82 free, high-quality past-paper-style questions on Wave model for IB Physics SL, each with a full mark scheme. Covers: The wave equation; Sound waves; Transverse & longitudinal; and more. For when you have finished the past papers, or want to practise one topic at a time.C.2 at Higher 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.374 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1C-1A-09The wave equationPaper 1AEasy1 mark
A sound wave has a frequency of 250 Hz and a wavelength of 1.36 m.
What is its speed?
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Step 1v = fλ = 250 × 1.36 = 340 m s⁻¹.
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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.
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.
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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.
3C-1A-13Transverse & longitudinalPaper 1AEasy1 mark
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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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.
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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.
7C-1A-66The electromagnetic spectrumPaper 1AEasy1 mark
A dental X-ray machine emits X-rays. A Wi-Fi router emits microwaves of frequency 5 GHz. Which row compares the X-rays with the microwaves when both travel through a vacuum?
Speed of the X-raysWavelength of the X-rays
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Step 1All electromagnetic waves travel through a vacuum at the same speed, c = 3.00 × 108 m s−1, whatever their frequency.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The microwaves have λ = c/f = 6 × 10−2 m, while X-rays have wavelengths of the order of 10−10 m, so the X-ray wavelength is much shorter.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis assumes that higher-frequency, more energetic radiation travels faster. All electromagnetic waves have the same speed c in a vacuum.
BCorrect: the speed is the same, c, and X-rays lie at the short-wavelength end of the spectrum.
CThe speed is right, but the spectrum has been reversed: X-ray wavelengths (about 10−10 m) are far shorter than microwave wavelengths (a few centimetres).
DThis reads v = fλ as if speed and wavelength both increase with frequency. In a vacuum v = c is fixed, so a higher frequency means a shorter wavelength.
8C-1A-71Mechanical & electromagnetic wavesPaper 1AEasy1 mark
A long-wave radio transmitter and the echo sounder of a ship both emit waves of frequency 200 kHz. The radio waves travel through air. The sound from the echo sounder travels through seawater, in which its speed is 1500 m s−1.
What is (wavelength of the radio waves)/(wavelength of the sound waves)?
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Step 1λ = v/f and both frequencies are the same, so the ratio of the wavelengths equals the ratio of the speeds.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Ratio = (3.00 × 108)/1500 = 2.0 × 105.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis is the inverse ratio, 1500/(3.00 × 108): the sound has the much shorter wavelength.
BThis assumes that waves of the same frequency have the same wavelength. The wavelength also depends on the speed, which is very different for an electromagnetic wave and a mechanical wave.
CCorrect: λradio/λsound = c/v = 2.0 × 105.
DThis uses the speed of sound in air, 340 m s−1, instead of in seawater: 3.00 × 108/340 = 8.8 × 105.
A medical ultrasound probe emits waves of frequency 5.0 MHz into soft tissue. The speed of sound in the 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) = 3.1×10−4 m (about 0.3 mm). This is small enough to resolve fine detail in tissue. 0.31 m comes from treating MHz as kHz. 3.2×103 m comes from calculating f/v. 7.7×109 m comes from multiplying v by f.
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Answer A
Answer: A
Common error: Mega means 10⁶ and kilo means 10³. Convert prefixes before you substitute, then check that the size of the answer makes sense.
10IBP2S-020-01-02The electromagnetic spectrumPaper 1AEasy1 mark
Which list places regions of the electromagnetic spectrum in order of increasing wavelength?
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Gamma rays have the shortest wavelengths. Ultraviolet comes next, then visible light, then infrared, then microwaves, which are longer still. The order gamma → UV → IR → microwave is therefore increasing wavelength. Option B is the order of increasing frequency, which is the reverse. Options C and D each swap a pair of neighbouring regions.
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Answer A
Answer: A
Exam tip: Learn one order, for example radio → gamma (decreasing wavelength). Reverse it when the question asks for frequency, because λ = c/f.
11IBP2S-020-01-06The electromagnetic spectrumPaper 1AEasy1 mark
A space probe sends radio waves to Earth, and a star emits visible light and X-rays. In a vacuum, which property is the same for all of these electromagnetic waves?
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All electromagnetic waves travel at the same speed in a vacuum, c = 3.00×108 m s−1. Frequency and wavelength differ from one region of the spectrum to another. Amplitude depends on the source and on the distance from it, not on the type of wave.
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Answer D
Answer: D
Exam tip: Because c is fixed in a vacuum, knowing either the frequency or the wavelength of an EM wave lets you find the other.
12IBP2S-020-02-02Transverse & longitudinalPaper 1AEasy1 mark
During an earthquake, primary (P) seismic waves travel through rock as longitudinal waves. Which statement about the motion of the rock particles as a P-wave passes is correct?
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In a longitudinal wave the particles oscillate about their equilibrium positions, parallel to the direction in which the wave transfers energy. This produces compressions and rarefactions. Perpendicular oscillation describes a transverse wave. A wave transfers energy, not matter, so the particles do not travel with the wave. The particles do move, because they oscillate.
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Answer B
Answer: B
Exam tip: When you define a wave type, always state what oscillates and compare its direction with the direction of energy transfer.
13IBP2S-020-02-08The electromagnetic spectrumPaper 1AEasy1 mark
A hospital uses X-rays for imaging, ultraviolet lamps for sterilisation, infrared cameras for thermal scans and microwaves in its kitchen ovens. Which list puts these radiations in order of increasing wavelength?
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From shortest to longest wavelength the order is X-ray, then ultraviolet, then (visible), then infrared, then microwave. Option A is the order of increasing frequency, which is the reverse. Option B swaps ultraviolet and infrared on either side of the visible region. Option D mixes up both pairs.
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Answer C
Answer: C
Common error: Check whether the question asks for wavelength or frequency. The two orders are opposite, because c = fλ.
14IBP2S-020-00-01Waves in a new mediumPaper 1AEasy1 mark
A swimmer floating just below the surface of a pool hears music played by a violinist standing at the poolside. The speed of sound is about 340 m s−1 in air and about 1480 m s−1 in water. How do the frequency and the wavelength of the sound in the water compare with those in the air?
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The frequency is set by the vibrating source and does not change when the wave enters a new medium. Because v = fλ and v increases in water while f is fixed, λ must increase, by a factor of about 1480/340 ≈ 4.4. Option A has the wavelength change in the wrong direction. Options C and D wrongly assume that the frequency changes at the boundary.
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Answer B
Answer: B
Common error: Do not assume that a 'faster' wave has a higher frequency. At a boundary only the speed and the wavelength change.
15IBP2S-020-00-02Transverse & longitudinalPaper 1AEasy1 mark
An ocean buoy anchored offshore bobs up and down as swell waves pass it. It does not drift towards the beach. What do the waves transfer towards the beach?
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A travelling wave transfers energy. The particles of the medium only oscillate about fixed equilibrium positions, so there is no net transport of water. This is why the buoy stays in place and only moves up and down. Options A and B confuse the wave speed with the movement of the medium. Option D ignores the energy that the wave carries.
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Answer C
Answer: C
Exam tip: A floating object that oscillates but does not drift is direct evidence that a wave transfers energy, not matter.
16IBP2S-020-00-10Mechanical & electromagnetic wavesPaper 1AEasy1 mark
Which property of electromagnetic waves provides evidence that they are transverse?
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Polarisation restricts the oscillations to a single plane perpendicular to the direction of travel. This is only possible for transverse waves, because longitudinal waves oscillate along the direction of travel. Diffraction happens for all wave types. Travelling through a vacuum and having a common speed are properties of EM waves, but neither shows whether the oscillations are transverse.
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Answer A
Answer: A
Exam tip: Sound cannot be polarised. Polarisation is the standard test that separates transverse waves from longitudinal ones.
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?
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Step 1From the graph, one complete cycle occupies 0.40 m, so λ = 0.40 m.
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Step 2v = λ/T = 0.40/0.020 = 20 m s−1.
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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.
A transverse wave travels with speed v along a stretched rope. A point on the rope takes a time t to move from a position of maximum displacement to its equilibrium position.
What is the wavelength of the wave?
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Step 1Moving from maximum displacement to the equilibrium position is one quarter of an oscillation, so the period is T = 4t.
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Step 2λ = vT = 4vt.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis divides by 4 instead of multiplying: the period is four times as long as the quarter-cycle time t, not a quarter of it.
BThis treats t as the period. From maximum displacement to equilibrium is only a quarter of a cycle.
CThis treats the motion from maximum displacement to equilibrium as half a cycle. Half a cycle takes the point from one maximum to the opposite maximum.
DCorrect: t is a quarter of the period, so T = 4t and λ = vT = 4vt.
A student sends two continuous transverse waves, P and Q, one after the other along the same long stretched spring. The tension in the spring is the same for both waves. The graph shows the displacement of the spring against distance for P (solid line) and for Q (dashed line), each at one instant.
What is (frequency of P)/(frequency of Q)?
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Step 1From the graph, λP = 0.60 m and λQ = 0.40 m.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Both waves travel along the same spring at the same tension, so they have the same speed. f = v/λ, so fP/fQ = λQ/λP = 0.40/0.60 = 2/3.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: at the same speed the frequency is inversely proportional to the wavelength, so the ratio is 0.40/0.60 = 2/3.
BThis assumes that the spring fixes the frequency. The medium fixes the speed; the frequency is set by the source, the student's hand.
CThis takes the frequency as proportional to the wavelength, 0.60/0.40. At constant speed a longer wavelength means a lower frequency.
DThis is the ratio of the amplitudes, 3.0 cm/1.5 cm. The amplitude has no effect on the frequency.
A wave machine in a science museum is a row of horizontal rods joined at their centres by a wire. A transverse wave travels along the machine from rod X towards rod Y, which is a quarter of a wavelength further along. In each graph the solid line shows the displacement y of the tip of X against time t. Which graph shows, as a dashed line, the displacement of the tip of Y?
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Step 1Y is a quarter of a wavelength further along the direction of travel, so the wave reaches Y a quarter of a period later: Y repeats the motion of X after a delay of T/4.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2So Y reaches each maximum T/4 after X does; at t = 0, when X is at zero displacement moving up, Y is at its lowest point. This is graph B.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AHere Y leads X by T/4. That would be correct only if the wave travelled from Y towards X.
BCorrect: Y does what X did a quarter of a period earlier, so its graph is shifted T/4 later in time.
CThis shows Y in antiphase with X, which needs a separation of half a wavelength, not a quarter.
DThis shows Y in phase with X, as for points a whole wavelength apart; it treats the whole machine as moving together.
