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Edexcel International A Level Physics · Unit 2 · Waves and Electricity
40 free, high-quality past-paper-style questions on Progressive waves for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which of the following is the wavelength of a progressive wave?
| Answer | Mark | Notes |
|---|---|---|
| The wavelength is the distance between two neighbouring points oscillating in phase — for example crest to crest, or compression to compression. | ✓ 1 | Answer D |
Answer: D
Exam tip: “In phase” is the key phrase: it makes the definition work for longitudinal waves too, where there are no crests.
Which of the following is a longitudinal wave?
| Answer | Mark | Notes |
|---|---|---|
| Sound and ultrasound are longitudinal: the particles of the medium oscillate parallel to the direction of energy transfer. Electromagnetic waves (microwaves, X-rays) and waves on strings are transverse. Answer: ultrasound in body tissue. | ✓ 1 | Answer A |
Answer: A
Exam tip: All electromagnetic waves are transverse. All sound waves, at any frequency, are longitudinal.
A transverse progressive wave travels along a rope. At one instant a particular point on the rope is at a crest.
Which statement about that point at this instant is correct?
| Answer | Mark | Notes |
|---|---|---|
| Each point oscillates at right angles to the rope. At a crest it has its maximum displacement and is momentarily at rest before moving back towards equilibrium. It never travels along the rope. | ✓ 1 | Answer D |
Answer: D
Exam tip: Separate the wave from the particles: the crest moves along at the wave speed, but each particle only moves up and down about its own position.
A music teacher strikes a tuning fork marked 440 Hz. The speed of sound in the air of the classroom is 343 m s−1.
State what is meant by the wavelength of a wave.
State what is meant by the frequency of a wave.
Calculate the wavelength of the sound from the tuning fork.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the distance between two neighbouring points oscillating in phase / the distance between neighbouring compressions | ✓ 1 | |
| Part (b) | ||
| the number of complete oscillations (of a particle / of the source) per unit time | ✓ 1 | |
| Part (c) | ||
| λ = v/f = 343 ÷ 440 | ✓ 1 | |
| = 0.780 m (0.7795 m) | ✓ 1 | |
Exam tip: A definition of frequency must contain “per unit time” or “per second”; “the number of waves” alone is not a frequency.
Waves can be classified as transverse or longitudinal.
Explain the difference between a transverse wave and a longitudinal wave.
Give one example of each type of wave.
State what is transferred from place to place by a progressive wave.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| transverse: the oscillations are perpendicular to the direction of energy transfer / of wave travel | ✓ 1 | |
| longitudinal: the oscillations are parallel to the direction of energy transfer / of wave travel | ✓ 1 | |
| Part (b) | ||
| transverse: any electromagnetic wave (light, radio, microwaves …) / a wave on a string / S-wave | ✓ 1 | |
| longitudinal: sound / ultrasound / P-wave / push–pull wave on a slinky | ✓ 1 | |
| Part (c) | ||
| energy (and information), with no net transfer of the medium | ✓ 1 | |
Common error: Writing “the wave moves at right angles”. It is the oscillation of the particles or fields that is perpendicular; the wave itself travels forwards.
A microphone connected to an oscilloscope picks up a steady note from a whistle. One complete cycle of the trace occupies 4.0 horizontal divisions with the time-base set to 0.20 ms div−1. The trace measures 3.0 divisions from its lowest point to its highest point, with the vertical sensitivity set to 0.50 V div−1. The speed of sound in air is 340 m s−1.
Determine the period of the sound.
Calculate the frequency of the sound.
Calculate the wavelength of the sound in air.
Determine the amplitude of the voltage signal shown on the oscilloscope.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 4.0 × 0.20 = 0.80 ms | ✓ 1 | |
| Part (b) | ||
| f = 1 ÷ 0.80 × 10−3 = 1250 Hz (1.25 kHz) | ✓ 1 | |
| Part (c) | ||
| λ = 340 ÷ 1250 = 0.27 m (0.272 m) | ✓ 1 | |
| Part (d) | ||
| 3.0 ÷ 2 × 0.50 = 0.75 V | ✓ 1 | |
Common error: Using the full lowest-to-highest height as the amplitude. The amplitude is half of it.
A transverse progressive wave travels along a stretched string. The graph shows the displacement of points on the string against distance along the string at one instant.
What is the wavelength of the wave?
| Answer | Mark | Notes |
|---|---|---|
| Neighbouring crests are at 10 cm and 50 cm, so λ = 50 − 10 = 40 cm. The curve also crosses zero going upwards at 0 and at 40 cm, one wavelength apart. | ✓ 1 | Answer D |
Answer: D
Common error: Reading the distance between neighbouring points of zero displacement (0 to 20 cm). Those points are only half a wavelength apart.
