sound energy is absorbed by the air or by the crowd.
—
• The ground or obstacles absorb or scatter some of the sound.
—
• The source is not truly a point source radiating uniformly; for example, some power is directed away from the listener.
—
Common error: Amplitude is proportional to 1/r, so the amplitude ratio equals the distance ratio, not its square. Squaring the distance ratio gives the intensity ratio instead.
49IBP2H-021-01-12Transverse and longitudinal wavesPaper 2Medium6 marks
An earthquake produces P-waves, which travel through rock at 6.0 km s−1, and S-waves, which travel at 3.5 km s−1. At a monitoring station, the S-waves arrive 42 s after the P-waves.
(a)
P-waves are longitudinal and S-waves are transverse. Explain why S-waves cannot pass through the liquid outer core of the Earth.
(2)
(b)
The P-waves have a frequency of 2.0 Hz. Calculate their wavelength.
(1)
(c)
Determine the distance from the station to the earthquake source.
(3)
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Part (a)
In a transverse wave, particles oscillate perpendicular to the direction of travel, which needs a shearing restoring force
✓ 1
A liquid cannot sustain shear forces, so the transverse oscillation cannot propagate
✓ 1
Part (b)
λ = 6000/2.0 = 3.0 × 103 m
✓ 1
Part (c)
d/3500 − d/6000 = 42
✓ 1
d = 42/(1/3500 − 1/6000)
✓ 1
d = 3.5 × 105 m (350 km)
✓ 1
Exam tip: Write each arrival time in terms of the unknown distance, then subtract. Keep the speeds in m s⁻¹ throughout.
50IBP2H-021-01-13Longitudinal wavesPaper 2Medium6 marks
A sound wave of frequency 680 Hz travels through air in the +x direction at 340 m s−1. Displacement is taken as positive in the +x direction. At time t = 0, the displacement s of air molecules at position x is s = (2.0 μm) sin(2πx/λ).
(a)
Calculate the wavelength λ.
(1)
(b)
Identify the position between x = 0 and x = 0.50 m where there is a compression at t = 0. Explain your answer.
(3)
(c)
Calculate the time taken for this compression to reach x = 0.40 m.
(2)
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Part (a)
λ = 340/680 = 0.50 m
✓ 1
Part (b)
x = 0.25 m (ECF from (a): x = λ/2)
✓ 1
Just left of 0.25 m, s is positive (molecules displaced in +x, towards 0.25 m)
✓ 1
Just right of 0.25 m, s is negative (molecules displaced in −x, towards 0.25 m), so molecules crowd together there
✓ 1
Part (c)
Distance = 0.40 − 0.25 = 0.15 m
✓ 1
t = 0.15/340 = 4.4 × 10−4 s (ECF from (b))
✓ 1
Common error: x = 0 is also a point of zero displacement, but it is a rarefaction: neighbouring molecules are displaced away from it.
51IBP3H-C2-20r34-11The wave equationPaper 2Medium7 marks
A medical ultrasound probe emits pulses of frequency 3.5 MHz into soft tissue, in which the speed of ultrasound is 1540 m s−1. In this question use unrounded values from earlier parts in later calculations.
(a)
Calculate the wavelength of the ultrasound in the tissue.
(2)
(b)
A pulse reflects from a tissue boundary and the echo returns to the probe 52 μs after the pulse was emitted. Determine the depth of the boundary below the probe, to 3 significant figures.
(3)
(c)
Calculate the number of wavelengths of the ultrasound in the distance from the probe to the boundary.
52IBP3H-C2-20r34-12Wave graphsPaper 2Medium8 marks
A transverse wave travels in the +x direction along a cable in a wave-energy research rig. At t = 0 the displacement–distance graph of the cable has a crest of displacement +2.5 cm at x = 0. The wavelength is 0.80 m and the wave speed is 6.0 m s−1.
(a)
Calculate the frequency of the wave.
(2)
(b)
Calculate the period of the wave, to 3 significant figures.
(1)
(c)
Sketch the displacement–time graph for the particle of the cable at x = 0 for two complete oscillations. Describe the sketch.
(3)
(d)
Calculate the phase difference between the oscillations of two points on the cable that are 0.20 m apart. Give your answer in radians to 3 significant figures.
