111 free, high-quality past-paper-style questions on Standing waves and resonance for IB Physics SL, each with a full mark scheme. Covers: Resonance & damping; Superposition; Standing vs travelling waves; and more. For when you have finished the past papers, or want to practise one topic at a time.C.4 at Higher Level →Read the C.4 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.516 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1C-1A-35Resonance & dampingPaper 1AEasy1 mark
Three oscillators with different degrees of damping are each displaced by the same amount and released at time t = 0. The graph shows how the displacement of each oscillator varies with time.
Which row identifies the critically damped oscillator and the heavily damped oscillator?
Graph drawn to scale
Critically dampedHeavily damped
Show mark scheme
Marking point
Mark
Notes
Step 1Q oscillates about equilibrium with decreasing amplitude, so it is lightly damped; of P and R, which do not oscillate, R returns to equilibrium sooner, so R is critically damped and P is heavily damped.
✓ 1
Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AP and R have been swapped: heavy damping slows the return, and critical damping returns to equilibrium in the shortest time without oscillating.
BQ oscillates about the equilibrium position, so it is lightly damped, not critically damped.
CCorrect: R returns to equilibrium in the shortest time without oscillating (critical damping), while P returns more slowly (heavy damping).
DQ oscillates, so it is lightly damped; and R, which returns fastest without oscillating, is critically rather than heavily damped.
Two waves that are identical except for their direction of travel move in opposite directions along a string and form a standing wave.
What is (amplitude of the standing wave at an antinode)/(amplitude of each travelling wave)?
Show mark scheme
Marking point
Mark
Notes
Step 1At an antinode the two waves always arrive in phase, so their displacements add: the amplitude is 2 × the amplitude of each wave, and the ratio is 2.
✓ 1
Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThis is the value at a node, where the two waves are always in antiphase and cancel.
BThis assumes the standing wave has the same amplitude as each travelling wave; at an antinode the two waves reinforce each other.
CThis adds the two amplitudes as though they were perpendicular vectors. Superposition adds displacements along the same line.
DCorrect: at an antinode the displacements of the two waves are always equal and in the same direction, so the amplitude doubles.
Which is an example of a useful application of resonance?
Show mark scheme
Marking point
Mark
Notes
Step 1Resonance is the large-amplitude oscillation of a system driven at its natural frequency. In a quartz watch this large, steady oscillation at one precise frequency is used to keep time.
✓ 1
Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AShock absorbers damp the oscillations of the springs so that the car stops bouncing quickly; oscillation at the natural frequency after every bump is exactly what they prevent.
BThis is resonance, but it is destructive: the large amplitude can damage the bridge.
CCorrect: the crystal is driven at its natural frequency, and its large, stable oscillation is used as the timing reference of the watch.
DThis is an effect of heavy damping, not of resonance; there is no periodic driving force.
A smart speaker emits a steady note towards a flat, hard wall. A standing wave forms in the air between the speaker and the wall.
Which statement explains how the standing wave forms?
Show mark scheme
Marking point
Mark
Notes
Step 1A standing wave is the superposition of two waves of the same frequency, wavelength and (similar) amplitude travelling in opposite directions.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The wall reflects the sound back towards the speaker without changing its frequency or speed, so the incident and reflected waves meet this condition.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: two waves of the same frequency and speed (so the same wavelength) travelling in opposite directions superpose; where they always cancel there are nodes, and where they always reinforce there are antinodes.
BReflection from a stationary wall does not change the frequency. Two waves of different frequencies would give a pattern that moves, not fixed nodes and antinodes.
CThe reflected wave travels back towards the speaker. Two waves travelling the same way combine into a travelling wave, which has no fixed nodes.
DA standing wave needs the sound to be reflected. If the wall absorbed the sound there would be no reflected wave to superpose with the incident wave.
In a vibration test, a circuit board on a flexible mount is shaken by a vibration table. The table exerts a periodic force of constant amplitude on the mount. The natural frequency of the board on its mount is 18 Hz, and the damping is light.
The table is set to 30 Hz. Once the motion of the board is steady, which row describes the frequency of oscillation of the board, and its amplitude compared with that when the table is set to 18 Hz?
Frequency of oscillationAmplitude compared with that at 18 Hz
Show mark scheme
Marking point
Mark
Notes
Step 1In the steady state a driven system oscillates at the driving frequency, 30 Hz.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The amplitude is greatest when the driving frequency equals the natural frequency, 18 Hz (resonance), so at 30 Hz it is smaller.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AOnce the motion is steady, a driven system oscillates at the driving frequency, not at its natural frequency.
BThe board does not settle at a frequency between its natural frequency and the driving frequency; in the steady state it follows the driver at 30 Hz.
CThe frequency is right, but the amplitude is largest when the driving frequency equals the natural frequency; at 30 Hz the board is far from resonance.
DCorrect: the board oscillates at the driving frequency, 30 Hz, with a much smaller amplitude than at resonance (18 Hz).
A diver jumps from the end of a springboard. The board then oscillates up and down with light damping until it comes to rest.
As the oscillations die away, which row describes the amplitude of the oscillations and the time taken for each oscillation?
AmplitudeTime for each oscillation
Show mark scheme
Marking point
Mark
Notes
Step 1The damping force removes energy from the board in every oscillation, so the amplitude decreases.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2With light damping the board keeps oscillating at (very nearly) its natural frequency, which does not depend on the amplitude, so the time for each oscillation stays the same.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: energy is removed in every oscillation, so the amplitude falls, while the period stays equal to the (lightly damped) natural period of the board.
BA smaller oscillation covers a shorter distance, but the board also moves more slowly, so the time for each oscillation does not fall.
CThe board moves more slowly as the amplitude falls, but it also has less distance to cover, so the time for each oscillation does not grow.
DThe damping force removes energy from the board in every oscillation, so the amplitude cannot stay the same.