The diagram shows a transverse wave at one instant. Four distances, P, Q, R and S, are shown.
Which distance is the amplitude of the wave?
| Answer | Mark | Notes |
|---|---|---|
| The amplitude is the maximum displacement from the equilibrium position: from the equilibrium line to a crest, which is S. P is one wavelength, Q is half a wavelength, and R, from a trough to a crest, is twice the amplitude. | ✓ 1 | Answer D |
Answer: D
Common error: Measuring the amplitude from a trough to a crest. That height is 2A.
The diagram shows a ripple tank viewed from above. A dipper vibrating at 4.0 Hz produces straight crests, shown as lines. The ruler shows the positions of the first and sixth crests.
What is the speed of the ripples?
| Answer | Mark | Notes |
|---|---|---|
| From the first to the sixth crest there are five wavelengths: λ = 15.0 ÷ 5 = 3.0 cm = 0.030 m. v = fλ = 4.0 × 0.030 = 0.12 m s−1. | ✓ 1 | Answer B |
Answer: B
Common error: Dividing by the number of crests. Six crests have five gaps between them, so they span five wavelengths.
A paddle at one end of a long laboratory wave flume makes water waves of frequency 1.5 Hz. The graph shows the displacement of the water surface against distance along the flume at one instant.
Use the graph to determine the amplitude of the waves.
Use the graph to determine the wavelength of the waves.
Calculate the speed of the waves.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| equilibrium to crest: 2.5 cm (0.025 m) | ✓ 1 | |
| Part (b) | ||
| crest to neighbouring crest, 0 to 0.80 m: 0.80 m (allow 0.78–0.82 m) | ✓ 1 | |
| Part (c) | ||
| v = fλ = 1.5 × 0.80 | ✓ 1 | |
| = 1.2 m s−1 (allow error carried forward) | ✓ 1 | |
Common error: Taking the distance between neighbouring points of zero displacement (0.20 m to 0.60 m) as the wavelength. That is only half a wavelength.
A progressive wave travels along a long rubber cord. The graph shows how the displacement of one point on the cord varies with time.
State the amplitude of the oscillation of the point.
Determine the period of the wave.
Calculate the frequency of the wave.
State one time shown on the graph at which the point is moving with its greatest speed. Give a reason for your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 6.0 mm | ✓ 1 | |
| Part (b) | ||
| maximum to next maximum, 0 to 40 ms: 40 ms (0.040 s; allow 39–41 ms) | ✓ 1 | |
| Part (c) | ||
| f = 1/T = 1 ÷ 0.040 = 25 Hz (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| any one of 10, 30, 50, 70 or 90 ms | ✓ 1 | |
| the displacement is zero / the point is passing through its equilibrium position, where the gradient of the graph is steepest | ✓ 1 | |
Exam tip: On a displacement–time graph the gradient is the velocity of the point: steepest where the displacement is zero, and zero at a maximum or minimum.
The figure represents a microwave travelling through air at one instant. The arrows show the direction and relative size of the electric field at points along the path of the wave. The speed of electromagnetic waves in air is 3.00 × 108 m s−1.
Explain how the figure shows that microwaves are transverse waves.
Determine the wavelength of the microwave.
Calculate the frequency of the microwave.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the electric field (arrows) oscillates at right angles to the direction of travel | ✓ 1 | |
| Part (b) | ||
| largest upward arrows at 0, 3.0 and 6.0 cm: λ = 3.0 cm | ✓ 1 | |
| Part (c) | ||
| f = c/λ = 3.00 × 108 ÷ 0.030 | ✓ 1 | |
| = 1.0 × 1010 Hz (allow error carried forward) | ✓ 1 | |
Common error: Leaving the wavelength in centimetres in f = c/λ. Convert 3.0 cm to 0.030 m first.
A sonar unit on a fishing boat emits sound of frequency 25 kHz into seawater, where the speed of sound is 1500 m s−1.
What is the distance between a compression and the nearest rarefaction in the seawater?
| Answer | Mark | Notes |
|---|---|---|
| λ = v/f = 1500 ÷ 25 000 = 0.060 m. A compression and the nearest rarefaction are half a wavelength apart: 0.030 m. | ✓ 1 | Answer B |
Answer: B
Common error: Giving the full wavelength. Compression to compression is λ; compression to the nearest rarefaction is λ/2.
A microphone is connected to an oscilloscope with the time-base set to 0.20 ms div−1. One complete cycle of the trace occupies 5.0 horizontal divisions.
What is the frequency of the sound?
| Answer | Mark | Notes |
|---|---|---|
| Period T = 5.0 div × 0.20 ms div−1 = 1.0 ms. f = 1/T = 1 ÷ 1.0 × 10−3 s = 1000 Hz = 1.0 kHz. | ✓ 1 | Answer C |
Answer: C
Exam tip: Multiply divisions by the time-base to get a time; check the units cancel (div × ms div−1 = ms).
