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IB Physics Higher Level · Topic C.4

C.4 Standing waves and resonance

87 free, high-quality past-paper-style questions on Standing waves and resonance for IB Physics HL, each with a full mark scheme. Covers: Forced oscillations and resonance; Wavelength of a standing wave; Formation of standing waves; and more. For when you have finished the past papers, or want to practise one topic at a time.C.4 at Standard Level →Read the C.4 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.383 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1C-1A-25Forced oscillations and resonance Paper 1AEasy1 mark

A lightly damped mass–spring system is driven by a periodic force. After some time the system oscillates with a constant amplitude.

Which statement about the steady oscillation is correct?

Show mark scheme
Marking pointMarkNotes
Step 1Damping removes energy from the oscillating system in every cycle.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The amplitude, and so the total energy ET = ½mω²x0², is constant, so energy must be supplied at the same rate as it is dissipated.—
Step 3That energy comes from the work done by the driving force: in each cycle the work done by the driver equals the energy dissipated.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: a constant amplitude means constant energy, so the energy supplied per cycle balances the energy dissipated per cycle.
  • BReturning to the same position does not mean zero work: the driving force is mostly in the direction of motion, so it does positive work each cycle.
  • COnce the transient has died away, a driven system oscillates at the driving frequency, not at its natural frequency.
  • DThe energy would fall only if the driver were removed; with a constant amplitude the energy is constant.
2C-1A-27Wavelength of a standing wave Paper 1AEasy1 mark

A string of length 1.2 m, fixed at both ends, vibrates at a frequency of 150 Hz in the mode shown.

What is the lowest frequency at which a standing wave can form on this string?

1.2 mNNNNAAA
The string vibrates in the mode shown; N = node, A = antinode.
Show mark scheme
Marking pointMarkNotes
Step 1The string vibrates in three loops: three half-wavelengths fit between the fixed ends, so this is the third harmonic.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2For a string fixed at both ends fn = nf1.—
Step 3f1 = 150/3 = 50 Hz.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • A150/4: this counts all four nodes, including the fixed ends, as the harmonic number.
  • BCorrect: three loops is the third harmonic, so f1 = 150/3 = 50 Hz.
  • C150/2: this counts only the two nodes between the ends as the harmonic number.
  • D150 × 3: the first harmonic is the lowest frequency, so the frequency must be divided by 3, not multiplied.
3C-1A-54Formation of standing waves Paper 1AEasy1 mark

Two waves of the same amplitude, frequency and wavelength λ travel in opposite directions along a string and superpose to form a standing wave.

Which statements about the standing wave are correct?

I. Its points oscillate at the same frequency as the travelling waves.

II. Adjacent nodes are a distance λ apart.

III. It transfers energy along the string from one end to the other.

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Marking pointMarkNotes
Step 1Every point is the superposition of two oscillations of frequency f, so it oscillates at f: I is correct.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2At a node the two waves always arrive in antiphase. Moving a distance x along the string changes the phase of one wave by +2πx/λ and of the other by −2πx/λ, so they are again in antiphase after λ/2: adjacent nodes are λ/2 apart and II is incorrect.—
Step 3The two travelling waves carry equal energy in opposite directions, so there is no net transfer of energy along a standing wave (no energy passes the nodes): III is incorrect.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: same frequency as the travelling waves; nodes are λ/2 apart; no net energy transfer.
  • BII is incorrect: adjacent nodes are half a wavelength apart.
  • CIII is incorrect: a standing wave stores energy between the nodes but does not transfer it along the string.
  • DII and III are both incorrect: nodes are λ/2 apart and there is no net energy transfer.
4C-1A-56Forced oscillations and resonance Paper 1AEasy1 mark

A model building on a shaking table has a natural frequency of sway of 2.5 Hz and is lightly damped. The table now moves the base sinusoidally at 2.0 Hz until a steady oscillation is reached.

Which row gives the frequency of the steady oscillation of the model and its amplitude compared with the amplitude when the table is driven at 2.5 Hz?

FrequencyAmplitude
Show mark scheme
Marking pointMarkNotes
Step 1Once the initial transient has died away, a driven system oscillates at the driving frequency, 2.0 Hz, not at its natural frequency.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The amplitude is largest (resonance) when the driving frequency is close to the natural frequency, 2.5 Hz.—
Step 3At 2.0 Hz the model is driven off resonance, so its amplitude is smaller.✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AThe model does not oscillate at its natural frequency when driven, and the amplitude depends on how close the driving frequency is to 2.5 Hz.
  • BThe amplitude comparison is right, but the steady oscillation follows the driver, at 2.0 Hz.
  • CFor light damping the amplitude is a maximum near the natural frequency; 2.0 Hz is below resonance, so the amplitude is smaller.
  • DCorrect: driven at 2.0 Hz, the model oscillates at 2.0 Hz with less than the resonant amplitude.
5IBP2H-025-00-01Pipes closed at both ends Paper 1AEasy1 mark

A narrow glass tube of length L is sealed at both ends and contains air. What is the wavelength of the sound wave in the fundamental (first harmonic) standing wave of the air column?

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Marking pointMarkNotes
A rigid sealed end prevents the air next to it from moving, so both ends are displacement nodes. The simplest pattern has one antinode between two nodes, which is half a wavelength. So L = λ/2 and λ = 2L. 4L is the result for a tube closed at one end only. L describes two half-loops, which is the second harmonic.✓ 1Answer C

Answer: C

Common error: A tube closed at both ends behaves like a string fixed at both ends, not like a tube closed at one end. Draw the nodes at the boundaries first, then fit the loops between them.

6IBP2H-025-00-08Wavelength of a standing wave Paper 1AEasy1 mark

A section of gas pipeline is sealed at both ends. The gas inside forms a standing sound wave in its fourth harmonic. How many displacement nodes are there along the section, including the ends?

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Marking pointMarkNotes
With nodes at both ends, the fourth harmonic has four half-wavelength loops along the pipe. Four loops need five nodes: one at each end and three between. Four is the number of antinodes, or the harmonic number. Nine is the total number of nodes plus antinodes.✓ 1Answer B

Answer: B

Exam tip: For a node–node system in harmonic n, there are n loops, n antinodes and n + 1 nodes. A quick sketch confirms this.

7IBP2H-025-01-01Standing waves and resonance Paper 1AEasy1 mark

A microwave oven with its turntable removed sets up a standing wave inside the cooking chamber. A flat bar of chocolate placed in the oven melts first at a row of spots. These spots mark adjacent antinodes and are 6.1 cm apart. What is the wavelength of the microwaves?

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Marking pointMarkNotes
Adjacent antinodes, like adjacent nodes, are half a wavelength apart. So λ = 2 × 6.1 cm = 12.2 cm. Option A uses node-to-antinode spacing (λ/4) the wrong way round. Option B treats the antinode spacing as a full wavelength. Option D multiplies by 4 instead of 2.✓ 1Answer C

Answer: C

Common error: Node to node, or antinode to antinode, is λ/2, not λ. Node to the nearest antinode is λ/4.

8IBP2H-025-01-08Formation of standing waves Paper 1AEasy1 mark

Which property belongs to a standing wave but not to a travelling wave?

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Marking pointMarkNotes
In a standing wave the amplitude changes with position: it is zero at nodes and largest at antinodes. In a travelling wave every point has the same amplitude (ignoring energy losses). Option A is true of both kinds of wave. Option C applies only to travelling waves; a standing wave has no net energy transfer. Option D applies to the component travelling waves in both cases.✓ 1Answer B

Answer: B

Common error: A standing wave does not transfer energy along its length. Energy is stored in, and exchanged between, the loops.

9IBP3H-C4-20r16-01Standing waves in pipes Paper 1AEasy1 mark

In a hospital acoustics test, an air-filled duct of length 0.85 m is sealed at both ends. The speed of sound in the air is 340 m s−1. What is the lowest frequency at which a standing wave can form in the duct?

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Marking pointMarkNotes
Both ends are displacement nodes, so the fundamental has λ = 2L = 1.70 m and f = v/λ = 340/1.70 = 200 Hz (B). 100 Hz comes from treating the duct as closed at one end only (λ = 4L). 400 Hz uses λ = L. 800 Hz uses λ = L/2.✓ 1Answer B

Answer: B

Common error: A pipe sealed at both ends has a node at each end, so its fundamental has half a wavelength along the pipe.

10IBP3H-C4-20r16-02Strings with free ends Paper 1AEasy1 mark

A beekeeper stretches a light cord of length 1.2 m between two vertical poles. A small ring at each end of the cord slides freely on its pole, so each end of the cord is free to move. The cord vibrates in its fundamental mode. What is the wavelength of the standing wave?

ring on smooth rodring on smooth rod1.2 m
Diagram not to scale
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Marking pointMarkNotes
Free ends are antinodes. The fundamental has an antinode at each end with one node between, so L = λ/2 and λ = 2.4 m (C). 0.60 m treats the length as λ/4. 1.2 m takes the length as one wavelength. 4.8 m uses λ = 4L, which is for one fixed and one free end.✓ 1Answer C

Answer: C

Exam tip: Count quarter-wavelength steps: antinode to adjacent antinode is half a wavelength.

11IBP3H-C4-20r16-03Formation of standing waves Paper 1AEasy1 mark

Fine powder in a horizontal glass tube settles into small heaps at points where the air is not moving. A loudspeaker sends sound of frequency 2000 Hz along the tube and the heaps form with a separation of 0.085 m between neighbouring heaps. What is the speed of sound in the tube?

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Marking pointMarkNotes
Heaps form at displacement nodes, which are half a wavelength apart, so λ = 0.17 m and v = fλ = 2000 × 0.17 = 340 m s−1 (B). 170 m s−1 takes the heap separation as λ. 680 m s−1 doubles the wavelength again. 85 m s−1 halves the separation.✓ 1Answer B

Answer: B

Common error: Adjacent nodes are λ/2 apart, not λ.

12IBP3H-C4-20r16-05Damping and resonance Paper 1AEasy1 mark

A wing of a Formula 1 car vibrates strongly when the engine runs at a particular speed. A tuned damper is fitted to the wing and the engine speed is swept through the same range. Compared with the undamped wing, the amplitude-frequency response of the damped wing has a peak that is

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Marking pointMarkNotes
Damping dissipates energy from the vibration, so the resonance peak is lower. The energy loss also means the system responds over a wider range of frequencies, so the peak is broader (B). A is the effect of reducing damping. C and D combine a correct and an incorrect feature.✓ 1Answer B

Answer: B

Common error: Damping lowers the peak and broadens it; it does not make the resonance sharper.

13IBP3H-C4-20r26-01Strings with free ends Paper 1AEasy1 mark

In a sports-science laboratory, a taut elastic cord of length 0.40 m is fixed at both ends. It vibrates in its fundamental mode at a frequency of 300 Hz. What is the speed of transverse waves on the cord?

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Marking pointMarkNotes
The fundamental wavelength is twice the length, λ = 0.80 m, so v = fλ = 300 × 0.80 = 240 m s-1. 120 m s-1 takes λ equal to the length of the cord, forgetting that only half a wavelength fits. 480 m s-1 uses λ = 4L, which applies to a pipe closed at one end. 750 m s-1 comes from dividing the frequency by the length.✓ 1Answer B

Answer: B

Common error: For a string fixed at both ends the fundamental fits half a wavelength, so λ = 2L. Do not use the closed-pipe relation λ = 4L.

14IBP3H-C4-20r26-02Standing waves in pipes Paper 1AEasy1 mark

A railway signal whistle is a pipe that is closed at one end and open at the other. Which row gives the ratio of the frequencies of its three lowest resonances?

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Marking pointMarkNotes
A closed pipe has a node at the closed end and an antinode at the open end, so the length is an odd number of quarter-wavelengths. The frequencies are f, 3f and 5f. The ratio 1 : 2 : 3 holds for a pipe open at both ends or a string fixed at both ends. 1 : 2 : 4 wrongly takes each resonance as double the previous one. 1 : 4 : 9 confuses the pattern with squares of the harmonic number.✓ 1Answer C

Answer: C

Exam tip: Count quarter-wavelengths in a closed pipe: L = λ/4, 3λ/4, 5λ/4. Only odd harmonics are present.

15IBP3H-C4-20r26-03Standing waves in pipes Paper 1AEasy1 mark

A rail-tunnel ventilation shaft acts as an air column open at both ends. It has length 4.25 m. The speed of sound in the air is 340 m s-1. What is the frequency of the third harmonic?

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Marking pointMarkNotes
For an open pipe f1 = v/2L = 340/8.50 = 40 Hz, and the third harmonic is 3 × 40 = 120 Hz. 40 Hz is the fundamental only. 60 Hz uses the closed-pipe formula 3v/4L. 80 Hz is the second harmonic (first overtone), reached by miscounting the harmonic number.✓ 1Answer D

Answer: D

Common error: The nth harmonic of an open pipe is n times the fundamental. The first overtone is the second harmonic, so the third harmonic is the second overtone.

16C-1A-08Standing waves in pipes Paper 1AMedium1 mark

A pipe closed at one end has a first-harmonic frequency of 100 Hz. A pipe of the same length open at both ends is sounded.

What is the frequency of its second harmonic?

Show mark scheme
Marking pointMarkNotes
Step 1Closed pipe: the first harmonic has a quarter wavelength in the pipe, f1 = v/4L = 100 Hz.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Open pipe of the same length: half a wavelength fits, f1 = v/2L = 200 Hz.—
Step 3The second harmonic of the open pipe is 2 × 200 = 400 Hz.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis is the first harmonic of the open pipe, not its second harmonic.
  • BThis is the next harmonic of the closed pipe (3f1).
  • CCorrect: open-pipe first harmonic 200 Hz, second harmonic 400 Hz.
  • DThis would be the third harmonic of the open pipe.
17C-1A-16Standing waves on strings Paper 1AMedium1 mark

A string of length L is fixed at both ends. Two consecutive resonant frequencies of the string are fA and fB, where fB > fA. The speed of transverse waves on the string is v.

What is L?

Show mark scheme
Marking pointMarkNotes
Step 1For a string fixed at both ends fn = nv/2L, so consecutive harmonics differ by v/2L, the first-harmonic frequency.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2fB − fA = v/(2L).—
Step 3L = v/(2(fB − fA)).✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: the spacing of consecutive resonances equals the first-harmonic frequency v/2L.
  • BThis uses v/4L (the lowest frequency of a string with one free end) as the spacing; the string here has two fixed ends, and its resonances are v/2L apart.
  • CThis matches the frequency difference to a wavelength equal to L, instead of 2L.
  • DThis assumes that fA is the first harmonic. Nothing in the data says so — only the difference between consecutive resonances gives v/2L.
18C-1A-19Standing waves in pipes Paper 1AMedium1 mark

The diagram shows the displacement nodes and antinodes of a standing wave in the air in a pipe that is open at both ends. The frequency of this standing wave is 850 Hz.