A longitudinal standing wave is set up in the air inside a tube. Which statement correctly describes a point that is a displacement node?
Show mark scheme
Marking point
Mark
Notes
At a displacement node the air layers on either side oscillate in antiphase. They move towards the node together, which forms a compression, and then away from it together, which forms a rarefaction. The pressure variation is therefore greatest at this point, so a displacement node is a pressure antinode. Option A wrongly assumes that no motion means no pressure change. Option C confuses zero displacement with zero pressure variation. Option D is wrong because an open end is a displacement antinode; displacement nodes occur at closed ends.
✓ 1
Answer B
Answer: B
Common error: Displacement nodes and pressure nodes are a quarter-wavelength apart. A displacement node is always a pressure antinode.
8IBP2S-024-03-02Standing vs travelling wavesPaper 1AEasy1 mark
An acoustic levitator traps tiny polystyrene beads at the nodes of a standing sound wave in air. Adjacent beads are 4.3 mm apart. The speed of sound in the air is 343 m s−1. What is the frequency of the sound?
Show mark scheme
Marking point
Mark
Notes
Adjacent nodes are half a wavelength apart, so λ = 2 × 4.3 mm = 8.6 × 10−3 m. Then f = v/λ = 343 / 8.6 × 10−3 ≈ 4.0 × 104 Hz. Option A takes the spacing as λ/4. Option C takes the spacing as a full wavelength. Option D takes the spacing as 2λ.
✓ 1
Answer B
Answer: B
Exam tip: Node to node is λ/2, and node to the nearest antinode is λ/4. Write this down before you substitute any numbers.
9IBP2S-024-03-05Standing waves in pipesPaper 1AEasy1 mark
An organ pipe is open at both ends and sounds its second harmonic. How many displacement nodes are there inside the pipe?
Show mark scheme
Marking point
Mark
Notes
Both open ends are displacement antinodes. For the second harmonic the length of the pipe equals one wavelength, so there is a third antinode at the centre. Between each pair of adjacent antinodes there is one node, at L/4 and at 3L/4. That gives 2 nodes. Option A describes the fundamental. Option C counts the antinodes instead of the nodes. Option D is a miscount.
✓ 1
Answer B
Answer: B
Exam tip: Sketch the displacement envelope quickly in the margin, starting from the boundary conditions at each end. Then count the nodes.
10IBP2S-024-01-01Standing waves in pipesPaper 1AEasy1 mark
A straight ventilation duct is open at both ends. Air in the duct resonates at its fundamental frequency. How many displacement nodes and displacement antinodes are there along the duct?
Show mark scheme
Marking point
Mark
Notes
Each open end is a displacement antinode. The fundamental is the simplest pattern that satisfies this: half a wavelength fits in the duct, with one displacement node at the centre. This gives 1 node and 2 antinodes. Two nodes and one antinode is the pattern for a string fixed at both ends. One node and one antinode is the fundamental of a pipe closed at one end.
✓ 1
Answer A
Answer: A
Common error: Do not copy the string pattern onto a pipe. Open ends are displacement antinodes. Closed ends are displacement nodes.
11IBP2S-024-01-02Standing waves in pipesPaper 1AEasy1 mark
The human ear canal can be modelled as a tube closed at one end by the eardrum. A particular ear canal has a fundamental resonant frequency of 3.0 kHz. Which of the following is also a resonant frequency of this ear canal?
Show mark scheme
Marking point
Mark
Notes
A tube closed at one end supports only odd harmonics: f1, 3f1, 5f1, and so on. Since 3 × 3.0 kHz = 9.0 kHz, this is a resonant frequency. The values 6.0 kHz and 12 kHz are even multiples of f1, which would only resonate in an open–open tube. A frequency of 1.5 kHz is below the fundamental.
✓ 1
Answer C
Answer: C
Exam tip: For a pipe closed at one end, divide the frequency by f₁. Only odd whole numbers give a resonance.
12IBP2S-024-01-03Standing waves in pipesPaper 1AEasy1 mark
A clarinet player warms up the instrument by blowing through it. The air inside becomes warmer, so the speed of sound in it increases. The length of the air column does not change. What happens to the wavelength and the frequency of the fundamental?
Show mark scheme
Marking point
Mark
Notes
The wavelength of a resonant mode is fixed by the length of the air column and the boundary conditions, so it does not change. From f = v/λ, a greater speed with the same wavelength gives a higher frequency. This is why wind instruments go sharp as they warm up.
✓ 1
Answer A
Answer: A
Common error: In a resonating pipe, the geometry fixes the wavelength. Any change in wave speed therefore shows up as a change in frequency.
13IBP2S-024-01-08Standing vs travelling wavesPaper 1AEasy1 mark
A standing sound wave is set up in a pipe. Which row correctly describes the open end of the pipe?
Show mark scheme
Marking point
Mark
Notes
At an open end, the air is free to move, so the displacement amplitude is a maximum: a displacement antinode. The pressure is held close to atmospheric pressure, so the pressure variation is a minimum: a pressure node. Displacement nodes and pressure nodes are always a quarter-wavelength apart, so they never coincide.
✓ 1
Answer C
Answer: C
Exam tip: A displacement node is always a pressure antinode, and a displacement antinode is always a pressure node. Always say which quantity you mean.
14IBP2S-024-00-01Resonance & dampingPaper 1AEasy1 mark
A mass–spring system is driven by an oscillator of variable frequency and a graph of amplitude against driving frequency is obtained. The experiment is repeated with a vane attached to the mass so that the damping is heavier. Compared with the first curve, the second curve has
Show mark scheme
Marking point
Mark
Notes
Heavier damping removes energy from the system faster. The maximum amplitude at resonance is therefore smaller, and the response is spread over a wider range of frequencies, so the peak is broader. The peak frequency moves very slightly lower, not higher. The peak height must change because the energy dissipated per cycle increases.