22IBP2S-020-01-03Mechanical & electromagnetic wavesPaper 1AMedium1 mark
A submarine's sonar pulse has its amplitude reduced to one third of its original value after passing through a layer of water containing many bubbles. By what factor has the intensity of the pulse changed?
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Intensity is proportional to amplitude squared (I ∝ A2). So the intensity changes by a factor of (1/3)2 = 1/9. Choosing 1/3 assumes I ∝ A. Choosing 1/√3 applies the square root the wrong way round. Choosing 1/27 cubes the factor instead of squaring it.
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Answer B
Answer: B
Common error: Going from amplitude to intensity, you square the factor. Going from intensity to amplitude, you take the square root.
23IBP2S-020-01-04Waves in a new mediumPaper 1AMedium1 mark
A lifeguard standing on a pier blows a whistle. Some of the sound travels from the air into the sea below. The speed of sound in water is greater than in air. What happens to the frequency and the wavelength of the sound as it enters the water?
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The source sets the frequency, so it does not change when the wave enters a new medium. From λ = v/f, a larger speed at the same frequency gives a larger wavelength, so A is correct. B and D wrongly let the frequency change. C keeps the frequency constant but reverses the change in wavelength.
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Answer A
Answer: A
Exam tip: At a boundary between media the frequency stays constant, so the wavelength changes in proportion to the wave speed.
24IBP2S-020-01-05Waves in a new mediumPaper 1AMedium1 mark
A small Wi-Fi router acts as a point source radiating equally in all directions. At a distance of 2.0 m from the router, the amplitude of the electric field of the wave is A. Absorption can be ignored. What is the amplitude at a distance of 6.0 m?
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Intensity follows the inverse-square law, so tripling the distance gives I/9. Since I ∝ A2, amplitude ∝ √I ∝ 1/r. The amplitude is therefore A/3. A/9 is the change in intensity, not amplitude. A/√3 applies the square root to the distance factor. A/6 confuses the new distance with the factor of change.
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Answer B
Answer: B
Common error: The inverse-square law applies to intensity. For a point source, amplitude falls only as 1/r.
25IBP2S-020-01-07Reading wave graphsPaper 1AMedium1 mark
A sensor clipped to a vibrating bridge cable records the displacement of one point on the cable against time. The shortest time between the point being at maximum positive displacement and next being at zero displacement is 2.0 ms. The wavelength of the wave on the cable is 0.60 m. What is the speed of the wave?
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Going from maximum displacement to zero displacement takes a quarter of a period. So T = 4 × 2.0 ms = 8.0 ms and f = 1/T = 125 Hz. Then v = fλ = 125 × 0.60 = 75 m s−1. 300 m s−1 treats 2.0 ms as the full period. 150 m s−1 treats it as half a period. 37.5 m s−1 uses T = 16 ms.
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Answer B
Answer: B
Common error: Count the fraction of a cycle carefully. Peak to zero is a quarter period, and peak to trough is a half period.
26IBP2S-020-01-09Intensity & amplitudePaper 1AMedium1 mark
A lamp used to sterilise hospital equipment emits electromagnetic radiation of frequency 1.0×1015 Hz. Which row gives the wavelength in a vacuum and the region of the spectrum?
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λ = c/f = 3.0×108 / 1.0×1015 = 3.0×10−7 m = 300 nm. This is shorter than the violet end of the visible range (about 400 nm), so the radiation is ultraviolet. Infrared has wavelengths longer than visible light. 3.3×106 m comes from calculating f/c.
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Answer A
Answer: A
Exam tip: Remember that visible light runs from about 400 nm to 700 nm. Use this as a reference point to place nearby wavelengths in the right region.
27IBP2S-020-02-03Reading wave graphsPaper 1AMedium1 mark
A sinusoidal transverse wave travels in the +x direction along a taut steel cable. At one instant, point P on the cable has zero displacement, and the displacement of the cable increases as x increases through P. What is the motion of P at this instant?
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Displacement increases with x through P. So the nearest crest lies a quarter wavelength ahead of P (at larger x), and the nearest trough lies a quarter wavelength behind it. The wave moves in +x, so the trough-side part of the waveform reaches P next, and P must be moving downward. On a sinusoidal wave, a point at zero displacement is passing through equilibrium, where its speed is greatest. 'Instantaneously at rest' applies only at a crest or trough. Particles of a transverse wave oscillate perpendicular to x and do not travel along x.
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Answer B
Answer: B
Exam tip: To find a particle's direction, imagine the whole waveform shifted a short distance in the direction of travel. Then see whether the point's displacement goes up or down.
28IBP2S-020-02-04Reading wave graphsPaper 1AMedium1 mark
A buoy on an offshore wave-energy converter moves up and down as ocean waves pass. Its displacement–time graph shows exactly 3 complete oscillations in 12 s. Satellite images show that adjacent wave crests are 18 m apart. What is the speed of the waves?
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The period is T = 12/3 = 4.0 s and the wavelength is λ = 18 m. So v = λ/T = 18/4.0 = 4.5 m s−1. 1.5 m s−1 divides by the total time instead of the period. 54 m s−1 multiplies λ by the number of oscillations. 72 m s−1 multiplies λ by the period.
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Answer B
Answer: B
Common error: A displacement–time graph gives the period, not the wavelength. Divide the total time by the number of complete cycles before using v = λ/T.
29IBP2S-020-02-05The electromagnetic spectrumPaper 1AMedium1 mark
A distant pulsar emits a burst of radio waves and a burst of X-rays at the same instant. Both travel through the vacuum of space to an observatory in orbit around Earth. Which statement is correct?
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All electromagnetic waves travel at the same speed, c, in a vacuum, so the two bursts arrive at the same time. X-rays have a much higher frequency than radio waves. Since c = fλ with c fixed, the X-rays must have the shorter wavelength. Options A and D wrongly link speed in a vacuum to frequency or wavelength.
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Answer B
Answer: B
Exam tip: In a vacuum, c is the same for every part of the EM spectrum, so frequency and wavelength are inversely proportional.
30IBP2S-020-02-06Waves in a new mediumPaper 1AMedium1 mark
A loudspeaker above a swimming pool produces a note. Sound waves pass from the air into the water, where the speed of sound is greater. What happens to the frequency and the wavelength of the sound as it enters the water?
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The frequency is set by the source, so it does not change when the wave crosses into a new medium. Because v = fλ and v increases while f stays constant, λ must increase. Option B assumes that waves always get 'squashed' in a denser medium. Options C and D wrongly allow the frequency to change.
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Answer A
Answer: A
Common error: Do not assume sound slows down in denser media the way light does. Sound is faster in water than in air.
31IBP2S-020-02-09Transverse & longitudinalPaper 1AMedium1 mark
An athlete training with 'battle ropes' moves one end of a long rope up and down 2.5 times per second. This sends transverse waves along the rope, and adjacent crests are 1.6 m apart. How long does a crest take to travel 8.0 m along the rope?
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The wave speed is v = fλ = 2.5 × 1.6 = 4.0 m s−1, so the time is t = d/v = 8.0/4.0 = 2.0 s. 0.50 s is v/d. 3.2 s divides the distance by the frequency. 5.0 s divides the distance by the wavelength.
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Answer B
Answer: B
Exam tip: A crest moves at the wave speed. Find v from v = fλ first, then use distance ÷ speed.
32IBP2S-020-00-03Reading wave graphsPaper 1AMedium1 mark
A loudspeaker emits a sinusoidal sound wave into air. Consider an air molecule located at the centre of a compression at a particular instant. What are the displacement of this molecule from its equilibrium position and its speed at that instant?
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At the centre of a compression, the molecules on either side are displaced towards that point, while the molecule at the centre itself sits at its equilibrium position. Its displacement is therefore zero. In simple harmonic oscillation, zero displacement corresponds to maximum speed. Option A describes a molecule midway between a compression and a rarefaction. That molecule has maximum displacement but is momentarily at rest. Options B and D combine the two quantities incorrectly.
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Answer C
Answer: C
Common error: A compression is a region of high density, not a place where the molecules have their largest displacement. The two occur a quarter of a wavelength apart.
33IBP2S-020-00-04Mechanical & electromagnetic wavesPaper 1AMedium1 mark
A fishing boat's echo-sounder sends a short ultrasound pulse vertically down to the sea bed. The reflected pulse returns 0.80 s after it was emitted. The speed of sound in sea water is 1500 m s−1. What is the depth of the sea bed below the boat?
Diagram not to scale
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In 0.80 s the pulse travels 1500 × 0.80 = 1200 m. This distance covers the trip down and back, so the depth is 1200/2 = 600 m. Option C forgets to halve the distance. Option A halves it twice. Option D divides the speed by the time instead of multiplying.
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Answer B
Answer: B
Exam tip: For any echo problem, write 2d = vt before substituting numbers.
34IBP2S-020-00-05Reading wave graphsPaper 1AMedium1 mark
A microphone records the steady hum of a drone. A displacement–time graph for one air molecule shows exactly 5 complete oscillations in 20 ms. A displacement–distance graph of the sound wave at one instant shows that adjacent points of maximum positive displacement are 1.36 m apart. What is the speed of the sound?
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The period is T = 20 ms / 5 = 4.0 ms, so f = 250 Hz. The wavelength, read from the distance graph, is λ = 1.36 m. Then v = fλ = 250 × 1.36 = 340 m s−1. Option B uses 20 ms as the period. Option C multiplies by the number of oscillations a second time. Option A divides λ by f.
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Answer D
Answer: D
Exam tip: Period comes from a displacement–time graph and wavelength comes from a displacement–distance graph. Check the horizontal axis before reading off a value.
35IBP2S-020-00-06The electromagnetic spectrumPaper 1AMedium1 mark
A home Wi-Fi router transmits electromagnetic waves of frequency 2.4 GHz. The speed of electromagnetic waves in air is 3.0×108 m s−1. What is the wavelength of these waves?
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λ = c/f = 3.0×108 / 2.4×109 = 0.125 m. Option D treats GHz as MHz, giving a value 1000 times too large. Option C calculates f/c instead of c/f. Option B is a decimal-place slip.
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Answer A
Answer: A
Common error: giga = 10⁹ and mega = 10⁶. Mixing up these prefixes changes the answer by a factor of 1000.
36IBP2S-020-00-07Transverse & longitudinalPaper 1AMedium1 mark
A plucked guitar string vibrates and produces a sound wave in the surrounding air. Which statement correctly compares the wave on the string with the sound wave in the air?
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The vibrating string forces the air to oscillate at the same rate, so the two waves have the same frequency. On the string, the particles move perpendicular to the length of the string, so that wave is transverse. In air, the molecules oscillate parallel to the direction of travel, so sound is longitudinal. The two speeds differ, so by v = fλ the wavelengths differ too. This rules out options A and B. Option C wrongly describes the string wave as longitudinal.