What is the first-harmonic frequency of the pipe?

ANANANAopen end at each side of the pipe
Displacement nodes (N) and antinodes (A) of the standing wave in a pipe open at both ends.
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Marking pointMarkNotes
Step 1Both open ends are antinodes. The pattern A–N–A–N–A–N–A has three nodes and fits three half-wavelengths into the pipe: L = 3λ/2.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2For a pipe open at both ends λn = 2L/n, so this is the third harmonic and its frequency is 3f1.—
Step 3f1 = 850/3 = 283 Hz.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • A850/4 = 213 Hz: this counts the four antinodes as the harmonic number.
  • BCorrect: three half-wavelengths in the pipe, so the third harmonic; f1 = 850/3 = 283 Hz.
  • C850/2 = 425 Hz: this counts only the two complete loops in the middle and ignores the half-loops at the open ends.
  • D850 × 3: the first harmonic is the lowest frequency, so the frequency of this mode must be divided by 3, not multiplied.
19C-1A-21Damping and resonance Paper 1AMedium1 mark

A mass on a spring is driven by a periodic force of variable frequency. The amplitude of the resulting oscillation is measured for a range of driving frequencies, first with light damping and then with the damping increased.

Which row gives the effect of increasing the damping on the resonance curve?

Maximum amplitudeFrequency at which the maximum occurs
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Marking pointMarkNotes
Step 1Damping removes energy from the oscillator each cycle, so at every driving frequency the steady amplitude is smaller: the peak is lower and broader.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The resonant peak also moves to a slightly lower frequency than the undamped natural frequency.—
Step 3So heavier damping gives a smaller maximum amplitude at a slightly lower frequency.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: heavier damping lowers and broadens the peak and shifts it to a slightly lower frequency.
  • BThe peak shifts to lower, not higher, frequency as the damping increases.
  • CThe maximum amplitude must fall — damping dissipates the energy fed in by the driver.
  • DDamping cannot increase the amplitude; it always removes energy from the oscillation.
20C-1A-30Phase along a standing wave Paper 1AMedium1 mark

A string of length 1.20 m fixed at both ends vibrates in its third harmonic, so that nodes are found at 0, 0.40 m, 0.80 m and 1.20 m from one end. P, Q and R are points on the string at 0.20 m, 0.30 m and 0.60 m from that end.

Which row gives the phase difference between the oscillations of P and Q, and between the oscillations of P and R?

PQR1.20 m
Diagram not to scale
P and QP and R
Show mark scheme
Marking pointMarkNotes
Step 1In a standing wave all points between two adjacent nodes (in the same loop) move up and down together — they are in phase, though with different amplitudes.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2P (0.20 m) and Q (0.30 m) both lie in the first loop (0 to 0.40 m), so their phase difference is 0.—
Step 3R (0.60 m) lies in the second loop (0.40 to 0.80 m); points in adjacent loops move in opposite directions at every instant, so P and R are in antiphase: phase difference π.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: same loop → in phase; adjacent loops → antiphase.
  • BThis is the reverse — P and Q are in the same loop, P and R are separated by a node.
  • CA phase difference of π/2 never occurs on a standing wave: points are either in phase or in antiphase (unlike a travelling wave).
  • DP and R are on opposite sides of the node at 0.40 m, so the difference is π, not π/2.
21C-1A-51Strings with free ends Paper 1AMedium1 mark

A light elastic cord of length 0.90 m is clamped at one end. The other end is tied to a light ring that can slide without friction along a smooth vertical rod, so that this end of the cord is a free end. The speed of transverse waves on the cord is 18 m s−1.

What is the second-lowest frequency at which a standing wave can form on the cord?

clampedring on smooth rod0.90 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Step 1The clamped end is a node and the free end is an antinode, so only an odd number of quarter-wavelengths fits on the cord: L = λ/4, 3λ/4, ….—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Lowest: λ = 4L = 3.6 m, f = 18/3.6 = 5.0 Hz; only odd multiples of this frequency occur.—
Step 3Second-lowest: λ = 4L/3 = 1.2 m, f = 3 × 5.0 = 15 Hz.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis doubles the lowest frequency (5.0 Hz); with a node at one end and an antinode at the other, only odd harmonics occur.
  • BCorrect: λ = 4L/3 = 1.2 m and f = 18/1.2 = 15 Hz.
  • CThis is the second harmonic of a cord with both ends fixed (λ = L): the free end has been treated as a node.
  • DThis applies the odd-harmonics rule to the first harmonic of a cord with two fixed ends (3 × 10 Hz) instead of to v/4L.
22C-1A-52Pipes closed at both ends Paper 1AMedium1 mark

A pipe open at both ends has a first-harmonic frequency f. One end of the pipe is closed with a cap, and then the other end is also closed with a cap. The speed of sound in the air in the pipe does not change.

What is the first-harmonic frequency of the pipe with both ends closed?

Show mark scheme
Marking pointMarkNotes
Step 1Open at both ends: an antinode at each end, L = λ/2 and f = v/2L.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Closed at one end: node to antinode, L = λ/4, so the first harmonic falls to v/4L = f/2.—
Step 3Closed at both ends: a node at each end, L = λ/2 again, so the first harmonic returns to v/2L = f.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis halves the frequency once for each cap, as if each closed end added a quarter-wavelength.
  • BThis is the first harmonic with only one end closed; the second cap makes both ends nodes.
  • CCorrect: node to node, like antinode to antinode, is half a wavelength.
  • DThis fits a whole wavelength between the two closed ends (λ = L); node to node is only half a wavelength.
23C-1A-53Phase along a standing wave Paper 1AMedium1 mark

A stretched wire vibrates in its second harmonic. A stroboscope photograph is taken at an instant when every point of the wire is at its equilibrium position, so the wire is straight.

Which statement describes the motion of the points of the wire at this instant?

Show mark scheme
Marking pointMarkNotes
Step 1Every point of a standing wave performs SHM with the same frequency but an amplitude that depends on position (zero at a node, maximum at an antinode).—All 3 steps must be completed — there is no mark for a part-answer.
Step 2At equilibrium each point has its maximum speed vmax = ωx0, so the speed is largest at the antinodes and zero at the nodes; the energy is all kinetic.—
Step 3Points in the same loop are in phase; points in adjacent loops (on either side of the central node) are in antiphase, so the two loops move in opposite directions.✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AAt the straight position the displacement is zero, but this is where each point moves fastest — a standing wave does not stop.
  • BThis treats the wire as having the same amplitude everywhere, like a travelling wave; the amplitude (and so the maximum speed) varies from node to antinode.
  • CThe amplitude pattern is right, but points on opposite sides of a node are in antiphase.
  • DCorrect: vmax ∝ amplitude, and adjacent loops are π out of phase.
24C-1A-55Strings with free ends Paper 1AMedium1 mark

Two elastic cords P and Q have the same length L. Both ends of P are tied to light rings that slide without friction on smooth vertical rods, so that both ends of P are free. One end of Q is clamped and its other end is free. The speed of transverse waves on Q is twice the speed of transverse waves on P.

What is (first-harmonic frequency of P)/(first-harmonic frequency of Q)?

Show mark scheme
Marking pointMarkNotes
Step 1Free ends are antinodes. The simplest pattern on P is A–N–A: half a wavelength on the cord, so λP = 2L and fP = v/(2L), where v is the wave speed on P.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2On Q the clamped end is a node and the free end is an antinode. The simplest pattern, N–A, is a quarter of a wavelength: λQ = 4L, and the wave speed is 2v, so fQ = 2v/(4L) = v/(2L).—
Step 3The ratio is fP/fQ = 1: the longer wavelength on Q is exactly compensated by its greater wave speed.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis swaps the two boundary conditions, taking λ = 4L for P (two free ends) and λ = 2L for Q: (v/4L)/(2v/2L) = 1/4.
  • BThis treats P as if one of its ends were fixed (λ = 4L): (v/4L)/(2v/4L) = 1/2. Two free ends are both antinodes, so half a wavelength fits.
  • CCorrect: fP = v/(2L) and fQ = 2v/(4L) = v/(2L).
  • DThis uses the right wavelengths (2L and 4L) but ignores the greater wave speed on Q: (v/2L)/(v/4L) = 2.
25C-1A-68Formation of standing waves Paper 1AMedium1 mark

Two loudspeakers face each other along a straight line and emit sound of frequency 680 Hz in phase. The speed of sound is 340 m s−1. A microphone moves at a constant speed of 0.85 m s−1 along the line between the loudspeakers, and the loudness that it detects rises and falls regularly.

What is the time between successive maxima of loudness?

Show mark scheme
Marking pointMarkNotes
Step 1The two waves travel in opposite directions and superpose to form a standing wave, with λ = 340/680 = 0.50 m.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The pattern of loud and quiet positions repeats every half wavelength (the distance between adjacent antinodes, or adjacent nodes): λ/2 = 0.25 m.—
Step 3Time = 0.25/0.85 = 0.29 s.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis is the period of the sound, 1/680 s; the loudness varies because the microphone moves through the standing-wave pattern, not at the frequency of the sound.
  • BThis takes the distance between a node and the adjacent antinode (a loud position and the next quiet one), λ/4 = 0.125 m: 0.125/0.85 = 0.15 s.
  • CCorrect: adjacent antinodes are λ/2 = 0.25 m apart, so the time is 0.25/0.85 = 0.29 s.
  • DThis takes adjacent maxima to be one wavelength apart, as for a travelling wave: 0.50/0.85 = 0.59 s.
26IBP2H-025-00-02Standing waves in pipes Paper 1AMedium1 mark

A sealed air duct in an aircraft cabin is 0.85 m long and is closed at both ends. The speed of sound in the air is 340 m s−1. Which frequency is a possible standing-wave frequency for the air in the duct?

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Marking pointMarkNotes
Both ends are displacement nodes, so λn = 2L/n. The fundamental is f1 = v/2L = 340/1.7 = 200 Hz. Every integer multiple of 200 Hz is allowed: 200, 400, 600 Hz and so on. 400 Hz is the second harmonic. The values 300, 500 and 700 Hz are odd multiples of 100 Hz. That pattern would only apply to a duct closed at one end, where f1 = v/4L.✓ 1Answer B

Answer: B

Exam tip: Find the fundamental first, then test each option by dividing by it. Only whole-number results are allowed for a pipe closed at both ends.

27IBP2H-025-00-03Strings with free ends Paper 1AMedium1 mark

In a model of a tethered buoy, a taut cable of length L is fixed at its lower end. Its upper end is attached to a light ring that slides without friction on a vertical pole. The upper end is therefore free to move transversely. Which wavelength is a possible standing wave on the cable?

clampedring on smooth rodL
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The fixed end is a node and the free (ring) end is an antinode. The length must therefore be an odd number of quarter-wavelengths: L = nλ/4 with n = 1, 3, 5, … This gives λ = 4L, 4L/3, 4L/5, … Only 4L/3 (with n = 3) appears. The values 2L, L and 2L/3 come from λ = 2L/n, which applies only when both ends are fixed.✓ 1Answer C

Answer: C

Common error: A free end of a string is an antinode, not a node. Using the fixed–fixed formula λ = 2L/n for a node–antinode system gives wavelengths that cannot occur.

28IBP2H-025-00-04Damping and resonance Paper 1AMedium1 mark

A vibration-testing rig for satellite components drives a mass–spring system over a range of frequencies. The amplitude of the mass is recorded at each frequency. The damping in the system is then increased and the test is repeated. Which describes the change to the amplitude–frequency curve?

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Marking pointMarkNotes
Greater damping removes more energy per cycle, so the steady amplitude at resonance is smaller. The response also falls off less sharply either side of the natural frequency, so the peak becomes broader (flatter). The peak shifts very slightly to a lower frequency, not a much higher one. Far from resonance the amplitude is barely affected, so the change is not the same factor at every frequency.✓ 1Answer A

Answer: A

Exam tip: When describing a damped resonance curve, give three features: lower peak, broader peak, and peak shifted slightly to lower frequency.

29IBP2H-025-00-06Forced oscillations and resonance Paper 1AMedium1 mark

As a washing machine's drum speeds up, the machine's frame vibrates violently at one particular spin rate. At higher spin rates the vibration becomes much smaller. What is the best explanation for the violent vibration?

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Marking pointMarkNotes
The off-balance drum applies a periodic driving force at its rotation frequency. When this driving frequency equals the natural frequency of the frame, resonance occurs. Energy is then transferred most efficiently and the amplitude is a maximum. The large vibration happens at an intermediate spin rate, not the maximum one. The damping and the natural frequency of the frame are properties of the frame and do not change with spin rate.✓ 1Answer D

Answer: D

Exam tip: Always name both frequencies when explaining resonance: the driving frequency of the external force and the natural frequency of the system being driven.

30IBP2H-025-01-02Phase along a standing wave Paper 1AMedium1 mark

A string of length L is fixed at both ends and vibrates in its third harmonic. Point X is 0.10L from one end, point Y is 0.25L from the same end, and point Z is 0.40L from the same end. What are the phase differences between X and Y, and between X and Z?

XYZL
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
In the third harmonic the nodes are at 0, L/3, 2L/3 and L. X (0.10L) and Y (0.25L) are both in the first loop, so they oscillate in phase (phase difference 0). Z (0.40L) is in the second loop, between L/3 and 2L/3. Adjacent loops move in antiphase, so the phase difference between X and Z is π. Option C works out phase from distance as if the wave were travelling. Option B assumes every point on a standing wave is in phase. Option D reverses the loop rule.✓ 1Answer A

Answer: A

Exam tip: On a standing wave, find the nodes first. Points in the same loop are in phase; points in adjacent loops are in antiphase. Distance within a loop does not change the phase.

31IBP2H-025-01-03Standing waves in pipes Paper 1AMedium1 mark

The human ear canal can be modelled as a tube of length 2.5 cm, open at the outer end and closed by the eardrum. The speed of sound in the canal is 340 m s−1. The lowest resonant frequency is 3400 Hz. What is the next higher resonant frequency?

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Marking pointMarkNotes
A tube closed at one end has a displacement node at the closed end and an antinode at the open end. Only odd harmonics can form: f1 = v/4L = 3400 Hz, then 3f1, 5f1 and so on. The next resonance is therefore 3 × 3400 = 10 200 Hz. Option A (2f1) treats the canal as open at both ends. Option C (4f1) assumes only even harmonics exist. Option D uses half-integer multiples.✓ 1Answer B

Answer: B

Common error: A pipe closed at one end has no second harmonic. The frequency after f₁ is 3f₁.

32IBP2H-025-01-04Damping and resonance Paper 1AMedium1 mark

A car's suspension is driven by bumps in the road and shows a resonance peak. Its shock absorbers wear out, so the damping is reduced. Compared with new shock absorbers, how does the graph of oscillation amplitude against driving frequency change?