On a displacement–distance graph of a progressive wave, the horizontal distance between a crest and the neighbouring trough is 0.30 m. Points X and Y on the wave are 0.20 m apart.
What is the phase difference between the oscillations of X and Y?
| Answer | Mark | Notes |
|---|---|---|
| Crest to neighbouring trough is half a wavelength, so λ = 0.60 m. Δφ = 2π × 0.20 ÷ 0.60 = 2π/3 = 2.1 rad. | ✓ 1 | Answer C |
Answer: C
Common error: Taking the crest-to-trough distance as the wavelength. Neighbouring crest and trough are λ/2 apart.
For a wave on a stretched elastic cord, a displacement–time graph for one point on the cord shows a period of 8.0 ms. A displacement–distance graph shows that 2.5 complete wavelengths occupy a length of 1.70 m.
What is the speed of the wave?
| Answer | Mark | Notes |
|---|---|---|
| λ = 1.70 ÷ 2.5 = 0.68 m; f = 1/T = 1 ÷ 8.0 × 10−3 = 125 Hz. v = fλ = 125 × 0.68 = 85 m s−1. | ✓ 1 | Answer B |
Answer: B
Exam tip: Take the wavelength from the distance graph and the period from the time graph, then v = λ/T.
In an experiment to measure the speed of sound in air, a signal generator set to 2.45 kHz drives a loudspeaker. A microphone starts at a position where the two oscilloscope traces are in phase. As the microphone is moved 0.700 m directly away from the loudspeaker, the traces come back into phase exactly 5 times, ending in phase.
What is the value obtained for the speed of sound?
| Answer | Mark | Notes |
|---|---|---|
| The traces return to being in phase once per wavelength moved, so λ = 0.700 ÷ 5 = 0.140 m. v = fλ = 2450 × 0.140 = 343 m s−1 ≈ 340 m s−1. | ✓ 1 | Answer A |
Answer: A
Common error: Counting the starting position as one of the in-phase positions. Five returns to being in phase means five wavelengths.
A data buoy floats on the open ocean. Its height above the sea bed is recorded and shows that it rises a total of 1.70 m from its lowest to its highest position, and that successive crests of the ocean swell pass the buoy every 9.0 s. Aerial photographs show that the crests are 120 m apart. The swell may be treated as a progressive transverse wave.
State the amplitude of the swell.
Calculate the frequency of the swell.
Calculate the speed of the swell.
A storm far away produced the swell. Calculate the time for a crest to travel 2.0 km towards the buoy.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1.70 ÷ 2 = 0.85 m | ✓ 1 | |
| Part (b) | ||
| f = 1/T = 1 ÷ 9.0 = 0.11 Hz (0.111 Hz) | ✓ 1 | |
| Part (c) | ||
| v = fλ (or λ/T) = 120 ÷ 9.0 | ✓ 1 | |
| = 13 m s−1 (13.3 m s−1) | ✓ 1 | |
| Part (d) | ||
| t = distance ÷ speed = 2000 ÷ 13.3 | ✓ 1 | |
| = 150 s (allow error carried forward) | ✓ 1 | |
Exam tip: Low-to-high movement is crest to trough, which is twice the amplitude.
A mechanical vibrator sends a transverse progressive wave along a long horizontal rope in the +x direction. The vibrator is at x = 0. A displacement–distance graph of the rope at one instant shows a sine curve of amplitude 0.12 m and wavelength 1.6 m. The frequency of the vibrator is 2.5 Hz.
Calculate the period of the wave.
Calculate the speed of the wave.
Determine the phase difference between the oscillations of the points at x = 0.20 m and x = 1.00 m.
Compare the displacement–time graph of the point at x = 0.40 m with that of the point at x = 0.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| T = 1/f = 1 ÷ 2.5 = 0.40 s | ✓ 1 | |
| Part (b) | ||
| v = fλ = 2.5 × 1.6 = 4.0 m s−1 | ✓ 1 | |
| Part (c) | ||
| separation 0.80 m = half a wavelength | ✓ 1 | |
| phase difference = π rad (180°), they are in antiphase | ✓ 1 | |
| Part (d) | ||
| same shape, same amplitude (0.12 m) and same period (0.40 s) | ✓ 1 | |
| shifted later in time by a quarter of a period, 0.10 s / phase difference π/2 rad, the point at 0.40 m lagging | ✓ 1 | |
Exam tip: Points further from the source repeat the motion of points nearer the source a little later: a distance Δx means a delay of Δx/v.
The cone of a loudspeaker vibrates backwards and forwards at a constant frequency of 850 Hz. A listener some distance away hears a steady note. The speed of sound in air is 340 m s−1.
