C.4 Standing Waves and Resonance Questions

1C-1A-08

Standing waves in pipes·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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 point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingC.4 — the nature and formation of standing waves in pipes; the boundary conditions for pipes open at one end and open at both ends; harmonics Command term: Determine

2C-1A-16

Standing waves on strings·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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

What is L?

Show mark scheme

Marking point	Mark	Notes
Step 1For a string fixed at both ends f_n = 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 2f_B − f_A = v/(2L).	—
Step 3L = v/(2(f_B − f_A)).	✓ 1	Answer 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 f_A is the first harmonic. Nothing in the data says so — only the difference between consecutive resonances gives v/2L.

Syllabus understandingC.4 — standing wave patterns in strings; the determination of the wavelength and the frequency of the nth harmonic given the length of the string and the speed of the wave Command term: Determine

3C-1A-19

Standing waves in pipes·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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?

Displacement nodes (N) and antinodes (A) of the standing wave in a pipe open at both ends.

Show mark scheme

Marking point	Mark	Notes
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 λ_n = 2L/n, so this is the third harmonic and its frequency is 3f₁.	—
Step 3f₁ = 850/3 = 283 Hz.	✓ 1	Answer 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.

Syllabus understandingC.4 — standing wave patterns in pipes; boundary conditions for air in pipes with two open ends; vibration modes in terms of displacement nodes and antinodes; the frequency of the nth harmonic Command term: Deduce

4C-1A-21

Damping and resonance·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

5C-1A-25

Forced oscillations and resonance·C.4 Standing waves and resonance

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

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 point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingC.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; the effects of light, critical and heavy damping on the system Command term: Identify

6C-1A-27

Wavelength of a standing wave·C.4 Standing waves and resonance

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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?

The string vibrates in the mode shown; N = node, A = antinode.

Show mark scheme

Marking point	Mark	Notes
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 f_n = nf₁.	—
Step 3f₁ = 150/3 = 50 Hz.	✓ 1	Answer 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.

Syllabus understandingC.4 — standing wave patterns in strings; nodes and antinodes; the frequency of the nth harmonic Command term: Deduce

7C-1A-30

Phase along a standing wave·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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?

P and QP and R

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingC.4 — nodes and antinodes, relative amplitude and phase difference of points along a standing wave; standing wave patterns on strings Command term: Deduce

8C-1A-51

Strings with free ends·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingC.4 — standing wave patterns in strings; boundary conditions for strings with one fixed and one free boundary; the frequency of the nth harmonic given the length of the string and the speed of the wave Command term: Determine

9C-1A-52

Pipes closed at both ends·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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 point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingC.4 — boundary conditions for air in pipes with two closed ends, one closed and one open end, and two open ends; the frequency of the nth harmonic Command term: Deduce

10C-1A-53

Phase along a standing wave·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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 point	Mark	Notes
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 v_max = 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.	✓ 1	Answer 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: v_max ∝ amplitude, and adjacent loops are π out of phase.

Syllabus understandingC.4 — nodes and antinodes, relative amplitude and phase difference of points along a standing wave; C.1 (HL) — v = ωx₀ cos(ωt + ϕ) Command term: Deduce

11C-1A-54

Formation of standing waves·C.4 Standing waves and resonance

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

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.

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

12C-1A-55

Strings with free ends·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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 point	Mark	Notes
Step 1Free ends are antinodes. The simplest pattern on P is A–N–A: half a wavelength on the cord, so λ_P = 2L and f_P = 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: λ_Q = 4L, and the wave speed is 2v, so f_Q = 2v/(4L) = v/(2L).	—
Step 3The ratio is f_P/f_Q = 1: the longer wavelength on Q is exactly compensated by its greater wave speed.	✓ 1	Answer 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: f_P = v/(2L) and f_Q = 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.

Syllabus understandingC.4 — standing waves patterns in strings and pipes (Guidance: boundary conditions for strings include two fixed boundaries, one fixed and one free boundary, and two free boundaries; a determination of the wavelength and the frequency of the nth harmonic given the length of the string or pipe and the speed of the wave is required) Command term: Determine

13C-1A-56

Forced oscillations and resonance·C.4 Standing waves and resonance

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksState

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 point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingC.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency Command term: State

14C-1A-68

Formation of standing waves·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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 point	Mark	Notes
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.	✓ 1	Answer 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.

15C-1A-111

Formation of standing waves·C.4 Standing waves and resonance

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Two waves X and Y of the same amplitude, frequency and wavelength travel in opposite directions along a long string. The graph shows the displacement y of each wave against position x at time t = 0, when the two waves coincide exactly.

The period of the waves is T. What is the resultant displacement of the string at t = T/4?

Waves X and Y at t = 0 (drawn to scale). The two curves coincide.

Show mark scheme

Marking point	Mark	Notes
Step 1In a quarter of a period each wave moves a quarter of a wavelength: X moves λ/4 to the right and Y moves λ/4 to the left.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The two waves are now displaced by λ/2 relative to each other, so at every point a crest of one meets a trough of the other.	—
Step 3The waves have equal amplitudes, so they cancel everywhere: the string is momentarily straight (every point is passing through its equilibrium position).	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: a relative shift of λ/2 puts the waves in antiphase at every point, so the resultant is zero everywhere at T/4.
BThis would be true only if the two waves moved together; travelling in opposite directions, they separate by λ/2 in T/4. The string next has this shape after a whole period T.
CThis is the shape at t = T/2, when each wave has moved λ/2 and the relative shift is a whole wavelength.
DThis treats the resultant as a travelling wave that moves with X. The resultant is a standing wave: its pattern does not move along the string.

Syllabus understandingC.4 — the nature and formation of standing waves in terms of superposition of two identical waves travelling in opposite directions; nodes and antinodes, relative amplitude and phase difference of points along a standing wave Command term: Deduce

16C-1A-112

Standing and travelling waves compared·C.4 Standing waves and resonance

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Which statements are true both for a standing wave on a string and for a travelling wave on a string?

I. All the points of the string that oscillate do so with the same frequency.

II. All the points of the string oscillate with the same amplitude.

III. Two points one wavelength apart oscillate in phase.

Show mark scheme

Marking point	Mark	Notes
Step 1Every point of either wave oscillates at the frequency of the source, so I is true for both.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2In a standing wave the amplitude varies from zero at a node to a maximum at an antinode, so II is true only for the travelling wave.	—
Step 3In a standing wave points one wavelength apart are separated by two nodes, so they are in the same phase of the pattern (and have the same amplitude); in a travelling wave they are 2π apart, i.e. in phase. III is true for both.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis rejects III, as if a point one wavelength away always lay in the adjacent loop of the standing wave. One wavelength spans two loops, so the phase reverses twice.
BCorrect: same frequency everywhere and points one wavelength apart in phase, for both waves; equal amplitudes only for the travelling wave.
CThis treats the standing wave as having the same amplitude everywhere, like a travelling wave, and also rejects III.
DII is false for the standing wave: the amplitude is zero at the nodes and largest at the antinodes.

Syllabus understandingC.4 — nodes and antinodes, relative amplitude and phase difference of points along a standing wave; the nature and formation of standing waves in terms of superposition of two identical waves travelling in opposite directions; C.2 — transverse travelling waves Command term: Deduce

17C-1A-113

Amplitude at an antinode·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A wave of amplitude A and frequency f travels along a string. When the same wave travels along the string in the opposite direction at the same time, a standing wave forms.

What is
(maximum speed of a point at an antinode of the standing wave) / (maximum speed of a point on the string when only one wave is present)?

Show mark scheme

Marking point	Mark	Notes
Step 1At an antinode the two waves always arrive in phase, so the amplitude there is 2A.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Every point oscillates with simple harmonic motion at the same angular frequency ω = 2πf, and the maximum speed is ωx₀.	—
Step 3The ratio is ω(2A)/(ωA) = 2.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis assumes the maximum speed depends only on the frequency, which is unchanged; it also depends on the amplitude, which doubles at an antinode.
BThis adds the energies of the two waves, giving an amplitude of √2A. The displacements, not the energies, superpose: the amplitude at an antinode is 2A.
CCorrect: the amplitude doubles at the antinode and ω is unchanged, so v_max = ωx₀ doubles.
DThis squares the amplitude ratio, as for the energy of oscillation; the maximum speed is proportional to the amplitude itself.

18C-1A-114

Identifying a pipe from its resonances·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A loudspeaker drives the air in a pipe. Its frequency is increased slowly from zero and the frequencies of the successive resonances of the air are recorded. The graph shows the frequency f of the k-th resonance against k, with a straight line through the points. The speed of sound is 340 m s⁻¹.