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Marking pointMarkNotes
Less damping means less energy is lost per cycle. The amplitude at resonance therefore grows larger, and the response falls away more quickly either side of the natural frequency, giving a sharper, narrower peak. The peak also moves very slightly closer to the natural frequency. Option B wrongly links low damping with a broad peak. Option C describes increased damping. Option D gets the size of the frequency shift wrong and ignores the change in amplitude.✓ 1Answer A

Answer: A

Exam tip: Light damping gives a tall, narrow peak. Heavy damping gives a low, broad peak whose maximum sits at a slightly lower frequency.

33IBP2H-025-01-06Standing waves on strings Paper 1AMedium1 mark

A 490 Hz tuning fork is held above a vertical tube whose lower end is in water. The water level is lowered. Resonance is first heard when the air column is 0.17 m long, and next heard when it is 0.52 m long. What is the speed of sound in the air?

tuning fork0.17 mfirst resonancetuning fork0.52 msecond resonance
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Successive resonance lengths differ by λ/2. This difference does not depend on the unknown end correction. λ = 2 × (0.52 − 0.17) = 0.70 m, so v = fλ = 490 × 0.70 = 343 m s−1. Option B uses λ = 4 × 0.17 m and ignores the end correction. Option A takes the difference in lengths as a full wavelength. Option D uses λ = 4 × 0.52 m.✓ 1Answer C

Answer: C

Exam tip: Use the difference between two resonance lengths. It removes the end correction and gives λ/2 directly.

34IBP3H-C4-20r16-04Pipes closed at both ends Paper 1AMedium1 mark

A tube closed at one end is held vertically and the water level is lowered. Sound from a loudspeaker of fixed frequency produces resonance when the air column is 0.160 m long and again when it is 0.500 m long. The speed of sound is 340 m s−1. What is the frequency of the sound?

tuning fork0.160 mfirst resonancetuning fork0.500 msecond resonance
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Successive resonances differ by λ/2, so λ = 2(0.500 − 0.160) = 0.680 m and f = 340/0.680 = 500 Hz (A). 531 Hz comes from λ = 4(0.160), ignoring the end effect. 510 Hz comes from λ = 4(0.500)/3. 1000 Hz uses λ = 0.340 m, treating the difference as a full wavelength.✓ 1Answer A

Answer: A

Exam tip: Using the difference between two resonance lengths removes any unknown end correction.

35IBP3H-C4-20r16-08Wavelength of a standing wave Paper 1AMedium1 mark

The beekeeper's cord (free ends, fundamental frequency 12 Hz) is now driven so that it vibrates in its fourth harmonic. Counting the end points, how many nodes and antinodes does the standing wave have, and what is its frequency?

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Marking pointMarkNotes
With an antinode at each end, the nth harmonic has n nodes and n + 1 antinodes. For n = 4 there are 4 nodes and 5 antinodes, and f = 4 × 12 = 48 Hz (A). B swaps nodes and antinodes (the pattern for fixed ends). C uses 3 × 12. D counts the pattern of the third harmonic.✓ 1Answer A

Answer: A

Exam tip: Compare the ends: free ends give antinodes at both ends, so the antinodes outnumber the nodes by one.

36IBP3H-C4-20r26-04Phase along a standing wave Paper 1AMedium1 mark

A standing wave is set up on a stretched string. Which statement correctly compares the motion of points on the string between two adjacent nodes?

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Marking pointMarkNotes
Between two adjacent nodes every point moves in phase with every other point, reaching maximum displacement at the same instant, but the amplitude varies from zero at the nodes to a maximum at the antinode. The statements with equal amplitudes are wrong because the amplitude depends on position. The statement with different phases between the nodes is wrong because the phase changes only when passing through a node. The phase difference of π applies to points in adjacent loops, not to points within one loop.✓ 1Answer A

Answer: A

Common error: In a standing wave the amplitude changes with position but the phase is constant within one loop. Adjacent loops are in antiphase.

37IBP3H-C4-20r26-05Strings with free ends Paper 1AMedium1 mark

A taut string of length L has one end fixed. The other end is attached to a light ring that slides without friction on a vertical rod, so that the end is free to move. The lowest resonant frequency of the string is f. What is the next-lowest resonant frequency of the string?

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Marking pointMarkNotes
The fixed end is a node and the free end is an antinode, so L is an odd number of quarter-wavelengths and the resonant frequencies are f, 3f, 5f. The next frequency is 3f. 2f and 4f are the harmonics of a string fixed at both ends, which are absent here. 5f is the second resonance after the fundamental, not the next one.✓ 1Answer B

Answer: B

Exam tip: A free end behaves like the open end of a pipe (antinode). With one fixed and one free end only odd multiples of the fundamental occur.

38IBP3H-C4-20r26-06Formation of standing waves Paper 1AMedium1 mark

A tuning fork of frequency 680 Hz is held over a vertical tube, closed at the lower end by water, whose level is lowered. Resonance is first heard when the air column is 0.115 m long and again when it is 0.365 m long. What is the speed of sound in the air?

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Marking pointMarkNotes
Successive resonances in a closed tube are half a wavelength apart: λ/2 = 0.365 − 0.115 = 0.250 m, so λ = 0.500 m and v = 680 × 0.500 = 340 m s-1. 170 m s-1 takes the spacing 0.250 m as the whole wavelength. 313 m s-1 assumes the first length is exactly λ/4 and ignores the end correction. 496 m s-1 takes the second length of 0.365 m as half a wavelength, ignoring the end correction.✓ 1Answer C

Answer: C

Exam tip: Use the difference between successive resonance lengths. This eliminates the unknown end correction of the tube.

39IBP3H-C4-20r26-07Damping and resonance Paper 1AMedium1 mark

A railway footbridge behaves as a driven oscillator with light damping, driven by regular footsteps of varying frequency. Extra damping is then added to the bridge. Which change to the graph of amplitude against driving frequency is correct?

Show mark scheme
Marking pointMarkNotes
More damping dissipates more energy per cycle, so the amplitude at resonance falls and the response spreads over a wider range of frequencies. A lower peak with a narrower curve is not possible, because damping flattens the response rather than sharpening it. A higher peak is the effect of reducing damping. Heavier damping does move the peak slightly to a lower frequency, but the height cannot stay the same.✓ 1Answer A

Answer: A

Common error: Damping reduces the resonance amplitude and flattens the curve. Light damping gives a tall, narrow peak.

40IBP2H-025-00-05Harmonics of a stretched string Paper 1AHard1 mark

A tennis racket string is stretched between two fixed points. A stringing machine increases the tension in the string by a factor of 4. The vibrating length of the string is also halved. The mass per unit length is unchanged. By what factor does the fundamental frequency change?

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Marking pointMarkNotes
Wave speed on a string is proportional to √(tension), so quadrupling the tension doubles v. The fundamental wavelength is 2L, so halving L halves λ. Since f = v/λ, the frequency changes by 2 ÷ ½ = ×4. The value ×8 comes from taking v ∝ T rather than √T. The value ×2 ignores the change in length. The value ×1 comes from wrongly thinking the two effects cancel.✓ 1Answer C

Answer: C

Common error: Wave speed on a string depends on the square root of tension. Treat the speed change and the wavelength change separately, then combine them using f = v/λ.

41IBP2H-025-00-07Phase along a standing wave Paper 1AHard1 mark

A string of length L is fixed at both ends and vibrates in its third harmonic. The maximum amplitude on the string is A. Point P is a distance L/12 from one end. Point Q is a distance L/2 from the same end. Which statement about the oscillations of P and Q is correct?

PQL
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
In the third harmonic the nodes are at 0, L/3, 2L/3 and L. P lies in the first loop, close to a node. Its amplitude is A sin(π/4) ≈ 0.71A. Q is at the antinode of the second loop, so its amplitude is A. In a standing wave, adjacent loops oscillate in antiphase, so P and Q are π out of phase. The value 5π/4 comes from wrongly applying the travelling-wave rule (phase difference = 2π × separation/λ) to a standing wave.✓ 1Answer B

Answer: B

Common error: In a standing wave, all points within one loop move in phase. Points in adjacent loops move in antiphase. Phase does not change gradually with distance as it does in a travelling wave.

42IBP2H-025-01-05Standing waves on strings Paper 1AHard1 mark

A violin string has a fundamental frequency of 440 Hz. A player presses the string so that its vibrating length becomes 2/3 of the original length. The tuning peg also slips, increasing the tension by 44%. The speed of waves on the string is proportional to the square root of the tension. What is the new fundamental frequency?

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Marking pointMarkNotes
f1 = v/2L. The tension rises by a factor of 1.44, so v rises by a factor of √1.44 = 1.2. The length falls by a factor of 2/3, so f rises by a factor of 1.5. New f = 440 × 1.2 × 1.5 = 792 Hz. Option D forgets the square root (440 × 1.44 × 1.5). Option A treats frequency as proportional to length. Option B includes only the change in tension.✓ 1Answer C

Answer: C

Common error: The frequency depends on the square root of the tension but on the reciprocal of the length. Apply each factor separately before multiplying.

43IBP2H-025-01-07Wavelength of a standing wave Paper 1AHard1 mark

An air duct in a building resonates at 450 Hz and at 750 Hz. There are no resonant frequencies between these two values. Which row correctly describes the duct and its fundamental frequency?

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Marking pointMarkNotes
The gap between successive resonances is 300 Hz. For an open–open pipe this gap would equal f1 = 300 Hz. But 450 Hz is not a whole-number multiple of 300 Hz, so the duct cannot be open at both ends. For a pipe closed at one end, successive resonances differ by 2f1, so f1 = 150 Hz. Then 450 Hz = 3f1 and 750 Hz = 5f1, which are consecutive odd harmonics. Option A assumes the gap equals f1 without checking. Option C chooses the right pipe but the wrong relationship. Option D fails because an open pipe with f1 = 150 Hz would also resonate at 600 Hz.✓ 1Answer B

Answer: B

Exam tip: Divide each resonant frequency by the gap between them. Whole-number ratios mean an open pipe. Ratios of odd halves (1.5, 2.5) mean a pipe closed at one end.

44IBP3H-C4-20r16-06Strings with free ends Paper 1AHard1 mark

A standing wave on a string has an amplitude of 6.0 mm at an antinode. What is the amplitude of vibration of a point on the string that is one twelfth of a wavelength from the nearest node?

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The amplitude at distance x from a node is A0 sin(2πx/λ). With x = λ/12 the phase is π/6, so the amplitude is 6.0 × sin 30° = 3.0 mm (B). 2.0 mm assumes amplitude increases linearly with distance from the node. 5.2 mm uses cos 30°. 1.5 mm divides by four.✓ 1Answer B

Answer: B

Exam tip: Amplitude varies sinusoidally with position along a standing wave, not linearly.

45IBP3H-C4-20r16-07Standing waves in pipes Paper 1AHard1 mark

A pipe of length L is closed at one end and open at the other, and its fundamental frequency is f. A second pipe of length 2L is closed at both ends and contains the same gas. Which statement describes the resonant frequencies of the second pipe?

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Marking pointMarkNotes
For the first pipe f = v/4L. The second pipe has nodes at both ends, so its resonances are fn = nv/(2×2L) = nv/4L = nf for n = 1, 2, 3, … (C). A is the series for a pipe closed at one end. B omits the fundamental. D would need a pipe four times as long.✓ 1Answer C

Answer: C

Common error: A pipe closed at both ends supports every harmonic; only a pipe with one open and one closed end is limited to odd harmonics.

46IBP3H-C4-20r26-08Wavelength of a standing wave Paper 1AHard1 mark

A string of length 1.20 m and mass per unit length 2.5 × 10-3 kg m-1 is fixed at both ends. It vibrates in its third harmonic at a frequency of 150 Hz. What is the tension in the string?

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Marking pointMarkNotes
For the third harmonic λ = 2L/3 = 0.80 m, so v = fλ = 150 × 0.80 = 120 m s-1. Then T = μv² = 2.5 × 10-3 × 120² = 36 N. 0.30 N results from using T = μv without squaring the speed. 9.0 N uses λ = L/3 = 0.40 m. About 20 N (20.25 N) comes from treating the third harmonic as having four loops (λ = L/2 = 0.60 m).✓ 1Answer D

Answer: D

Common error: Count the loops: the nth harmonic has n loops, so L = nλ/2. Remember that tension depends on v².

47C-1B-22Strings with free ends Paper 1BEasy7 marks

An elastic cord of length 1.20 m lies along a horizontal bench. The end at x = 0 is attached to a vibration generator of very small amplitude, so that it behaves almost as a fixed end. The end at x = 1.20 m is tied to a light ring that slides freely on a smooth vertical rod. The generator frequency is 15.0 Hz and a standing wave forms.

The amplitude A of the cord is measured at different distances x from the generator with an uncertainty of ±1 mm. The graph shows the data with a smooth curve.

x / m0.000.150.300.450.600.750.901.051.20
A / mm114232518592125
0.000.150.300.450.600.750.901.051.20x / m051015202530A / mm
Amplitude A of the cord against distance x from the generator (drawn to scale).
(a)
(i)

Use the graph to determine the wavelength of the waves on the cord.

(2)
(b)
(i)

Deduce the harmonic number of this standing wave and calculate the speed of the waves.

(2)
(c)
(i)

Determine the frequency of the next harmonic above 15.0 Hz at which a standing wave forms on this cord.

(2)
(d)
(i)

At 15.0 Hz, state the phase difference between the oscillations of the points at x = 0.30 m and x = 1.05 m.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Nodes at x = 0 and x ≈ 0.80 m are half a wavelength apart✓ 1Or node at 0.80 m to antinode at 1.20 m = λ/4.
λ = 1.60 m✓ 1Accept 1.56–1.64 m.
Part (b)(i)
1.20 m = 3λ/4 (node at the fixed end, antinode at the free end): third harmonic✓ 1Allow ECF from (a).
v = fλ = 15.0 × 1.60 = 24.0 m s−1✓ 1Accept 23.4–24.6 m s−1. Allow ECF from (a).
Part (c)(i)
Only odd harmonics occur, so the next is the fifth: 1.20 = 5λ/4, λ = 0.96 m✓ 1Award [0] for 20.0 Hz (fourth harmonic, not possible here).
f = 24.0/0.96 = 25.0 Hz✓ 1Or 15.0 × 5/3. Allow ECF from (b).
Part (d)(i)
π rad: the points are on opposite sides of the node at 0.80 m✓ 1Allow ECF from the node position in (a).

Answers: (a)(i) 1.60 m  ·  (b)(i) third; 24.0 m s−1  ·  (c)(i) 25.0 Hz  ·  (d)(i) π rad (the remaining parts are explanations — see the table above)

48C-1B-08Harmonics of a stretched string Paper 1BMedium7 marks

A string passes over a pulley at one end of a bench and carries a hanging load of mass M. A vibration generator near the other end drives the string, and the vibrating length between the generator and the pulley is 0.800 m. For each load the student finds the frequency f1 of the first harmonic.