✓ 1
Answer B
Answer: B
Common error: Damping never raises the resonant frequency. If anything, the peak shifts slightly to a lower frequency.
15IBP2S-024-00-02Tension & wave speedPaper 1AEasy1 mark
A guitar string is tuned by turning a peg. The length of the vibrating string stays the same, but the tension is increased to four times its original value. What happens to the frequency of the fundamental?
Show mark scheme
Marking point
Mark
Notes
The wave speed is v = √(T/μ), so quadrupling T doubles v. The wavelength of the fundamental is fixed at 2L, so f = v/2L also doubles. Option D assumes f ∝ T. Option B wrongly assumes the frequency depends only on length.
✓ 1
Answer C
Answer: C
Exam tip: For a fixed length, the wavelength of each harmonic is fixed. Any change in wave speed therefore appears directly as a change in frequency.
16IBP2S-024-00-09Resonance & dampingPaper 1AEasy1 mark
Which statement about a lightly damped system at resonance is correct?
Show mark scheme
Marking point
Mark
Notes
Resonance occurs when the driving frequency equals (or is very close to) the natural frequency. Energy is then transferred to the system most efficiently, so the amplitude is greatest. In any real system damping is present, so the amplitude stays finite. Damping does not disappear at resonance. A driven system oscillates at the driving frequency, not at its own natural frequency.
✓ 1
Answer A
Answer: A
Common error: Do not describe resonance simply as 'maximum amplitude'. State the frequency condition and the maximum energy transfer as well.
17IBP2S-024-02-01Standing vs travelling wavesPaper 1AEasy1 mark
A standing wave is set up on a string. Point X lies in one loop and point Y lies in the adjacent loop, on the other side of a node. What is the phase difference between the oscillations of X and Y?
Show mark scheme
Marking point
Mark
Notes
In a standing wave, all points between two adjacent nodes oscillate in phase. Points in adjacent loops oscillate in antiphase, so the phase difference is π rad. This is true wherever X and Y sit inside their loops, so the phase difference does not depend on their separation. That rules out D, which is the rule for a travelling wave. A treats the whole wave as one loop. B is the travelling-wave phase difference for a quarter-wavelength separation.
✓ 1
Answer C
Answer: C
Common error: Do not use Δφ = 2πΔx/λ for a standing wave. Phase changes only in jumps of π at each node.
18IBP2S-024-02-02Standing vs travelling wavesPaper 1AEasy1 mark
Which row correctly compares a travelling wave with a standing wave on a string?
Travelling wave / Standing wave
Show mark scheme
Marking point
Mark
Notes
A travelling wave carries energy in its direction of propagation. A standing wave is the superposition of two equal waves travelling in opposite directions, so there is no net energy flow; energy stays trapped between the nodes. B reverses this. C swaps the amplitude properties: amplitude is constant in a travelling wave but varies from zero at nodes to a maximum at antinodes in a standing wave. D swaps the phase properties.
✓ 1
Answer A
Answer: A
Exam tip: Learn the three comparisons as a set: energy transfer, amplitude and phase.
19IBP2S-024-02-09Standing waves on stringsPaper 1AEasy1 mark
A sitar string is fixed at both ends and vibrates in its fourth harmonic. How many nodes are there on the string, including the two ends?
Show mark scheme
Marking point
Mark
Notes
The fourth harmonic has four loops. Four loops are separated by three internal nodes, and there is a node at each fixed end, giving 3 + 2 = 5 nodes. Option A counts only the internal nodes. Option B confuses nodes with loops, which equal the number of antinodes. Option D doubles the harmonic number.
✓ 1
Answer C
Answer: C
Exam tip: For a string fixed at both ends in harmonic n, there are n antinodes and n + 1 nodes.
21C-1A-31Standing waves in pipesPaper 1AMedium1 mark
A pipe open at both ends has a first harmonic of frequency fopen. One end of the pipe is then closed, and the new first harmonic has frequency fclosed. The speed of sound is unchanged.
What is fclosed/fopen?
Show mark scheme
Marking point
Mark
Notes
Step 1Open at both ends: antinodes at both ends, so λ1 = 2L. Closed at one end: node to antinode, so λ1 = 4L.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2f = v/λ, so fclosed/fopen = 2L/4L = 1/2.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis takes the open pipe's first harmonic as one whole wavelength (λ = L); from antinode to antinode is only half a wavelength.
BCorrect: the first-harmonic wavelength doubles from 2L to 4L, so the frequency halves.
CThe length is unchanged, but the boundary condition at the closed end changes which wavelengths fit.
DThis is the inverse ratio. Closing one end lengthens the first-harmonic wavelength, which lowers the frequency.
The tension in a string fixed at both ends is increased to four times its original value. The length of the string does not change. The speed of waves on the string is proportional to the square root of the tension.
What is (new first-harmonic frequency)/(original first-harmonic frequency)?
Show mark scheme
Marking point
Mark
Notes
Step 1v ∝ √(tension), so the wave speed is multiplied by √4 = 2.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The first-harmonic wavelength, 2L, is unchanged, so f = v/2L is multiplied by 2.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis is the inverse ratio: a tighter string carries faster waves, so it has a higher first-harmonic frequency.
BThe wavelength is fixed by the length of the string, but the wave speed changes, so the frequency must change too.
CCorrect: the wave speed doubles and the wavelength of the first harmonic is unchanged, so the frequency doubles.
DThis applies the factor of 4 to the speed directly; the wave speed depends on the square root of the tension.
A system is made to oscillate by a periodic driving force whose frequency can be varied. The damping of the system is then increased.
Which statement describes the change to the graph of amplitude against driving frequency?
Show mark scheme
Marking point
Mark
Notes
Step 1Heavier damping removes energy from the system faster, so the amplitude at resonance is smaller.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The response is spread over a wider range of driving frequencies (a broader peak), and the frequency of maximum amplitude moves slightly below the natural frequency.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the maximum amplitude falls, the peak broadens, and the resonant frequency moves slightly lower.