Calculate the distance between neighbouring compressions in the sound wave.
Explain how the sound is transmitted through the air from the loudspeaker to the listener. Your answer should refer to the motion of the air molecules and to the pressure of the air.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| compression to compression is one wavelength: λ = 340 ÷ 850 = 0.40 m | ✓ 1 | |
| Part (b) | ||
| Indicative content: | — | |
| as the cone moves forwards it pushes neighbouring air molecules closer together, forming a compression | ✓ 1 | |
| as the cone moves back the molecules spread out, forming a rarefaction | ✓ 1 | |
| each layer of molecules pushes on the next, so compressions and rarefactions travel outwards and energy is transferred to the listener | ✓ 1 | |
| the molecules oscillate about fixed equilibrium positions, with no net movement of air from the loudspeaker to the listener | ✓ 1 | |
| the oscillations are parallel to the direction of energy transfer, so the wave is longitudinal | ✓ 1 | |
| a compression is a region of pressure above atmospheric and a rarefaction is a region below it; the pressure at the listener's ear varies at the frequency of the cone | ✓ 1 | |
Exam tip: In an extended answer, make each sentence carry one physics idea, and use the key words: compression, rarefaction, oscillate, equilibrium position, parallel, energy, pressure.
A student determines the speed of sound in air using a signal generator, a loudspeaker, a microphone, a metre rule and a two-beam oscilloscope. The signal generator is set to a frequency of 2.50 kHz.
Describe how the oscilloscope and microphone are used to determine the wavelength of the sound.
The first in-phase position is at the 12.0 cm mark and the sixth in-phase position is at the 80.4 cm mark. Calculate the speed of sound.
Explain why the student measures across several wavelengths rather than one.
The student estimates that the uncertainty in the 68.4 cm distance is ± 0.3 cm. Calculate the percentage uncertainty in this distance.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| signal generator output to one channel and microphone output to the other channel of the oscilloscope | ✓ 1 | |
| move the microphone along the metre rule, directly away from the loudspeaker, and note positions where the two traces are in phase | ✓ 1 | |
| measure the distance moved across several in-phase positions and divide by the number of wavelengths; the distance between neighbouring in-phase positions is one wavelength | ✓ 1 | |
| Part (b) | ||
| five wavelengths in 80.4 − 12.0 = 68.4 cm, so λ = 13.68 cm | ✓ 1 | |
| v = fλ = 2500 × 0.1368 = 342 m s−1 | ✓ 1 | |
| Part (c) | ||
| the uncertainty in judging each in-phase position is the same, so it is a smaller percentage of a larger distance / reduces the percentage uncertainty in the wavelength | ✓ 1 | |
| Part (d) | ||
| 0.3 ÷ 68.4 × 100 = 0.44% | ✓ 1 | |
Common error: Dividing by six because six in-phase positions were recorded. From the first to the sixth position is five wavelengths.
A sound wave passes through air. The graph shows how the displacement of one air molecule from its equilibrium position varies with time.
What is the frequency of the sound?
| Answer | Mark | Notes |
|---|---|---|
| Neighbouring maxima are at 0 and 2.5 ms, so T = 2.5 ms. f = 1/T = 1 ÷ (2.5 × 10−3 s) = 400 Hz. | ✓ 1 | Answer C |
Answer: C
Exam tip: Read the period between two points at the same stage of the cycle, such as neighbouring maxima, and convert ms to s before using f = 1/T.
The diagram shows the positions of layers of air molecules at one instant as a sound wave travels through the air.
Which labelled position is at the centre of a rarefaction?
| Answer | Mark | Notes |
|---|---|---|
| A rarefaction is where the layers are furthest apart, so the pressure is lowest. The layers are most widely spaced around W. Y is the centre of a compression; at X and Z the spacing is normal. | ✓ 1 | Answer A |
Answer: A
Exam tip: Compression: layers closest together, pressure above normal. Rarefaction: layers furthest apart, pressure below normal. The two are half a wavelength apart.
The diagram shows a transverse wave on a rope at one instant. The wave is travelling to the left.
Which point on the rope is moving upwards at this instant?
| Answer | Mark | Notes |
|---|---|---|
| A moment later the whole wave shape has moved slightly to the left, so each point takes the displacement that the rope just to its right has now. Just to the right of K the rope is above the equilibrium line, so K is moving upwards (at its maximum speed). M is moving downwards; L and N, at a crest and a trough, are momentarily at rest. | ✓ 1 | Answer A |
Answer: A
Exam tip: Sketch the wave shifted a little in its direction of travel. Each point moves towards its position on the shifted curve.
A student sends transverse waves along a long stretched spring and changes the frequency of the vibrator. For each frequency she measures the wavelength λ and the period T of the waves. The graph shows her results and the line of best fit.