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Answer D
Answer: D
Exam tip: When one wave drives another, the frequency carries over unchanged. Speed and wavelength depend on each medium.
37IBP2S-020-00-08Mechanical & electromagnetic wavesPaper 1AMedium1 mark
A railway engineer strikes a steel rail with a hammer. A colleague 1020 m away along the track hears the sound twice: once through the rail and once through the air. The speed of sound is 5100 m s−1 in steel and 340 m s−1 in air. What is the time interval between the two sounds?
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Through the air the sound takes 1020/340 = 3.0 s. Through the steel it takes 1020/5100 = 0.20 s. The interval between them is 3.0 − 0.20 = 2.8 s. Options A and B are the individual travel times. Option D adds the two times instead of subtracting.
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Answer C
Answer: C
Exam tip: Sound travels much faster in solids than in gases, because the particles are closely bound and pass on disturbances quickly.
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?
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Velocity of PVelocity of Q
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Step 1P is at maximum displacement, so it is momentarily at rest.
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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.
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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.
In a computer simulation, a sound wave travels through air in the positive x-direction. The graph shows how the displacement s of one air molecule from its undisturbed position varies with time t. A displacement in the positive x-direction is positive.
At which time is the molecule at the centre of a compression?
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Step 1At the centre of a compression the molecule has zero displacement and the air on each side is displaced towards it, so the answer is 1.0 ms or 2.0 ms.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The wave travels in the +x-direction, so the molecule next takes the displacement that the air just behind it has now. At a compression that air is displaced forwards, so the molecule is moving in the +x-direction.
—
Step 3s = 0 with s increasing at t = 2.0 ms. (At 1.0 ms s = 0 but is decreasing: that is the centre of a rarefaction.)
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis takes the largest forward displacement as a compression. At a maximum of displacement the neighbouring air is displaced by almost the same amount, so the density there is normal.
BThe displacement is zero, but the molecule is moving in the −x-direction, against the wave. The air on each side is displaced away from it: this is the centre of a rarefaction.
CThis takes the largest backward displacement as a compression. As at 0.5 ms, the density is normal where the displacement is a maximum.
DCorrect: zero displacement while moving in the direction of travel means that the air on both sides is displaced towards the molecule.
Water waves travel past a post at the end of a pier. A student starts a stopwatch as one crest passes the post, counting it as crest 1, and stops the stopwatch as crest n passes. The stopwatch reads t. In a photograph of the waves, the horizontal distance between a crest and the adjacent trough is d.
What is the speed of the waves?
Show mark scheme
Marking point
Mark
Notes
Step 1From crest 1 to crest n there are n − 1 complete oscillations, so T = t/(n − 1).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2A crest and the adjacent trough are half a wavelength apart, so λ = 2d.
—
Step 3v = λ/T = 2(n − 1)d/t.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis takes d as the wavelength. A crest and the next trough are only half a wavelength apart.
BThis counts n periods in time t. Between the first and the nth crest there are only n − 1.
CThis multiplies the wavelength by the period. Speed is wavelength divided by period.
DCorrect: λ = 2d and T = t/(n − 1), so v = λ/T = 2(n − 1)d/t.
42IBP2S-020-01-08Intensity & amplitudePaper 1AHard1 mark
A point source of sound has power P. At distance r from the source, the sound has amplitude A. The power of the source is doubled and the listener moves to a distance 2r. Absorption can be ignored. What is the new amplitude?
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Notes
I = P/(4πr2). The new intensity is (2P)/(4π(2r)2) = (2/4) × P/(4πr2) = I/2. Since A ∝ √I, the new amplitude is A/√2. A/4 and A/2 come from forgetting to double the power, or from applying the square root incorrectly. A comes from wrongly assuming the two changes cancel.
✓ 1
Answer C
Answer: C
Exam tip: Find the factor for intensity first. Then take its square root to get the factor for amplitude.
43IBP2S-020-01-10Transverse & longitudinalPaper 1AHard1 mark
A longitudinal sound wave travels to the right along a pipe. A displacement is taken as positive when it is to the right. At one instant, a point P in the air is at the centre of a compression. Which statement describes the air near P at that instant?
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Notes
At the centre of a compression, the air layers on both sides have moved towards P. The layer at P itself has zero displacement. Air on the left has moved right (positive) and air on the right has moved left (negative), which bunches the layers together. Option C describes the centre of a rarefaction. At points of maximum displacement (options A and D), the spacing between layers is normal, so the density is normal.
✓ 1
Answer B
Answer: B
Common error: Compressions and rarefactions are found where the displacement is zero, not where it is a maximum. Look at the displacement of the neighbouring layers to decide which one you have.
44IBP2S-020-02-07Reading wave graphsPaper 1AHard1 mark
A longitudinal wave travels along a long spring used to test vibration dampers. On a displacement–distance graph, positive displacement means displacement in the +x direction. At one instant the displacement is zero at x = 0.20 m. It is positive for points just below x = 0.20 m and negative for points just above x = 0.20 m. The displacement is a positive maximum at x = 0.10 m. Which statement is correct at this instant?
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Notes
Coils just to the left of x = 0.20 m are displaced to the right (+x). Coils just to the right are displaced to the left (−x). Both groups have moved towards x = 0.20 m, so the coils are bunched there: a compression. A rarefaction would need the neighbouring coils to move apart. At a displacement maximum (x = 0.10 m) the neighbouring coils are displaced by almost the same amount, so the spacing there is normal. Zero displacement does not mean normal spacing.
✓ 1
Answer A
Answer: A
Exam tip: For a longitudinal wave, compressions and rarefactions occur at points of zero displacement. Check which way the neighbouring particles have moved to decide which one it is.
45IBP2S-020-02-10Reading wave graphsPaper 1AHard1 mark
A drone films a transverse wave travelling along an overhead power line. In the frame taken at t = 0, a crest is at x = 0.30 m. In the frame taken at t = 0.050 s, the same crest is at x = 0.45 m. The wave has moved less than one wavelength in this time, and the wavelength is 0.80 m. What is the frequency of the wave?
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Notes
The crest moves 0.15 m in 0.050 s, so v = 3.0 m s−1. Then f = v/λ = 3.0/0.80 = 3.75 Hz ≈ 3.8 Hz. 0.27 Hz is the period (0.27 s) written as a frequency. 2.4 Hz is v × λ. 20 Hz wrongly treats the time between frames as the period.
✓ 1
Answer C
Answer: C
Common error: The time between two snapshots is not the period unless the wave has moved exactly one wavelength in that time.
A student uses a data handbook to investigate how the speed of sound v in a gas depends on the molar mass M of the gas. To make a fair comparison, the student chooses five noble gases and uses the values listed for a temperature of 0 °C. The handbook states that each speed is accurate to ±1 %.
The student's hypothesis is that v = KMn, where K is a constant and n = −0.5. The graph shows log v against log M; the error bars are too small to show.
gas
M / g mol−1
v / m s−1 ± 1 %
log( M / g mol−1 )
log( v / m s−1 )
helium
4.00
972
0.602
2.988
neon
20.18
433
1.305
2.636
argon
39.95
308
1.602
2.489
krypton
83.80
213
xenon
131.29
170
2.118
2.230
Graph drawn to scale
(a)
A tuning fork of frequency 440 Hz is sounded in xenon at 0 °C. Calculate the wavelength of the sound.
(1)
(b)
Complete the table for krypton.
(2)
(c)
Show that the uncertainty in log v is about ±0.004.
(1)
(d)
Draw the line of best fit on the graph.
(1)
(e)
Determine the gradient of the line of best fit.
(2)
(f)
State whether the data support the hypothesis.
(1)
(g)
A second handbook lists the speeds at 20 °C, where each speed is greater by the same factor of 1.036. State and explain the effect, if any, on the gradient and on the intercept of the graph if these values were used.
(2)
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Marking point
Mark
Notes
Part (a)
λ = v/f = 170/440 = 0.386 m
✓ 1
Accept 0.39 m.
Part (b)
log( M / g mol−1 ) = log 83.80 = 1.923
✓ 1
Accept 2 or 4 decimal places.
log( v / m s−1 ) = log 213 = 2.328
✓ 1
Accept 2 or 4 decimal places.
Part (c)
log(1.01v) − log v = log 1.01 = 0.0043 «≈ 0.004»
✓ 1
Must see the calculation (or log 1.01 or log 0.99) AND an answer to at least 2 s.f.
Part (d)
Single straight line with points evenly distributed about it
✓ 1
Do not accept a curve.
Part (e)
Uses a triangle covering more than half the line
✓ 1
gradient = −0.50
✓ 1
Accept −0.48 to −0.52. No unit.
Part (f)
Yes: the points lie on a straight line whose gradient equals −0.5 within the precision of the graph, so v ∝ M−0.5
✓ 1
Allow ECF from (e): a gradient outside −0.47 to −0.53 leads to 'not supported'.
Part (g)
Every value of log v increases by the same amount, log 1.036 = 0.015
✓ 1
OWTTE.
so the line moves up parallel to itself: the gradient is unchanged and the intercept increases «by about 0.015»
✓ 1
Both effects needed. A factor common to every reading is a systematic shift, not a change of gradient.
Answers: (a) λ ≈ 0.386 m · (b) 1.923; 2.328 · (c) 0.0043 · (e) −0.50 (the remaining parts are explanations — see the table above)
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)
Show mark scheme
Marking point
Mark
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)
A student measures the wavelength of ultrasound in air. An ultrasonic transmitter, driven by a signal generator set to a frequency of (40.0 ± 0.1) kHz, faces a receiver that slides along a metre rule. A dual-trace oscilloscope shows the signal sent to the transmitter and the signal from the receiver.
The student moves the receiver slowly away from the transmitter, starting at a position where the two traces are in phase (n = 0). The traces come back into phase again and again; the student records the position x of the receiver after every fifth return to phase. Each position has an uncertainty of ±0.2 cm. The air temperature is 22 °C. The error bars are too small to be shown on the graph.
n
x / cm ± 0.2 cm
0
14.6
5
18.9
10
23.2
15
27.5
20
31.8
25
36.1
30
40.6
Graph drawn to scale
(a)
Outline why the two traces are in phase again each time the receiver moves one more wavelength away from the transmitter.
(1)
(b)
Suggest why the student records the position only after every fifth return to phase.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the wavelength λ of the ultrasound, in m, from the graph.
(2)
(e)
The uncertainty in λ may be estimated as the sum of the uncertainties in the first and the last positions, divided by 30. Determine the speed of the ultrasound and its absolute uncertainty.
(3)
(f)
The accepted speed of sound in air at 22 °C is 344 m s−1. Comment on the student's result.