BHeavier damping flattens the curve, spreading the response over a wider range of driving frequencies, so the peak becomes broader, not narrower.
CThe damping removes more energy per cycle, so the maximum amplitude must fall.
DThe height and width are right, but the shift is the wrong way: increased damping moves the peak slightly below the natural frequency.
As the drum of a washing machine speeds up, the whole machine vibrates violently at one particular drum speed. At higher drum speeds the vibration becomes much smaller.
Which statement explains this?
Show mark scheme
Marking point
Mark
Notes
Step 1The rotating drum is a periodic driving force; the amplitude of the forced oscillation is greatest when the driving frequency equals the natural frequency of the machine on its supports — resonance.
✓ 1
Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: at resonance energy is transferred from the driver to the machine most effectively, so the amplitude is at its largest.
BThe out-of-balance force from the drum grows as the drum speeds up; the large amplitude comes from matching frequencies, not from a larger force.
CThe natural frequency is fixed by the mass and stiffness of the machine and its supports; it is the driving frequency that changes.
DDamping does not disappear at any speed; it limits the amplitude at resonance to a finite value.
26C-1A-55Standing waves in pipesPaper 1AMedium1 mark
The air in a pipe that is open at both ends can form standing waves of two successive harmonics with frequencies 540 Hz and 720 Hz. The speed of sound in the air is v.
What is the length of the pipe?
Show mark scheme
Marking point
Mark
Notes
Step 1A pipe open at both ends has harmonics at every whole-number multiple of the first harmonic, so successive harmonics differ by f1: f1 = 720 − 540 = 180 Hz (these are the third and fourth harmonics).
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The first harmonic has λ = 2L, so f1 = v/2L and L = v/(2 × 180 Hz) = v/(360 Hz).
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis takes the first-harmonic wavelength as L. Between two antinodes there is only half a wavelength, so λ1 = 2L.
BCorrect: the difference between successive harmonics is the first-harmonic frequency, 180 Hz, and L = v/2f1.
CThis uses f1 = v/4L, which is the first harmonic of a pipe closed at one end; this pipe is open at both ends.
DThis treats 540 Hz as the first harmonic. Since 540 Hz and 720 Hz are successive harmonics, f1 is their difference, 180 Hz.
The air in a pipe that is closed at one end and open at the other vibrates in its fifth harmonic.
Which row gives the number of displacement nodes and the number of displacement antinodes in the air column? Include any at the ends of the pipe.
Displacement nodesDisplacement antinodes
Show mark scheme
Marking point
Mark
Notes
Step 1There is a displacement node at the closed end and an antinode at the open end, so the pipe holds an odd number of quarter-wavelengths; the fifth harmonic has L = 5λ/4.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Along the pipe the pattern is N A N A N A: 3 nodes and 3 antinodes.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: five quarter-wavelengths from a node at the closed end to an antinode at the open end give N A N A N A.
BThe node at the closed end has been left out. Air cannot move at a closed end, so it is always a displacement node.
CThis is the fifth harmonic of a pipe open at both ends (five half-wavelengths, antinodes at both ends). Here one end is closed.
DThis takes the harmonic number as the number of nodes and of antinodes. A closed pipe's fifth harmonic holds only five quarter-wavelengths.
A string is fixed at both ends and vibrates in its second harmonic.
Which graph shows how the amplitude A of oscillation of the points on the string varies with the distance x from one end of the string?
Sketch graphs, not drawn to scale. Each x-axis runs from one fixed end to the other.Show mark scheme
Marking point
Mark
Notes
Step 1A fixed end cannot move, so the amplitude is zero at both ends.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The second harmonic has two loops: a third node at the centre and antinodes at a quarter and three quarters of the length. Amplitude is a maximum displacement, so it is never negative: graph D.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis is the first harmonic: one loop with a single antinode at the centre. The second harmonic has two loops.
BThis is the displacement of the string at one instant of maximum displacement. The amplitude of a point is the size of its maximum displacement, so it cannot be negative.
CThis puts antinodes at the two ends. A fixed end is a node; this is the pattern of the second harmonic of a pipe open at both ends.
DCorrect: zero amplitude at the two fixed ends and at the centre, with maximum amplitude midway between them — two equal loops.
29C-1A-85Standing waves in pipesPaper 1AMedium1 mark
The air in a tube of length L that is closed at both ends vibrates in its first harmonic with frequency f.
One end of the tube is then opened. Later the other end is opened as well. The speed of sound in the air does not change.
Which row gives the frequency of the first harmonic of the air when one end is open and when both ends are open?
One end openBoth ends open
Show mark scheme
Marking point
Mark
Notes
Step 1Closed at both ends: a displacement node at each end, so half a wavelength fits in the tube: λ = 2L and f = v/2L.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2One end open: a node at the closed end and an antinode at the open end, so λ = 4L and the frequency is f/2. Both ends open: an antinode at each end, so λ = 2L again and the frequency is f.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis keeps λ = 2L when one end is opened. A node at one end and an antinode at the other fit only a quarter of a wavelength, so λ = 4L and the frequency halves.
BThis treats the tube closed at both ends like a tube closed at one end, with λ = 4L. With a node at each end, half a wavelength fits, so λ = 2L.
CCorrect: the same type of point at both ends (two nodes, or two antinodes) gives λ = 2L; a node at one end and an antinode at the other gives λ = 4L. So the frequency falls to f/2 and then returns to f.
DThe frequency with one end open is right, but with both ends open there is an antinode at each end, so λ = 2L again, not 4L: the frequency returns to f.
In an atomic force microscope, a tiny cantilever is driven at its natural frequency of 300 kHz and oscillates with a constant amplitude. Air resistance removes energy from the cantilever at a rate of 9.0 × 10−12 W.