What is the speed of the waves on the spring?
| Answer | Mark | Notes |
|---|---|---|
| Since v = fλ = λ/T, λ = vT: the gradient of the graph is the wave speed. The line passes through the origin and (0.50 s, 1.20 m), so v = 1.20 ÷ 0.50 = 2.4 m s−1. | ✓ 1 | Answer D |
Answer: D
Exam tip: Write the equation in the form of the graph (λ = v × T, like y = mx) to see what the gradient means. Use a large triangle on the line, not one data point.
Two waves, A and B, travel along the same stretched cord at the same speed. The graph shows the displacement–distance graph of each wave at one instant.
What is the ratio (frequency of A) ÷ (frequency of B)?
| Answer | Mark | Notes |
|---|---|---|
| From the graph, λA = 0.75 m (A crosses zero going upwards at 0 and at 0.75 m) and λB = 0.50 m. At the same speed f = v/λ, so fA/fB = λB/λA = 0.50 ÷ 0.75 = 0.67. | ✓ 1 | Answer B |
Answer: B
Exam tip: The medium sets the speed. At a fixed speed, frequency and wavelength are inversely proportional; the amplitudes play no part.
A transverse wave travels along a long rope. The graph shows the shape of the rope at time t = 0 (solid line) and at t = 0.025 s (dashed line). In this time the wave moves less than half a wavelength. P and Q are points on the rope.
State the direction in which the wave is travelling. Explain how you can tell from the graph.
Determine the speed of the wave.
Calculate the frequency of the wave.
State the direction in which point P is moving at t = 0.
State the phase difference between the oscillations of P and Q, in radians.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| to the right (increasing distance), because each crest is further along the rope on the dashed curve, e.g. the crest at 0.30 m has moved to 0.45 m | ✓ 1 | |
| Part (b) | ||
| a crest moves 0.45 − 0.30 = 0.15 m in 0.025 s | ✓ 1 | |
| v = 0.15 ÷ 0.025 = 6.0 m s−1 | ✓ 1 | |
| Part (c) | ||
| λ = 1.2 m from the graph; f = v/λ = 6.0 ÷ 1.2 = 5.0 Hz (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| upwards — its displacement has become positive on the dashed curve | ✓ 1 | |
| Part (e) | ||
| separation 0.30 m = λ/4, so π/2 rad (1.57 rad) | ✓ 1 | |
Exam tip: Follow one feature, such as a crest, from one snapshot to the other. The distance it moves divided by the time is the wave speed; the rope itself only moves up and down.
A loudspeaker sends a sound wave along the x-axis through air. The graph shows the displacement of air molecules from their equilibrium positions against distance x at one instant; a positive displacement is in the +x direction. The diagram below the graph shows the corresponding positions of layers of air, with the displacements greatly exaggerated. The speed of sound in air is 340 m s−1.
State which of the points P, Q, R and S is at the centre of a compression. Use the graph to explain your answer.
Determine the wavelength of the sound.
Calculate the frequency of the sound.
State how the air pressure at P compares with atmospheric pressure. Explain your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Q | ✓ 1 | |
| air just to the left of Q has positive displacement (moved towards Q) and air just to the right has negative displacement (also moved towards Q), so the layers crowd together at Q | ✓ 1 | |
| Part (b) | ||
| 0.50 m, e.g. from x = 0 to S, or compression at 0.25 m to the next at 0.75 m | ✓ 1 | |
| Part (c) | ||
| f = v/λ = 340 ÷ 0.50 = 680 Hz (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| it is equal to (normal) atmospheric pressure | ✓ 1 | |
| the molecules on either side of P are displaced by almost the same amount in the same direction, so their spacing is normal, even though the displacement at P is a maximum | ✓ 1 | |
Common error: Placing a compression where the displacement is largest. On a displacement–distance graph, compressions and rarefactions are at points of zero displacement.
A student measures the wavelength λ of sound in air at the frequencies 1.0, 1.5, 2.0, 2.5, 3.0 and 4.0 kHz, using a signal generator, a loudspeaker, a microphone, a two-beam oscilloscope and a metre rule. She plots a graph of λ against 1/f and draws the line of best fit shown, which passes through the origin.
Describe how the student uses the oscilloscope traces to measure the wavelength at one frequency.
Explain why the graph of λ against 1/f is expected to be a straight line through the origin.
Use the line of best fit to determine the speed of sound in air.
Identify the anomalous result and state how the student should deal with it.