The speed of waves on a string is v = √(F/μ), where F is the tension and μ the mass per unit length. The student plots f1² against M.

M / kg0.1000.2000.3000.4000.500
f1 / Hz17.724.730.535.139.2
f1² / Hz²31361093012321537
0.00.10.20.30.40.50.6M / kg0.00.30.60.91.21.51.8f₁² / 10³ Hz²
f1² against M with the line of best fit (drawn to scale).
(a)
(i)

Show that f1² = (g/4L²μ)M, where L is the vibrating length.

(2)
(b)
(i)

Determine the gradient of the graph, including its unit.

(2)
(c)
(i)

Determine μ.

(1)
(d)
(i)

A 2.000 m sample of the string has a mass of 2.40 g. Compare this with your answer to (c) and suggest a reason for the difference.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
First harmonic: λ = 2L, so f1 = v/2L; the load is in equilibrium so F = Mg✓ 1
f1² = v²/4L² = Mg/(4L²μ)✓ 1
Part (b)(i)
Large triangle, e.g. (0, 3) to (0.500, 1538)✓ 1
Gradient = 3.07 × 103 Hz² kg−1✓ 1Accept 3000–3150 Hz² kg−1. Unit required.
Part (c)(i)
μ = g/(4L² × gradient) = 9.81/(4 × 0.800² × 3.07 × 103) = 1.25 × 10−3 kg m−1✓ 1Accept 1.22–1.28 × 10−3 kg m−1. Allow ECF from (a) and (b).
Part (d)(i)
μ = 2.40 × 10−3/2.000 = 1.20 × 10−3 kg m−1: the graph value is about 4 % larger✓ 1Allow ECF from (c).
Friction at the pulley means the tension in the vibrating part is less than Mg, so each f1 is lower, the gradient smaller and μ from the graph too large✓ 1The direction of the effect must match the difference. Accept: the string stretches under tension (lower μ), only if the direction is argued consistently.

Answers: (b)(i) 3.07 × 103 Hz² kg−1  ·  (c)(i) 1.25 × 10−3 kg m−1  ·  (d)(i) 1.20 × 10−3 kg m−1; graph value ≈ 4 % larger (the remaining parts are explanations — see the table above)

49C-1B-20Forced oscillations and resonance Paper 1BMedium7 marks

A metal strip is clamped at one end so that it can oscillate horizontally; an aluminium vane is fixed to its free end. The strip is plucked and its free oscillation is recorded with a motion sensor (Figure 1). The clamp is then attached to a vibration generator, and for each driving frequency f the student records the steady amplitude A of the vane, taking readings every 0.5 Hz (curve X, Figure 2).

Two strong magnets are then placed on either side of the vane, without touching it, and the measurements are repeated (curve Y).

0.00.20.40.60.81.01.21.4t / s-25-20-15-10-50510152025x / mm
Figure 1: free oscillation of the strip (drawn to scale).
5.56.06.57.07.58.08.59.09.5f / Hz05101520253035A / mmXY
Figure 2: amplitude A against driving frequency f without (X) and with (Y) the magnets (drawn to scale).
(a)
(i)

Use Figure 1 to determine the natural frequency of the strip.

(2)
(b)
(i)

State the frequency at which curve X has its maximum and explain whether it is consistent with (a).

(1)
(c)
(i)

Explain, with reference to the vane, why curve Y has a much smaller maximum than curve X.

(2)
(d)
(i)

Suggest how the student could use the same signal generator to determine the resonant frequency of X more precisely, and estimate the uncertainty of the present value.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
10 complete cycles between the peaks at t = 0 and t ≈ 1.32 s✓ 1Fewer than 5 cycles used: [0] for this mark.
f0 = 10/1.32 = 7.6 Hz✓ 1Accept 7.5–7.7 Hz.
Part (b)(i)
Peak of X at ≈ 7.6 Hz; this equals the natural frequency from (a), as expected for resonance with light damping✓ 1Accept 7.5–7.7 Hz. Allow ECF from (a).
Part (c)(i)
The vane moves through the magnetic field, so an emf is induced and eddy currents flow in the aluminium✓ 1
By Lenz's law these currents produce a force opposing the motion; energy is dissipated as internal energy in the vane (greater damping), so the maximum amplitude is smaller (and the peak broader / at slightly lower frequency)✓ 1"More friction" without induction: [0].
Part (d)(i)
With readings every 0.5 Hz the peak is only located to about ±0.25 Hz (the largest measured amplitude is at 7.5 Hz, with the true peak between 7.5 and 8.0 Hz)✓ 1Allow ECF from (b).
Take readings at small intervals (e.g. 0.05–0.1 Hz) between about 7.0 Hz and 8.0 Hz, close to the peak✓ 1Readings more finely spaced only near the peak. "Use a better generator": [0].

Answers: (a)(i) 7.6 Hz  ·  (b)(i) 7.6 Hz; consistent  ·  (d)(i) ±0.25 Hz (the remaining parts are explanations — see the table above)

50C-1B-03Resonance tubes Paper 1BHard7 marks

A long horizontal tube is surrounded by a water jacket whose temperature θ is read on a thermometer. A loudspeaker at the open end emits sound of frequency 1500 Hz. A piston inside the tube is moved slowly away from the loudspeaker and a microphone detects successive resonances. The distance Δx between the piston positions of two adjacent resonances is measured with a metre rule.

Kinetic theory predicts that the square of the speed of sound v is proportional to the absolute temperature of the air. The graph shows v² against θ with the line of best fit.

θ / °C1530456075
Δx / mm113.6115.9119.2121.4124.3
v² / 104 m² s−211.6112.0912.7913.26
-300-250-200-150-100-50050100θ / °C03691215v² / 10⁴ m² s⁻²
v² against θ with the line of best fit (drawn to scale). The θ axis extends to −300 °C.
(a)
(i)

Show that v² at 75 °C is about 13.9 × 104 m² s−2.

(1)
(b)
(i)

Determine the gradient of the line, including its unit.

(2)
(c)
(i)

By extrapolating the line, determine the temperature at which v² would be zero.

(2)
(d)
(i)

Compare your answer to (c) with the prediction of kinetic theory, and suggest one reason for the difference.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Adjacent resonances are half a wavelength apart: λ = 2Δx = 0.2486 m, v = fλ = 1500 × 0.2486 = 372.9 m s−1, so v² = 13.91 × 104 m² s−2✓ 1The factor 2 and a value to at least 3 s.f. (or full substitution) must be seen.
Part (b)(i)
Large triangle on the line, e.g. using (15, 11.58) and (75, 13.89)✓ 1The 75 °C point uses the value from (a): allow ECF from (a).
Gradient = 0.0385 × 104 m² s−2 °C−1 (= 385 m² s−2 K−1)✓ 1Accept 0.0375–0.0393 × 104. Unit required (°C−1 or K−1).
Part (c)(i)
Intercept on the v² axis at θ = 0 is 11.00 × 104 m² s−2✓ 1Accept 10.9–11.1. ALT: extend the line to the θ axis.
θ = −intercept/gradient = −11.00/0.0385 = −286 °C✓ 1Accept −275 to −300 °C. Allow ECF from (b).
Part (d)(i)
Theory: v² ∝ T, so v² = 0 at absolute zero, −273 °C; the experimental value is about 13 °C lower / 5 % different✓ 1Allow ECF from (c): the comparison must use their value.
Reason: a long extrapolation (from 15 °C to below −270 °C) magnifies any small error in the gradient; OR the air inside the tube is cooler than the water jacket at the higher temperatures, so the measured gradient is too small and the intercept on the θ axis is too negative✓ 1Accept any reason that is consistent with the direction of their difference. Do not accept "human error" or "inaccurate thermometer" without a mechanism.

Answers: (b)(i) 385 m² s−2 K−1  ·  (c)(i) −286 °C (the remaining parts are explanations — see the table above)

51C-2-06Standing waves on strings Paper 2Easy10 marks

A cello string has a vibrating length of 0.690 m between its two fixed ends and a mass per unit length μ of 1.50 × 10−3 kg m−1. The speed of transverse waves on the string is v = √(T/μ), where T is the tension. The string is wound around a peg of radius 4.0 mm, which is turned to tune it.

(a)
(i)

The first harmonic of the string is tuned to 220 Hz. Show that the tension in the string is about 140 N.

(2)
(ii)

Determine the torque that friction at the peg must provide to hold the string in tune.

(2)
(iii)

Before tuning, the first harmonic was 212 Hz. Determine the percentage increase in tension that was needed.

(2)
(b)

The string now vibrates in its third harmonic. Point P is 0.115 m and point Q is 0.345 m from one end of the string.

(i)

Compare the amplitudes and the phases of the oscillations of P and Q, and calculate the wavelength in air of the sound produced. The speed of sound in air is 343 m s−1.

(3)
(ii)

Explain why the wavelength of the sound in air is different from the wavelength of the wave on the string.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
λ = 2L = 1.38 m, so v = fλ = 220 × 1.38 = 303.6 m s−1✓ 1
T = μv² = 1.50 × 10−3 × 303.6² = 138 N✓ 1Answer to at least 3 s.f. or full substitution required.
Part (a)(ii)
Torque of the tension about the axis of the peg = Tr = 138 × 4.0 × 10−3 = 0.55 N m✓ 1Allow ECF from (a)(i).
In rotational equilibrium the friction torque is equal in size and opposite in direction: 0.55 N m✓ 1
Part (a)(iii)
f ∝ √T, so Tnew/Told = (220/212)²✓ 1
= 1.077, an increase of 7.7 %✓ 1Accept 7.7–7.8 %. 3.8 % (ratio not squared): [1 max].
Part (b)(i)
λ = 2L/3 = 0.460 m: nodes at 0, 0.230, 0.460 and 0.690 m, so P and Q are both at antinodes and have the same (maximum) amplitude✓ 1
P and Q are in adjacent loops, separated by the node at 0.230 m, so they oscillate in antiphase (phase difference π rad)✓ 1
f3 = 3 × 220 = 660 Hz, so λair = 343/660 = 0.520 m✓ 1Allow ECF from (a)(i).
Part (b)(ii)
The frequency is the same (set by the vibrating string) but the wave speeds differ, so λ = v/f differs✓ 1Reference to the same frequency and different speeds needed.

Answers: (a)(i) 138 N  ·  (a)(ii) 0.55 N m  ·  (a)(iii) 7.7 %  ·  (b)(i) equal; antiphase; 0.520 m (the remaining parts are explanations — see the table above)

52C-2-35Forced oscillations and resonance Paper 2Easy13 marks

A thin wine glass rings at a frequency of 880 Hz when it is tapped. A small loudspeaker connected to a signal generator is placed 0.30 m from the glass. The speed of sound in air is 340 m s−1.

(a)
(i)

Calculate the wavelength in air of sound of frequency 880 Hz.

(1)
(ii)

The output of the signal generator is kept constant while its frequency is increased slowly from 800 Hz to 950 Hz. Describe and explain how the amplitude of vibration of the rim of the glass changes.

(3)
(b)

At resonance the points of the rim with the largest amplitude vibrate with simple harmonic motion of amplitude 0.15 mm.

(i)

Calculate the maximum acceleration of the rim.

(2)
(ii)

Calculate the maximum speed of the rim.

(1)
(iii)

Viewed from above, the rim vibrates at resonance as a standing wave with four nodes and four antinodes, equally spaced around it. The circumference of the rim is 0.24 m. Determine the speed of the waves that travel around the rim to form this standing wave.

(2)
(c)

The loudspeaker emits sound energy at a rate of 2.0 W. Assume that it radiates uniformly in all directions and that no energy is absorbed by the air.

(i)

Calculate the intensity of the sound at the glass.

(2)
(d)
(i)

Water is poured into the glass. When the glass is tapped, it now rings at a lower frequency. Suggest why, by modelling the rim of the glass as a mass–spring oscillator.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
λ = 340/880 = 0.386 m✓ 1Accept 0.39 m.
Part (a)(ii)
The amplitude increases to a large maximum at (or very close to) 880 Hz and then decreases again✓ 1A sharp peak should be described.
880 Hz is the natural frequency of the glass, so the glass is then driven at resonance✓ 1"Resonance" alone, with no reference to the natural frequency, is not enough.
At resonance the driving force transfers the most energy to the glass in each cycle (it acts in phase with the velocity of the rim), so energy builds up until the energy dissipated per cycle equals that supplied✓ 1Accept "energy transfer is greatest at resonance".
Part (b)(i)
ω = 2π × 880 = 5.53 × 103 rad s−1✓ 1
amax = ω²x0 = (5.53 × 103)² × 1.5 × 10−4 = 4.6 × 103 m s−2✓ 1Accept (4.5–4.6) × 103 m s−2. Allow ECF for ω.
Part (b)(ii)
vmax = ωx0 = 5.53 × 103 × 1.5 × 10−4 = 0.83 m s−1✓ 1Allow ECF from (b)(i).
Part (b)(iii)
Adjacent nodes are half a wavelength apart, so the circumference is 4 × λ/2 = 2λ: λ = 0.12 m✓ 1Award [0] for this mark for λ = 0.24 m or 0.06 m.
v = fλ = 880 × 0.12 = 1.1 × 102 m s−1✓ 1Accept 106 m s−1. Allow ECF for λ.
Part (c)(i)
I = P/(4πr²)✓ 1
I = 2.0/(4π × 0.30²) = 1.77 W m−2✓ 1Accept 1.8 W m−2.
Part (d)(i)
The water next to the wall of the glass has to move with it, so the effective oscillating mass increases while the stiffness (spring constant) of the glass is unchanged✓ 1
T = 2π√(m/k), so a larger mass gives a longer period and a lower natural frequency✓ 1"The water damps the glass" alone scores [0]: light damping hardly changes the frequency.

Answers: (a)(i) 0.386 m  ·  (b)(i) 4.6 × 103 m s−2  ·  (b)(ii) 0.83 m s−1  ·  (b)(iii) 1.1 × 102 m s−1  ·  (c)(i) 1.77 W m−2 (the remaining parts are explanations — see the table above)

53IBP2H-025-00-09Wavelength of a standing wave Paper 2Easy5 marks

A steel stay cable on a pedestrian footbridge is 24 m long. It is fixed at both ends. In strong wind, engineers observe a standing wave on the cable with three loops between the ends. The cable oscillates at a frequency of 1.8 Hz.

24 m
Diagram not to scale
(a)

State the number of nodes on the cable, including the ends.

(1)
(b)

Calculate the wavelength of the standing wave.

(2)
(c)

Calculate the speed of the transverse waves on the cable.