Which row describes the pipe?

Frequency f of the k-th resonance against k (drawn to scale).

PipeLength of pipe

Show mark scheme

Marking point	Mark	Notes
Step 1The line does not pass through the origin: the resonances are 170, 510, 850, 1190 Hz, in the ratio 1 : 3 : 5 : 7. Only odd harmonics occur, so the pipe is closed at one end and open at the other.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2For this pipe the first harmonic is v/4L, so L = 340/(4 × 170) = 0.50 m.	—
Step 3Check: successive resonances are 2 × 170 = 340 Hz apart, the gradient of the graph, which equals v/2L = 340/(2 × 0.50) = 340 Hz.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis uses only the spacing of the resonances, 340 Hz = v/2L, and ignores the intercept: for a pipe open at both ends the line would pass through the origin.
BCorrect: odd harmonics only (the intercept is −170 Hz, not zero), and v/4L = 170 Hz gives 0.50 m.
CThis identifies the pipe correctly but takes the spacing of the resonances, 340 Hz, as the first-harmonic frequency v/4L.
DThis takes the lowest resonance, 170 Hz, as v/2L, the first harmonic of a pipe open at both ends; that pipe would then have its next resonance at 340 Hz, not 510 Hz.

Syllabus understandingC.4 — standing waves patterns in strings and pipes; Guidance: boundary conditions for air in pipes include two closed ends, one closed and one open end, and two open ends; Guidance: a determination of the wavelength and the frequency of the nth harmonic given the length of the string or pipe and the speed of the wave is required Command term: Deduce

19C-1A-115

Changing the length of a string·C.4 Standing waves and resonance

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksCalculate

A guitar string, fixed at both ends, has a first-harmonic frequency of 240 Hz. A guitarist presses the string against a fret so that the vibrating length of the string becomes three-quarters of its original value. The tension in the string does not change.

What is the new first-harmonic frequency?

Show mark scheme

Marking point	Mark	Notes
Step 1The tension and the mass per unit length are unchanged, so the wave speed v is unchanged.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2For the first harmonic λ = 2L, so f₁ = v/2L ∝ 1/L.	—
Step 3New frequency = 240 × 4/3 = 320 Hz.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis uses f ∝ L²: 240 × (3/4)² = 135 Hz.
BThis uses f ∝ L: 240 × 3/4 = 180 Hz. A shorter string has a shorter wavelength and so a higher frequency.
CThis uses f ∝ 1/√L, confusing the dependence on the length with the dependence on the tension: 240 × √(4/3) = 277 Hz.
DCorrect: f ∝ 1/L at constant wave speed, so the frequency rises by a factor 4/3 to 320 Hz.

Syllabus understandingC.4 — standing waves patterns in strings and pipes; Guidance: a determination of the wavelength and the frequency of the nth harmonic given the length of the string or pipe and the speed of the wave is required; C.2 — v = fλ Command term: Calculate

20C-1A-116

Effect of wave speed on the mode·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A string fixed at both ends is driven at a constant frequency, and a standing wave with six loops forms on it. The speed of transverse waves on the string is v = √(T/μ), where T is the tension and μ is the mass per unit length.

The tension is increased by a factor of 2.25. The frequency is not changed. How many loops does the standing wave on the string now have?

Show mark scheme

Marking point	Mark	Notes
Step 1The wave speed increases by a factor √2.25 = 1.5; at the same frequency the wavelength λ = v/f also increases by a factor 1.5.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The number of loops is L/(λ/2) = 2L/λ, which is inversely proportional to λ.	—
Step 3New number of loops = 6/1.5 = 4, a whole number, so a standing wave forms.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the wavelength is 1.5 times longer, so 6/1.5 = 4 loops fit on the string.
BThis assumes the pattern is fixed by the frequency alone. The number of loops depends on the wavelength, which changes when the wave speed changes.
CThis multiplies by 1.5 instead of dividing: a faster wave at the same frequency has a longer, not a shorter, wavelength.
DThis takes v ∝ T, so that the wavelength increases by 2.25 and 6/2.25 = 2.67 loops would be needed. The speed depends on √T.

21C-1A-117

Harmonics with a forced node·C.4 Standing waves and resonance

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A string fixed at both ends has a first-harmonic frequency of 110 Hz. A player touches the string lightly at a point one-third of the length of the string from one end. This forces a node at that point, but both sections of the string continue to vibrate.

What is the lowest frequency of the standing waves that can now form on the string?

Show mark scheme

Marking point	Mark	Notes
Step 1The whole string still vibrates with nodes at both ends, so only its harmonics, n × 110 Hz with λ_n = 2L/n, can occur.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2There must also be a node at L/3, so a whole number of half-wavelengths must fit into L/3. Since λ/2 = L/n, the number of half-wavelengths in L/3 is n/3, a whole number only if n is a multiple of 3.	—
Step 3The lowest allowed harmonic is the third: 3 × 110 = 330 Hz.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis assumes that touching the string lightly does not change the lowest mode. The first harmonic has an antinode, not a node, at L/3.
BThis lets the longer section (2L/3) vibrate in its own first harmonic, 110 × 3/2 = 165 Hz; the shorter section would then need L/3 to be a whole number of half-wavelengths of 4L/3, which is impossible.
CThis takes the next harmonic, 220 Hz; the second harmonic has its middle node at L/2, not at L/3.
DCorrect: a node at L/3 needs λ/2 = L/3, L/6, …, so the lowest frequency is the third harmonic, 330 Hz.

Syllabus understandingC.4 — nodes and antinodes, relative amplitude and phase difference of points along a standing wave; standing waves patterns in strings and pipes; Guidance: a determination of the wavelength and the frequency of the nth harmonic given the length of the string or pipe and the speed of the wave is required Command term: Determine

22C-1A-118

Critical and heavy damping·C.4 Standing waves and resonance

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

The needle of an analogue meter is critically damped. When the current is switched off, the needle returns to zero. The damping of the needle is then increased so that it is heavily damped.

Which row describes the motion of the heavily damped needle when the current is switched off, compared with the critically damped needle?

Time to return to zeroNeedle overshoots zero

Show mark scheme

Marking point	Mark	Notes
Step 1Critical damping returns a displaced system to equilibrium in the shortest possible time without oscillating.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2With heavy damping the resistive force is larger, so the needle creeps back more slowly: it takes longer to return to zero.	—
Step 3A heavily damped system does not oscillate, so the needle does not overshoot zero.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis assumes that more damping always gives a quicker return. Beyond critical damping, extra damping slows the return.
BCorrect: heavy damping gives a slower return with no overshoot.
CThis describes reducing the damping below critical: a lightly damped needle first reaches zero sooner and then overshoots.
DThis assumes that heavy damping makes the needle oscillate slowly about zero; a heavily damped system returns without oscillating.

Syllabus understandingC.4 — the effects of light, critical and heavy damping on the system Command term: Identify

23C-1A-119

Strings with two free ends·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Both ends of an elastic cord of length L are tied to light rings that slide without friction on smooth vertical rods, so that both ends of the cord are free. The speed of transverse waves on the cord is v. A standing wave forms that has n nodes.

What is the frequency of the standing wave?

Show mark scheme

Marking point	Mark	Notes
Step 1A free end is a displacement antinode, so the pattern starts and ends with an antinode: A–N–A–N–…–A.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Adjacent antinodes are half a wavelength apart, and with n nodes there are n + 1 antinodes and n half-wavelengths between the two ends: L = nλ/2.	—
Step 3f = v/λ = nv/(2L).	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis counts the n + 1 antinodes as the number of half-wavelengths; the number of half-wavelengths between the first and last antinode is one fewer.
BThis treats the ends as nodes, as for a string fixed at both ends, so that the n nodes enclose only n − 1 loops.
CThis uses the result for a cord with one fixed and one free end, where only odd numbers of quarter-wavelengths fit.
DCorrect: n nodes between two antinode ends means n half-wavelengths, so λ = 2L/n and f = nv/(2L).

Syllabus understandingC.4 — standing waves patterns in strings and pipes; Guidance: boundary conditions for strings include two fixed boundaries, one fixed and one free boundary, and two free boundaries; Guidance: a determination of the wavelength and the frequency of the nth harmonic given the length of the string or pipe and the speed of the wave is required Command term: Determine

24C-1A-120

Tuning a driven oscillator to resonance·C.4 Standing waves and resonance

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A trolley of mass m is held between two stretched springs on a horizontal track. A motor drives the trolley with a periodic force of frequency f, and the trolley oscillates with a very large amplitude. The frequency of the motor is then halved.