(1)
(g)
The student did not measure the distance between the transmitter and the receiver, only the position of the receiver on the rule. Explain why this does not affect the value of λ.
(1)
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Marking point
Mark
Notes
Part (a)
moving one wavelength further adds one wavelength to the path, so the sound takes one more whole period to arrive «the extra delay is T = λ/v»; a delay of a whole period leaves the received trace in phase with the transmitted one
✓ 1
OWTTE. Do not accept 'the wave repeats' alone.
Part (b)
the uncertainty in each position «±0.2 cm» is then a smaller fraction of the distance between readings «5λ ≈ 4.3 cm rather than 0.86 cm», so the percentage uncertainty in λ is smaller
✓ 1
Accept 'one wavelength is too short to measure precisely with the rule'.
Part (c)
single straight line with the points evenly distributed about it, not forced through the origin
✓ 1
Part (d)
gradient from a large triangle = 0.864 cm
✓ 1
Accept 0.851–0.877 cm. The gradient is the distance per return to phase, which is one wavelength.
λ = 8.64 × 10−3 m
✓ 1
Must convert to m. Allow ECF from the gradient.
Part (e)
v = fλ = 40.0 × 103 × 8.64 × 10−3 = 346 m s−1
✓ 1
Allow ECF from (d).
Δλ/λ = (0.4/30)/0.864 = 1.5 % AND Δf/f = 0.1/40.0 = 0.25 %, so Δv/v = 1.8 %
✓ 1
Fractional uncertainties are added for a product.
v = (346 ± 6) m s−1
✓ 1
Accept ±5 to ±7 m s−1. MP3 is for matching the precision of the value and the uncertainty, with the unit.
Part (f)
344 m s−1 lies within the range 340 to 352 m s−1, so the result agrees with the accepted value «within its uncertainty»
✓ 1
Allow ECF from (e). Must refer to the uncertainty.
Part (g)
an unknown distance from the transmitter to the zero of the rule adds the same amount to every x; it changes only the intercept of the line, not its gradient, which gives λ
✓ 1
OWTTE.
Answers: (d) λ ≈ 8.64 × 10−3 m · (e) v = (346 ± 6) m s−1(the remaining parts are explanations — see the table above)
A student investigates how the speed of sound in air depends on temperature. Two ultrasonic transducers are fixed (0.500 ± 0.001) m apart inside a sealed, insulated duct that contains a heater. A timer measures the time t for a short pulse to travel from one transducer to the other, with an uncertainty of ±0.5 μs. The temperature θ of the air is measured with a thermometer, with an uncertainty of ±0.5 °C.
The student's hypothesis is that the speed v is proportional to √T, where T is the absolute temperature. The error bars are too small to be shown on the graph.
θ / °C ± 0.5 °C
t / μs ± 0.5 μs
v / m s−1
v² / 105 m2 s−2
10.0
1481
337.6
1.140
22.0
1452
344.4
1.186
35.0
1421
351.9
1.238
47.0
1394
1.287
60.0
1366
366.0
1.340
73.0
1341
372.9
1.390
85.0
1318
379.4
1.439
Graph drawn to scale
(a)
Show that v for θ = 47.0 °C is about 359 m s−1.
(1)
(b)
Calculate the percentage uncertainty in v² for θ = 47.0 °C.
(2)
(c)
Explain why, if the hypothesis is correct, the graph of v² against θ is a straight line that would meet the θ axis at −273 °C.
(2)
(d)
Draw the line of best fit on the graph.
(1)
(e)
Determine the gradient of the line and its intercept on the v² axis.
(2)
(f)
Hence estimate the temperature at which the speed of sound would be zero, and comment on the hypothesis.
(2)
(g)
The thermometer is later found to read 1.5 °C too high at every temperature. State and explain the effect of this error on the gradient and on your estimate in (f).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = 0.500/(1394 × 10−6) = 358.7 «m s−1»
✓ 1
Must see full substitution OR answer to at least 4 s.f.
Part (b)
%Δv = ΔL/L + Δt/t = 0.20 % + 0.036 % = 0.24 %
✓ 1
Accept fractional uncertainties.
%Δ(v²) = 2 × 0.24 % = 0.47 %
✓ 1
Accept 0.42–0.52 %. Award [1 max] for 0.24 % «not doubled». Award [2] for CNA.
Part (c)
v² ∝ T = θ + 273, so v² = k(θ + 273), which is of the form y = mx + c with gradient k: a straight line «not through the origin»
✓ 1
Accept v² = kθ + 273k.
v² = 0 when T = 0 K, that is, when θ = −273 °C
✓ 1
OWTTE.
Part (d)
single straight line with the points evenly distributed about it
✓ 1
Do not accept a line forced through a point on the axis.
Part (e)
gradient = 4.00 × 102 m2 s−2 °C−1
✓ 1
Accept 388–412 m2 s−2 °C−1. Unit required; accept K−1 for °C−1.
this is close to −273 °C, especially for so long an extrapolation, so the data support the hypothesis «v ∝ √T»
✓ 1
Allow ECF: a value differing from −273 °C by more than about 15 °C leads to 'not supported'.
Part (g)
every point moves the same distance «1.5 °C» along the θ axis, so the gradient is unchanged
✓ 1
OWTTE.
the whole line, and its intercept on the θ axis, is shifted by +1.5 °C, so the estimate of absolute zero is 1.5 °C too high «the true estimate is 1.5 °C lower»
✓ 1
Award MP2 for 'the estimate is too high by the same amount' or equivalent.
Answers: (a) v = 358.7 m s−1 · (b) 0.47 % · (e) gradient ≈ 4.00 × 102 m2 s−2 °C−1, intercept ≈ 1.099 × 105 m2 s−2 · (f) ≈ −275 °C (the remaining parts are explanations — see the table above)
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)
Show mark scheme
Marking point
Mark
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)
A student investigates whether the speed of ripples in a ripple tank depends on their frequency. The water has a uniform depth h = (5.0 ± 0.2) mm. A lamp above the tank casts a shadow pattern of the ripples onto a screen below it. For each frequency f of the dipper, set with a stroboscope to ±0.1 Hz, the student measures the distance D on the screen across five wavelengths, with an uncertainty of ±0.2 cm, and calculates the speed v of the pattern on the screen and its uncertainty Δv.
f / Hz ± 0.1 Hz
D / cm ± 0.2 cm
v / m s−1
Δv / m s−1
6.0
22.0
0.264
0.007
9.0
14.8
0.266
0.007
12.0
11.2
0.269
0.007
15.0
9.2
18.0
7.9
0.284
0.009
21.0
6.9
0.290
0.010
24.0
6.1
0.293
0.011
Graph drawn to scale; the point for f = 15.0 Hz has not been plotted
(a)
Calculate v and Δv for f = 15.0 Hz, and plot this point, with its error bar, on the graph.
(3)
(b)
Draw the line or curve of best fit on the graph.
(1)
(c)
Determine whether the data support the hypothesis that the speed of the ripples does not depend on their frequency.
(2)
(d)
The shadow pattern on the screen is 1.20 times the size of the pattern of ripples on the water.
(i)
Calculate the speed of the ripples on the water for f = 6.0 Hz.
(1)
(ii)
State and explain whether this magnification affects your answer to (c).
(1)
(e)
A simple model predicts that the speed of water waves in shallow water is v = √(gh). Determine whether your answer to (d)(i) agrees with this model.
(3)
(f)
Suggest one improvement to the method that would reduce the uncertainty in v at the highest frequencies.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = f × D/5 = 15.0 × (9.2 × 10−2/5) = 0.276 m s−1
✓ 1
Accept 2 or 3 s.f.
Δv/v = 0.2/9.2 + 0.1/15.0 = 0.0284, so Δv = 0.008 m s−1
✓ 1
Accept 0.007–0.009 m s−1. Allow ECF for v.
point plotted at (15.0 Hz, 0.276 m s−1) with an error bar from about 0.268 to 0.284 m s−1
✓ 1
Accept ± half a small square «0.002 m s−1». Allow ECF from MP1 and MP2.
Part (b)
smooth line or curve through all the error bars, roughly level at low frequencies and rising at higher frequencies
✓ 1
Accept a single straight line through all the error bars.
Part (c)
a horizontal line cannot pass through all the error bars: e.g. the top of the 6.0 Hz bar «0.271 m s−1» is below the bottom of the 24.0 Hz bar «0.282 m s−1»
✓ 1
Must refer to the error bars or uncertainties.
so the hypothesis is not supported over this range: the speed increases with frequency «above about 12 Hz»
✓ 1
Accept 'supported only at low frequencies (up to about 12 Hz)'. MP2 only scores if MP1 scores. «At these short wavelengths surface tension also affects the speed.»
Part (d)(i)
0.264/1.20 = 0.220 m s−1
✓ 1
Accept 0.22 m s−1.
Part (d)(ii)
no: every speed is divided by the same factor, so the trend in v with f, and the conclusion, are unchanged
✓ 1
OWTTE. Accept 'it shifts all the points down by the same fraction'.
Part (e)
√(gh) = √(9.81 × 0.0050) = 0.221 m s−1
✓ 1
uncertainty in the model value = ½ × 4 % = 2 %, so 0.221 ± 0.004 m s−1; measured value 0.220 ± 0.006 m s−1
✓ 1
Δv for (d)(i) = 0.0068/1.20. Allow ECF from (d)(i).
the ranges overlap «the values agree within their uncertainties», so the measurement agrees with the model
✓ 1
MP3 only scores if at least one uncertainty has been considered.
Part (f)
measure across more wavelengths «e.g. ten» at high frequency, where five wavelengths span only a few centimetres, so the percentage uncertainty in D is smaller
✓ 1
Accept 'photograph the frozen pattern beside a scale and measure many wavelengths'. Do not accept 'repeat the readings' alone.
Answers: (a) v = 0.276 m s−1, Δv = 0.008 m s−1 · (d)(i) 0.220 m s−1 · (e) √(gh) = 0.221 ± 0.004 m s−1(the remaining parts are explanations — see the table above)
52C-2-06Mechanical & electromagnetic wavesPaper 2Easy10 marks
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)
Show mark scheme
Marking point
Mark
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)
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)
An ultrasound probe used in a medical scan sends pulses of ultrasound of frequency 3.50 MHz into soft tissue, in which the speed of sound is 1540 m s−1. Figure 1 shows, at one instant, the displacement of particles of the tissue from their undisturbed positions against distance along the beam, for part of a pulse. A displacement in the direction in which the pulse travels is positive.
Figure 1 — drawn to scale
(a)
State what is meant by a longitudinal wave.
(1)
(b)
(i)
Show that the wavelength of the ultrasound in the tissue is 0.44 mm.