(2)
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Part (a)
f = v/λ = 6.0 / 0.80
✓ 1
f = 7.5 Hz
✓ 1
Part (b)
T = 1/f = 0.133 s
✓ 1
Part (c)
the curve starts at its maximum displacement of +2.5 cm at t = 0 and follows a cosine shape
✓ 1
the period is 0.133 s, so two complete oscillations end at t = 0.267 s
✓ 1
the axes are labelled with the amplitude 2.5 cm (and −2.5 cm minimum) and the time values
✓ 1
Part (d)
Δφ = 2π × 0.20 / 0.80
✓ 1
Δφ = 1.57 rad (π/2)
✓ 1
Model sketch
Model sketch: two complete oscillations
Common error: A displacement–time graph for one particle repeats with the period, not with the wavelength; check the horizontal axis before reading off a value.
53IBP3H-C2-20r34-13Electromagnetic and mechanical wavesPaper 2Medium7 marks
A loudspeaker in a shop produces a tone of frequency 680 Hz. The distance between the centres of successive compressions in the air is 0.50 m.
(a)
Describe how the vibration of the loudspeaker cone produces a longitudinal sound wave in the air.
(3)
(b)
Calculate the speed of the sound wave.
(2)
(c)
A microphone is placed 12.0 m from the loudspeaker. Calculate the time taken for the sound to reach it, to 3 significant figures.
(2)
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Part (a)
the cone moves back and forth, pushing the adjacent layers of air together and then apart
✓ 1
this creates alternate regions of high pressure (compressions) and low pressure (rarefactions)
✓ 1
the air particles oscillate parallel to the direction of travel and pass the disturbance along, so energy moves outwards while the air itself does not travel with the wave
✓ 1
Part (b)
λ = 0.50 m, since successive compression centres are one wavelength apart
✓ 1
v = fλ = 680 × 0.50 = 340 m s−1
✓ 1
Part (c)
t = d/v = 12.0 / 340
✓ 1
t = 0.0353 s
✓ 1
Exam tip: In a description of a sound wave, mention both the pressure variation and the direction of oscillation relative to the direction of travel.
54IBP3H-C2-20r34-14Electromagnetic and mechanical wavesPaper 2Medium7 marks
An ultraviolet-C lamp in a hospital ward emits electromagnetic waves of wavelength 254 nm in a vacuum. The speed of electromagnetic waves is 3.00 × 108 m s−1 in a vacuum and 2.00 × 108 m s−1 in the fused quartz window of the lamp.
(a)
Calculate the frequency of the radiation, to 3 significant figures.
(2)
(b)
Outline what oscillates in an electromagnetic wave and how this relates to the direction of energy transfer.
(2)
(c)
Determine the wavelength of the radiation in the quartz, to 3 significant figures.
(3)
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Part (a)
f = c/λ = 3.00 × 108 / 254 × 10−9
✓ 1
f = 1.18 × 1015 Hz
✓ 1
Part (b)
the electric field and the magnetic field oscillate
✓ 1
the oscillations are perpendicular to each other and to the direction of energy transfer, so the wave is transverse
✓ 1
Part (c)
the frequency is unchanged when the wave enters the quartz
✓ 1
λ = v/f = 2.00 × 108 / 1.18 × 1015
✓ 1
λ = 1.69 × 10−7 m (169 nm)
✓ 1
Common error: When a wave enters a new medium, the frequency stays the same; do not use the wavelength from the first medium with the new speed.
55IBP2H-021-00-12Longitudinal wavesPaper 2Hard8 marks
A longitudinal wave of wavelength 0.40 m travels in the +x direction along a stretched spring at 2.0 m s−1. Displacement to the right (+x) is positive. At t = 0, the displacement s of the coil whose undisturbed position is x is given by \(s = (6.0\text{ mm})\sin\left(\frac{2\pi x}{0.40}\right)\), with x in metres.
(a)
State the displacements at t = 0 of the coils at x = 0.10 m and x = 0.30 m.
(1)
(b)
Determine where, between x = 0 and x = 0.40 m, the centre of a compression and the centre of a rarefaction are located at t = 0.
(3)
(c)
Calculate the earliest time after t = 0 at which the coil at x = 0.40 m is at the centre of a compression.
(2)
(d)
Explain the direction of motion of the coil at x = 0.20 m at t = 0.
(2)
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Notes
Part (a)
x = 0.10 m: +6.0 mm (to the right); x = 0.30 m: −6.0 mm (to the left). Both are needed for the mark.
✓ 1
Part (b)
Near x = 0.20 m, coils on the left are displaced right and coils on the right are displaced left, so they converge
✓ 1
The centre of the compression is at x = 0.20 m
✓ 1
Coils diverge from x = 0 (and x = 0.40 m), so a rarefaction is centred there
✓ 1
Part (c)
The compression must travel from x = 0.20 m to x = 0.40 m, a distance of 0.20 m
✓ 1
t = 0.20/2.0 = 0.10 s. ECF from (b).