What mass must be added to the trolley so that it again oscillates with a very large amplitude?

Show mark scheme

Marking point	Mark	Notes
Step 1The large amplitude is resonance: the driving frequency f equals the natural frequency (1/2π)√(k/m).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2For resonance at f/2 the natural frequency must halve. The natural frequency is proportional to 1/√(mass), so the total mass must become 4m.	—
Step 3The mass to be added is 4m − m = 3m.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis takes the natural frequency to be proportional to 1/(mass), so that the total mass must double to 2m, an added m.
BThis takes the natural frequency to be proportional to 1/(mass) and then gives the new total mass, 2m, rather than the mass added.
CCorrect: f ∝ 1/√(mass), so the total mass must be 4m, and 3m must be added.
DThis is the new total mass, 4m; the trolley already has mass m, so only 3m must be added.

25C-1A-121

Reducing destructive resonance·C.4 Standing waves and resonance

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

A tall building has a natural frequency of sway of 2.0 Hz. Earthquakes in the region shake the ground most strongly at frequencies close to 1.5 Hz.

Which changes would reduce the amplitude of sway of the building during an earthquake?

I. fitting dampers that dissipate energy as the building sways

II. stiffening the frame, so that the natural frequency rises well above 2.0 Hz

III. adding heavy equipment to the upper floors without changing the stiffness

Show mark scheme

Marking point	Mark	Notes
Step 1Dampers remove energy in each cycle, so the amplitude at and near resonance is smaller (I reduces the sway).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The amplitude is largest when the driving frequency is close to the natural frequency. A stiffer frame raises the natural frequency further from 1.5 Hz (II reduces the sway).	—
Step 3Extra mass at the same stiffness lowers the natural frequency from 2.0 Hz towards 1.5 Hz, closer to resonance (III increases the sway).	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: more damping and a natural frequency further from the driving frequency both reduce the amplitude; added mass moves the building towards resonance.
BThis rejects II, as if a stiffer building simply followed the ground more closely, and accepts III because of the extra inertia, ignoring the change in natural frequency.
CThis rejects I, as if dampers acted only after the shaking stops; damping lowers the amplitude of a driven system too.
DIII is wrong: extra mass lowers the natural frequency (f ∝ 1/√m) towards 1.5 Hz, closer to resonance.

Syllabus understandingC.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; the effect of damping on the maximum amplitude and resonant frequency of oscillation; Guidance: knowledge of the useful and destructive effects of resonance is required Command term: Identify

26C-1A-122

Amplitude and phase along a standing wave·C.4 Standing waves and resonance

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A string fixed at both ends vibrates in its fifth harmonic, as shown. The amplitude at each antinode is 6.0 mm. Point P, in the loop nearest the left-hand end, oscillates with an amplitude of 3.0 mm.

How many points on the string, including P, oscillate with an amplitude of 3.0 mm and in phase with P?

The string vibrating in its fifth harmonic; the solid and dashed curves show the extreme positions.

Show mark scheme

Marking point	Mark	Notes
Step 1Within each loop the amplitude rises smoothly from zero at one node to 6.0 mm at the antinode and falls back to zero, so each loop contains exactly two points with amplitude 3.0 mm: 5 × 2 = 10 such points.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Points in the same loop are in phase; points in adjacent loops are in antiphase. The loops in phase with P are the first, third and fifth.	—
Step 3Points with amplitude 3.0 mm in phase with P: 3 loops × 2 = 6 (including P).	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis counts one point per loop that is in phase with P; each loop has a 3.0 mm point on each side of its antinode.
BThis counts one point in each loop and ignores the phase: the second and fourth loops move in antiphase with P.
CCorrect: two 3.0 mm points in each of the three loops (first, third, fifth) that move in phase with P.
DThis counts every point with amplitude 3.0 mm but ignores the phase; half of the loops are in antiphase with P.

Syllabus understandingC.4 — nodes and antinodes, relative amplitude and phase difference of points along a standing wave; standing waves patterns in strings and pipes Command term: Deduce

27C-1A-123

Speed of a point on a standing wave·C.4 Standing waves and resonance

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A standing wave of frequency 50 Hz forms on a string fixed at both ends. The graph shows the displacement y of the string against position x at an instant when every point is at its maximum displacement.

What is the maximum speed of point P?

Shape of the string at an instant of maximum displacement (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1At this instant every point is at an extreme of its oscillation, so the magnitude of the displacement of P, 4.0 mm, is its amplitude.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2P oscillates with simple harmonic motion of angular frequency ω = 2π × 50 = 314 rad s⁻¹.	—
Step 3v_max = ωx₀ = 314 × 4.0 × 10⁻³ = 1.3 m s⁻¹.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis uses fx₀ instead of 2πfx₀: 50 × 4.0 × 10⁻³ = 0.20 m s⁻¹.
BCorrect: amplitude 4.0 mm (read at the instant of maximum displacement) and v_max = 2πfx₀ = 1.3 m s⁻¹.
CThis uses the antinode amplitude, 8.0 mm, for P, as if every point of a standing wave had the same amplitude.
DThis uses the peak-to-peak range at the antinodes, 16 mm, as the amplitude.

Syllabus understandingC.4 — nodes and antinodes, relative amplitude and phase difference of points along a standing wave; C.1 (HL) — v = ωx₀ cos(ωt + ϕ) Command term: Determine

28C-1B-03

Resonance tubes·C.4 Standing waves and resonance

Paper 1BHard7 marks

Data-based question5 steps to full marksDetermine

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 / °C	15	30	45	60	75
Δx / mm	113.6	115.9	119.2	121.4	124.3
v² / 10⁴ m² s⁻²	11.61	12.09	12.79	13.26

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 point	Mark	Notes
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	✓ 1	The 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)	✓ 1	The 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)	✓ 1	Accept 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	✓ 1	Accept 10.9–11.1. ALT: extend the line to the θ axis.
θ = −intercept/gradient = −11.00/0.0385 = −286 °C	✓ 1	Accept −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	✓ 1	Allow 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	✓ 1	Accept 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)

Syllabus understandingC.4 — standing waves in pipes; the relationship between the length of the pipe and the wavelength of a resonant mode; B.3 — absolute temperature, kinetic model of an ideal gas; Tools — extrapolation to an intercept, gradient with its unit, comparison with a theoretical value Command term: Determine

29C-1B-08

Harmonics of a stretched string·C.4 Standing waves and resonance

Paper 1BMedium7 marks

Data-based question5 steps to full marksDetermine

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 / kg	0.100	0.200	0.300	0.400	0.500
f₁ / Hz	17.7	24.7	30.5	35.1	39.2
f₁² / Hz²	313	610	930	1232	1537

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 point	Mark	Notes
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	✓ 1	Accept 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	✓ 1	Accept 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	✓ 1	Allow 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	✓ 1	The 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)

Syllabus understandingC.4 — standing waves on strings with two fixed ends; the wavelength and frequency of the nth harmonic; A.2 — tension and the equilibrium of the hanging load; Tools — linearising, gradient with its unit, comparison with an independent value Command term: Determine

30C-1B-20

Forced oscillations and resonance·C.4 Standing waves and resonance

Paper 1BMedium7 marks

Data-based question5 steps to full marksDetermine

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).

Figure 1: free oscillation of the strip (drawn to scale).