(1)
(d)

Determine the fundamental frequency of the cable.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
4 nodes✓ 1
Part (b)
Three loops means 3 × λ/2 = 24 m✓ 1
λ = 16 m✓ 1
Part (c)
v = fλ = 1.8 × 16 = 29 m s−1 (accept 28.8 m s−1; allow error carried forward from the calculated wavelength)✓ 1
Part (d)
The observed wave is the third harmonic, so f1 = 1.8/3 = 0.60 Hz (also accept f1 = v/2L, allowing error carried forward from the calculated wave speed)✓ 1

Exam tip: Count loops, not nodes, to find the harmonic number on a string fixed at both ends. Each loop is half a wavelength.

54IBP2H-025-01-09Formation of standing waves Paper 2Easy5 marks

An acoustic levitator suspends small droplets of a medical sample in air without touching them. An ultrasonic transducer emits sound towards a flat reflector placed directly opposite it. The droplets settle at a line of positions, each a node of the standing wave, spaced 4.3 mm apart. The speed of sound in the air is 344 m s−1.

transducerreflector4.3 mm
Diagram not to scale; beads at nodes of the standing wave
(a)

Outline how a standing wave forms between the transducer and the reflector.

(2)
(b)

State the distance between adjacent nodes in terms of the wavelength λ.

(1)
(c)

Calculate the frequency of the ultrasound.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Sound from the transducer is reflected by the reflector back towards the transducer ; the incident and reflected waves have the same frequency (and similar amplitude), travel in opposite directions and superpose to form a standing wave .✓ 2
Part (b)
λ/2✓ 1
Part (c)
λ = 2 × 4.3 mm = 8.6×10−3 m ; f = v/λ = 344/8.6×10−3 = 4.0×104 Hz . Allow ECF from (b).✓ 2

Exam tip: To explain how a standing wave forms, mention three things: reflection, two waves of the same frequency travelling in opposite directions, and superposition.

55IBP3H-C4-20r16-09Standing waves in pipes Paper 2Easy5 marks

A hospital sound laboratory uses an air-filled pipe of length 0.60 m that is sealed at both ends. Take the speed of sound in air as 340 m s−1.

(a)

State the type of displacement point (node or antinode) at each sealed end.

(1)
(b)

Calculate the frequency of the fundamental standing wave in the pipe. Give your answer to three significant figures.

(2)
(c)

Determine the frequency of the third harmonic. Use the unrounded value from (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Node at each end, since the air cannot move at a closed end✓ 1
Part (b)
M1 \(\lambda=2L=1.20\) m and \(f=v/\lambda\)✓ 1
A1 283 Hz✓ 1
Part (c)
M1 \(f_3=3f_1\)✓ 1
A1 850 Hz✓ 1

Common error: A sealed end is a displacement node, even though it is a pressure antinode.

56IBP3H-C4-20r16-10Formation of standing waves Paper 2Easy5 marks

A tuning fork of frequency 680 Hz is held above a tube closed at one end. The water level in the tube is lowered until the first resonance is heard. Take the speed of sound as 340 m s−1 and neglect end corrections.

(a)

Calculate the wavelength of the sound.

(2)
(b)

Determine the length of the air column at the first resonance.

(1)
(c)

Determine the length of the air column at the second resonance.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\lambda=v/f\)✓ 1
A1 0.50 m✓ 1
Part (b)
A1 \(\lambda/4=0.125\) m✓ 1
Part (c)
M1 length \(=\dfrac{3\lambda}{4}\)✓ 1
A1 0.375 m✓ 1

Exam tip: Resonant lengths of a tube closed at one end are \(\lambda/4\), \(3\lambda/4\), \(5\lambda/4\), …

57IBP3H-C4-20r26-09Strings with free ends Paper 2Easy7 marks

A sports-science team studies a battle rope of length 9.0 m. Both ends of the rope are effectively fixed and act as nodes. Transverse waves travel along the rope at 13.5 m s-1.

(a)

Outline how a standing wave forms on the rope.

(3)
(b)

Calculate the frequency of the fundamental standing wave.

(2)
(c)

Determine the frequency of the fourth harmonic and the number of nodes along the rope, including the two ends.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Incident wave travels along the rope and is reflected at the fixed end✓ 1
The reflected and incident waves (of same frequency and speed) travel in opposite directions and superpose✓ 1
Superposition gives nodes (destructive) and antinodes (constructive) at fixed positions, so the pattern does not travel✓ 1
Part (b)
M1 λ = 2L = 18 m, or f = v/2L✓ 1
A1 f = 13.5/18 = 0.75 Hz✓ 1
Part (c)
A1 f4 = 4 × 0.75 = 3.0 Hz✓ 1
A1 5 nodes (ECF from (b) for frequency)✓ 1

Exam tip: The nth harmonic has n antinodes and n + 1 nodes, counting the two fixed ends.

58IBP3H-C4-20r26-10Standing waves in pipes Paper 2Easy8 marks

A train horn is a pipe of length 0.60 m, closed at one end and open at the other. Take the speed of sound in air as 340 m s-1 and ignore end corrections.

(a)

State the displacement of the air at the closed end and at the open end of the pipe in the fundamental mode.

(2)
(b)

Calculate the frequency of the fundamental note.

(2)
(c)

Calculate the frequencies of the next two resonances. Use the unrounded fundamental frequency.

(2)
(d)

Explain why a frequency of 2f is not produced by this pipe.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Closed end is a displacement node✓ 1
Open end is a displacement antinode✓ 1
Part (b)
M1 λ = 4L = 2.4 m, or f = v/4L✓ 1
A1 f = 340/2.4 = 142 Hz (141.7 Hz)✓ 1
Part (c)
A1 3f = 425 Hz✓ 1
A1 5f = 708 Hz✓ 1
Part (d)
A resonance requires a node at the closed end and an antinode at the open end✓ 1
At 2f the length would be a whole number of half-wavelengths, which would give a node (or antinode) at both ends, so this is not possible✓ 1

Common error: Do not use the string or open-pipe sequence f, 2f, 3f for a closed pipe. Closed-pipe resonances are f, 3f, 5f.

59C-2-14Standing waves and resonance Paper 2Medium12 marks

A microwave oven produces electromagnetic waves of frequency 2.45 GHz. The waves are reflected by the metal walls of the oven. A student removes the rotating turntable and heats a flat plate of cheese for a short time. The cheese melts only in small spots arranged in lines.

(a)
(i)

Calculate the wavelength of the microwaves.

(1)
(ii)

Explain why the cheese melts only in spots.

(2)
(iii)

Determine the distance between adjacent melted spots along a line.

(1)
(b)
(i)

The student measures the separation of adjacent spots as (6.3 ± 0.3) cm. Determine, using the frequency stated on the oven, a value for the speed of light with its absolute uncertainty, and comment on your answer.

(3)
(c)
(i)

Calculate the energy of a microwave photon in eV, and explain why microwaves cannot ionise the molecules of the cheese however intense the radiation.

(2)
(ii)

The water in a cup of mass 0.25 kg absorbs microwave energy at a rate of 800 W. Determine the number of photons absorbed per second and the time needed to raise the temperature of the water by 20 K. The specific heat capacity of water is 4200 J kg−1 K−1.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
λ = c/f = 3.00 × 108/2.45 × 109 = 0.122 m✓ 1
Part (a)(ii)
Waves reflected from the walls superpose with the incident waves (of the same frequency, travelling in opposite directions) to form a standing wave✓ 1
At antinodes the oscillations of the field are largest, so energy is transferred to the cheese fastest; at nodes there is (almost) no oscillation, so the cheese there stays solid✓ 1
Part (a)(iii)
Adjacent antinodes are λ/2 apart: 0.061 m✓ 1Allow ECF from (a)(i). 0.122 m (a whole wavelength): [0].
Part (b)(i)
c = fλ = 2 × 0.063 × 2.45 × 109 = 3.1 × 108 m s−1✓ 1The factor 2 (spot spacing = λ/2) is needed. Allow ECF from (a)(iii).
Fractional uncertainty = 0.3/6.3 = 0.048, so Δc = ±1.5 × 107 m s−1✓ 1Accept ±1 × 107 to ±2 × 107 m s−1.
The accepted value 3.00 × 108 m s−1 lies within the range, so the result is consistent✓ 1Conclusion must agree with the candidate's own range.
Part (c)(i)
E = hf = 6.63 × 10−34 × 2.45 × 109 = 1.62 × 10−24 J = 1.02 × 10−5 eV✓ 1Unit eV required.
Ionisation needs several eV given to one electron by one photon; each photon carries far too little energy, and a greater intensity only gives more photons, not more energetic ones✓ 1
Part (c)(ii)
Number per second = 800/1.62 × 10−24 = 4.9 × 1026 s−1✓ 1Allow ECF from (c)(i).
Q = mcΔT = 0.25 × 4200 × 20 = 2.1 × 104 J✓ 1
t = Q/P = 2.1 × 104/800 = 26 s✓ 1Accept 26 s.

Answers: (a)(i) 0.122 m  ·  (a)(iii) 0.061 m  ·  (b)(i) (3.09 ± 0.15) × 108 m s−1  ·  (c)(i) 1.02 × 10−5 eV  ·  (c)(ii) 4.9 × 1026 s−1; 26 s (the remaining parts are explanations — see the table above)

60C-2-30Pipes closed at both ends Paper 2Medium12 marks

A gas sensor uses a rigid tube of length 0.180 m sealed at both ends, as shown. A small loudspeaker in one cap drives the gas; a microphone in the other cap detects the amplitude of the sound. The frequency of the loudspeaker is swept and the frequencies at which the microphone signal is a maximum are recorded.

P and Q are points on the axis of the tube, 0.045 m and 0.135 m from the left cap.

loudspeakermicrophonePQ0.180 mrigid caprigid cap
The sealed tube (not to scale vertically). P is 0.045 m and Q is 0.135 m from the left cap.
(a)
(i)

Describe the standing wave in the air when it vibrates in its first harmonic.

(1)
(ii)

The speed of sound in air is 343 m s−1. Show that the first-harmonic frequency is about 950 Hz.

(2)
(b)
(i)

Compare the amplitudes and the phases of the oscillations of the air at P and at Q in the first harmonic.

(2)
(ii)

The amplitude at a distance d from a displacement node is Amax sin(2πd/λ). The air now vibrates in its third harmonic. Determine the amplitudes at P and at Q as fractions of Amax.

(2)
(c)

The tube is filled with a mixture of hydrogen and air at 293 K. The lowest frequency at which the microphone signal is a maximum is now 1520 Hz. The speed of sound in a gas is v = √(γRT/M), where M is the molar mass; γ = 1.40 for hydrogen, for air and for any mixture of the two.

(i)

Determine the molar mass of the mixture.

(3)
(ii)

The molar masses of hydrogen and air are 2.0 g mol−1 and 29.0 g mol−1. Deduce the fraction of the molecules in the mixture that are hydrogen.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Displacement nodes at both caps and a single displacement antinode at the centre: half a wavelength fits in the tube✓ 1
Part (a)(ii)
λ = 2 × 0.180 = 0.360 m✓ 1
f = 343/0.360 = 953 Hz✓ 1Answer to at least 3 s.f. required.
Part (b)(i)
Equal amplitudes (P and Q are symmetrical about the central antinode), each smaller than at the centre✓ 1
In phase: both lie in the same loop, between the same two nodes✓ 1
Part (b)(ii)
λ = 2L/3 = 0.120 m: nodes at 0, 0.060, 0.120 and 0.180 m, so P and Q are each 0.015 m from a node✓ 1
Amplitude = sin(2π × 0.015/0.120) = sin(π/4) = 0.71Amax at both✓ 1Accept 0.7.
Part (c)(i)
First harmonic: v = 2Lf = 2 × 0.180 × 1520 = 547 m s−1✓ 1
M = γRT/v² = 1.40 × 8.31 × 293/547²✓ 1Allow ECF for the speed.
M = 1.14 × 10−2 kg mol−1✓ 1Accept 11.4 g mol−1.
Part (c)(ii)
11.4 = x × 2.0 + (1 − x) × 29.0, with x the fraction of hydrogen molecules✓ 1The hidden step: the molar mass of a mixture is the number-weighted mean.
x = (29.0 − 11.4)/27.0 = 0.65✓ 1Allow ECF from (c)(i).

Answers: (a)(ii) 953 Hz  ·  (b)(ii) 0.71Amax at both  ·  (c)(i) 1.14 × 10−2 kg mol−1  ·  (c)(ii) 0.65 (the remaining parts are explanations — see the table above)

61C-2-31Damping and resonance Paper 2Medium9 marks

A test rig for car shock absorbers holds a mass of 300 kg on a spring. The mass is pulled down 50 mm and released, and the graph shows its displacement x against time t for three different shock absorbers P, Q and R.

0.00.40.81.21.62.02.4t / s-50-40-30-20-1001020304050x / mmQRP
Displacement of the mass against time for shock absorbers P, Q and R (drawn to scale).
(a)
(i)

Identify the shock absorber that gives critical damping, giving a reason.

(1)
(ii)

Use the graph to determine the spring constant of the spring.

(2)
(iii)

For shock absorber Q, determine the fraction of the energy of the oscillation that is dissipated during the first cycle.

(2)
(b)
(i)

In use, shock absorber R dissipates all the energy stored in the spring when it is compressed by 50 mm, twice every second for 5.0 minutes. Estimate the rise in temperature of the 0.40 kg of oil in the shock absorber. The specific heat capacity of the oil is 1900 J kg−1 K−1. State an assumption you make.

(3)
(ii)

State which type of damping is suitable for a car suspension.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
R: it returns the mass to equilibrium in the shortest time without oscillating✓ 1Reason required.
Part (a)(ii)
Period of Q = 0.80 s✓ 1Accept 0.78–0.82 s; reading over several cycles is better.
k = 4π²m/T² = 4π² × 300/0.80² = 1.85 × 104 N m−1✓ 1Accept 1.8–1.9 × 104 N m−1. Allow ECF from the period read.
Part (a)(iii)
Amplitude after one cycle = 35 mm (from 50 mm)✓ 1Accept 34–36 mm.
Energy ∝ amplitude², so the fraction dissipated = 1 − (35/50)² = 0.51✓ 1Accept 0.48–0.54. 0.30 (amplitude ratio used): [1 max].
Part (b)(i)
Energy stored each time = ½kx² = ½ × 1.85 × 104 × 0.050² = 23.1 J✓ 1Allow ECF from (a)(ii).
Total energy = 23.1 × 2 × 300 = 1.4 × 104 J✓ 1
ΔT = 1.4 × 104/(0.40 × 1900) = 18 K, assuming no energy is lost from the oil to the surroundings✓ 1Accept 18–19 K. Value and assumption both needed for this mark.
Part (b)(ii)
Critical (or slightly less than critical), so that the car settles quickly after a bump without bouncing✓ 1

Answers: (a)(i) R  ·  (a)(ii) 1.85 × 104 N m−1  ·  (a)(iii) 0.51  ·  (b)(i) 18 K (the remaining parts are explanations — see the table above)

62IBP2H-025-00-10Pipes closed at both ends Paper 2Medium9 marks

A cryogenic transfer line contains a straight cylindrical section of length 1.40 m. The section is sealed with rigid metal caps at both ends and is filled with helium gas at 300 K. A small speaker and microphone inside the section detect acoustic resonances at 365 Hz, 730 Hz and 1095 Hz. There is no resonance below 365 Hz.