(1)
(ii)
Calculate the period of the ultrasound.
(1)
(c)
The pulse travels in the positive direction. Sketch, on Figure 1, the displacement of the tissue particles a quarter of a period after the instant shown.
(2)
(d)
The pulse is reflected at the boundary of an organ and the echo reaches the probe 70.0 μs after the pulse was emitted. Calculate the depth of the boundary below the probe.
(2)
(e)
Images of bones are often made with X-rays instead. State two differences between ultrasound and X-rays as waves.
(2)
(f)
The pulse enters a bone, in which the speed of sound is 4080 m s−1.
(i)
State the frequency of the ultrasound in the bone.
(1)
(ii)
Calculate the wavelength of the ultrasound in the bone.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A wave in which the particles of the medium oscillate parallel to the direction of travel of the wave «direction of energy transfer»
Must see full substitution OR answer to at least 3 s.f. Accept the wavelength read from Figure 1 with the calculation.
Part (b)(ii)
T = 1/f = 2.86 × 10−7 s
✓ 1
Accept 0.286 μs.
Part (c)
Sine curve of the same amplitude «30 nm» and wavelength «0.44 mm»
✓ 1
Shifted a quarter of a wavelength «0.11 mm» in the positive direction: crests at 0.22 mm, 0.66 mm and 1.10 mm
✓ 1
Do not award MP2 for a shift in the negative direction. Accept ±0.02 mm.
Part (d)
The pulse travels to the boundary and back: distance = vt/2
✓ 1
Award [1 max] for 0.108 m.
depth = 1540 × 70.0 × 10−6/2 = 0.0539 m
✓ 1
Award [2] for CNA.
Part (e)
Ultrasound is a mechanical wave that needs a medium; X-rays are electromagnetic and can travel through a vacuum
✓ 1
Award [1] per correct difference, [2 max].
Ultrasound is longitudinal; X-rays are transverse OR ultrasound travels far more slowly «≈ 103 m s−1 compared with ≈ 3 × 108 m s−1» OR X-rays have far shorter wavelengths / higher frequencies
✓ 1
Each difference must state both waves, OWTTE.
Part (f)(i)
3.50 MHz «the frequency is set by the source and does not change»
✓ 1
Part (f)(ii)
λ = 4080/(3.50 × 106) = 1.17 × 10−3 m
✓ 1
Accept 1.17 mm. Allow ECF from (f)(i).
Answers: (b)(i) 0.440 mm · (b)(ii) 2.86 × 10−7 s · (d) 0.0539 m · (f)(i) 3.50 MHz · (f)(ii) 1.17 × 10−3 m (the remaining parts are explanations — see the table above)
A small robotic helicopter flies in the thin carbon dioxide atmosphere of Mars. A microphone on a rover 80 m away records the sound of the helicopter's rotor blades, which has a frequency of 84 Hz. The speed of sound in the Martian atmosphere is 240 m s−1. The speed of sound in air on the Earth is 340 m s−1.
(a)
Describe how the sound of the rotor travels through the Martian atmosphere to the microphone.
(2)
(b)
Calculate the wavelength of the sound in the Martian atmosphere.
(1)
(c)
Calculate the time taken for the sound to travel from the helicopter to the microphone.
(1)
(d)
The rover sends the recording to the Earth using electromagnetic waves of frequency 8.4 GHz.
(i)
Calculate the wavelength of these waves.
(1)
(ii)
State the region of the electromagnetic spectrum to which these waves belong.
(1)
(e)
On the Earth, the recording is played through a loudspeaker.
(i)
State the frequency of the sound of the rotor produced by the loudspeaker.
(1)
(ii)
Calculate the wavelength of this sound in air.
(1)
(iii)
A student says that the rotor will sound lower in pitch on the Earth than on Mars because its wavelength in air is longer. Explain why the student is wrong.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the blades make the molecules of the gas oscillate «about fixed positions» parallel to the direction in which the sound travels «a longitudinal wave»
✓ 1
Do not accept 'the gas moves to the microphone'.
forming compressions and rarefactions «regions of higher and lower pressure» that travel outwards OR energy is passed on by collisions between molecules, with no net flow of gas
✓ 1
Accept either point for MP2. OWTTE.
Part (b)
λ = 240/84 = 2.9 m
✓ 1
Accept 2.86 m.
Part (c)
t = 80/240 = 0.33 s
✓ 1
Part (d)(i)
λ = 3.00 × 108/(8.4 × 109) = 3.6 × 10−2 m
✓ 1
Accept 0.036 m or 3.6 cm.
Part (d)(ii)
microwaves
✓ 1
Allow ECF from (d)(i): the region must match the candidate's wavelength «data booklet orders of magnitude».
Part (e)(i)
84 Hz
✓ 1
Do not accept a different value.
Part (e)(ii)
λ = 340/84 = 4.0 m
✓ 1
Allow ECF from (e)(i).
Part (e)(iii)
pitch depends on the frequency, not on the wavelength
✓ 1
the frequency is set by the source «the recording reproduces the same 84 oscillations per second»; the wavelength is longer only because sound travels faster in the Earth's air «λ = v/f»
✓ 1
OWTTE.
Answers: (b) 2.9 m · (c) 0.33 s · (d)(i) 3.6 × 10−2 m · (d)(ii) microwaves · (e)(i) 84 Hz · (e)(ii) 4.0 m (the remaining parts are explanations — see the table above)
56C-2-55Transverse & longitudinalPaper 2Easy10 marks
Spectators in a sports stadium make a crowd wave. Each spectator stands up and then sits down again as the wave reaches them, so that the wave travels along the rows of seats. In one stand the seats are in a straight row 150 m long. A video shows that a crowd wave travels the 150 m in 12.5 s and that each spectator takes 0.80 s to stand up and sit down again. Treat the motion of each spectator's head as vertical.
Figure 1 — for part (d)
(a)
State and explain whether the crowd wave is a model of a transverse wave or of a longitudinal wave.
(2)
(b)
Show that the speed of the crowd wave is about 12 m s−1.
(1)
(c)
Calculate the length of the row in which spectators are out of their seats at any one instant.
(1)
(d)
The crowd wave starts at one end of the row at time t = 0. Spectator S sits 6.0 m from that end. When standing, the head of S is 0.60 m above its seated position. On Figure 1, sketch a graph to show how the height h of the head of S above its seated position varies with time from t = 0 to t = 2.0 s.
(2)
(e)
A cheerleader then starts a new crowd wave every 4.0 s, so that a series of crowd waves travels along the row at the same speed.
(i)
Calculate the frequency of the waves.
(1)
(ii)
Calculate the wavelength of the waves.
(1)
(f)
Suggest one way in which a crowd wave is a good model of a mechanical wave and one way in which it is not.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
transverse
✓ 1
Award MP1 only with a reason.
each head moves up and down, perpendicular to the direction in which the wave travels along the row
✓ 1
Do not accept 'the spectators move up and down' without a reference to the direction of travel.
Part (b)
v = 150/12.5 = 12.0 «m s−1»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
12 × 0.80 = 9.6 m
✓ 1
Allow ECF from (b).
Part (d)
h = 0 until t = 0.50 s «= 6.0/12» and again after t = 1.30 s
✓ 1
Accept ±0.05 s for both times.
a single smooth hump between these times with a maximum of 0.60 m «at about 0.9 s»
✓ 1
Do not accept a curve that goes below h = 0 or a repeating pattern.
Part (e)(i)
f = 1/4.0 = 0.25 Hz
✓ 1
Part (e)(ii)
λ = v/f = 12/0.25 = 48 m
✓ 1
Allow ECF from (b) and (e)(i).
Part (f)
good: the disturbance travels along the row while each spectator only moves up and down about a fixed position «no spectator travels along the row»
✓ 1
OWTTE.
not good: in a mechanical wave each particle is made to move by forces from its neighbours, which pass energy along the medium; each spectator uses their own energy and nothing is passed from one to the next OR the speed depends on the reactions of the people, not on the properties of a medium
✓ 1
Accept any one valid limitation. OWTTE.
Answers: (b) 12.0 m s−1 · (c) 9.6 m · (e)(i) 0.25 Hz · (e)(ii) 48 m (the remaining parts are explanations — see the table above)
57IBP2S-020-01-11Transverse & longitudinalPaper 2Easy4 marks
A small instrument buoy floats freely in the open ocean. Swell waves pass the buoy. The time between successive crests passing the buoy is 8.0 s, and the distance between successive crests is 96 m.
(a)
Calculate the frequency of the waves.
(1)
(b)
Calculate the speed of the waves.
(2)
(c)
Outline why the buoy moves up and down but is not carried along with the wave.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
f = 1/T = 1/8.0 = 0.125 Hz (accept 0.13 Hz)
✓ 1
Part (b)
Uses v = fλ (or v = λ/T)
✓ 1
v = 0.125 × 96 = 12 m s−1
✓ 1
Allow error carried forward from the frequency calculated in the first part.
—
Part (c)
The wave transfers energy, not matter; the water, and so the buoy, only oscillates about a fixed equilibrium position with no net displacement
✓ 1
Exam tip: "Successive crests" means one full wavelength in space and one full period in time. Use the distance for λ and the time for T.
58IBP2S-020-01-12The electromagnetic spectrumPaper 2Easy4 marks
A 5G mobile-phone network uses electromagnetic waves of frequency 28 GHz.
(a)
State the speed of these waves in a vacuum.
(1)
(b)
Calculate the wavelength of the waves.
(2)
(c)
State the region of the electromagnetic spectrum to which these waves belong.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
3.00×108 m s−1
✓ 1
Part (b)
λ = c/f = 3.00×108 / 28×109
✓ 1
λ = 1.07×10−2 m (accept 1.1×10−2 m or 0.011 m)
✓ 1
Part (c)
Microwave
✓ 1
Common error: Giga means 10⁹. Getting the prefix wrong gives a wavelength 1000 times too large or too small, which can also put the wave in the wrong region.
59IBP2S-020-02-11The electromagnetic spectrumPaper 2Easy4 marks
A speed gun at a tennis tournament measures the speed of a serve. It emits microwaves of frequency 24 GHz towards the ball.
(a)
State the speed of microwaves in air.
(1)
(b)
Outline one difference between the microwaves from the speed gun and the sound of the racket hitting the ball.
(1)
(c)
Calculate the wavelength of the microwaves.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
3.0×108 m s−1 (essentially the same as the speed of all EM waves in a vacuum)
✓ 1
Part (b)
Any one of: microwaves are transverse but sound is longitudinal; microwaves need no medium but sound needs a medium; microwaves travel much faster than sound
✓ 1
Part (c)
λ = c/f = 3.0×108 / 24×109
✓ 1
λ = 1.25×10−2 m (accept 1.3×10−2 m or 1.25 cm)
✓ 1
Exam tip: Write out the GHz conversion (×10⁹) in your working. A wrong final answer can then still earn the method mark.