How much energy does the driving force transfer to the cantilever during each oscillation?
Show mark scheme
Marking point
Mark
Notes
Step 1The amplitude is constant, so in each oscillation the driver supplies as much energy as air resistance removes.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2T = 1/f = 1/(3.00 × 105) = 3.33 × 10−6 s, so the energy per oscillation = PT = 9.0 × 10−12 × 3.33 × 10−6 = 3.0 × 10−17 J.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis multiplies the power by the frequency instead of by the period (energy = power × time).
BThis takes the period as 2π/f. The period is 1/f, so this answer is 2π times too large.
CCorrect: with a constant amplitude the energy supplied per cycle equals the energy dissipated per cycle, P/f = 3.0 × 10−17 J.
DThis divides the power by the angular frequency ω = 2πf instead of by f, so the time used is 2π times too short.
The shock absorbers of a car normally give damping that is close to critical. A garage replaces them with shock absorbers that give much heavier damping.
The car then goes over a single bump. Which row describes the motion of the car body afterwards, and the time it takes to settle at its equilibrium position, compared with the original shock absorbers?
Motion of the car bodyTime to settle
Show mark scheme
Marking point
Mark
Notes
Step 1Critical damping returns a system to equilibrium in the shortest possible time without oscillating.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Heavier damping also prevents oscillation, but the body creeps back to equilibrium more slowly, so it takes longer to settle.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AHeavy damping does prevent oscillation, but it slows the return: a heavily damped system takes longer than a critically damped one to reach equilibrium.
BCorrect: with heavy damping the body does not oscillate, but it returns to equilibrium more slowly than with critical damping.
COscillation with decreasing amplitude is light damping. Damping heavier than critical prevents any oscillation.
DThis describes light damping, and even then the body would take longer to settle because it keeps oscillating.
32IBP2S-024-03-03Standing vs travelling wavesPaper 1AMedium1 mark
Two air particles, P and Q, lie on opposite sides of the same displacement node in a longitudinal standing wave. Each particle is a distance λ/8 from the node. What is the phase difference between the oscillations of P and Q?
Show mark scheme
Marking point
Mark
Notes
In a standing wave, all particles between two adjacent nodes oscillate in phase. Particles on opposite sides of a node oscillate in antiphase. P and Q are separated by one node, so their phase difference is π rad. Option C applies the travelling-wave rule to a standing wave, since a separation of λ/4 gives π/2 for a travelling wave. Option B uses λ/8 as the separation in that same rule. Option A would be correct only if P and Q were in the same loop.
✓ 1
Answer D
Answer: D
Common error: Do not use phase difference = 2πx/λ for a standing wave. In a standing wave the phase difference is either 0 or π. It is 0 if the particles are in the same loop or separated by an even number of nodes, and π if they are separated by an odd number of nodes.
33IBP2S-024-03-04Standing waves in pipesPaper 1AMedium1 mark
The human ear canal can be modelled as a tube that is open at the outer end and closed by the eardrum. Its lowest resonant frequency is 3.4 kHz. Which of the following is not a resonant frequency of this model?
Show mark scheme
Marking point
Mark
Notes
A tube closed at one end supports only odd harmonics: f, 3f, 5f, 7f… This gives 3.4, 10.2, 17.0 and 23.8 kHz. The frequency 6.8 kHz is 2f, an even harmonic. An even harmonic would need a displacement antinode at the closed end, which is impossible. Options B, C and D are the 3rd, 5th and 7th harmonics.
✓ 1
Answer A
Answer: A
Exam tip: To identify allowed frequencies quickly, divide each frequency by the fundamental. For a tube closed at one end, the result must be an odd integer.
34IBP2S-024-03-07Standing waves in pipesPaper 1AMedium1 mark
A narrow ventilation duct in a building resonates in its fundamental mode. The air temperature rises, which increases the speed of sound. The duct length does not change. What happens to the fundamental frequency and to the wavelength of the fundamental?
Show mark scheme
Marking point
Mark
Notes
The length of the duct and the conditions at its ends fix the positions of the nodes and antinodes. This means the wavelength of each harmonic is fixed by the geometry. Since f = v/λ, a larger v with the same λ gives a higher frequency. Option B assumes the frequency is fixed, as it would be for a source with a set frequency. Option C assumes both quantities change. Option D reverses the dependence of f on v.
✓ 1
Answer A
Answer: A
Exam tip: For resonance in a pipe, the geometry fixes λ and the medium fixes v. The frequency f then follows from f = v/λ.
35IBP2S-024-03-08Standing waves in pipesPaper 1AMedium1 mark
A rocket combustion chamber of length 0.50 m behaves acoustically as a tube closed at both ends. A longitudinal instability is observed in its lowest mode at a frequency of 1.2 kHz. What is the speed of sound in the hot gas?
Show mark scheme
Marking point
Mark
Notes
Both closed ends are displacement nodes. The lowest mode has a single antinode at the centre, so L = λ/2 and λ = 1.0 m. Then v = fλ = 1200 × 1.0 = 1200 m s−1. Option B uses λ = L. Option D uses λ = 4L, which is correct only for a tube closed at one end. Option A uses λ = L/2.
✓ 1
Answer C
Answer: C
Common error: A tube closed at both ends has the same wavelength pattern as a tube open at both ends (λ = 2L/n). The only difference is that the nodes and antinodes are interchanged.
36IBP2S-024-03-10Standing vs travelling wavesPaper 1AMedium1 mark
A longitudinal standing wave is set up in a long spring used in a laboratory demonstration. Consider these statements. I. The amplitude of oscillation varies with position along the spring. II. There is a net transfer of energy along the spring. III. All coils between two adjacent nodes oscillate in phase. Which statements are correct?