(a)

Explain why both ends of the section are displacement nodes for the sound wave.

(2)
(b)

Outline how the data are consistent with a tube closed at both ends rather than a tube closed at one end only.

(1)
(c)

Determine the speed of sound in the helium.

(2)
(d)

For the 730 Hz resonance, describe the positions of the displacement antinodes and the pressure nodes along the section.

(2)
(e)

The helium is cooled to 75 K. The speed of sound in the gas is proportional to the square root of its absolute temperature. Calculate the new lowest resonant frequency. Assume the length of the section is unchanged.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The caps are rigid, so the gas molecules in contact with them cannot oscillate along the tube✓ 1
So the displacement amplitude at each end is zero, which is a node✓ 1
Part (b)
The resonances are 1, 2 and 3 times 365 Hz, so even as well as odd multiples of the fundamental occur, which fits f, 2f, 3f… for a tube closed at both ends. A tube closed at one end would give only odd multiples (f, 3f, 5f…)✓ 1
Part (c)
The fundamental is 365 Hz, with λ = 2L = 2.80 m✓ 1
v = fλ = 365 × 2.80 = 1.02 × 103 m s−1✓ 1
Part (d)
This is the second harmonic, so there are displacement nodes at both ends and at the centre, and displacement antinodes at 0.35 m and 1.05 m from one end✓ 1
Pressure nodes coincide with the displacement antinodes, so they are also at 0.35 m and 1.05 m from one end✓ 1
Part (e)
The speed is multiplied by √(75/300) = 0.50, and the wavelength is unchanged, so the frequency is multiplied by 0.50 too✓ 1
f1 = 365 × 0.50 = 183 Hz (accept 182 Hz to 183 Hz)✓ 1

Common error: Displacement nodes and pressure nodes are not at the same places. At a rigid closed end, displacement is a node but pressure is an antinode.

63IBP2H-025-00-11Forced oscillations and resonance Paper 2Medium7 marks

An unsealed desert road has developed regular ridges, called corrugations, spaced 0.75 m apart. Each wheel assembly of a vehicle, together with its tyre and suspension, has a natural vertical oscillation frequency of 12 Hz.

(a)

State what is meant by the natural frequency of the wheel assembly.

(1)
(b)

Calculate the speed of the vehicle at which the wheel assemblies resonate.

(2)
(c)

Worn shock absorbers are replaced with new ones that provide greater damping. Explain the effect of this on the wheel vibration when the vehicle travels at the resonant speed.

(2)
(d)

Explain why driving at a speed much higher than the resonant speed reduces the vibration of the wheels.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The frequency at which the assembly oscillates when it is displaced and then left to vibrate freely, with no periodic driving force✓ 1
Part (b)
Resonance requires the driving frequency f = v/d to equal the natural frequency of 12 Hz✓ 1
v = fd = 12 × 0.75 = 9.0 m s−1✓ 1
Part (c)
Greater damping dissipates more energy from the oscillation in each cycle✓ 1
So the amplitude of the wheel vibration at resonance is smaller✓ 1
Part (d)
The driving frequency (v/d) becomes much greater than the natural frequency, so the system is far from resonance✓ 1
The wheel cannot respond to the rapidly changing force, so energy transfer to it is inefficient and the amplitude is small✓ 1

Exam tip: For a vehicle on a periodic surface, the driving frequency equals speed divided by ridge spacing. Set this equal to the natural frequency to find the resonant speed.

64IBP2H-025-00-13Wavelength of a standing wave Paper 2Medium8 marks

Vibrating-wire strain gauges are embedded in a concrete dam to monitor stress. Each gauge contains a steel wire fixed at both ends. The wire is plucked electromagnetically and vibrates at its fundamental frequency. In a calibration test, the tension is kept constant and the length L of the wire is varied. The fundamental frequency f is measured for each length:

L = 0.100 m, f = 1495 Hz
L = 0.125 m, f = 1204 Hz
L = 0.150 m, f = 998 Hz
L = 0.200 m, f = 752 Hz

(a)

Show that the data are consistent with f being inversely proportional to L.

(2)
(b)

Determine the speed of transverse waves on the wire.

(2)
(c)

In the dam, the concrete around a gauge stretches slightly, increasing the tension in the wire. Explain the effect on the measured frequency.

(2)
(d)

The vibrating 0.150 m wire produces a faint sound in the surrounding air. The speed of sound in air is 340 m s−1. Calculate the wavelength of this sound in air.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Calculates fL for at least three data pairs, e.g. 149.5, 150.5, 149.7, 150.4 Hz m✓ 1
The product is constant to within about 1%, so f ∝ 1/L✓ 1
Part (b)
For the fundamental, λ = 2L, so v = 2fL✓ 1
Mean fL ≈ 150 Hz m, so v ≈ 300 m s−1✓ 1
Part (c)
Greater tension increases the wave speed on the wire✓ 1
The wavelength of the fundamental stays about 2L, so f = v/λ increases✓ 1
Part (d)
The frequency of the sound equals the wire's frequency, 998 Hz✓ 1
λ = 340/998 = 0.341 m✓ 1

Common error: A wave keeps the same frequency when it passes from the wire into the air. Its wavelength in air is not 2L; use the speed of sound in air instead.

65IBP2H-025-00-15Formation of standing waves Paper 2Medium7 marks

In pharmaceutical research, an ultrasonic levitator holds small droplets in mid-air, so that no container can contaminate them. An ultrasound transducer faces a flat reflector, and a standing sound wave forms in the air between them. Six tiny polystyrene beads become trapped at six consecutive pressure nodes. The distance between the first and sixth bead is 21.4 mm. The speed of sound in air is 343 m s−1.

transducerreflector21.4 mm
Diagram not to scale; beads at nodes of the standing wave
(a)

Explain how the standing wave is formed between the transducer and the reflector.

(2)
(b)

State the distance between adjacent pressure nodes in terms of the wavelength λ.

(1)
(c)

Determine the wavelength of the ultrasound.

(2)
(d)

Calculate the frequency of the ultrasound.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Sound waves from the transducer reflect from the reflector and travel back, giving two waves of the same frequency and speed travelling in opposite directions✓ 1
These waves superpose, producing fixed positions of minimum and maximum amplitude (nodes and antinodes)✓ 1
Part (b)
λ/2✓ 1
Part (c)
Six beads span five node-to-node gaps, so 5 × λ/2 = 21.4 mm✓ 1
λ = 8.56 mm✓ 1
Part (d)
f = v/λ = 343/(8.56 × 10−3)✓ 1
f = 4.0 × 104 Hz (accept 4.01 × 104 Hz; allow error carried forward from the wavelength calculated above)✓ 1

Common error: Count the gaps, not the beads. Six objects at consecutive nodes are separated by five half-wavelengths, not six.

66IBP2H-025-01-10Forced oscillations and resonance Paper 2Medium8 marks

An overhead power cable is fixed at two towers 60.0 m apart. Wind makes the cable vibrate. Sensors record the frequencies of several standing-wave modes, where n is the harmonic number.

n: 2, 3, 5, 7
f / Hz: 1.90, 2.85, 4.75, 6.65

(a)

Show that the fundamental frequency of the cable is about 0.95 Hz.

(2)
(b)

Calculate the speed of transverse waves on the cable.

(2)
(c)

In a steady wind, air vortices push the cable sideways at a frequency of 4.75 Hz. Explain why the cable oscillates with a large amplitude.

(2)
(d)

Engineers clamp a damper to the cable 1.0 m from one tower rather than at the midpoint. Outline why this position is better.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
f/n = 0.95 Hz for every row, e.g. 4.75/5 = 0.95 ; the harmonics are whole-number multiples of f1, so f1 = 0.95 Hz .✓ 2
Part (b)
λ1 = 2L = 120 m ; v = f1λ1 = 0.95 × 120 = 114 m s−1 . ECF from (a).✓ 2
Part (c)
The driving frequency equals the natural frequency of the fifth harmonic ; this is resonance, so energy is transferred efficiently every cycle and the amplitude builds up until it is limited only by damping .✓ 2
Part (d)
The midpoint is a node for every even harmonic, so a damper there would not move in those modes and would remove no energy from them ; 1.0 m from the end is not a node for any of the low harmonics (the nodes nearest the end are at 60/n m), so the damper moves in every likely mode and dissipates energy from each .✓ 2

Exam tip: A damper can only remove energy where the medium moves. Check whether the chosen point is a node for the harmonic in question.

67IBP2H-025-01-11Resonance tubes Paper 2Medium7 marks

A sound source of unknown frequency is held above a vertical tube. The lower end of the tube is in water, and the water level is slowly lowered. The first resonance occurs when the air column is 0.142 m long, and the second when it is 0.462 m long. The speed of sound on that day is 346 m s−1.

tuning fork0.142 mfirst resonancetuning fork0.462 msecond resonance
Diagram not to scale
(a)

Determine the frequency of the source.

(3)
(b)

The antinode at the open top of the tube lies a short distance above the end of the tube. This distance is the end correction. Determine the end correction.

(2)
(c)

Predict the length of the air column at the third resonance.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The difference between successive resonance lengths is λ/2 ; λ = 2 × (0.462 − 0.142) = 0.640 m ; f = 346/0.640 = 541 Hz .✓ 3
Part (b)
First resonance: λ/4 = 0.160 m ; end correction = 0.160 − 0.142 = 0.018 m . ECF from (a).✓ 2
Part (c)
The third resonance is at 5λ/4 minus the end correction, or equivalently a further λ/2 beyond the second resonance ; length = 0.462 + 0.320 = 0.782 m . ECF from (a)/(b).✓ 2

Common error: Calculating λ = 4 × (first length) ignores the end correction. Use the difference between two resonance lengths instead.

68IBP2H-025-01-16Standing waves in pipes Paper 2Medium8 marks

A simple flute is a pipe 0.330 m long, open at both ends. On Earth the speed of sound is 343 m s−1.

(a)

Describe the standing wave in the pipe when it sounds its fundamental.

(2)
(b)

Calculate the fundamental frequency on Earth.

(2)
(c)

One end of the pipe is now closed. Determine the new fundamental frequency and state which harmonics can now be produced.

(2)
(d)

An astronaut on the surface of Mars wants an open–open pipe that sounds the same note as in (b). The speed of sound in the Martian atmosphere is 240 m s−1. Determine the length of pipe needed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
There is a displacement antinode at each open end ; there is a single displacement node at the centre, so L = λ/2 .✓ 2
Part (b)
λ = 2 × 0.330 = 0.660 m ; f = 343/0.660 = 520 Hz .✓ 2
Part (c)
f1 = v/4L = 343/(4 × 0.330) = 260 Hz ; only odd harmonics (1st, 3rd, 5th, …) .✓ 2
Part (d)
L = v/2f = 240/(2 × 520) ; L = 0.231 m . Allow ECF from (b).✓ 2

Exam tip: The frequency sets the note, so it does not change in (d). A lower wave speed gives a shorter wavelength, so the pipe must be shorter.

69IBP3H-C4-20r16-11Strings with free ends Paper 2Medium6 marks

A light elastic cord of length 1.8 m has a smooth ring at each end that slides on a vertical rod, so both ends of the cord are free. Transverse waves travel along the cord at 8.4 m s−1.

ring on smooth rodring on smooth rod1.8 m
Diagram not to scale
(a)

Describe, in words, the positions of the nodes and antinodes for the fundamental standing wave on the cord.

(2)
(b)

Calculate the frequency of the fundamental. Give your answer to three significant figures.

(2)
(c)

Determine the frequency of the third harmonic, and state how many nodes are present. Use the unrounded value from (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Antinodes at both ends of the cord✓ 1
One node at the midpoint (0.90 m from each end)✓ 1
Part (b)
M1 \(\lambda=2L=3.6\) m, \(f=v/\lambda\)✓ 1
A1 2.33 Hz✓ 1
Part (c)
A1 7.00 Hz✓ 1
A1 3 nodes✓ 1

Common error: For free ends the harmonic frequencies are whole-number multiples of the fundamental, as for two fixed ends.

70IBP3H-C4-20r16-13Standing waves in pipes Paper 2Medium6 marks

A student measures the speed of sound using a tube closed at one end and a tuning fork of frequency 425 Hz. The first resonance occurs when the air column is 0.185 m long and the second when it is 0.585 m long.

tuning fork0.185 mfirst resonancetuning fork0.585 msecond resonance
Diagram not to scale
(a)

Use the two lengths to calculate the wavelength of the sound.

(2)
(b)

Calculate the speed of sound.

(1)
(c)

The first resonance does not occur at exactly \(\lambda/4\). Calculate the difference between \(\lambda/4\) and the first resonant length.

(2)
(d)

Suggest why using the difference between the two resonant lengths gives a more reliable value of \(\lambda\) than using the first length alone.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\lambda/2=0.585-0.185\)✓ 1
A1 0.800 m✓ 1
Part (b)
A1 \(425\times0.800=340\) m s−1✓ 1
Part (c)
M1 \(\lambda/4=0.200\) m✓ 1
A1 0.015 m✓ 1
Part (d)
The antinode forms slightly beyond the open end, so the first length is shorter than \(\lambda/4\); the difference between lengths cancels this unknown shift✓ 1

Exam tip: The antinode at the open end lies a little beyond the tube, so the measured length is shorter than the ideal \(\lambda/4\).

71IBP3H-C4-20r16-14Standing waves in pipes Paper 2Medium8 marks

A pipe of length 0.85 m is sealed at both ends. A researcher sets up the third harmonic standing wave. Take the speed of sound as 340 m s−1.

(a)

Calculate the frequency of the third harmonic.

(2)
(b)

Determine the separation of adjacent displacement nodes. Give your answer to three significant figures.

(2)
(c)

A pressure sensor is placed at a displacement node inside the pipe. State and explain the size of the pressure variation there compared with the rest of the pipe.

(2)
(d)

One end of the pipe is now opened. Explain whether a standing wave of frequency 600 Hz can still form.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(f_3=3v/2L\)✓ 1
A1 600 Hz✓ 1
Part (b)
M1 separation \(=\lambda/2=v/2f\)✓ 1
A1 0.283 m✓ 1
Part (c)
Greatest pressure variation (pressure antinode)✓ 1
Air at a displacement node cannot move, so it is alternately compressed and rarefied by the neighbouring air✓ 1
Part (d)
With one open end the resonances are odd multiples of \(v/4L=100\) Hz: 100, 300, 500, 700 Hz✓ 1
600 Hz is not among them, so no standing wave forms at 600 Hz✓ 1

Common error: Displacement nodes are pressure antinodes and displacement antinodes are pressure nodes.