60IBP2S-020-02-12Transverse & longitudinalPaper 2Easy5 marks
A railway engineer strikes a steel rail with a hammer. A sensor 1.20 km further along the track detects two sounds: one that travelled through the rail and one that travelled through the air. The table gives the speed of sound in some materials.
Material — speed of sound / m s−1 Air — 340 Water — 1480 Steel — 5900
Diagram not to scale
(a)
State the type of wave by which sound travels through the rail, and describe the motion of the steel particles as the wave passes.
(2)
(b)
Calculate the time interval between the two sounds arriving at the sensor.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Longitudinal wave
✓ 1
The particles oscillate about their fixed positions, parallel to the direction of energy transfer, forming compressions and rarefactions
✓ 1
Part (b)
tair = 1200/340 = 3.53 s
✓ 1
tsteel = 1200/5900 = 0.203 s
✓ 1
Δt = 3.3 s
✓ 1
Common error: Convert 1.20 km to 1200 m before dividing. The speeds in the table are in m s⁻¹.
A medical ultrasound transducer emits pulses of frequency 3.5 MHz into a patient's abdomen. The speed of ultrasound in soft tissue is 1540 m s−1.
(a)
Outline what is meant by a longitudinal wave.
(2)
(b)
Calculate the wavelength of the ultrasound in soft tissue.
(2)
(c)
A pulse passes from soft tissue into bone, where the speed of ultrasound is 4080 m s−1. State the frequency of the ultrasound in bone and determine its wavelength there.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The particles of the medium oscillate parallel to the direction of energy transfer / wave travel . This produces a series of compressions and rarefactions .
✓ 2
Part (b)
λ = v/f = 1540 / 3.5×106 . λ = 4.4×10−4 m .
✓ 2
Part (c)
The frequency stays 3.5 MHz because it is set by the source . λ = 4080 / 3.5×106 = 1.17×10−3 m (accept 1.2×10−3 m) .
✓ 2
Common error: Do not calculate a new frequency when the wave enters bone. Keep f fixed and find the new λ from v = fλ.
62IBP2S-020-00-12Transverse & longitudinalPaper 2Easy5 marks
At a football stadium, spectators create a 'crowd wave'. Each person stands up and sits down again as the wave reaches them, and the disturbance moves horizontally along the stand. The crest of the crowd wave travels 72 m along the stand in 6.0 s.
(a)
Explain whether the crowd wave is a transverse or a longitudinal wave.
(2)
(b)
Calculate the speed of the crowd wave.
(1)
(c)
Each spectator takes 1.5 s to complete one stand-and-sit cycle. Calculate the wavelength of the crowd wave.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The people move vertically, up and down . This is perpendicular to the horizontal direction in which the wave travels (the direction of energy transfer), so the wave is transverse .
✓ 2
Part (b)
v = 72/6.0 = 12 m s−1 .
✓ 1
Part (c)
Uses λ = vT, or f = 1/1.5 = 0.67 Hz with λ = v/f . λ = 12 × 1.5 = 18 m . Allow error carried forward from the speed calculated in the previous part.
✓ 2
Exam tip: When the period is given directly, λ = vT is quicker than finding f first. It also avoids rounding f too early.
63C-2-04Transverse & longitudinalPaper 2Medium12 marks
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)
Show mark scheme
Marking point
Mark
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)
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)
Show mark scheme
Marking point
Mark
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)
65C-2-20Intensity & amplitudePaper 2Medium14 marks
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)
66C-2-21Transverse & longitudinalPaper 2Medium12 marks
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
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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)
67C-2-56Mechanical & electromagnetic wavesPaper 2Medium12 marks
A self-driving car uses three types of sensor to measure the distance to nearby objects: ultrasonic parking sensors that emit pulses of sound of frequency 48 kHz, a radar that emits microwaves of frequency 77 GHz, and a lidar that emits pulses of infrared light of wavelength 905 nm. Each sensor measures the time between emitting a pulse and detecting its echo.
Speed of sound in air at 20 °C = 343 m s−1
Figure 1 — time axis drawn to scale; the oscillations within the pulse are not drawn individually
(a)
Calculate the wavelength of the ultrasound in air at 20 °C.
(1)
(b)
(i)
Calculate the wavelength of the radar waves.
(1)
(ii)
Calculate the frequency of the lidar light.
(1)
(c)
The frequency of the radar waves is more than a million times that of the ultrasound. Explain why their wavelengths are nevertheless of the same order of magnitude.
(2)
(d)
Figure 1 shows the signal at the ultrasonic transducer as it emits a pulse at t = 0. The pulse is reflected by a wall 1.20 m from the sensor. The air is at 20 °C. On Figure 1, draw the echo detected by the transducer.
(2)
(e)
The pulse in Figure 1 lasts 0.25 ms. An air molecule M is 0.60 m in front of the sensor, on the line between the sensor and the wall. Describe the motion of M between t = 0 and t = 3.0 ms.
(2)
(f)
The electronics of the ultrasonic sensor always assume a speed of 343 m s−1. On a hot day the air is at 40 °C, and the speed of sound is 355 m s−1.
(i)
Determine the distance displayed by the sensor for the wall 1.20 m away.
(2)
(ii)
Explain why the distances measured by the lidar are not affected by the temperature of the air in this way.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
λ = 343/(48 × 103) = 7.15 × 10−3 m
✓ 1
Accept 7.1 mm.
Part (b)(i)
λ = 3.00 × 108/(77 × 109) = 3.9 × 10−3 m
✓ 1
Part (b)(ii)
f = 3.00 × 108/(905 × 10−9) = 3.31 × 1014 Hz
✓ 1
Part (c)
λ = v/f: the radar frequency is about 1.6 × 106 times greater
✓ 1
but the radar waves are electromagnetic and travel at «about» c, about 8.7 × 105 times the speed of sound «a mechanical wave», so the two factors almost cancel
✓ 1
Must compare the speeds of the two waves. OWTTE.
Part (d)
echo begins at t = 2 × 1.20/343 = 7.0 ms
✓ 1
Accept 6.8–7.2 ms on the graph.
echo of the same duration «about 0.25 ms» and smaller amplitude than the emitted pulse
✓ 1
Do not award MP2 for an echo with an amplitude equal to or greater than that of the pulse.
Part (e)
M is at rest until the pulse reaches it at t = 0.60/343 = 1.75 ms; it then oscillates until t = 2.0 ms «12 oscillations at 48 kHz»
✓ 1
Accept 1.7–1.8 ms for the start. Both times needed.
it oscillates about its fixed equilibrium position, parallel to the line from the sensor to the wall «longitudinal», and is at rest at its original position again after the pulse has passed «no net displacement»
✓ 1
Do not accept 'M moves towards the wall with the pulse'. The echo reaches M only at about 5.2 ms. OWTTE.
Part (f)(i)
echo time = 2 × 1.20/355 = 6.76 ms
✓ 1
ALTERNATIVE: displayed distance = 1.20 × 343/355.
displayed distance = 343 × 6.76 × 10−3/2 = 1.16 m
✓ 1
Accept 1.159 m. Award [2] for CNA. Award [1 max] for 1.24 m.
Part (f)(ii)
light is an electromagnetic wave whose speed in air is «almost exactly» c, whatever the air temperature, whereas the speed of sound depends on the motion of the air molecules «and so on the temperature»
✓ 1
OWTTE.
Answers: (a) 7.15 × 10−3 m · (b)(i) 3.9 × 10−3 m · (b)(ii) 3.31 × 1014 Hz · (d) echo at 7.0 ms · (e) oscillates from 1.75 ms to 2.0 ms · (f)(i) 1.16 m (the remaining parts are explanations — see the table above)
68C-2-57Transverse & longitudinalPaper 2Medium12 marks
In a fitness exercise an athlete holds one end of a heavy rope. The other end is tied to a post 15.0 m away. The athlete moves the end of the rope up and down with a frequency of 2.0 Hz, sending waves along the rope towards the post. Figure 1 shows, drawn to scale, the first 6.0 m of the rope at time t = 0. Over this length the rope is clear of the floor and the decrease in amplitude is negligible.
Figure 1 — drawn to scale
(a)
Explain why the waves on the rope are transverse.
(1)
(b)
(i)
State the wavelength of the waves.
(1)
(ii)
Show that the speed of the waves is about 6 m s−1.
(1)
(iii)
Calculate the time taken for a crest to travel from the athlete to the post.
(1)
(c)
On Figure 1, draw the shape of the rope at t = 0.125 s.
(2)
(d)
The athlete now moves the end of the rope with a frequency of 3.0 Hz. The tension in the rope does not change. State and explain the new wavelength.
(2)
(e)
Beyond 6.0 m the rope strikes the floor in every oscillation, and 12.0 m from the athlete the amplitude of the waves is only 0.10 m. Outline what this suggests about the energy carried by the waves as they travel along the rope.
(2)
(f)
During a 30 s set, the athlete's arms do work on the rope at an average rate of 180 W. The muscles have an efficiency of 22 %. Calculate the energy transferred from the athlete's food store during the set.
(2)
Show mark scheme
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Mark
Notes
Part (a)
each part of the rope oscillates «up and down» perpendicular to the direction in which the wave travels «and transfers energy» along the rope
✓ 1
OWTTE.
Part (b)(i)
3.0 m
✓ 1
Accept 2.9–3.1 m.
Part (b)(ii)
v = fλ = 2.0 × 3.0 = 6.0 «m s−1»
✓ 1
Must see full substitution OR answer to 3 s.f. Allow ECF from (b)(i).
Part (b)(iii)
t = 15.0/6.0 = 2.5 s
✓ 1
Allow ECF from (b)(ii).
Part (c)
sine curve of the same amplitude «0.25 m» and wavelength «3.0 m»
✓ 1
shifted 0.75 m «a quarter of a wavelength» towards the post: crests at 1.5 m and 4.5 m, with the rope at −0.25 m at the athlete's hands
✓ 1
Accept ±0.1 m. Do not award MP2 for a shift towards the athlete.
Part (d)
the speed depends only on the rope «the medium and its tension», so it stays 6.0 m s−1
✓ 1
Allow ECF from (b)(ii).
λ = 6.0/3.0 = 2.0 m
✓ 1
MP2 may be awarded without MP1.
Part (e)
the waves transfer energy along the rope from the athlete towards the post
✓ 1
some of this energy is dissipated along the way «as thermal energy, by friction between the rope and the floor, air resistance or internal friction in the rope», so less energy reaches the post and the amplitude is smaller
✓ 1
OWTTE. Do not accept 'the energy is used up'.