✓ 1
Part (d)
The pattern moves in +x, so this coil next takes the displacement of coils just to its left, which are displaced in +x
✓ 1
So the coil moves in the +x direction, at maximum speed because its displacement is zero
✓ 1
Exam tip: For a longitudinal wave, look at the neighbours of a point. Displacements pointing towards it mean a compression; displacements pointing away mean a rarefaction.
56IBP2H-021-00-14Transverse and longitudinal wavesPaper 2Hard8 marks
Three seismic stations record P-waves and S-waves from the same earthquake. Within the crust, P-waves travel at 6.0 km s−1 and S-waves at 3.5 km s−1. The table gives the delay between the arrival of the P-wave and the arrival of the S-wave at each station.
Station: A, B, C S − P delay / s: 40.0, 25.0, 60.0
(a)
Outline the difference between P-waves and S-waves in terms of the motion of the rock particles.
(2)
(b)
Explain why S-waves are not detected after passing through the Earth's liquid outer core.
(2)
(c)
Show that station A is about 340 km from the earthquake.
(3)
(d)
Determine the distance of station C from the earthquake.
(1)
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Part (a)
P-waves are longitudinal: particles oscillate parallel to the direction of propagation
✓ 1
S-waves are transverse: particles oscillate perpendicular to the direction of propagation
✓ 1
Part (b)
Transverse mechanical waves need the medium to provide a restoring force against shear (sideways) deformation
✓ 1
A liquid cannot sustain shear stress, so S-waves cannot propagate through it
✓ 1
Part (c)
Write the delay as d/vS − d/vP = 40.0 s
✓ 1
d(1/3.5 − 1/6.0) = 40.0, so d × 0.119 = 40.0
✓ 1
d = 336 km ≈ 340 km
✓ 1
Part (d)
The distance is proportional to the delay: d = 336 × 60.0/40.0 = 504 km (about 500 km). ECF from (c).
✓ 1
Common error: The delay is the difference between two travel times. Do not divide the distance by the difference in speeds.
57IBP2H-021-00-16Electromagnetic and mechanical wavesPaper 2Hard9 marks
A railway engineer strikes a steel rail with a hammer. A colleague further along the track, with one ear against the rail, hears two sounds: one through the rail and one through the air. The second sound arrives 0.55 s after the first. The speed of sound in the steel rail is 5100 m s−1 and in air it is 340 m s−1.
(a)
Outline why sound can travel through the rail as both longitudinal and transverse waves, but through air only as longitudinal waves.
(2)
(b)
The blow produces a component of frequency 1.7 kHz. Calculate the wavelength of this component in the steel and in the air.
(2)
(c)
Determine the distance between the engineer and the colleague.
(3)
(d)
The colleague can measure the delay only to ±0.05 s. Determine the absolute uncertainty in the distance found in (c).
(2)
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Part (a)
Steel is a solid that resists both compression and shear, so it provides restoring forces for both kinds of oscillation
✓ 1
Air resists compression but cannot sustain shear stress, so only longitudinal waves propagate
✓ 1
Part (b)
Steel: λ = 5100/1700 = 3.0 m
✓ 1
Air: λ = 340/1700 = 0.20 m
✓ 1
Part (c)
d/340 − d/5100 = 0.55
✓ 1
d = 0.55 × (340 × 5100)/(5100 − 340) = 0.55 × 364
✓ 1
d ≈ 200 m
✓ 1
Part (d)
d is proportional to the delay, so Δd = 364 × 0.05 (or the fractional uncertainty is 0.05/0.55 = 9.1%)
✓ 1
Δd ≈ ±18 m (accept ±20 m). ECF from (c).
✓ 1
Exam tip: When two waves cover the same distance at different speeds, set up an equation for the difference in travel times and solve it for the distance.
58IBP2H-021-01-14Intensity and amplitudePaper 2Hard8 marks
A deep-space probe is 1.8 × 1013 m from Earth. It transmits radio signals of frequency 8.4 GHz with a transmitter power of 23 W.
(a)
Calculate the wavelength of the signal.
(1)
(b)
Calculate the time taken for the signal to reach Earth.
(1)
(c)
Assume the probe radiates uniformly in all directions. Calculate the intensity of the signal at Earth.
(2)
(d)
The probe's dish actually concentrates the signal. The intensity at Earth is therefore 4.0 × 104 times the uniform-radiation intensity you calculated in the previous part. A receiving dish of diameter 70 m faces the probe. Determine the power collected by the receiving dish.
(2)
(e)
Later, the probe is 2.4 × 1013 m from Earth, with the same transmitted power and beam. Determine the ratio of the new received wave amplitude to the earlier amplitude.