72IBP3H-C4-20r16-16Formation of standing waves Paper 2Medium7 marks

Two identical loudspeakers face each other 2.40 m apart in a long corridor and emit sound in phase at 425 Hz. A microphone is moved along the line joining them. Take the speed of sound as 340 m s−1.

(a)

Calculate the wavelength of the sound.

(1)
(b)

Determine the distance between adjacent nodes of the standing wave.

(1)
(c)

Show that the midpoint between the loudspeakers is an antinode.

(2)
(d)

Determine the number of nodes between the two loudspeakers.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(340/425=0.800\) m✓ 1
Part (b)
A1 \(\lambda/2=0.400\) m✓ 1
Part (c)
Both waves travel 1.20 m to the midpoint so the path difference is zero✓ 1
The waves arrive in phase and interfere constructively, giving an antinode✓ 1
Part (d)
M1 nodes are \(\lambda/4=0.200\) m either side of the midpoint antinode✓ 1
M1 nodes at 0.2, 0.6, 1.0, 1.4, 1.8 and 2.2 m from one loudspeaker✓ 1
A1 6 nodes✓ 1

Exam tip: Nodes lie midway between adjacent antinodes, so start from the known antinode at the midpoint.

73IBP3H-C4-20r26-11Pipes closed at both ends Paper 2Medium9 marks

In a simple model of a pulsating star, the gas pulsates radially as a standing sound wave. The centre of the star is a displacement node and the surface is a displacement antinode. The star has radius R = 3.6 × 1010 m and the average speed of sound inside it is v = 3.0 × 105 m s-1.

(a)

Explain why the centre is a node and the surface is an antinode.

(2)
(b)

Show that the period of the fundamental pulsation is T = 4R/v.

(2)
(c)

Calculate the period of the fundamental pulsation, in seconds and in days, to three significant figures.

(3)
(d)

Calculate the period of the first overtone that fits this model. Use the unrounded value of T from (c) and give the answer in days to three significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
At the centre the gas cannot move in any direction (symmetry), so displacement is zero✓ 1
At the surface the gas is free to move (no restoring layer beyond it), so amplitude is a maximum✓ 1
Part (b)
For the fundamental, R = λ/4 so λ = 4R✓ 1
T = λ/v = 4R/v✓ 1
Part (c)
M1 T = 4 × 3.6 × 1010 / 3.0 × 105✓ 1
A1 T = 4.8 × 105 s✓ 1
A1 T = 5.56 days✓ 1
Part (d)
M1 first overtone is the third harmonic, so R = 3λ/4 and period is T/3✓ 1
A1 1.60 × 105 s = 1.85 days✓ 1

Exam tip: A node at one end and an antinode at the other means the fundamental has a quarter of a wavelength in the length.

74IBP3H-C4-20r26-12Strings with free ends Paper 2Medium9 marks

A cable in a climbing-wall test rig has length 0.75 m. One end of the cable is fixed. The other end is attached to a light ring that can slide freely on a vertical rod. The cable has a tension of 28.8 N and a mass per unit length of 2.0 × 10-3 kg m-1.

clampedring on smooth rod0.75 m
Diagram not to scale
(a)

Explain why there is an antinode at the ring.

(2)
(b)

Calculate the speed of transverse waves on the cable.

(2)
(c)

Determine the wavelength and frequency of the fundamental mode.

(2)
(d)

State the frequencies of the next two resonances.

(1)
(e)

Calculate the fundamental frequency if both ends of the same cable were fixed, with the same tension.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The ring is free to move, so the end cannot exert a transverse force on the cable✓ 1
A wave is reflected without a phase change, so incident and reflected waves reinforce and the amplitude is a maximum✓ 1
Part (b)
M1 v = √(T/μ) = √(28.8 / 2.0 × 10-3)✓ 1
A1 v = 120 m s-1✓ 1
Part (c)
A1 λ = 4L = 3.0 m✓ 1
A1 f = 120/3.0 = 40 Hz✓ 1
Part (d)
120 Hz and 200 Hz (both required)✓ 1
Part (e)
M1 λ = 2L = 1.5 m✓ 1
A1 f = 120/1.5 = 80 Hz✓ 1

Common error: Do not treat a free end as a node. A free end is an antinode, so the fundamental is λ = 4L, not 2L.

75IBP3H-C4-20r26-13Standing waves in pipes Paper 2Medium8 marks

A rail technician measures the speed of sound using a tuning fork of frequency 500 Hz held above a vertical tube partly filled with water. The water level is lowered. The first resonance is heard when the air column has length 0.158 m and the second when the length is 0.498 m.

tuning fork0.158 mfirst resonancetuning fork0.498 msecond resonance
Diagram not to scale
(a)

Determine the wavelength of the sound in the tube.

(2)
(b)

Calculate the speed of sound in air.

(2)
(c)

Determine the end correction of the tube.

(2)
(d)

Explain why using the difference between the two lengths gives a better estimate of the wavelength than using the first length alone.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 λ/2 = 0.498 − 0.158 = 0.340 m✓ 1
A1 λ = 0.680 m✓ 1
Part (b)
M1 v = fλ = 500 × 0.680✓ 1
A1 v = 340 m s-1✓ 1
Part (c)
M1 first resonance L + e = λ/4 = 0.170 m✓ 1
A1 e = 0.170 − 0.158 = 0.012 m✓ 1
Part (d)
The air column resonates at a length slightly longer than the tube length because the antinode lies just beyond the open end✓ 1
This end correction is the same for both resonances, so it cancels when the lengths are subtracted✓ 1

Exam tip: If the two resonance lengths are known, use their difference. The end correction cancels and no extra assumption is needed.

76IBP3H-C4-20r26-16Standing waves in pipes Paper 2Medium9 marks

A sports-arena air horn contains an open pipe of length 0.42 m, open at both ends. The speed of sound in air at temperature θ (in °C) is v = 331 + 0.6θ m s-1. Ignore end corrections and any change in length of the pipe with temperature.

(a)

Calculate the fundamental frequency of the pipe at 20 °C.

(2)
(b)

A second horn uses a pipe closed at one end and has the same fundamental frequency at 20 °C. Determine its length.

(2)
(c)

State the frequency of the second lowest resonance of each pipe at 20 °C. Use the unrounded fundamental frequency.

(2)
(d)

Calculate the fundamental frequency of the open pipe at −5 °C, in a cold stadium. Give your answer to three significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 v = 331 + 0.6 × 20 = 343 m s-1 and f = v/2L✓ 1
A1 f = 343/0.84 = 408 Hz (408.3 Hz)✓ 1
Part (b)
M1 f = v/4L for the closed pipe✓ 1
A1 L = 343/(4 × 408.3) = 0.21 m✓ 1
Part (c)
Open pipe 2f = 817 Hz✓ 1
Closed pipe 3f = 1225 Hz✓ 1
Part (d)
M1 v = 331 − 3 = 328 m s-1✓ 1
M1 f = 328/(2 × 0.42)✓ 1
A1 f = 390 Hz✓ 1

Common error: The length of the pipe is the same, so a change in the fundamental frequency comes only from the change in the speed of sound.

77C-2-03Standing waves in pipes Paper 2Hard9 marks

A simple model of the human vocal tract is a pipe of length 17.0 cm, closed at the vocal cords and open at the lips. Sound from the vibrating vocal cords is strongest at frequencies close to the resonant frequencies of the air in the tract.

The speed of sound in a gas at kelvin temperature T is v = √(γRT/M), where M is the molar mass and γ is a constant for the gas. For air, γ = 1.40 and M = 29.0 g mol−1. At 20 °C the speed of sound in air is 343 m s−1.

(a)
(i)

Calculate the speed of sound in the air in the vocal tract at 37 °C.

(1)
(ii)

Show that the lowest resonant frequency of the tract is about 520 Hz.

(2)
(iii)

State the frequencies of the next two resonances.

(1)
(b)

A diver breathes heliox, a mixture of helium and oxygen for which γ = 1.59 and M = 9.6 g mol−1, at the same temperature.

(i)

Deduce, without numerical substitution, an expression for the ratio of a resonant frequency of the tract in heliox to the same resonance in air.

(2)
(ii)

Calculate the lowest resonant frequency of the tract in heliox. Hence explain why the voice sounds higher although the vocal cords vibrate at the same rate.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
v = 343 × √(310/293) = 353 m s−1✓ 1Use of Celsius temperatures: [0].
Part (a)(ii)
Closed–open pipe: a quarter wavelength fits in the tract, λ = 4L = 0.680 m✓ 1
f = 352.8/0.680 = 519 Hz✓ 1Answer to at least 3 s.f. required. Allow ECF from (a)(i).
Part (a)(iii)
Only odd harmonics: 1557 Hz and 2594 Hz✓ 1Accept 1560 Hz and 2600 Hz. Allow ECF from (a)(ii).
Part (b)(i)
The length and the boundary conditions are unchanged, so each resonant wavelength is unchanged and f ∝ v✓ 1
At the same T, v ∝ √(γ/M), so fheliox/fair = √(γhMair/γairMh)✓ 1Equivalent forms accepted.
Part (b)(ii)
Ratio = √(1.59 × 29.0/(1.40 × 9.6)) = 1.85, so f = 961 Hz✓ 1Accept 955–965 Hz. Allow ECF from (a)(ii) and (b)(i).
The frequency of the sound produced by the vocal cords (the source) depends on the cords, not on the gas, so the basic note is unchanged✓ 1
The tract resonances, which decide which frequency components are made loud, move up by a factor of about 1.85, so the higher-frequency components are emphasised✓ 1Dependent on the idea that the source frequency is unchanged.

Answers: (a)(i) 353 m s−1  ·  (a)(ii) 519 Hz  ·  (a)(iii) 1557 Hz, 2594 Hz  ·  (b)(ii) 961 Hz (the remaining parts are explanations — see the table above)

78C-2-42Forced oscillations and resonance Paper 2Hard18 marks

A wind turbine has three blades, each 40 m long, mounted on a tall tower. As each blade passes in front of the tower, the tower receives a small sideways push. A slight imbalance of the rotor also pushes the tower sideways once per revolution.

The graph shows the amplitude A of the sideways vibration of the top of the tower as the rotation rate of the rotor is slowly increased.

2468101214161820rotor rate / revolutions per minute04812162024A / mm
Amplitude of vibration of the top of the tower against the rotation rate of the rotor (drawn to scale).
(a)
(i)

Explain why the graph has two peaks.

(2)
(ii)

Deduce the natural frequency of the sideways vibration of the tower.

(1)
(b)
(i)

The top of the tower behaves as a mass–spring system with an effective mass of 2.0 × 105 kg. Determine its effective spring constant.

(2)
(ii)

Both peaks occur when the tower is driven at the same frequency. Suggest why the peak at 6 revolutions per minute is much higher than the peak at 18 revolutions per minute.

(2)
(c)

In normal operation the rotor turns at 15 revolutions per minute.

(i)

Calculate the angular velocity of the rotor. Give an appropriate unit.

(1)
(ii)

Calculate the speed of the tip of a blade.

(1)
(iii)

The tip of each blade emits a whistling sound of frequency 600 Hz. A distant observer stands in the plane of the rotor, so that the tip of a blade at the top or bottom of its circle moves directly towards or away from the observer. The speed of sound is 340 m s−1. Determine the highest frequency heard by the observer.

(2)
(iv)

Determine the time between successive instants at which the observer hears this highest frequency.

(2)
(d)

The moment of inertia of the rotor about its axis is 1.2 × 107 kg m2.

(i)

Calculate the rotational kinetic energy of the rotor in normal operation.

(2)
(ii)

The wind stops and the rotor slows down under a frictional torque of about 1.5 × 105 N m. Estimate the time it takes to stop, and state one assumption you make.

(2)
(iii)

Suggest why the tower may vibrate strongly near the end of this spin-down.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
The tower is driven at two frequencies: three pushes per revolution from the blades passing, and one push per revolution from the imbalance✓ 1The hidden modelling step.
Each driving frequency produces resonance when it equals the natural frequency of the tower: at 6 revolutions per minute for the blade pushes and at 18 revolutions per minute for the imbalance✓ 1
Part (a)(ii)
18 revolutions per minute = 0.30 revolutions per second, or 3 × 6.0/60 = 0.30, so f0 = 0.30 Hz✓ 1Allow ECF from (a)(i). 0.10 Hz (3-blade factor missed): [0].
Part (b)(i)
k = m(2πf0)² = 2.0 × 105 × (2π × 0.30)²✓ 1Allow ECF from (a)(ii).
k = 7.1 × 105 N m−1✓ 1
Part (b)(ii)
Both peaks are resonances at the same driving frequency (0.30 Hz) with the same damping, so the amplitude at each peak depends on the size of the periodic driving force✓ 1Allow ECF from (a).
So the sideways push from the blades passing the tower is larger (about 3 times) than the push from the imbalance of the rotor✓ 1Do not accept "more pushes per revolution" alone: the driving frequency is the same at both peaks.
Part (c)(i)
ω = 15 × 2π/60 = 1.57 rad s−1✓ 1Unit rad s−1 required for this mark.
Part (c)(ii)
u = ωr = 1.57 × 40 = 62.8 m s−1✓ 1Allow ECF from (c)(i).
Part (c)(iii)
The highest frequency is heard from sound emitted when a tip moves directly towards the observer, as a moving source: f′ = fv/(v − u)✓ 1
f′ = 600 × 340/(340 − 62.8) = 736 Hz✓ 1Accept 735–737 Hz. Allow ECF from (c)(ii).
Part (c)(iv)
One revolution takes 60/15 = 4.0 s, and each tip moves directly towards the observer once per revolution✓ 1
There are three blades, so the highest frequency is heard every 4.0/3 = 1.3 s✓ 1Accept 1.33 s. 4.0 s (one blade only): [1 max].
Part (d)(i)
Ek = ½Iω² = ½ × 1.2 × 107 × 1.57²✓ 1Allow ECF from (c)(i).
Ek = 1.5 × 107 J✓ 1
Part (d)(ii)
α = τ/I = 1.5 × 105/1.2 × 107 = 0.0125 rad s−2, so t = ω/α = 126 s✓ 1Allow ECF from (c)(i). Energy route (Ek = τ × angle) also accepted.
Assumption: the frictional torque stays constant (and the wind exerts no torque)✓ 1
Part (d)(iii)
As the rotor slows it passes through 6 revolutions per minute, where the blade pushes drive the tower at its natural frequency (resonance)✓ 1Allow ECF from (a).