At 2.0 kHz the student's value of λ is 17.1 cm, with an uncertainty of ± 0.5 cm. Calculate the percentage uncertainty in this value.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| signal generator to one input and microphone to the other; move the microphone directly away from the loudspeaker and record the positions where the two traces are in phase | ✓ 1 | |
| neighbouring in-phase positions are one wavelength apart — measure the distance across several of them and divide by the number of wavelengths | ✓ 1 | |
| Part (b) | ||
| λ = v × (1/f) with v constant, so λ is proportional to 1/f, the form y = mx | ✓ 1 | |
| Part (c) | ||
| gradient from a large triangle, e.g. 34.0 cm ÷ 1.00 ms = 0.340 m ÷ 1.00 × 10−3 s | ✓ 1 | |
| v = 340 m s−1 (allow 330–350 m s−1) | ✓ 1 | |
| Part (d) | ||
| the point at 1/f = 0.40 ms (2.5 kHz), well above the line; repeat that measurement / leave it out when drawing the line | ✓ 1 | |
| Part (e) | ||
| 0.5 ÷ 17.1 × 100 = 2.9% | ✓ 1 | |
Exam tip: Convert both axis units to SI before quoting a gradient as a speed: here cm to m and ms to s.
The figure shows two waves sent along the same long spring, each at one instant. In (a) the end of the spring is moved from side to side; in (b) it is moved backwards and forwards along the length of the spring. Each ellipse shows the position of one coil.
Use the scale to determine the wavelength of the wave in (b).
Compare the motion of the coils in the two waves and explain how each wave transfers energy along the spring. Refer to the figure in your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| neighbouring compressions centred at 0.30 m and 0.90 m: 0.60 m (allow 0.55–0.65 m) | ✓ 1 | |
| Part (b) | ||
| Indicative content: | — | |
| in (a) each coil oscillates at right angles to the length of the spring / to the direction of travel, so the wave is transverse | ✓ 1 | |
| in (b) each coil oscillates along the spring, parallel to the direction of travel, so the wave is longitudinal | ✓ 1 | |
| in (b) the coils are crowded together in some places (compressions, e.g. at 0.30 m) and spread apart in others (rarefactions, e.g. at 0.60 m) | ✓ 1 | |
| in both waves each coil oscillates about its own equilibrium position, with no net movement of coils along the spring | ✓ 1 | |
| each coil exerts forces on its neighbours and sets the next coil oscillating slightly later, so energy is passed from coil to coil | ✓ 1 | |
| the pattern (crests and troughs in (a), compressions and rarefactions in (b)) travels along the spring at the wave speed, carrying energy away from the hand | ✓ 1 | |
Exam tip: In a compare question, make each point about both waves (“in (a) … whereas in (b) …”) rather than describing one wave and leaving the comparison implied.
A sound wave travels in the +x direction through air. At one instant, a graph of the displacement of air molecules against distance x is a sine curve: displacement is zero at x = 0, a positive maximum at x = 0.25 m, zero at x = 0.50 m, a negative maximum at x = 0.75 m and zero at x = 1.00 m. A positive displacement means a displacement in the +x direction.
At which position is there a compression at this instant?
| Answer | Mark | Notes |
|---|---|---|
| Just before x = 0.50 m molecules are displaced forwards (+x); just after it they are displaced backwards. Both sets move towards x = 0.50 m, so there is a compression at x = 0.50 m, where the displacement itself is zero. | ✓ 1 | Answer B |
Answer: B
Exam tip: Look at the molecules on either side of a point. If both are displaced towards it, it is a compression; if both are displaced away from it, it is a rarefaction.
A hammer strikes one end of a long steel rail, producing sound that travels both through the air above the rail and through the steel. The speed of sound is 343 m s−1 in air and 5150 m s−1 in steel.
What is the ratio (wavelength of the sound in the steel) ÷ (wavelength of the sound in the air) for one frequency component of the sound?
| Answer | Mark | Notes |
|---|---|---|
| The frequency is set by the source and is the same in both media. Since λ = v/f, the ratio of wavelengths equals the ratio of speeds: 5150 ÷ 343 = 15. | ✓ 1 | Answer C |
Answer: C
Exam tip: When a wave changes medium, keep f fixed and let v and λ change in the same ratio.
In a variation of the speed-of-sound experiment, the frequency is 3.20 kHz. A student measures the distance d from the front of the loudspeaker to the microphone at successive positions where the oscilloscope traces are in phase, numbering them n = 1, 2, 3 … Her results are:
n = 1: 0.149 m; n = 2: 0.257 m; n = 3: 0.363 m; n = 4: 0.471 m; n = 5: 0.577 m; n = 6: 0.685 m.
She plots a graph of d against n and draws a straight line of best fit.
Explain why the gradient of the graph is equal to the wavelength.
The line passes through the first and last points. Use the gradient to determine the wavelength.
Calculate the speed of sound.