Show mark scheme
Marking point
Mark
Notes
In a standing wave the amplitude varies from zero at the nodes to a maximum at the antinodes, so I is correct. All particles within one loop oscillate in phase, so III is correct. Statement II is wrong. A standing wave is the superposition of two equal waves travelling in opposite directions, so there is no net energy transfer. Energy is stored in each loop and does not pass through the nodes. The options containing II therefore reflect the misconception that a standing wave behaves like a travelling wave.
✓ 1
Answer B
Answer: B
Common error: Do not describe a standing wave as 'transferring energy'. That description belongs to travelling waves.
37IBP2S-024-01-04SuperpositionPaper 1AMedium1 mark
Noise-cancelling headphones detect an ambient sound wave of pressure amplitude 4.0 mPa. They emit a wave of the same frequency with amplitude 3.0 mPa, exactly in antiphase with it. What is the intensity of the resulting sound as a fraction of the intensity of the ambient sound?
Show mark scheme
Marking point
Mark
Notes
By superposition, the displacements of waves in antiphase subtract. The resultant amplitude is 4.0 − 3.0 = 1.0 mPa. Intensity is proportional to amplitude squared, so the fraction is (1.0/4.0)2 = 1/16. The option 1/4 is the amplitude ratio. The option 7/16 comes from subtracting intensities instead of amplitudes. The option 9/16 is the intensity of the cancelling wave alone.
✓ 1
Answer B
Answer: B
Common error: Superpose the amplitudes (displacements) first, then square the result. Intensities do not add or subtract directly.
38IBP2S-024-01-05Standing waves in pipesPaper 1AMedium1 mark
A sound engineer needs two pipes with the same fundamental frequency. Pipe P is open at both ends. Pipe Q is closed at one end. What is the ratio of their lengths, LP/LQ?
Show mark scheme
Marking point
Mark
Notes
For the open–open pipe, λ = 2LP. For the closed pipe, λ = 4LQ. The same frequency in the same air means the same wavelength. So 2LP = 4LQ, which gives LP/LQ = 2. The option 1/2 inverts the ratio. The option 4 compares 4L with L.
✓ 1
Answer C
Answer: C
Exam tip: Write λ in terms of L for each pipe, then set the two wavelengths equal.
39IBP2S-024-01-06Standing waves in pipesPaper 1AMedium1 mark
A smartphone speaker emits a steady 850 Hz tone above a tall measuring cylinder. The cylinder is slowly filled with water, which shortens the air column. Loud resonances are heard at successive air-column lengths that differ by 0.20 m. What is the speed of sound in the air?
Show mark scheme
Marking point
Mark
Notes
Successive resonances of a closed air column occur at lengths λ/4, 3λ/4, 5λ/4, and so on. These differ by λ/2, so λ = 0.40 m. Then v = fλ = 850 × 0.40 = 340 m s−1. The option 170 m s−1 treats 0.20 m as a whole wavelength. The option 680 m s−1 treats it as λ/4.
✓ 1
Answer C
Answer: C
Exam tip: Using the difference between successive resonance lengths cancels the end correction, which is why it is the preferred method.
40IBP2S-024-01-07Standing waves in pipesPaper 1AMedium1 mark
A tube of length L is closed at one end. The air inside vibrates at the second-lowest frequency at which resonance is possible. At which distances from the closed end are the displacement nodes?
Show mark scheme
Marking point
Mark
Notes
The second-lowest resonance of a closed tube is the third harmonic, so L = 3λ/4 and λ = 4L/3. The closed end is a displacement node, and nodes are spaced λ/2 = 2L/3 apart. The nodes are therefore at 0 and 2L/3. The open end at L is an antinode. The option L/3 and L places nodes at antinode positions. The option 0 and L/2 assumes the second harmonic, which cannot exist in a closed tube.
✓ 1
Answer A
Answer: A
Common error: The second resonance of a closed pipe is the third harmonic, not the second.
41IBP2S-024-01-09Resonance & dampingPaper 1AMedium1 mark
A building's air-conditioning duct hums loudly because a fan drives one of its resonant modes. Engineers line the inside of the duct with sound-absorbing foam. Compared with the unlined duct, how does the resonance curve (amplitude against driving frequency) for this mode change?
Show mark scheme
Marking point
Mark
Notes
The foam increases damping: it removes energy from the sound wave on each reflection. Greater damping lowers the maximum amplitude at resonance and makes the resonance peak broader (less sharp). The peak frequency stays close to the natural frequency, shifting only slightly lower. A narrower peak would mean less damping.
✓ 1
Answer A
Answer: A
Common error: Damping does not remove resonance. It reduces the peak amplitude and broadens the peak.
42IBP2S-024-00-03Resonance & dampingPaper 1AMedium1 mark
The suspension of a car uses springs together with shock absorbers. The shock absorbers are designed to give approximately critical damping. What is the main purpose of this?
Show mark scheme
Marking point
Mark
Notes
A critically damped system returns to equilibrium in the minimum possible time, with no overshoot or oscillation. This keeps the ride comfortable and keeps the tyres in contact with the road. Option A describes resonance, which is exactly what the design avoids. The springs still supply the restoring force. Constant amplitude would mean no damping at all.
✓ 1
Answer B
Answer: B
Exam tip: Learn the three cases: light damping (oscillates, amplitude decays), critical (fastest return with no oscillation) and heavy (no oscillation, slow return).
43IBP2S-024-00-04Tension & wave speedPaper 1AMedium1 mark
A string on a stringed instrument has fundamental frequency f. It is replaced by a string of the same material, length and tension, but with twice the diameter. What is the new fundamental frequency?
Show mark scheme
Marking point
Mark
Notes
Mass per unit length μ is proportional to the cross-sectional area, so it is proportional to d². Doubling d makes μ four times larger. Then v = √(T/μ) halves, and with λ = 2L unchanged, f halves. Option C assumes μ ∝ d. Option A forgets the square root in the wave speed equation.