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 point	Mark	Notes
Part (a)(i)
10 complete cycles between the peaks at t = 0 and t ≈ 1.32 s	✓ 1	Fewer than 5 cycles used: [0] for this mark.
f₀ = 10/1.32 = 7.6 Hz	✓ 1	Accept 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	✓ 1	Accept 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)	✓ 1	Allow 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	✓ 1	Readings 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)

Syllabus understandingC.4 — natural frequency, forced oscillations and resonance; the effect of damping on the maximum amplitude and resonant frequency (qualitative); D.4 (HL) — induced currents and Lenz's law; Tools — timing by counting cycles on a trace, choosing the interval of the independent variable Command term: Determine

31C-1B-22

Strings with free ends·C.4 Standing waves and resonance

Paper 1BEasy7 marks

Data-based question5 steps to full marksDeduce

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 / m	0.00	0.15	0.30	0.45	0.60	0.75	0.90	1.05	1.20
A / mm	1	14	23	25	18	5	9	21	25

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 point	Mark	Notes
Part (a)(i)
Nodes at x = 0 and x ≈ 0.80 m are half a wavelength apart	✓ 1	Or node at 0.80 m to antinode at 1.20 m = λ/4.
λ = 1.60 m	✓ 1	Accept 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	✓ 1	Allow ECF from (a).
v = fλ = 15.0 × 1.60 = 24.0 m s⁻¹	✓ 1	Accept 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	✓ 1	Award [0] for 20.0 Hz (fourth harmonic, not possible here).
f = 24.0/0.96 = 25.0 Hz	✓ 1	Or 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	✓ 1	Allow 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)

Syllabus understandingC.4 — standing wave patterns on strings with one fixed and one free boundary; nodes and antinodes; the wavelength and frequency of the nth harmonic; C.2 — v = fλ Command term: Deduce

32C-1B-35

Wave speed from standing-wave modes·C.4 Standing waves and resonance

Paper 1BEasy7 marks

Data-based question5 steps to full marksDetermine

An elastic cord is stretched horizontally between a vibration generator and a fixed clamp 1.50 m away. The generator moves its end of the cord with a very small amplitude, so both ends behave as nodes. A student increases the frequency f of the generator and records the frequency at which a steady standing wave with n loops is clearest, for n = 1 to 6. The uncertainty in each frequency is ±1.0 Hz.

The graph shows f against n with error bars and the line of best fit.

n	1	2	3	4	5	6
f / Hz	5.8	12.2	18.4	24.7	29.5	35.7

Frequency f against number of loops n, with error bars and the line of best fit (drawn to scale).

(a)

(i)

Show that f = (v/2L)n, where L is the length of the cord and v is the speed of the waves on it.

(1)

(b)

(i)

Determine the gradient of the line of best fit, including its unit, and hence the speed of the waves on the cord.

(2)

(c)

(i)

Draw the steepest and shallowest lines that pass through all the error bars. Use them to determine the absolute uncertainty in the speed of the waves.

(2)

(d)

(i)

The tension in the cord is 4.0 N and a 1.50 m length of the cord has a mass of 18.5 g. The speed of transverse waves on a cord is v = √(T/μ), where T is the tension and μ is the mass per unit length. Determine whether this prediction is consistent with your answers to (b) and (c).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
n loops fill the cord, each half a wavelength long: L = nλ/2, so λ = 2L/n and f = v/λ = (v/2L)n	✓ 1	Both the half-wavelength per loop and f = v/λ must be seen.
Part (b)(i)
Gradient from a large triangle, e.g. (41.8 − 0.3)/(7 − 0) = 5.93 Hz	✓ 1	Accept 5.8–6.1. Unit Hz (or s⁻¹) required; accept "Hz per loop".
v = 2L × gradient = 3.00 × 5.93 = 17.8 m s⁻¹	✓ 1	Accept 17.4–18.3 m s⁻¹. Allow ECF from the gradient.
Part (c)(i)
Steepest ≈ 6.38 Hz (e.g. through the bottom of the error bar at n = 2 and the top of the bar at n = 6) and shallowest ≈ 5.63 Hz (e.g. through the top of the bar at n = 1 and the bottom of the bar at n = 4)	✓ 1	Accept 6.2–6.5 and 5.5–5.8. Lines drawn simply through the opposite ends of the bars at n = 1 and n = 6 miss some error bars and are not accepted.
v ranges from 16.9 to 19.1 m s⁻¹, so Δv = ±1.1 m s⁻¹	✓ 1	Accept ±0.9 to ±1.5 m s⁻¹ (half the range). Allow ECF from (b).
Part (d)(i)
μ = 0.0185/1.50 = 1.23 × 10⁻² kg m⁻¹; v = √(4.0/1.23 × 10⁻²) = 18.0 m s⁻¹	✓ 1
18.0 m s⁻¹ lies within 17.8 ± 1.1 m s⁻¹, so the prediction is consistent with the experiment	✓ 1	The conclusion must agree with the candidate's own range. Allow ECF from (b) and (c).

Answers: (b)(i) 5.93 Hz; 17.8 m s⁻¹ · (c)(i) ±1.1 m s⁻¹ · (d)(i) 18.0 m s⁻¹; consistent (the remaining parts are explanations — see the table above)

33C-1B-36

Natural frequency from resonance·C.4 Standing waves and resonance

Paper 1BMedium7 marks

Data-based question6 steps to full marksDetermine

A spring hangs from a vibration generator, and a hanger carrying slotted masses is attached to its lower end; m is the total mass of the hanger and slotted masses. For each m, the student slowly varies the frequency of the generator and records the frequency f₀ at which the masses oscillate with the greatest amplitude. The damping is light.

Theory predicts that f₀ = (1/2π)√(k/(m + m_e)), where k is the spring constant and m_e is a constant that allows for the mass of the spring. The student plots f₀⁻² against m.

m / kg	f₀ / Hz	f₀⁻² / s²
0.050	2.37	0.178
0.100	1.91	0.274
0.150	1.63	0.376
0.200	1.45
0.250	1.31	0.583
0.300	1.22	0.672

f₀⁻² against m with the line of best fit (drawn to scale).

(a)

(i)

Calculate the missing value of f₀⁻².

(1)

(b)

(i)

Determine the gradient of the line, including its unit.

(2)

(ii)

Hence determine k.

(1)

(c)

(i)

Determine m_e.

(2)

(d)

(i)

Explain why the damping of the oscillating masses must be kept light in this experiment.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
1/1.45² = 0.476 s²	✓ 1	Accept 0.476 or 0.48.
Part (b)(i)
Large triangle on the line, e.g. (0.776 − 0.077)/(0.35 − 0)	✓ 1	Two adjacent data points only: [0] for this mark.
Gradient = 2.00 s² kg⁻¹	✓ 1	Accept 1.92–2.08. Unit required.
Part (b)(ii)
f₀⁻² = (4π²/k)(m + m_e), so gradient = 4π²/k and k = 4π²/2.00 = 19.8 N m⁻¹	✓ 1	Accept 19.0–20.6 N m⁻¹. Allow ECF from (b)(i).
Part (c)(i)
Intercept on the f₀⁻² axis = 0.077 s²	✓ 1	Accept 0.065–0.090 s². ALT: the intercept on the m axis is −m_e.
Intercept = 4π²m_e/k, so m_e = intercept/gradient = 0.077/2.00 = 0.038 kg	✓ 1	Accept 0.032–0.046 kg. Allow ECF from (b).
Part (d)(i)
Only for light damping is the frequency of maximum amplitude (almost) equal to the natural frequency; greater damping gives a lower, broader peak at a slightly lower frequency, so f₀ would be harder to locate and systematically below (1/2π)√(k/(m + m_e))	✓ 1	Reference to the shift of the peak to a lower frequency, or to the broader peak being harder to locate, is needed. "To get accurate results" alone: [0].

Answers: (a)(i) 0.476 s² · (b)(i) 2.00 s² kg⁻¹ · (b)(ii) 19.8 N m⁻¹ · (c)(i) 0.038 kg (the remaining parts are explanations — see the table above)

Syllabus understandingC.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; the effect of damping on the maximum amplitude and resonant frequency of oscillation; Guidance: only a qualitative analysis is required concerning the impact of damping on the frequency response of a driven oscillator; C.1 — the time period of a mass–spring system; Tools — linearising a relationship, gradient and intercept with units, systematic error Command term: Determine

34C-1B-37

Tension reduced by upthrust·C.4 Standing waves and resonance

Paper 1BHard7 marks

Data-based question5 steps to full marksDetermine

A steel wire is fixed at one end, passes over two bridges 0.600 m apart and then over a pulley, and supports a solid metal cylinder of mass M = 2.000 kg. The cylinder hangs fully immersed in a liquid of density ρ and does not touch the container. The wire between the bridges is plucked and the frequency f₁ of its first harmonic is measured. The experiment is repeated in air (ρ taken as zero) and in four liquids.

The speed of transverse waves on the wire is v = √(T/μ), where T is the tension and μ is the mass per unit length of the wire. The graph shows f₁² against ρ.

Medium	ρ / kg m⁻³	f₁ / Hz	f₁² / 10³ Hz²
air	0	73.8	5.45
paraffin	800	62.1	3.86
water	1000	58.4	3.41
salt solution	1200	55.2	3.05
glycerol	1260	53.9	2.91

f₁² against the density ρ of the liquid, with the line of best fit over the range of the data (drawn to scale).

(a)

(i)

Show that f₁² = g(M − ρV)/(4L²μ), where V is the volume of the cylinder and L is the distance between the bridges.