The line does not pass through the origin. Explain why this does not affect the value of the speed of sound obtained from the graph.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| each increase of 1 in n means the microphone has moved one more wavelength from the loudspeaker, so Δd/Δn = λ | ✓ 1 | |
| Part (b) | ||
| gradient = (0.685 − 0.149) ÷ (6 − 1) | ✓ 1 | |
| = 0.107 m (0.1072 m) | ✓ 1 | |
| Part (c) | ||
| v = fλ = 3200 × 0.1072 = 343 m s−1 (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| the intercept arises because the sound does not start at the front of the loudspeaker / the in-phase positions are not at whole numbers of wavelengths from the measured zero / it is a systematic (zero) error | ✓ 1 | |
| a constant offset adds the same amount to every d, so it changes the intercept but not the gradient, and the speed depends only on the gradient | ✓ 1 | |
Exam tip: A gradient method cancels any constant error in where distances are measured from. Say so explicitly when asked about an intercept.
A small motor-driven dipper at the edge of a still pond produces a continuous train of ripples that spread outwards. Two corks, A and B, float 0.10 m apart on a line through the source, with B further from the source. Each cork bobs up and down with a period of 0.50 s. Cork B reaches its highest point 0.125 s after cork A does. The corks are less than one wavelength apart.
Determine the phase difference between the oscillations of A and B, in radians.
Calculate the wavelength of the ripples.
Calculate the speed of the ripples.
Explain why the corks do not drift away from the source as the ripples pass them.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| fraction of a cycle = 0.125 ÷ 0.50 = 0.25 | ✓ 1 | |
| phase difference = 0.25 × 2π = π/2 rad (1.57 rad) | ✓ 1 | |
| Part (b) | ||
| 0.10 m is a quarter of a wavelength, so λ = 4 × 0.10 = 0.40 m | ✓ 1 | |
| Part (c) | ||
| v = λ/T = 0.40 ÷ 0.50 = 0.80 m s−1 (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| a progressive wave transfers energy but not the medium; the water at each place only oscillates about its equilibrium position | ✓ 1 | |
Exam tip: A time delay as a fraction of the period equals the separation as a fraction of the wavelength — both are the same fraction of a cycle.
An earthquake produces two types of wave that travel through the Earth's crust: P-waves at 6.0 km s−1 and S-waves at 3.5 km s−1. P-waves are longitudinal and S-waves are transverse. At a seismometer station the S-waves arrive 24 s after the P-waves. Assume both travel the same straight path at constant speed.
Describe how the motion of the rock as a P-wave passes differs from its motion as an S-wave passes.
Calculate the distance from the earthquake to the station.
The S-waves have a frequency of 2.5 Hz. Calculate their wavelength.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| P-wave: rock oscillates parallel to the direction of energy transfer / backwards and forwards along the path | ✓ 1 | |
| S-wave: rock oscillates perpendicular to the direction of energy transfer | ✓ 1 | |
| Part (b) | ||
| d/3.5 − d/6.0 = 24 (with d in km) | ✓ 1 | |
| d × 0.119 = 24 | ✓ 1 | |
| d = 2.0 × 102 km (202 km) | ✓ 1 | |
| Part (c) | ||
| λ = v/f = 3500 ÷ 2.5 = 1.4 × 103 m (1.4 km) | ✓ 1 | |
Exam tip: With two waves over the same distance, write the time for each as distance ÷ speed and set the difference equal to the measured delay.
The graph shows the displacement against distance for a progressive wave at one instant. X and Y are two points on the wave.
What is the phase difference between the oscillations of X and Y?
| Answer | Mark | Notes |
|---|---|---|
| Crests at 0.20 m and 1.00 m give λ = 0.80 m. X is at 0.10 m and Y at 0.40 m, so Δx = 0.30 m. Δφ = 2πΔx/λ = 2π × 0.30 ÷ 0.80 = 3π/4 = 2.4 rad. | ✓ 1 | Answer C |
Answer: C
Common error: Using the displacements of X and Y. The phase difference depends on their separation along the wave, not on how far each is from equilibrium.
In an experiment to measure the speed of sound, a signal generator drives a loudspeaker and is also connected to one input of a two-beam oscilloscope. A microphone in front of the loudspeaker is connected to the other input. The figure shows the screen. The speed of sound in the air is 340 m s−1.
The microphone is moved slowly, directly away from the loudspeaker. What is the smallest distance it must move for the two traces to be exactly in phase?
| Answer | Mark | Notes |
|---|---|---|
| Period = 5.0 div × 0.10 ms div−1 = 0.50 ms, so f = 2000 Hz and λ = 340 ÷ 2000 = 0.170 m. The microphone trace reaches each maximum 1.0 div after the signal trace, so it lags by 1/5 of a cycle. Moving away increases the lag, and the traces are next in phase when the lag is a whole cycle: a further 4/5 of a wavelength, 0.8 × 0.170 = 0.14 m (0.136 m). | ✓ 1 | Answer C |
Answer: C
Exam tip: Moving the microphone further from the loudspeaker always increases the delay of its trace. Work out how much more delay is needed to reach the next whole cycle.