✓ 1
Answer B
Answer: B
Common error: Mass per unit length depends on area (∝ d²), not on diameter alone.
44IBP2S-024-00-05Resonance & dampingPaper 1AMedium1 mark
A tall building has a natural frequency of sway of 0.8 Hz. During an earthquake the ground oscillates steadily at 0.5 Hz. Once steady oscillations are established, the building sways with a frequency of
Show mark scheme
Marking point
Mark
Notes
In forced oscillation, the steady-state motion always takes place at the driving frequency, here 0.5 Hz. The natural frequency affects only the amplitude of the response, which becomes large when the two frequencies are close. Options C and D wrongly average or subtract the two frequencies.
✓ 1
Answer A
Answer: A
Exam tip: A driven system oscillates at the driver's frequency. Its natural frequency controls how large the amplitude is.
45IBP2S-024-00-06Tension & wave speedPaper 1AMedium1 mark
A stay cable on a cable-stayed bridge is fixed at both ends. Gusting wind makes it vibrate strongly in its third harmonic. How many antinodes and nodes, including the end points, does the standing wave on the cable have?
Show mark scheme
Marking point
Mark
Notes
The third harmonic on a string fixed at both ends fits three half-wavelengths into the length L. This gives 3 antinodes. Nodes sit at both fixed ends and between each pair of loops, giving 4 nodes in total. Option A misses the end nodes. Option D treats the third harmonic as three full wavelengths.
✓ 1
Answer B
Answer: B
Common error: The fixed ends of a string are always nodes, so they must be counted.
46IBP2S-024-00-07Tension & wave speedPaper 1AMedium1 mark
A piano wire sounds a fundamental of 440 Hz. Over a hot day the technician finds the wire tension has risen by 10% while the length and mass per unit length are unchanged. The new fundamental frequency is approximately
Show mark scheme
Marking point
Mark
Notes
f ∝ √T when L and μ are fixed. The new frequency is therefore 440 × √1.10 ≈ 440 × 1.05 ≈ 462 Hz. Option C assumes f ∝ T. Option A assumes the length alone sets the frequency. Option D has the relationship inverted.
✓ 1
Answer B
Answer: B
Exam tip: For small changes, √(1 + x) ≈ 1 + x/2. A 10% rise in tension therefore gives about a 5% rise in frequency.
47IBP2S-024-00-10Resonance & dampingPaper 1AMedium1 mark
A diver jumps off the end of a springboard. The board then oscillates with light damping caused by air resistance and internal friction. Which describes the subsequent motion of the board?
Show mark scheme
Marking point
Mark
Notes
With light damping the amplitude decays gradually, by roughly the same fraction each cycle. The period stays very close to the undamped natural period. The period does not depend on amplitude for this kind of oscillation. A linear decrease to zero (option C) is not the behaviour of a damped oscillator of this type.
✓ 1
Answer A
Answer: A
Exam tip: Light damping changes the amplitude, not (to a good approximation) the frequency.
48IBP2S-024-02-03Standing waves on stringsPaper 1AMedium1 mark
A tennis racket string has a vibrating length of 0.30 m between its fixed ends. It vibrates in its third harmonic. What is the wavelength of the standing wave?
Show mark scheme
Marking point
Mark
Notes
The third harmonic on a string fixed at both ends has three loops. Each loop is half a wavelength, so 3(λ/2) = 0.30 m, giving λ = 0.20 m. Option A, 0.10 m, is the node-to-node distance (half a wavelength). Option C assumes the length equals one wavelength. Option D multiplies the length by the harmonic number instead of dividing.
✓ 1
Answer B
Answer: B
Common error: One loop is half a wavelength, not a whole wavelength.
49IBP2S-024-02-04Tension & wave speedPaper 1AMedium1 mark
A cellist presses a string so that its vibrating length becomes three-quarters of its original length. The wave speed on the string is unchanged. What is the ratio (new fundamental frequency) / (original fundamental frequency)?
Show mark scheme
Marking point
Mark
Notes
For the fundamental, λ = 2L and f = v/2L. With v constant, f is proportional to 1/L. Reducing L to 3L/4 multiplies f by 4/3. Option A assumes f is proportional to L. Options C and D wrongly square the ratio.
✓ 1
Answer B
Answer: B
Exam tip: A shorter string gives a shorter wavelength, and at fixed speed that means a higher frequency.
50IBP2S-024-02-05Standing vs travelling wavesPaper 1AMedium1 mark
Which statement correctly describes the motion of the points on a string that lie between two adjacent nodes of a standing wave?
Show mark scheme
Marking point
Mark
Notes
Within one loop, every point reaches its maximum displacement at the same instant, so all points are in phase. The amplitude rises from zero at each node to a maximum at the antinode. Option A describes a travelling wave. Option B ignores the variation of amplitude. Option D mixes up behaviour within one loop with behaviour across adjacent loops.
✓ 1
Answer C
Answer: C
Common error: Saying the amplitude is the same everywhere describes a travelling wave, not a standing wave.
51IBP2S-024-02-06Standing waves in pipesPaper 1AMedium1 mark
An engineer drives a guy wire on a radio mast with a variable-frequency shaker. Both ends of the wire are fixed. Two successive resonant frequencies are 240 Hz and 300 Hz. What is the fundamental frequency of the wire?
Show mark scheme
Marking point
Mark
Notes
For a string fixed at both ends, the resonant frequencies are nf₁ with n = 1, 2, 3, … Successive harmonics therefore differ by f₁, so f₁ = 300 − 240 = 60 Hz. This is consistent: 240 Hz = 4 × 60 Hz and 300 Hz = 5 × 60 Hz. Option A halves the difference, a rule that applies to a pipe closed at one end, where only odd harmonics exist. Option D assumes the lower frequency is the fundamental.
✓ 1
Answer B
Answer: B
Exam tip: For a string fixed at both ends, the gap between successive resonances equals the fundamental frequency.