(2)

(b)

(i)

By extending the line, determine the density of the metal of the cylinder. Explain your method.

(2)

(c)

(i)

Determine μ.

(2)

(d)

(i)

A student suggests adding a point closer to the intercept by using a liquid denser than the cylinder. Explain why this cannot work.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The cylinder is in equilibrium: T + ρVg = Mg, so T = g(M − ρV)	✓ 1	The upthrust ρVg must be seen.
First harmonic: λ = 2L, so f₁ = v/(2L) and f₁² = T/(4L²μ) = g(M − ρV)/(4L²μ)	✓ 1
Part (b)(i)
f₁² = 0 when the tension is zero, i.e. when ρV = M, so the intercept on the ρ axis is M/V, the density of the cylinder	✓ 1	The link between zero tension and equal densities is needed. Allow ECF from (a).
Intercept ≈ 2.7 × 10³ kg m⁻³	✓ 1	Accept 2.55 × 10³ to 2.85 × 10³ kg m⁻³.
Part (c)(i)
Intercept on the f₁² axis = 5.45 × 10³ Hz² = gM/(4L²μ)	✓ 1	Accept 5.40–5.50 × 10³ Hz²; the air value read from the table is acceptable.
μ = 9.81 × 2.000/(4 × 0.600² × 5450) = 2.50 × 10⁻³ kg m⁻¹	✓ 1	Accept 2.46–2.54 × 10⁻³ kg m⁻¹. Allow ECF from (a).
Part (d)(i)
The upthrust on the fully immersed cylinder would be greater than its weight, so the cylinder would float up: the wire would go slack (the tension cannot become negative), so no standing wave forms / no point beyond the intercept exists	✓ 1	Accept "the cylinder floats, so the wire is not under tension".

Answers: (b)(i) 2.7 × 10³ kg m⁻³ · (c)(i) 2.50 × 10⁻³ kg m⁻¹ (the remaining parts are explanations — see the table above)

35C-2-03

Standing waves in pipes·C.4 Standing waves and resonance

Paper 2Hard9 marks

Short answer & extended response6 steps to full marksExplain

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 point	Mark	Notes
Part (a)(i)
v = 343 × √(310/293) = 353 m s⁻¹	✓ 1	Use 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	✓ 1	Answer to at least 3 s.f. required. Allow ECF from (a)(i).
Part (a)(iii)
Only odd harmonics: 1557 Hz and 2594 Hz	✓ 1	Accept 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 f_heliox/f_air = √(γ_hM_air/γ_airM_h)	✓ 1	Equivalent forms accepted.
Part (b)(ii)
Ratio = √(1.59 × 29.0/(1.40 × 9.6)) = 1.85, so f = 961 Hz	✓ 1	Accept 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	✓ 1	Dependent 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)

Syllabus understandingC.4 — standing wave patterns in pipes with one closed and one open end; the nature of resonance; C.2 — v = fλ; B.3 — the kelvin temperature and the molar mass of a gas Command term: Explain

36C-2-06

Standing waves on strings·C.4 Standing waves and resonance

Paper 2Easy10 marks

Short answer & extended response6 steps to full marksDetermine

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 point	Mark	Notes
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	✓ 1	Answer 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	✓ 1	Allow 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 T_new/T_old = (220/212)²	✓ 1
= 1.077, an increase of 7.7 %	✓ 1	Accept 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 λ_air = 343/660 = 0.520 m	✓ 1	Allow 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	✓ 1	Reference 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)

Syllabus understandingC.4 — standing wave patterns in strings with two fixed boundaries; nodes and antinodes, relative amplitude and phase difference of points along a standing wave; C.2 — v = fλ; A.4 (HL) — torque and rotational equilibrium Command term: Determine

37C-2-14

Standing waves and resonance·C.4 Standing waves and resonance

Paper 2Medium12 marks

Short answer & extended response7 steps to full marksExplain

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 point	Mark	Notes
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	✓ 1	Allow 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	✓ 1	The 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	✓ 1	Accept ±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	✓ 1	Conclusion 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	✓ 1	Unit 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	✓ 1	Allow ECF from (c)(i).
Q = mcΔT = 0.25 × 4200 × 20 = 2.1 × 10⁴ J	✓ 1
t = Q/P = 2.1 × 10⁴/800 = 26 s	✓ 1	Accept 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)

Syllabus understandingC.4 — the nature and formation of standing waves in terms of superposition of two identical waves travelling in opposite directions; nodes and antinodes; C.2 — the nature of electromagnetic waves; Tools — propagation of uncertainty; E.1 — photon energy E = hf; B.1 — specific heat capacity Command term: Explain

38C-2-17

Forced oscillations and resonance·C.4 Standing waves and resonance

Paper 2Medium12 marks

Short answer & extended response7 steps to full marksDetermine

The deck of a footbridge can sway sideways. Engineers drive the deck sideways with a periodic force of constant size and variable frequency f, and measure the amplitude A of the sway. The graph shows the results before and after a damper is fitted.

When a person walks, each foot pushes sideways on the deck, alternately to the left and to the right. A typical pedestrian takes steps 0.75 m long.

Amplitude of the sideways sway of the deck against driving frequency, without and with the damper (drawn to scale).

(a)

(i)

State the natural frequency of the sideways oscillation of the deck.

(1)

(ii)

Determine the walking speed at which pedestrians in step with each other would drive the deck at resonance.

(2)

(b)

(i)

The deck behaves as a mass–spring system with an effective mass of 2.4 × 10⁴ kg. Determine its effective spring constant.

(2)

(ii)

Design guidance requires the maximum sideways acceleration of the deck to stay below 0.30 m s⁻². Deduce whether this requirement is met without and with the damper when the deck is driven at resonance by the force used in the test.

(3)

(iii)

Without the damper, a pedestrian of mass 70 kg stands still on the deck while it sways at resonance with the amplitude shown. Determine the maximum sideways friction force on the pedestrian and the minimum coefficient of static friction needed to stop the pedestrian sliding.

(2)

(c)

(i)

Describe two effects of the damper shown by the graph, and explain one of them.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
1.00 Hz (the frequency of the peak without the damper)	✓ 1	Accept 0.98–1.02 Hz.
Part (a)(ii)
The sideways push repeats every two steps, so the sideways driving frequency is half the step frequency: steps at 2.0 Hz are needed	✓ 1	This is the hidden step. Using 1.0 steps per second (0.75 m s⁻¹): [1 max].
v = 0.75 × 2.0 = 1.5 m s⁻¹	✓ 1	Allow ECF from (a)(i).
Part (b)(i)
k = m(2πf)²	✓ 1
k = 2.4 × 10⁴ × (2π × 1.00)² = 9.5 × 10⁵ N m⁻¹	✓ 1	Allow ECF from (a)(i).
Part (b)(ii)
Without the damper: a_max = ω²x₀ = (2π × 1.00)² × 0.040 = 1.58 m s⁻²	✓ 1	Allow ECF from (a)(i).
With the damper: amplitude read as about 6.7 mm, so a_max ≈ 0.26 m s⁻²	✓ 1	Accept 6–7.5 mm and 0.24–0.30 m s⁻².
The requirement is met only with the damper	✓ 1	Conclusion must agree with the candidate's own values.
Part (b)(iii)
F = ma_max = 70 × 1.58 = 111 N	✓ 1	Allow ECF from (b)(ii).
μ ≥ F/(mg) = 1.58/9.81 = 0.16	✓ 1
Part (c)(i)
The peak amplitude is much smaller (and the peak broader), because the damper dissipates energy each cycle so a balance between energy supplied and energy dissipated is reached at a smaller amplitude	✓ 1
The peak occurs at a slightly lower frequency than the natural frequency	✓ 1	A description of the shift is sufficient; no explanation is required for this mark.

Answers: (a)(i) 1.00 Hz · (a)(ii) 1.5 m s⁻¹ · (b)(i) 9.5 × 10⁵ N m⁻¹ · (b)(ii) 1.58 m s⁻²; 0.26 m s⁻² · (b)(iii) 111 N; 0.16 (the remaining parts are explanations — see the table above)

Syllabus understandingC.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; the effect of damping on the maximum amplitude and resonant frequency; useful and destructive effects of resonance; C.1 — a = −ω²x and the period of a mass–spring system; A.1 — speed; A.2 — Newton's second law and friction Command term: Determine

39C-2-30

Pipes closed at both ends·C.4 Standing waves and resonance

Paper 2Medium12 marks

Short answer & extended response7 steps to full marksExplain

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.