A progressive wave travels along a stretched string. Figure 1(a) shows the displacement–distance graph of the string at time t = 0. Figure 1(b) shows how the displacement of the point at x = 0.30 m varies with time from t = 0.
State the wavelength of the wave.
Determine the frequency of the wave.
Calculate the speed of the wave.
Deduce the direction in which the wave is travelling along the string. Explain your reasoning.
State the phase difference between the oscillations of the points at x = 0.30 m and x = 0.45 m.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0.60 m | ✓ 1 | |
| Part (b) | ||
| T = 0.20 s from Figure 1(b), so f = 1 ÷ 0.20 = 5.0 Hz | ✓ 1 | |
| Part (c) | ||
| v = fλ = 5.0 × 0.60 = 3.0 m s−1 (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| Figure 1(b) shows that just after t = 0 the point at x = 0.30 m moves to negative displacement / downwards | ✓ 1 | |
| in Figure 1(a) the string just to the right of x = 0.30 m is below the equilibrium line, so that part of the shape must be arriving from the right: the wave travels in the −x direction (towards x = 0) | ✓ 1 | |
| Part (e) | ||
| separation 0.15 m = λ/4, so π/2 rad (1.57 rad; accept 3π/2 rad) | ✓ 1 | |
Exam tip: A displacement–distance graph alone cannot show which way a wave travels. Use the extra information (how one point moves next), then imagine the curve shifted slightly each way and see which shift matches.
In an experiment to measure the speed of sound in air, a signal generator drives a loudspeaker and is also connected to one input of a two-beam oscilloscope. A microphone placed in front of the loudspeaker is connected to the other input. The figure shows the oscilloscope screen.
Determine the frequency of the sound.
Determine the phase difference between the two traces, in radians.
The student moves the microphone slowly, directly away from the loudspeaker. The traces are first exactly in phase when the microphone has moved 8.5 cm. Calculate the speed of sound.
The microphone trace becomes smaller as the microphone moves away. Explain why this does not affect the measurement of the wavelength.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| period = 8.0 div × 0.050 ms div−1 = 0.40 ms | ✓ 1 | |
| f = 1 ÷ (0.40 × 10−3 s) = 2500 Hz (2.5 kHz) | ✓ 1 | |
| Part (b) | ||
| the microphone maximum is 3.0 div after the signal maximum: 3.0/8.0 of a cycle | ✓ 1 | |
| Δφ = 3/8 × 2π = 3π/4 = 2.4 rad (2.36 rad; allow 5π/4 = 3.9 rad, the microphone trace taken as leading by 5/8 of a cycle) | ✓ 1 | |
| Part (c) | ||
| the lag grows from 3/8 of a cycle to a whole cycle, so 8.5 cm = 5/8 λ and λ = 13.6 cm | ✓ 1 | |
| v = fλ = 2500 × 0.136 = 340 m s−1 (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| the in-phase positions depend only on the time delay (phase) of the sound reaching the microphone, which depends on distance and wavelength, not on the amplitude | ✓ 1 | |
Common error: Taking the 8.5 cm move as one whole wavelength. The microphone started 3/8 of a cycle behind, so it only had to move a further 5/8 of a wavelength.
A vibrator at x = 0 sends a transverse wave along a light rope that is joined at x = 1.2 m to a heavier rope. The graph shows the displacement of both ropes at one instant. The wave travels at 3.2 m s−1 along the light rope. Ignore any wave reflected at the join.
Explain why the frequency of the wave is the same on both ropes.
Use the graph to determine the wavelength on each rope.
Calculate the frequency of the wave and its speed on the heavy rope.
Calculate the time taken for a crest to travel from the vibrator to x = 2.2 m.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the frequency is set by the vibrator; the heavy rope is driven by the oscillations of the light rope at the join, so it oscillates at the same rate | ✓ 1 | |
| Part (b) | ||
| light rope: 3 waves in 1.2 m, λ1 = 0.40 m | ✓ 1 | |
| heavy rope: 4 waves in 1.0 m, λ2 = 0.25 m | ✓ 1 | |
| Part (c) | ||
| f = 3.2 ÷ 0.40 = 8.0 Hz | ✓ 1 | |
| v2 = 8.0 × 0.25 = 2.0 m s−1 (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| t = 1.2 ÷ 3.2 + 1.0 ÷ 2.0 = 0.375 + 0.50 | ✓ 1 | |
| = 0.88 s (0.875 s) | ✓ 1 | |
Exam tip: When a wave passes into a new medium its frequency stays the same; its speed and wavelength change in the same proportion, since v = fλ.
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