Answers: (a)(ii) 0.30 Hz  ·  (b)(i) 7.1 × 105 N m−1  ·  (c)(i) 1.57 rad s−1  ·  (c)(ii) 62.8 m s−1  ·  (c)(iii) 736 Hz  ·  (c)(iv) 1.3 s  ·  (d)(i) 1.5 × 107 J  ·  (d)(ii) 126 s (the remaining parts are explanations — see the table above)

79IBP2H-025-00-14Forced oscillations and resonance Paper 2Hard7 marks

A tall office tower sways in its fundamental mode with a natural period of 6.0 s. After a single strong gust, the amplitude of sway at the top decreases from 0.40 m to 0.10 m over 10 complete oscillations. The damping is light, so the period can be taken as constant.

(a)

Calculate the natural frequency of the tower.

(1)
(b)

Determine the fraction of the initial oscillation energy that remains after the 10 oscillations.

(2)
(c)

On some days, eddies shed by the wind push the tower periodically at a frequency close to 0.17 Hz. Explain why fitting a system that increases the damping reduces the danger to the tower.

(2)
(d)

Assume the amplitude decreases by the same fraction in every oscillation. Estimate the percentage of the oscillation energy lost in each cycle.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
f = 1/6.0 = 0.17 Hz✓ 1
Part (b)
Energy is proportional to (amplitude)2✓ 1
Fraction = (0.10/0.40)2 = 0.063 (accept 1/16)✓ 1
Part (c)
The driving frequency is close to the natural frequency, so resonance occurs and the amplitude builds up✓ 1
The steady amplitude is reached when the energy lost per cycle equals the energy supplied per cycle. Greater damping makes this balance occur at a smaller amplitude✓ 1
Part (d)
Either amplitude ratio per cycle = (0.25)1/10 = 0.871, so energy ratio per cycle = 0.8712 = 0.758; or energy ratio per cycle = (0.0625)1/10 = 0.758✓ 1
Percentage lost per cycle ≈ 24% (allow error carried forward from the energy fraction found earlier)✓ 1

Common error: The energy lost per cycle is not the total fractional loss divided by 10. Repeated losses multiply, so take the tenth root of the overall ratio.

80IBP2H-025-00-16Wavelength of a standing wave Paper 2Hard9 marks

A concert harp string has a vibrating length of 0.65 m between its fixed ends. Its fundamental frequency is 196 Hz. The speed of sound in air is 343 m s−1.

(a)

Calculate the speed of transverse waves on the string.

(2)
(b)

The harpist lightly touches the string at a point one-third of its length from one end, then plucks it. The touch forces a node at that point. Explain why the lowest frequency that can now be sustained is three times the fundamental, and determine this frequency.

(3)
(c)

When the string is plucked exactly at its midpoint without being touched, the second harmonic is absent from the sound. Explain this observation.

(2)
(d)

A decorative air-filled tube, closed at both ends, sits near the harp. Determine the shortest length of such a tube that would resonate with the 588 Hz sound produced when the string is touched at one-third of its length.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
λ1 = 2L = 2 × 0.65 = 1.30 m✓ 1
v = fλ = 196 × 1.30 = 255 m s−1✓ 1
Part (b)
The standing wave must have nodes at both fixed ends and at L/3, so the segment of length L/3 must contain a whole number of half-wavelengths✓ 1
The longest wavelength that fits is λ = 2L/3. This gives three loops on the string, which is the third harmonic✓ 1
f = 3 × 196 = 588 Hz✓ 1
Part (c)
Plucking displaces the midpoint, so the vibration has a large amplitude there and the midpoint behaves as an antinode, not a node✓ 1
The second harmonic has a node at the midpoint, so it cannot be excited by plucking there✓ 1
Part (d)
The shortest tube resonates in its fundamental, with displacement nodes at both ends, so L = λ/2, where λ = 343/588 = 0.583 m✓ 1
L = 0.29 m (allow error carried forward from the frequency found earlier)✓ 1

Exam tip: A forced node or a pluck point acts as a filter. Only harmonics that have a node at the touched point survive. Harmonics that have a node at the pluck point are not excited.

81IBP2S-024-02-16Phase along a standing wave Paper 2Hard7 marks

A scale-model bridge stay cable is tested in a wind tunnel. The cable vibrates in a standing wave at frequency 50 Hz. Adjacent nodes are 0.40 m apart, and the amplitude at an antinode is 6.0 mm. The amplitude A of a point a distance x from a node is given by A = Amax sin(2πx/λ), where Amax is the antinode amplitude and the angle is in radians.

(a)

The standing wave is formed by two identical travelling waves moving in opposite directions. State the amplitude of each travelling wave.

(1)
(b)

Calculate the speed of the travelling waves on the cable.

(2)
(c)

Point P on the cable is 0.10 m from a node. Determine the amplitude of oscillation of P.

(2)
(d)

Calculate the maximum speed of P.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
3.0 mm, because the two waves add in phase at an antinode to give 6.0 mm✓ 1
Part (b)
λ = 2 × 0.40 = 0.80 m✓ 1
v = fλ = 50 × 0.80 = 40 m s−1✓ 1
Part (c)
Uses λ from (b) to obtain 2πx/λ = 2π × 0.10/0.80 = π/4 rad. Equivalent working such as x = λ/8 or 45° earns this mark. Allow ECF from (b).✓ 1
A = 6.0 sin(π/4) = 4.2 mm (4.24 mm)✓ 1
Part (d)
ω = 2πf = 2π × 50 = 314 rad s−1✓ 1
vmax = ωA = 314 × 4.24×10−3 = 1.3 m s−1 (1.33 m s−1). Allow ECF from (c).✓ 1

Common error: Do not confuse the wave speed along the cable (40 m s⁻¹) with the oscillation speed of a point on the cable. The maximum speed of P comes from ωA, not from fλ.

82IBP2H-025-01-12Standing waves in pipes Paper 2Hard8 marks

In speech science, the vocal tract is modelled as a tube 17.0 cm long. It is closed at the vocal folds and open at the lips. The resonances of this tube, called formants, strengthen certain frequencies in the sound produced by the vocal folds. The speed of sound in warm, humid exhaled air is 350 m s−1.

(a)

Explain why only odd harmonics form in this tube.

(2)
(b)

Calculate the three lowest resonant frequencies of the tube.

(2)
(c)

A diver breathes heliox, a helium–oxygen mixture in which the speed of sound is 610 m s−1. Calculate the new lowest resonant frequency of the vocal tract.

(2)
(d)

The vibration frequency of the vocal folds does not change when heliox is breathed. Explain why the diver's voice nevertheless sounds different.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
There is a displacement node at the closed end and a displacement antinode at the open end ; the tube length must therefore be an odd number of quarter-wavelengths, L = (2n − 1)λ/4, so only odd multiples of f1 are possible .✓ 2
Part (b)
f1 = 350/(4 × 0.170) = 515 Hz ; 3f1 ≈ 1540 Hz and 5f1 ≈ 2570 Hz . Accept 1545 Hz and 2575 Hz from the rounded value of f1.✓ 2
Part (c)
Uses f1 = v/4L = 610/(4 × 0.170), or scales the answer to (b) by 610/350 ; f1 ≈ 897 Hz .✓ 2
Part (d)
The basic pitch is set by the vocal-fold frequency, so it stays the same ; the resonant frequencies of the tract rise, so different, higher harmonics of the vocal-fold sound are strengthened, which changes the quality (timbre) of the voice .✓ 2

Exam tip: Separate the source frequency (set by the vibrating object) from the resonator frequencies (set by the tube length and the wave speed). A change of gas affects only the resonator.

83IBP2H-025-01-13Phase along a standing wave Paper 2Hard8 marks

A string 0.90 m long is fixed at both ends and vibrates in its third harmonic at 240 Hz. The amplitude at each antinode is 4.0 mm. Points P, Q and R on the string are 0.10 m, 0.25 m and 0.50 m from the same end. The amplitude at a point a distance x from a node varies as Amax|sin(2πx/λ)|.

PQR0.90 m
Diagram not to scale
(a)

Calculate the speed of waves on the string.

(2)
(b)

State the phase difference between the oscillations of P and Q, and between the oscillations of P and R.

(2)
(c)

Determine the amplitude of oscillation of P.

(2)
(d)

P moves in simple harmonic motion. Determine its maximum speed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
λ = 2L/3 = 0.60 m ; v = fλ = 240 × 0.60 = 144 m s−1 .✓ 2
Part (b)
The nodes are at 0, 0.30, 0.60 and 0.90 m. P and Q lie in the same loop, so their phase difference is 0 . R lies in the adjacent loop, so the phase difference between P and R is π rad (180°) .✓ 2
Part (c)
AP = 4.0 sin(2π × 0.10/0.60) = 4.0 sin(60°) ; AP = 3.46 mm ≈ 3.5 mm . Allow ECF from (a) for λ.✓ 2
Part (d)
ω = 2π × 240 = 1.51×103 rad s−1 ; vmax = ωAP = 1.51×103 × 3.46×10−3 = 5.2 m s−1 . Accept 5.3 m s−1 from using AP = 3.5 mm. Allow ECF from (c).✓ 2

Common error: The wave speed (144 m s⁻¹) is not the speed of the particles on the string. Find a particle's speed from SHM, using the local amplitude at that point.

84IBP2H-025-01-15Standing waves on strings Paper 2Hard8 marks

Bridge engineers check the tension in a stay cable by measuring how it vibrates. The cable is 50.0 m long between fixed anchorages and has a mass per unit length of 60 kg m−1. Its fundamental frequency is measured as 2.20 Hz. The speed v of transverse waves on the cable is given by v = √(T/μ), where T is the tension and μ is the mass per unit length.

(a)

Show that the wave speed on the cable is about 220 m s−1.

(2)
(b)

Calculate the tension in the cable.

(2)
(c)

A vibration sensor fixed at the midpoint of the cable never records the second harmonic. Explain why.

(2)
(d)

Years later the fundamental frequency is measured as 2.09 Hz. The length and mass per unit length are unchanged. Determine the percentage reduction in tension.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
λ1 = 2L = 100 m ; v = fλ = 2.20 × 100 = 220 m s−1 .✓ 2
Part (b)
T = μv2 = 60 × 2202 ; = 2.9×106 N . ECF from (a).✓ 2
Part (c)
The midpoint is a node of the second harmonic ; the displacement there is always zero, so the sensor detects no oscillation at that frequency .✓ 2
Part (d)
f ∝ v ∝ √T, so T ∝ f2, giving Tnew/Told = (2.09/2.20)2 = 0.9025 ; reduction = 9.8% .✓ 2

Common error: A 5% fall in frequency is not a 5% fall in tension. Because T ∝ f², the tension falls by almost 10%.

85IBP3H-C4-20r16-15Forced oscillations and resonance Paper 2Hard9 marks

Bees in a hive beat their wings at about 230 Hz. A wooden entrance tube of length 0.370 m is closed at one end and open at the other. Take the speed of sound as 340 m s−1 and neglect end corrections.

(a)

Calculate the fundamental frequency of the tube.

(2)
(b)

Determine the highest resonant frequency of the tube below 800 Hz. Give your answer to three significant figures.

(2)
(c)

Explain why the amplitude of the air vibration in the tube becomes large when the bees' wing frequency is close to 230 Hz.

(2)
(d)

The tube walls are lined with soft fibre which increases the damping. State and explain how the response curve of amplitude against driving frequency changes.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(f=v/4L\)✓ 1
A1 230 Hz (229.7 Hz)✓ 1
Part (b)
M1 odd multiples of the fundamental: 3 × 229.7✓ 1
A1 689 Hz✓ 1
Part (c)
The wings drive the air at a frequency equal to the natural frequency of the air column✓ 1
Energy is transferred efficiently on each cycle so the amplitude builds up (resonance)✓ 1
Part (d)
Peak amplitude decreases✓ 1
Peak becomes broader, so a wider range of frequencies produces a response✓ 1
More energy is dissipated each cycle by the fibre✓ 1

Exam tip: For a resonance peak, comment on both height and width when damping changes.

86IBP3H-C4-20r26-14Standing waves on strings Paper 2Hard9 marks

In a sports-science project, an archery bowstring of length 1.40 m, fixed at both ends, is plucked. It vibrates in its second harmonic with a frequency of 100 Hz. The mass per unit length of the string is 4.0 × 10-3 kg m-1.

(a)

Calculate the speed of transverse waves on the string.

(2)
(b)

Determine the tension in the string.

(2)
(c)

The tension is adjusted so that the fundamental frequency becomes 60 Hz. Calculate the new tension. Give your answer to three significant figures.

(3)
(d)

The string is plucked at its midpoint. Explain why the second harmonic is not excited.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 second harmonic has λ = L = 1.40 m✓ 1
A1 v = 100 × 1.40 = 140 m s-1✓ 1
Part (b)
M1 T = μv2✓ 1
A1 T = 4.0 × 10-3 × 1402 = 78.4 N✓ 1
Part (c)
The original fundamental frequency is 50 Hz✓ 1
M1 f ∝ √T, so T′ = T(60/50)2✓ 1
A1 T′ = 78.4 × 1.44 = 113 N (112.9 N)✓ 1
Part (d)
Plucking at the midpoint produces an antinode (maximum displacement) at the centre✓ 1
The second harmonic has a node at the centre, so it cannot be excited✓ 1

Common error: Tension depends on v², so frequency is proportional to √T. Doubling the frequency needs four times the tension.

87IBP3H-C4-20r26-15Forced oscillations and resonance Paper 2Hard10 marks

A railway carriage on its suspension has a natural vertical frequency of 1.25 Hz. The rails are joined with a regular gap every 18.0 m, and each time a wheel crosses a gap the carriage receives a small impulse.

(a)

Explain why the carriage vibrates with large amplitude at certain speeds.

(2)
(b)

Calculate the speed of the train, in m s-1 and in km h-1, at which resonance occurs.

(3)
(c)

Describe the effect on the resonance of adding dampers to the suspension.

(2)
(d)

Describe a sketch of the graph of amplitude against driving frequency for a lightly damped and a heavily damped suspension.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
The impulses from the rail joints provide a periodic driving force with a frequency that depends on the speed✓ 1
When the driving frequency equals the natural frequency there is resonance, with maximum energy transfer and large amplitude✓ 1
Part (b)
M1 driving frequency = v/18.0 = 1.25 Hz✓ 1
A1 v = 22.5 m s-1✓ 1
A1 22.5 × 3.6 = 81 km h-1✓ 1
Part (c)
The amplitude at resonance decreases✓ 1
The resonance peak becomes broader and, for heavy damping, shifts to a slightly lower frequency✓ 1
Part (d)
Both curves have a peak near the natural frequency (1.25 Hz)✓ 1
The lightly damped curve has a taller, narrower peak✓ 1
The heavily damped curve has a lower, broader peak and lies below the first curve near resonance✓ 1

Exam tip: For resonance, equate the driving frequency to the natural frequency. For a train, the driving frequency is speed divided by the spacing of the joints.