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 A_max sin(2πd/λ). The air now vibrates in its third harmonic. Determine the amplitudes at P and at Q as fractions of A_max.

(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 point	Mark	Notes
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	✓ 1	Answer 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) = 0.71A_max at both	✓ 1	Accept 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	✓ 1	Allow ECF for the speed.
M = 1.14 × 10⁻² kg mol⁻¹	✓ 1	Accept 11.4 g mol⁻¹.
Part (c)(ii)
11.4 = x × 2.0 + (1 − x) × 29.0, with x the fraction of hydrogen molecules	✓ 1	The hidden step: the molar mass of a mixture is the number-weighted mean.
x = (29.0 − 11.4)/27.0 = 0.65	✓ 1	Allow ECF from (c)(i).

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

Syllabus understandingC.4 — standing wave patterns in pipes; boundary conditions for air in pipes with two closed ends; displacement nodes and antinodes; relative amplitude and phase difference of points along a standing wave; the nature of resonance; B.3 — molar mass and the kelvin temperature Command term: Explain

40C-2-31

Damping and resonance·C.4 Standing waves and resonance

Paper 2Medium9 marks

Short answer & extended response6 steps to full marksDetermine

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.

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 point	Mark	Notes
Part (a)(i)
R: it returns the mass to equilibrium in the shortest time without oscillating	✓ 1	Reason required.
Part (a)(ii)
Period of Q = 0.80 s	✓ 1	Accept 0.78–0.82 s; reading over several cycles is better.
k = 4π²m/T² = 4π² × 300/0.80² = 1.85 × 10⁴ N m⁻¹	✓ 1	Accept 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)	✓ 1	Accept 34–36 mm.
Energy ∝ amplitude², so the fraction dissipated = 1 − (35/50)² = 0.51	✓ 1	Accept 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	✓ 1	Allow 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	✓ 1	Accept 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)

Syllabus understandingC.4 — the effects of light, critical and heavy damping on the system; C.1 — the time period of a mass–spring system; C.1 (HL) — E_T = ½mω²x₀²; B.1 — specific heat capacity Q = mc ΔT Command term: Determine

41C-2-35

Forced oscillations and resonance·C.4 Standing waves and resonance

Paper 2Easy13 marks

Short answer & extended response8 steps to full marksExplain

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 point	Mark	Notes
Part (a)(i)
λ = 340/880 = 0.386 m	✓ 1	Accept 0.39 m.
Part (a)(ii)
The amplitude increases to a large maximum at (or very close to) 880 Hz and then decreases again	✓ 1	A 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	✓ 1	Accept "energy transfer is greatest at resonance".
Part (b)(i)
ω = 2π × 880 = 5.53 × 10³ rad s⁻¹	✓ 1
a_max = ω²x₀ = (5.53 × 10³)² × 1.5 × 10⁻⁴ = 4.6 × 10³ m s⁻²	✓ 1	Accept (4.5–4.6) × 10³ m s⁻². Allow ECF for ω.
Part (b)(ii)
v_max = ωx₀ = 5.53 × 10³ × 1.5 × 10⁻⁴ = 0.83 m s⁻¹	✓ 1	Allow 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	✓ 1	Award [0] for this mark for λ = 0.24 m or 0.06 m.
v = fλ = 880 × 0.12 = 1.1 × 10² m s⁻¹	✓ 1	Accept 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	✓ 1	Accept 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)

Syllabus understandingC.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; the nature and formation of standing waves in terms of superposition of two identical waves travelling in opposite directions; nodes and antinodes; C.1 — the time period of a mass–spring system T = 2π√(m/k); C.1 (HL) — v = ωx₀ cos(ωt + φ), a = −ω²x; C.2 — v = fλ; B.1 — apparent brightness and the inverse-square law b = L/4πd², applied here to sound Command term: Explain

42C-2-55

Wind-driven vibration of a power line·C.4 Standing waves and resonance

Paper 2Easy12 marks

Short answer & extended response8 steps to full marksExplain

An overhead power-line conductor is clamped to two towers 300 m apart; the clamps act as fixed ends. The conductor has a mass per unit length of 1.60 kg m⁻¹, and transverse waves travel along it at 90.0 m s⁻¹.

A steady wind blowing across a conductor of diameter d sheds swirling eddies that push the conductor alternately up and down with frequency f = 0.20U/d, where U is the wind speed. The diameter of this conductor is 30.0 mm.

(a)

(i)

Show that the first-harmonic frequency of the conductor is about 0.15 Hz.

(1)

(ii)

The speed of transverse waves on the conductor is v = √(T/μ), where T is the tension and μ is the mass per unit length. Calculate the tension in the conductor.

(1)

(b)

The wind speed is 4.5 m s⁻¹.

(i)

Calculate the frequency of the up-and-down push of the wind.

(1)

(ii)

The conductor vibrates as a standing wave at this frequency. Deduce the harmonic number of the standing wave and the distance between adjacent nodes.

(3)

(iii)

Explain why the conductor can vibrate with a large amplitude for almost any steady wind speed.

(2)

(c)

To reduce the vibration, a damper (a heavy mass on a short, flexible steel cable) is clamped to the conductor. The damper dissipates energy when the point to which it is attached moves.

(i)

Explain why a damper attached at a node of the standing wave would not reduce its amplitude, and suggest why dampers are clamped about 1 m from a tower rather than at the clamp itself.

(2)

(ii)

At the antinodes the amplitude of the conductor is 15 mm. Calculate the maximum acceleration of the conductor at an antinode and express it as a multiple of g.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
λ = 2L = 600 m, so f₁ = v/λ = 90.0/600 = 0.150 Hz	✓ 1	Full substitution or an answer to at least 3 s.f. required.
Part (a)(ii)
T = μv² = 1.60 × 90.0² = 1.30 × 10⁴ N	✓ 1
Part (b)(i)
f = 0.20 × 4.5/0.0300 = 30.0 Hz	✓ 1	Diameter not converted to metres: [0].
Part (b)(ii)
Harmonic number n = f/f₁ = 30.0/0.150 = 200	✓ 1	Allow ECF from (a)(i) and (b)(i).
λ = v/f = 90.0/30.0 = 3.00 m	✓ 1	ALT: λ = 2L/n = 600/200.
Adjacent nodes are λ/2 apart: 1.50 m	✓ 1	Allow ECF for λ. 3.00 m (a whole wavelength): [0] for this mark.
Part (b)(iii)
The natural frequencies of the conductor are nf₁, only 0.15 Hz apart, so the driving frequency is always very close to one of them	✓ 1	Allow ECF from (a)(i).
The wind then drives the conductor at (almost) a natural frequency: resonance, and with little damping the amplitude becomes large	✓ 1	The idea of driving frequency equal to a natural frequency is needed; "resonance" alone: [0] for this mark.
Part (c)(i)
At a node the conductor does not move, so no work is done on the damper and no energy is removed from that standing wave	✓ 1
The clamp is always a node; 1 m from it is less than the node separation (1.50 m), so the damper is between a node and an antinode and moves	✓ 1	Allow ECF from (b)(ii). Accept "close to the clamp but not at a node for wind-driven waves of this wavelength".
Part (c)(ii)
ω = 2π × 30.0 = 188.5 rad s⁻¹	✓ 1	Allow ECF from (b)(i).
a_max = ω²x₀ = 188.5² × 0.015 = 533 m s⁻² ≈ 54g	✓ 1	Accept 530–535 m s⁻² and 54g.

Answers: (a)(i) 0.150 Hz · (a)(ii) 1.30 × 10⁴ N · (b)(i) 30.0 Hz · (b)(ii) 200; 1.50 m · (c)(ii) 533 m s⁻² ≈ 54g (the remaining parts are explanations — see the table above)

Syllabus understandingC.4 — standing waves patterns in strings and pipes; the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; Guidance: a determination of the wavelength and the frequency of the nth harmonic given the length of the string or pipe and the speed of the wave is required; Guidance: knowledge of the useful and destructive effects of resonance is required; C.1 (HL) — a = −ω²x Command term: Explain

43C-2-56

Pitch of a filling bottle·C.4 Standing waves and resonance

Paper 2Medium12 marks

Short answer & extended response8 steps to full marksDetermine

A cylindrical bottle has an internal height of 0.240 m and an internal diameter of 6.2 cm. It stands upright and is filled with water from a tap at a constant volume flow rate, starting empty at t = 0. Treat the bottle as a uniform tube, open at the top.

The air above the water vibrates as the water runs in, and a microphone records the frequency f of the first harmonic of this air column. The graph shows how f varies with t. The speed of sound in the air is 343 m s⁻¹.