(2)
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Part (a)
λ = 3.00 × 108/8.4 × 109 = 3.6 × 10−2 m
✓ 1
Part (b)
t = 1.8 × 1013/3.00 × 108 = 6.0 × 104 s
✓ 1
Part (c)
I = 23/(4π(1.8 × 1013)2)
✓ 1
I = 5.6 × 10−27 W m−2
✓ 1
Part (d)
I = 4.0 × 104 × 5.6 × 10−27 = 2.3 × 10−22 W m−2 (allow error carried forward from the previous part)
✓ 1
P = I × π(35)2 = 8.7 × 10−19 W (accept 8.6–8.9 × 10−19 W, depending on rounding)
✓ 1
Part (e)
Intensity ratio = (1.8/2.4)2 = 0.56
✓ 1
Intensity ∝ amplitude2, so amplitude ratio = √0.5625 = 0.75
✓ 1
Exam tip: Focusing the beam changes the constant in the inverse-square law, but not the 1/r² dependence. Ratios therefore follow from the distances alone, and amplitude scales as 1/r.
59IBP2H-021-01-15Intensity and amplitudePaper 2Hard7 marks
A student investigates how the speed of sound in air depends on absolute temperature T.
T / K: 250, 280, 300, 330, 360 v / m s−1: 317, 335, 347, 364, 380
(a)
Show that the data are consistent with v ∝ √T.
(2)
(b)
Explain, in terms of molecular motion, why sound travels faster at higher temperature.
(2)
(c)
A car's ultrasonic parking sensor converts echo time into distance. It always assumes v = 347 m s−1, the value at 300 K. On a day when the air is at 260 K, an obstacle is actually 1.50 m away. Determine the distance displayed by the sensor.
(3)
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Part (a)
Calculates v/√T for at least three values spread across the range (or all five), e.g. 317/√250 = 20.05, 347/√300 = 20.03, 380/√360 = 20.03
✓ 1
States that the ratio is constant to within experimental uncertainty (≈ 20.0 m s−1 K−1/2), so v ∝ √T
✓ 1
Part (b)
At higher T the mean kinetic energy, and hence mean speed, of the molecules is greater
✓ 1
So molecules reach and collide with their neighbours sooner, passing the disturbance (compression) on more quickly
✓ 1
Part (c)
v at 260 K = 20.0 × √260 ≈ 323 m s−1 (allow ECF from the ratio found in part 1, or 347 × √(260/300))
✓ 1
Echo time = 2 × 1.50/323 = 9.28 × 10−3 s, or uses displayed distance ∝ assumed speed: d = 1.50 × 347/323
✓ 1
Displayed distance = 347 × 9.28 × 10−3/2 = 1.61 m (accept 1.6 m)
✓ 1
Common error: Remember that the echo path is there and back (2d), although the factor of 2 cancels if you use the ratio of speeds. The colder, slower air makes the sensor over-read the distance.
60IBP3H-C2-20r34-15Electromagnetic and mechanical wavesPaper 2Hard10 marks
The nacelle at the top of a wind turbine contains a gearbox that vibrates at a frequency of 120 Hz. The vibration travels 80 m down the steel tower as a longitudinal wave of speed 5.0 × 103 m s−1, and also through the air along the same 80 m to a sensor at the base, where the speed of sound is 340 m s−1.
(a)
Calculate the wavelength of the vibration in the steel and in the air, each to 3 significant figures.
(3)
(b)
Explain why the frequency of the vibration is the same in the steel and in the air, although the wavelengths are different.
(2)
(c)
The gearbox emits a single burst of vibration lasting 0.050 s. Show that the burst arriving through the steel has ended before the burst arriving through the air begins.
(3)
(d)
Calculate the longest burst duration for which the two bursts do not overlap at the sensor, to 3 significant figures.
(2)
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Part (a)
λ = v/f for each medium
✓ 1
steel: 5000 / 120 = 41.7 m
✓ 1
air: 340 / 120 = 2.83 m
✓ 1
Part (b)
the frequency is determined by the source, which delivers 120 oscillations each second to both paths
✓ 1
since v = fλ, the same f with different speeds gives different wavelengths
✓ 1
Part (c)
time through steel = 80 / 5000 = 0.016 s
✓ 1
time through air = 80 / 340 = 0.235 s
✓ 1
the steel burst ends at 0.016 + 0.050 = 0.066 s, which is before 0.235 s, so the bursts do not overlap
✓ 1
Part (d)
Δt = 80/340 − 80/5000
✓ 1
Δt = 0.219 s
✓ 1
Exam tip: For two paths of equal length, the arrival-time difference is the length divided by each speed, subtracted; do not subtract the speeds first.
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