Part (f)
work done on the rope = 180 × 30 = 5400 J
✓ 1
energy from food = 5400/0.22 = 2.5 × 104 J
✓ 1
Award [1 max] for 1.2 × 103 J «multiplied by the efficiency». Award [2] for CNA.
Answers: (b)(i) 3.0 m · (b)(iii) 2.5 s · (d) 2.0 m · (f) 2.5 × 104 J (the remaining parts are explanations — see the table above)
69IBP2S-020-01-13The wave equationPaper 2Medium7 marks
A student uses a vibrating dipper to produce water waves of different frequencies f in a ripple tank of constant depth. The student measures the wavelength λ for each frequency.
71IBP2S-020-02-13The wave equationPaper 2Medium6 marks
A data buoy floats on the open ocean. Its displacement–time graph is sinusoidal. The buoy reaches its maximum upward displacement of +1.2 m at t = 0. It next reaches its maximum downward displacement of −1.2 m at t = 3.5 s. The ocean waves travel at 11 m s−1.
(a)
State the amplitude of the waves.
(1)
(b)
Determine the frequency of the waves.
(2)
(c)
Outline why the buoy does not travel along with the waves.
(1)
(d)
Calculate the wavelength of the waves.
(2)
Show mark scheme
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Mark
Notes
Part (a)
1.2 m
✓ 1
Part (b)
Going from a crest to a trough takes half a period, so T = 2 × 3.5 = 7.0 s
✓ 1
f = 1/7.0 = 0.14 Hz (accept 0.143 Hz)
✓ 1
Part (c)
A wave transfers energy, not matter: the water and the buoy only oscillate about a fixed equilibrium position
✓ 1
Part (d)
Uses λ = v/f or λ = vT, e.g. 11 × 7.0
✓ 1
λ = 77 m (accept 77–79 m if the rounded f = 0.14 Hz is used)
✓ 1
Allow error carried forward from the frequency found in the earlier part.
—
Common error: The time from a maximum to the next minimum is half a period, not a full period.
72IBP2S-020-02-14Mechanical & electromagnetic wavesPaper 2Medium5 marks
A rover on Mars sends data to Earth using radio waves of frequency 8.4 GHz. At the time of transmission, the distance between Earth and Mars is 2.25×1011 m.
(a)
State two properties that all electromagnetic waves have when travelling through a vacuum.
(2)
(b)
Calculate the wavelength of the radio waves.
(1)
(c)
The rover detects a hazard and signals Earth. Controllers reply immediately with a radio command. Determine the minimum time between the rover detecting the hazard and receiving the command.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Acceptable properties: they all travel at the same speed, c = 3.0×108 m s−1; they are transverse waves; they consist of oscillating electric and magnetic fields perpendicular to each other and to the direction of travel; they can be polarised.
—
First correct property
✓ 1
Second, different correct property
✓ 1
Part (b)
λ = c / f = 3.0×108 / 8.4×109 = 3.6×10−2 m
✓ 1
Part (c)
One-way time = 2.25×1011 / 3.0×108 = 750 s
✓ 1
Round trip = 2 × 750 = 1500 s, which is about 25 minutes
✓ 1
Exam tip: Read whether the signal makes a one-way trip or a round trip. A reply means the signal crosses the distance twice.
73IBP2S-020-02-15Reading wave graphsPaper 2Medium6 marks
A longitudinal wave of frequency 15 Hz travels in the +x direction along a long spring used to test a vehicle suspension. The table shows the displacement s of coils at positions x along the spring at one instant. Positive s means displacement in the +x direction.
State the wavelength and the amplitude of the wave.
(2)
(b)
Identify a position of a compression. Explain your answer.
(2)
(c)
Determine the time taken for the compression at x = 20 cm to reach x = 60 cm.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Wavelength = 40 cm
✓ 1
Amplitude = 4 mm
✓ 1
Part (b)
x = 20 cm (or 60 cm)
✓ 1
The coils to the left of this point are displaced in the +x direction and the coils to the right are displaced in the −x direction, so both are displaced towards it and the coils are bunched together
✓ 1
Part (c)
v = fλ = 15 × 0.40 = 6.0 m s−1
✓ 1
t = 0.40 / 6.0 = 0.067 s
✓ 1
Alternative: the distance of 40 cm is one wavelength, so t = T = 1/15 = 0.067 s (2)
—
Allow error carried forward from the wavelength in the first part.
—
Exam tip: In a longitudinal wave, compressions and rarefactions occur where the displacement is zero, not where it is a maximum. Check which way the neighbouring coils are displaced to tell them apart.
74IBP2S-020-00-13Mechanical & electromagnetic wavesPaper 2Medium8 marks
A student investigates how the speed of sound in dry air depends on temperature. The results are shown below.
Temperature / °C: 0, 10, 20, 30, 40 Speed of sound / m s−1: 331, 337, 343, 349, 355
A tuning fork of frequency 512 Hz is sounded in the air.
Data from the question
(a)
Describe the relationship between the speed of sound and the temperature over this range.
(2)
(b)
Determine the wavelength of the sound from the tuning fork at 20 °C.
(2)
(c)
Explain why the frequency of the sound in the air does not change when the air temperature changes, even though the speed does.
(2)
(d)
Determine the change in wavelength of the sound when the air temperature rises from 0 °C to 40 °C.
(2)
Show mark scheme
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Mark
Notes
Part (a)
The speed increases as the temperature increases . The relationship is approximately linear, with an increase of about 0.6 m s−1 per °C .
✓ 2
Part (b)
λ = v/f = 343/512 . λ = 0.670 m .
✓ 2
Part (c)
The frequency is determined by the vibration of the source (the tuning fork), which does not depend on the air . The wavelength changes instead, according to λ = v/f .
✓ 2
Part (d)
λ0 = 331/512 = 0.6465 m and λ40 = 355/512 = 0.6934 m . Change = 0.047 m, an increase .
✓ 2
Exam tip: For 'determine the change', subtract the speeds first and divide by f once. This reduces rounding errors.
75IBP2S-020-00-15Mechanical & electromagnetic wavesPaper 2Medium6 marks
A rover on Mars sends data to Earth using radio waves of frequency 8.4 GHz. At the time of transmission, Mars is 2.25×1011 m from Earth. The speed of electromagnetic waves is 3.00×108 m s−1.
(a)
Describe the nature of an electromagnetic wave.
(2)
(b)
Calculate the wavelength of the radio waves.
(1)
(c)
Astronauts standing close together on the Moon cannot hear each other speak, but they can communicate by radio. Explain why.
(2)
(d)
Calculate the time taken for the signal to travel from Mars to Earth.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
It consists of oscillating electric and magnetic fields that are perpendicular to each other . Both fields are perpendicular to the direction of travel, so the wave is transverse and does not need a medium .
✓ 2
Part (b)
λ = 3.00×108 / 8.4×109 = 3.6×10−2 m .
✓ 1
Part (c)
Sound is a mechanical wave that needs particles of a medium to oscillate, and there is no air on the Moon . Radio waves are electromagnetic and can travel through a vacuum .
✓ 2
Part (d)
t = 2.25×1011 / 3.00×108 = 750 s, about 12.5 minutes .
✓ 1
Exam tip: When explaining sound in space, say that sound needs a medium because it is a mechanical wave. Saying only 'there is no air' gives the situation but not the physics.
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
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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)
A sonic anemometer at a weather station measures the wind speed with ultrasound. Two transducers, P and Q, face each other 0.150 m apart on an east–west line, with P to the west. P sends a short pulse of ultrasound to Q and then Q sends a pulse to P. The wind blows from west to east along the line PQ.
The speed of a pulse relative to the ground is the speed of sound in still air plus or minus the wind speed. In one measurement the time taken from P to Q is 433.5 μs and the time taken from Q to P is 449.1 μs.
Diagram not to scale
(a)
State the direction in which the molecules of the air oscillate as the pulse travels from P to Q.
(1)
(b)
(i)
Show that the speed of the pulse from P to Q, relative to the ground, is about 346 m s−1.
(1)
(ii)
Determine the wind speed and the speed of sound in still air.
(4)
(c)
The speed of sound v in air is given by v = 20.0√T, where T is the temperature in kelvin. Calculate the temperature of the air in °C.
(2)
(d)
Explain, using the molecular model of a gas, why the speed of sound increases when the temperature of the air increases.
(2)
(e)
Each time is measured with an uncertainty of ±0.1 μs. Determine whether the anemometer can give the wind speed with an uncertainty no greater than ±0.1 m s−1.
(3)
(f)
Suggest one advantage of timing pulses in both directions rather than timing a pulse in one direction only.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Parallel to PQ «east–west, the direction in which the pulse travels»
✓ 1
Accept 'backwards and forwards along PQ'.
Part (b)(i)
0.150/(433.5 × 10−6) = 346.0 «m s−1»
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
Speed from Q to P = 0.150/(449.1 × 10−6) = 334.0 m s−1
✓ 1
v + u = 346.0 AND v − u = 334.0
✓ 1
Award MP2 for the two equations in any form.
u = (346.0 − 334.0)/2 = 6.0 m s−1
✓ 1
Award [1 max] for 12 m s−1.
v = (346.0 + 334.0)/2 = 340 m s−1
✓ 1
Award [4] for CNA for both.
Part (c)
T = (340/20.0)2 = 289 K
✓ 1
Allow ECF from (b)(ii).
16 °C
✓ 1
Accept 15–16 °C.
Part (d)
The mean «random» kinetic energy of the molecules is proportional to the «kelvin» temperature, so at a higher temperature the molecules move faster
✓ 1
Accept reference to Ek = (3/2)kBT.
The disturbance «compression» is passed from molecule to molecule by collisions, which now happen sooner, so the wave travels faster
✓ 1
OWTTE.
Part (e)
Uncertainty in each speed: Δv = vΔt/t = 346.0 × 0.1/433.5 = 0.080 m s−1 AND 334.0 × 0.1/449.1 = 0.074 m s−1
✓ 1
Accept a fractional uncertainty of 0.023 % for each.
u = (v1 − v2)/2, so Δu = (0.080 + 0.074)/2 = 0.077 m s−1
✓ 1
ALTERNATIVE: largest and smallest possible values of u from the extreme times.
0.08 m s−1 < 0.1 m s−1, so yes
✓ 1
MP3 only if MP2 is scored. Do not award MP3 for 0.15 m s−1 «halving omitted» and 'no'.
Part (f)
The speed of sound in still air «and so the temperature» does not need to be known: it cancels when the two speeds are subtracted
✓ 1
OWTTE.