Frequency f of the first harmonic of the air column against time t (drawn to scale).

(a)

(i)

Explain why the wavelength of the first harmonic is four times the length of the air column.

(1)

(ii)

Show that the frequency at t = 0 is about 360 Hz.

(2)

(b)

(i)

Use the graph to determine the speed at which the water surface rises.

(3)

(ii)

Calculate the volume flow rate of the water.

(1)

(c)

(i)

Show that the time taken for the frequency to rise from any value f to 2f is v/(8fu), where u is the speed at which the water surface rises.

(2)

(ii)

Hence explain the shape of the graph, and calculate the time taken for the frequency to rise from 1000 Hz to 2000 Hz.

(2)

(d)

(i)

State the frequency of the next resonance of the air column above its first harmonic at t = 0.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The water surface is a closed end (displacement node) and the top is an open end (displacement antinode); the shortest such pattern is node to adjacent antinode, a quarter of a wavelength	✓ 1	Both boundary conditions are needed.
Part (a)(ii)
λ = 4 × 0.240 = 0.960 m	✓ 1
f = 343/0.960 = 357 Hz	✓ 1	Answer to at least 3 s.f. required.
Part (b)(i)
Read a point on the curve, e.g. f = 714 Hz (double the initial value) at t ≈ 24 s	✓ 1	Accept 22–26 s for 714 Hz; any correctly read point.
Length of the air column L = v/(4f) = 343/(4 × 714) = 0.120 m, so the water has risen 0.240 − 0.120 = 0.120 m	✓ 1	ALT: doubling f halves L. Allow ECF from (a).
u = 0.120/24 = 5.0 × 10⁻³ m s⁻¹	✓ 1	Accept 4.6–5.5 × 10⁻³ m s⁻¹.
Part (b)(ii)
Rate = π(0.031)² × 5.0 × 10⁻³ = 1.5 × 10⁻⁵ m³ s⁻¹	✓ 1	Allow ECF from (b)(i). Accept 15 cm³ s⁻¹.
Part (c)(i)
At f the air column is v/(4f); at 2f it is v/(8f), so the water rises by v/(4f) − v/(8f) = v/(8f)	✓ 1
Time = distance/u = v/(8fu)	✓ 1
Part (c)(ii)
The time for each doubling is proportional to 1/f, so as f rises each doubling takes less time: f rises ever more rapidly (the curve gets steeper)	✓ 1
Time = 343/(8 × 1000 × 5.0 × 10⁻³) = 8.6 s	✓ 1	Accept 7.8–9.3 s. Allow ECF from (b)(i).
Part (d)(i)
Only odd harmonics occur: 3 × 357 = 1.07 × 10³ Hz	✓ 1	714 Hz: [0]. Allow ECF from (a)(ii).

Answers: (a)(ii) 357 Hz · (b)(i) 5.0 × 10⁻³ m s⁻¹ · (b)(ii) 1.5 × 10⁻⁵ m³ s⁻¹ · (c)(ii) 8.6 s · (d)(i) 1.07 × 10³ Hz (the remaining parts are explanations — see the table above)

44C-2-57

Resonant vibration energy harvester·C.4 Standing waves and resonance

Paper 2Medium12 marks

Short answer & extended response7 steps to full marksDiscuss

A slight imbalance of the rotor of an industrial motor makes the casing of the motor vibrate at the rotation frequency of the rotor. The rotor turns at 3000 revolutions per minute.

An energy harvester is fixed to the casing. It is a thin springy strip with a mass at its tip, and can be modelled as a mass–spring oscillator of spring constant 2.40 × 10³ N m⁻¹; the mass of the strip can be ignored. As the tip oscillates, a layer on the strip converts some of the energy of oscillation into electrical energy.

(a)

(i)

Show that the casing vibrates with a frequency of 50 Hz.

(1)

(ii)

Determine the mass at the tip for which the harvester is driven at resonance.

(2)

(b)

(i)

Explain, with reference to energy, why the amplitude of the tip is largest when its natural frequency equals the driving frequency.

(2)

(c)

At resonance the amplitude of the tip is 1.5 mm.

(i)

Show that the energy of oscillation of the tip is about 3 mJ.

(2)

(ii)

In each cycle 4.0 % of the energy of oscillation is converted into electrical energy. Determine the average electrical power delivered by the harvester.

(2)

(d)

(i)

In use, the rotation rate of the motor varies between 2850 and 3150 revolutions per minute. Discuss whether a lightly damped harvester or a more heavily damped harvester would deliver more energy over a long time.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
3000 revolutions in 60 s: f = 3000/60 = 50.0 Hz	✓ 1
Part (a)(ii)
Resonance: natural frequency = driving frequency, (1/2π)√(k/m) = 50.0 Hz	✓ 1	Allow ECF from (a)(i).
m = k/(4π²f²) = 2.40 × 10³/(4π² × 50.0²) = 2.43 × 10⁻² kg	✓ 1	Accept 24 g.
Part (b)(i)
At resonance the driving force keeps in step with the motion of the tip, so the energy transferred to the oscillator in each cycle is greatest	✓ 1	Accept "the driving force is in phase with the velocity".
The amplitude grows until the energy removed in each cycle (damping and electrical conversion) equals the energy supplied in each cycle; this balance is reached at the largest amplitude at resonance	✓ 1
Part (c)(i)
E_T = ½mω²x₀² = ½ × 2.43 × 10⁻² × (2π × 50.0)² × (1.5 × 10⁻³)²	✓ 1	ALT: ½kx₀² = ½ × 2400 × (1.5 × 10⁻³)². Allow ECF from (a)(ii).
= 2.70 × 10⁻³ J	✓ 1	Answer to at least 2 s.f. required.
Part (c)(ii)
Energy converted per cycle = 0.040 × 2.70 × 10⁻³ = 1.08 × 10⁻⁴ J	✓ 1	Allow ECF from (c)(i).
Power = 1.08 × 10⁻⁴ × 50.0 = 5.4 × 10⁻³ W	✓ 1	Accept 5.4 mW. Allow ECF from (a)(i).
Part (d)(i)
The driving frequency varies between 47.5 Hz and 52.5 Hz	✓ 1	Allow ECF from (a)(i).
Light damping gives a tall, narrow resonance peak: the amplitude is very large only when the driving frequency is very close to 50 Hz and falls sharply away from it	✓ 1
Heavier damping gives a lower but broader peak, so the amplitude stays moderate over the whole range of driving frequencies: the more heavily damped harvester is likely to deliver more energy on average	✓ 1	Accept a conclusion for light damping only if it is argued that the speed stays close to 3000 revolutions per minute most of the time. [3 max]; also accept: the peak of the heavily damped harvester is at a slightly lower frequency.

Answers: (a)(i) 50.0 Hz · (a)(ii) 2.43 × 10⁻² kg · (c)(i) 2.70 × 10⁻³ J · (c)(ii) 5.4 × 10⁻³ W (the remaining parts are explanations — see the table above)

Syllabus understandingC.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; the effect of damping on the maximum amplitude and resonant frequency of oscillation; Guidance: only a qualitative analysis is required concerning the impact of damping on the frequency response of a driven oscillator; Guidance: knowledge of the useful and destructive effects of resonance is required; C.1 — the time period of a mass–spring system; C.1 (HL) — E_T = ½mω²x₀²; A.3 — power as the rate of energy transfer Command term: Discuss

45C-2-58

Flute and clarinet as pipes·C.4 Standing waves and resonance

Paper 2Medium12 marks

Short answer & extended response8 steps to full marksDetermine

In a simple model, a flute is a pipe open at both ends, and a clarinet is a pipe closed at one end (by the reed and the player's mouth) and open at the other. In this question both instruments are modelled as air columns of length 0.600 m when all their finger holes are covered.

The speed of sound in the air in both instruments is 343 m s⁻¹ at 20 °C.

(a)

(i)

Show that the lowest note of the flute has a frequency of about 290 Hz.

(2)

(ii)

Hence calculate the frequency of the lowest note of the clarinet.

(1)

(b)

By blowing harder, a player can make the air column resonate at the next higher resonant frequency instead of the lowest.

(i)

Deduce, without numerical substitution, the ratio (next higher resonant frequency)/(lowest resonant frequency) for the flute.

(1)

(ii)

Deduce, without numerical substitution, the same ratio for the clarinet.

(1)

(c)

On the flute, uncovering a finger hole makes the air column end at the hole, which acts as an open end.

(i)

Determine the distance from the mouth end of the flute at which a hole must be uncovered for the lowest note to be 392 Hz.

(2)

(d)

(i)

A flautist on a parade float plays the lowest note of the flute while the float moves at a constant speed of 6.0 m s⁻¹ along a straight road. Determine the difference between the frequencies heard by a stationary observer at the roadside as the float approaches and after it has passed.

(3)

(e)

(i)

During playing, the air in the flute warms to 32 °C. The speed of sound in air is proportional to √T, where T is the kelvin temperature. Determine the frequency of the lowest note of the flute at this temperature.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Antinode at each open end: λ = 2L = 1.20 m	✓ 1
f = 343/1.20 = 286 Hz	✓ 1	Answer to at least 3 s.f. required.
Part (a)(ii)
Node to antinode is a quarter of a wavelength: λ = 4L = 2.40 m, twice the flute wavelength, so f = 286/2 = 143 Hz	✓ 1	Allow ECF from (a)(i).
Part (b)(i)
Open at both ends: all harmonics nf₁ occur, so the next is 2f₁: ratio 2	✓ 1
Part (b)(ii)
Closed at one end: only odd numbers of quarter-wavelengths fit (L = λ/4, 3λ/4, …), so only odd harmonics occur and the next is 3f₁: ratio 3	✓ 1	Ratio 2: [0].
Part (c)(i)
λ = 343/392 = 0.875 m	✓ 1
Length = λ/2 = 0.438 m	✓ 1	Accept 0.44 m. Using λ/4 (0.219 m): [1 max].
Part (d)(i)
Approaching: f′ = 286 × 343/(343 − 6.0) = 290.9 Hz	✓ 1	Allow ECF from (a)(i).
Receding: f′ = 286 × 343/(343 + 6.0) = 280.9 Hz	✓ 1
Difference = 10.0 Hz	✓ 1	Accept 9.9–10.1 Hz. One correct shifted frequency only: [2 max].
Part (e)(i)
The length and the boundary conditions are unchanged, so λ is unchanged and f ∝ v ∝ √T: ratio = √(305/293) = 1.020	✓ 1	Celsius temperatures used: [0] for this mark.
f = 286 × 1.020 = 292 Hz	✓ 1	Allow ECF from (a)(i).

Answers: (a)(i) 286 Hz · (a)(ii) 143 Hz · (b)(i) 2 · (b)(ii) 3 · (c)(i) 0.438 m · (d)(i) 10.0 Hz · (e)(i) 292 Hz (the remaining parts are explanations — see the table above)

46C-2-59

Wire driven by a magnetic force·C.4 Standing waves and resonance

Paper 2Hard14 marks

Short answer & extended response8 steps to full marksExplain

A copper wire of mass per unit length 1.13 × 10⁻³ kg m⁻¹ is stretched horizontally between two rigid clamps 0.800 m apart. A horseshoe magnet produces a uniform horizontal magnetic field of 0.090 T, perpendicular to the wire, over a 0.050 m length of the wire centred on its midpoint. The ends of the wire are connected to a signal generator that passes an alternating current of amplitude 1.5 A and variable frequency through the wire.

The speed of transverse waves on the wire is v = √(T/μ), where T is the tension and μ is the mass per unit length.

The wire, the magnet at its midpoint and the signal generator (not to scale).

(a)

(i)

Explain why the wire vibrates, and state the frequency of the vibration.

(2)

(b)

(i)

Calculate the maximum force on the wire.

(1)

(ii)

As the frequency is increased slowly from 20 Hz, the wire vibrates with a large amplitude at 62.5 Hz and at 187.5 Hz, but not at 125 Hz. Explain these observations.

(2)

(iii)

Determine the tension in the wire.

(2)

(c)

(i)

The magnet is moved so that the field region is centred 0.200 m from one clamp. Deduce which of the first six harmonics can no longer be driven.

(3)

(d)

The signal generator is replaced by an oscilloscope and the magnet is returned to the midpoint. The wire is plucked so that it vibrates in its first harmonic with an amplitude of 2.0 mm at its midpoint.

(i)

Treating the 0.050 m length in the field as a straight conductor moving with the speed of the midpoint, determine the maximum emf shown on the oscilloscope.

(2)

(ii)

The ends of the wire are now joined by a thick lead of negligible resistance. Explain why the vibration of the plucked wire now dies away more quickly.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The current in the field experiences a force (F = BIL) perpendicular to both the wire and the field, i.e. vertically	✓ 1
The force reverses each time the current reverses, so it is a periodic driving force and the wire vibrates at the frequency of the alternating current	✓ 1	Accept "forced oscillation at the generator frequency".
Part (b)(i)
F = BIL = 0.090 × 1.5 × 0.050 = 6.8 × 10⁻³ N	✓ 1
Part (b)(ii)
62.5 Hz, 125 Hz and 187.5 Hz are the first, second and third harmonics; the amplitude is large when the driving frequency equals a natural frequency (resonance)	✓ 1
The midpoint is an antinode of the first and third harmonics but a node of the second; the force acts at a point that does not move in the second harmonic, so it does no work on it and cannot drive it	✓ 1	The idea that a node does not move / no energy transfer is needed.
Part (b)(iii)
First harmonic: v = 2Lf₁ = 2 × 0.800 × 62.5 = 100 m s⁻¹	✓ 1	Allow ECF from the harmonic identified in (b)(ii).
T = μv² = 1.13 × 10⁻³ × 100² = 11.3 N	✓ 1
Part (c)(i)
The n-th harmonic has nodes at distances j × 0.800/n from a clamp, j = 1, 2, …, n − 1	✓ 1
0.200 m = L/4 is a node only when n is a multiple of 4	✓ 1
So only the fourth harmonic (250 Hz) cannot be driven; the second harmonic (node at L/2) can now be driven	✓ 1	Second harmonic still listed as impossible: [2 max].
Part (d)(i)
v_max = 2πfx₀ = 2π × 62.5 × 2.0 × 10⁻³ = 0.785 m s⁻¹	✓ 1	Allow ECF from the first-harmonic frequency in (b).
ε = BvL = 0.090 × 0.785 × 0.050 = 3.5 × 10⁻³ V	✓ 1	Accept 3.5 mV.
Part (d)(ii)
The induced emf now drives a current through the closed circuit, so there is a force BIL on the moving wire	✓ 1
By Lenz's law this force opposes the motion of the wire; energy of oscillation is transferred to internal energy in the circuit, so the damping is greater	✓ 1	"The magnet slows it down" without induced current: [0].

Answers: (b)(i) 6.8 × 10⁻³ N · (b)(iii) 11.3 N · (c)(i) 4th harmonic · (d)(i) 3.5 × 10⁻³ V (the remaining parts are explanations — see the table above)

Syllabus understandingC.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; nodes and antinodes, relative amplitude and phase difference of points along a standing wave; standing waves patterns in strings and pipes; Guidance: a determination of the wavelength and the frequency of the nth harmonic given the length of the string or pipe and the speed of the wave is required; D.3 — F = BIL sin θ; D.4 (HL) — ε = BvL for a straight conductor moving at right angles to a magnetic field, Lenz's law; C.1 (HL) — v = ωx₀ cos(ωt + ϕ) Command term: Explain

47C-2-67

Forced oscillations and resonance·C.4 Standing waves and resonance

Paper 2Hard18 marks

Short answer & extended response13 steps to full marksExplain

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.

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 point	Mark	Notes
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	✓ 1	The 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	✓ 1	Allow 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	✓ 1	Allow 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	✓ 1	Allow 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	✓ 1	Do 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	✓ 1	Unit rad s⁻¹ required for this mark.
Part (c)(ii)
u = ωr = 1.57 × 40 = 62.8 m s⁻¹	✓ 1	Allow 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	✓ 1	Accept 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	✓ 1	Accept 1.33 s. 4.0 s (one blade only): [1 max].
Part (d)(i)
E_k = ½Iω² = ½ × 1.2 × 10⁷ × 1.57²	✓ 1	Allow ECF from (c)(i).
E_k = 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	✓ 1	Allow ECF from (c)(i). Energy route (E_k = 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)	✓ 1	Allow 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)

Syllabus understandingC.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; C.1 — the time period of a mass–spring system; C.5 (HL) — the observed frequency for a moving source as given by f′ = fv/(v ± us); A.4 (HL) — angular velocity, rotational kinetic energy ½Iω², Newton's second law for rotation τ = Iα; Tools — estimation and units Command term: Explain

C.4 Standing waves and resonance: IB Physics HL exam-style questions

24 practice questions on C.4 Standing waves and resonance

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