Answers: (b)(i) 346.0 m s−1 · (b)(ii) u = 6.0 m s−1; v = 340 m s−1 · (c) 16 °C · (e) ±0.077 m s−1: yes (the remaining parts are explanations — see the table above)
78C-2-58The electromagnetic spectrumPaper 2Hard13 marks
A satellite in a circular orbit carries a laser altimeter to measure the height of ice sheets. The satellite is 500 km above the Earth's surface and its orbital period is 94.5 minutes. The laser emits short pulses of green light of wavelength 532 nm at a rate of 10 000 pulses per second. Each pulse reflects from the ground directly below and a telescope on the satellite detects the reflected light. The altimeter measures the time between the emission of each pulse and the detection of its reflection.
Radius of the Earth = 6.37 × 106 m. Assume that the light travels at c along the whole of its path.
(a)
Calculate the frequency of the laser light.
(1)
(b)
Show that the time between the emission of a pulse and the detection of its reflection is about 3.3 ms.
(1)
(c)
Determine whether each pulse is detected before the next pulse is emitted. If it is not, estimate the number of pulses that are travelling between the satellite and the ground at any instant.
(3)
(d)
The altimeter must measure the height of the ice to within ±3.0 cm. Calculate the uncertainty in the time measurement that this requires.
(2)
(e)
Ignore the rotation of the Earth.
(i)
Show that the speed of the satellite is about 7.6 km s−1.
(1)
(ii)
Determine the distance between the points on the ground directly below the satellite at the instants when two successive pulses are emitted.
(2)
(f)
Each pulse has an energy of 0.20 mJ. Calculate the number of photons in each pulse.
(2)
(g)
Only about ten photons from each pulse are detected by the telescope. Suggest one reason why so few are detected.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
f = 3.00 × 108/(532 × 10−9) = 5.64 × 1014 Hz
✓ 1
Part (b)
t = 2 × 5.0 × 105/(3.00 × 108) = 3.33 × 10−3 «s»
✓ 1
Must see the factor of 2 AND full substitution OR answer to 3 s.f.
Part (c)
time between pulses = 1/10 000 = 1.0 × 10−4 s
✓ 1
3.3 × 10−3 s ≫ 1.0 × 10−4 s, so each pulse is NOT detected before the next is emitted
✓ 1
MP2 only scores if the two times are compared.
number in flight ≈ 3.3 × 10−3/1.0 × 10−4 ≈ 33
✓ 1
Accept 33 or 34. Allow ECF from (b).
Part (d)
the light travels to the ice and back, so an error of 3.0 cm in height changes the path by 6.0 cm
✓ 1
Δt = 0.060/(3.00 × 108) = 2.0 × 10−10 s
✓ 1
Award [1 max] for 1.0 × 10−10 s. Award [2] for CNA.
the light reflected from the ground spreads out in many directions, so only a tiny fraction of it enters the small telescope 500 km away OR some of the light is absorbed or scattered by the atmosphere or absorbed by the surface
✓ 1
Accept any one sensible reason. OWTTE.
Answers: (a) 5.64 × 1014 Hz · (b) 3.33 × 10−3 s · (c) no; about 33 pulses · (d) 2.0 × 10−10 s · (e)(i) 7.61 × 103 m s−1 · (e)(ii) 0.71 m · (f) 5.3 × 1014(the remaining parts are explanations — see the table above)
79IBP2S-020-01-16Transverse & longitudinalPaper 2Hard8 marks
An earthquake produces P-waves, which travel through the crust at 6.0 km s−1, and S-waves, which travel at 3.5 km s−1. At a seismometer station, the first S-waves arrive 48 s after the first P-waves. The amplitude of the P-waves recorded at this station is 0.40 mm.
(a)
The P-waves are longitudinal and the S-waves are transverse. State the difference between a longitudinal and a transverse wave.
(1)
(b)
S-waves cannot pass through a liquid. Outline why.
(1)
(c)
Determine the distance from the earthquake to the station.
(3)
(d)
The P-waves have a frequency of 2.0 Hz. Calculate their wavelength.
(1)
(e)
A second station records P-waves of amplitude 0.25 mm. Assume the waves spread out from a point source with no absorption. Determine the distance of the second station from the earthquake.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
In a longitudinal wave the particles oscillate parallel to the direction of energy transfer; in a transverse wave they oscillate perpendicular to it
✓ 1
Part (b)
A transverse wave needs a medium that resists shear (sideways) deformation; a liquid cannot provide a shear restoring force between layers
✓ 1
Part (c)
d/3.5 − d/6.0 = 48, with d in km
✓ 1
d × (0.2857 − 0.1667) = 48, so d = 48/0.1190
✓ 1
d ≈ 403 km, equivalent to 4.0×102 km to 2 s.f.; either value scores
✓ 1
Part (d)
λ = v/f = 6.0×103/2.0 = 3.0×103 m
✓ 1
Part (e)
I ∝ A2 and I ∝ 1/r2, so A ∝ 1/r
✓ 1
r = 403 × 0.40/0.25 ≈ 645 km, equivalent to 6.5×102 km to 2 s.f.; either value scores
✓ 1
Allow error carried forward from the distance calculated in the earlier part.
—
Exam tip: For arrival-time differences, write one equation that uses the same unknown distance for both waves. Do not try to find each travel time separately.
80IBP2S-020-02-16Reading wave graphsPaper 2Hard7 marks
Engineers monitor a transverse wave travelling in the +x direction along the haul cable of a ski lift. The amplitude of the wave is 0.15 m. In a displacement–distance snapshot at t = 0, there is a crest at x = 2.0 m and the adjacent trough is at x = 5.0 m. In a second snapshot at t = 0.20 s, that crest is at x = 3.2 m. The wave moved less than one wavelength between the snapshots.
(a)
Determine the wavelength of the wave.
(1)
(b)
Calculate the speed of the wave.
(1)
(c)
Determine the frequency of the wave.
(1)
(d)
At t = 0, a point P on the cable at x = 3.5 m has zero displacement. State and explain the direction in which P is moving at this instant.
(2)
(e)
Determine the displacement of P at t = 0.25 s.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A crest and the adjacent trough are half a wavelength apart, so λ = 2 × 3.0 = 6.0 m
✓ 1
Part (b)
v = (3.2 − 2.0)/0.20 = 6.0 m s−1
✓ 1
Part (c)
f = v/λ = 6.0/6.0 = 1.0 Hz
✓ 1
ECF from (a) and (b)
—
Part (d)
P is moving upward (towards positive displacement)
✓ 1
The crest at x = 2.0 m is behind P and travels towards it in the +x direction, so P's displacement is about to increase
✓ 1
Part (e)
In 0.25 s the crest moves 6.0 × 0.25 = 1.5 m, from x = 2.0 m to x = 3.5 m. Alternatively, 0.25 s is a quarter of the period
✓ 1
So the displacement of P is +0.15 m
✓ 1
ECF from (b) and (c)
—
Exam tip: To find a later displacement, shift the whole waveform by vt and read off the new displacement at the point.
81IBP2S-020-00-14Transverse & longitudinalPaper 2Hard8 marks
An earthquake produces P-waves and S-waves that travel through the Earth's crust. In the crust, P-waves travel at 6.0 km s−1 and S-waves travel at 3.5 km s−1. A seismometer records the first P-wave 40 s before the first S-wave.
(a)
State which type of seismic wave is longitudinal and which is transverse.
(1)
(b)
S-waves are not detected at seismometers on the opposite side of the Earth from an earthquake. Explain how this provides evidence that part of the Earth's core is liquid.
(2)
(c)
An S-wave in the crust has a period of 0.50 s. Calculate its wavelength.
(2)
(d)
Determine the distance from the seismometer to the source of the earthquake.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
P-waves are longitudinal and S-waves are transverse .
✓ 1
Part (b)
S-waves are transverse, so they need a medium that resists shear (sideways) deformation; a liquid cannot support shear, so transverse waves cannot travel through it . Waves reaching the opposite side must pass through the core, so the absence of S-waves there shows that part of the core (the outer core) blocks them and must be liquid .
✓ 2
Part (c)
λ = vT = 3.5×103 × 0.50 . λ = 1.75×103 m ≈ 1.8×103 m; either value is accepted .
✓ 2
Part (d)
The difference in travel times equals the delay: d/vS − d/vP = 40 s . d = 40 / (1/3500 − 1/6000) . d ≈ 3.4×105 m, i.e. about 340 km .
✓ 3
Common error: Do not divide the time delay by the difference in speeds. The difference in travel times equals (1/vS − 1/vP)d, so d is not equal to 40/(vP − vS).
82IBP2S-020-00-16Reading wave graphsPaper 2Hard8 marks
In a gym, an athlete moves one end of a long, heavy 'battle rope' up and down with a frequency of 2.5 Hz and an amplitude of 0.15 m. A crest is observed to travel 6.0 m along the rope in 0.80 s. Ignore energy losses.
(a)
Calculate the speed of the wave on the rope.
(1)
(b)
Calculate the wavelength of the wave on the rope.
(1)
(c)
The athlete starts moving the end upwards from rest at t = 0. Describe the displacement–time graph, from t = 0 to t = 1.0 s, for a point on the rope 1.5 m from the athlete's end.
(2)
(d)
The athlete doubles the frequency to 5.0 Hz without changing the tension in the rope. State and explain the effect on the wavelength.
(2)
(e)
The rope is now joined to a heavier rope, in which waves travel at 5.0 m s−1. The athlete again moves the end at 2.5 Hz. Determine the wavelength in the heavier rope.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = 6.0/0.80 = 7.5 m s−1
✓ 1
Part (b)
λ = v/f = 7.5/2.5 = 3.0 m . Allow error carried forward from the wave speed.
✓ 1
Part (c)
The displacement stays zero until the wave arrives at t = 1.5/7.5 = 0.20 s, and the point then begins by moving upwards . From 0.20 s to 1.0 s the graph is sinusoidal, with amplitude 0.15 m and period 0.40 s, giving two complete cycles that end at zero displacement at t = 1.0 s . Allow error carried forward from the wave speed. Accept equivalent statements, e.g. 'the same shape as the motion of the end, delayed by 0.20 s (half a period)'.
✓ 2
Part (d)
The wave speed depends only on the rope (the medium) and its tension, so it is unchanged . Since λ = v/f, the wavelength halves, to 7.5/5.0 = 1.5 m .
✓ 2
Part (e)
The frequency stays at 2.5 Hz when the wave enters the heavier rope, because it is set by the source . λ = 5.0/2.5 = 2.0 m .
✓ 2
Common error: Increasing the frequency does not make the wave travel faster. In a given medium at fixed tension the speed is fixed, so the wavelength changes instead.
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