52IBP2S-024-02-08SuperpositionPaper 1AMedium1 mark
A long tether connects two parts of a spacecraft and is held taut, with both ends fixed. A small pulse sent along the tether takes time t to travel from one end to the other and back again. What is the fundamental frequency of standing waves on the tether?
Show mark scheme
Marking point
Mark
Notes
The round trip covers 2L, so t = 2L/v. The fundamental has λ = 2L, so f₁ = v/2L = 1/t. Option B uses the one-way distance with the round-trip time. Option A is the result for a tether fixed at only one end. Option D doubles instead of halving.
✓ 1
Answer C
Answer: C
Exam tip: The period of the fundamental equals the time for a wave to travel to the far end and back.
53C-1A-30Standing waves in pipesPaper 1AHard1 mark
A pipe of length L is closed at one end and open at the other. The speed of sound in the air in the pipe is v.
What is the frequency of the standing wave with the lowest frequency above that of the first harmonic?
Show mark scheme
Marking point
Mark
Notes
Step 1There is a displacement node at the closed end and an antinode at the open end, so the pipe holds an odd number of quarter-wavelengths: f1 = v/4L, and only odd multiples of f1 occur.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The next standing wave is therefore the third harmonic, 3v/4L.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis is the first harmonic itself, not the next standing wave above it.
BThis is 2f1. A pipe closed at one end cannot have an antinode at the closed end, so even harmonics do not occur.
CCorrect: only odd harmonics fit a pipe closed at one end, so the next one above v/4L is 3v/4L.
DThis is the second harmonic of a pipe open at both ends, 2 × v/2L; here one end is closed.
A standing wave is set up on a string fixed at both ends. The diagram shows the string at one instant of maximum displacement (solid line) and half a period later (dashed line). P, Q and R are points on the string. The distance PQ is λ/4 and the distance PR is 3λ/4, where λ is the wavelength.
Which row gives the phase difference between the oscillations of P and Q, and between those of P and R?
Diagram drawn to scale
Between P and QBetween P and R
Show mark scheme
Marking point
Mark
Notes
Step 1P and Q lie between the same pair of adjacent nodes, so they oscillate in phase: phase difference 0.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2R lies in the neighbouring loop, on the other side of a node, so it oscillates in antiphase with P: phase difference π.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis applies the travelling-wave rule, phase difference = 2π × separation/λ, to both pairs; in a standing wave the phase does not change steadily with position.
BCorrect: points in the same loop are in phase, and points in adjacent loops are in antiphase.
CP and R are right, but P and Q have been given the travelling-wave phase difference for points λ/4 apart; they are in the same loop, so they are in phase.
DP and Q are right, but P and R have been given the travelling-wave phase difference for points 3λ/4 apart; one node separates them, so they are in antiphase.
55C-1A-33Standing waves on stringsPaper 1AHard1 mark
A string is clamped at end X. Its other end, Y, is tied to a light ring that slides without friction on a vertical rod, so that Y is a free end.
Which diagram shows the third harmonic of the string? Each diagram shows the two extreme positions of the string.
Diagrams NOT accurately drawnShow mark scheme
Marking point
Mark
Notes
Step 1The clamped end X is a node and the free end Y is an antinode, so the string holds an odd number of quarter-wavelengths; the first harmonic is a quarter-wavelength.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The third harmonic has three times the first-harmonic frequency, so three quarter-wavelengths fit on the string: node at X, one node between, antinode at Y.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: a node at X, an antinode at Y and three quarter-wavelengths on the string, so the frequency is three times that of the first harmonic.
BThe boundary conditions are reversed: this has an antinode at the clamped end X and a node at the free end Y.
CThis has five quarter-wavelengths on the string: it is the third possible standing wave, but its frequency is 5 times that of the first harmonic, so it is the fifth harmonic.
DThis is the third harmonic of a string fixed at both ends; it puts a node at Y, but a free end is an antinode.
One end of a long, heavy rope is tied to a wall in a gym. An athlete moves the other end up and down so that waves of frequency 2.5 Hz and wavelength 1.8 m travel along the rope. The waves are reflected at the wall and a standing wave forms.
Which row gives the time between successive instants at which the rope is straight and the distance between adjacent nodes?
Time between straight instantsDistance between adjacent nodes
Show mark scheme
Marking point
Mark
Notes
Step 1T = 1/f = 1/2.5 = 0.40 s. All points of a standing wave pass through equilibrium together twice in each cycle, so the rope is straight every T/2 = 0.20 s.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Adjacent nodes are half a wavelength apart: 1.8/2 = 0.90 m.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
A0.40 s is the period. The rope is straight twice in every cycle — once as the loops move one way through equilibrium and once as they move back — so the interval is half a period.
BCorrect: the rope is straight every half period, 0.40/2 = 0.20 s, and adjacent nodes are λ/2 = 0.90 m apart.
CThe time is right, but adjacent nodes are half a wavelength apart, not a whole wavelength.
DThese are quarter values: T/4 is the time from maximum displacement to straight, and λ/4 is the distance from a node to the nearest antinode.
57IBP2S-024-03-09Standing waves in pipesPaper 1AHard1 mark
A pipe of length L is closed at one end and open at the other. It resonates at its third harmonic. Point X lies on the axis of the pipe, a distance L/3 from the closed end. Which row describes X?
Show mark scheme
Marking point
Mark
Notes
For the third harmonic, L = 3λ/4, so λ = 4L/3. Measured from the closed end (which is a displacement node), the displacement nodes are at 0 and λ/2 = 2L/3. The displacement antinodes are at λ/4 = L/3 and at L. X is therefore a displacement antinode. Displacement antinodes are pressure nodes. Option B swaps the two descriptions. Options C and D are impossible, because displacement and pressure nodes are always offset by λ/4.
✓ 1
Answer A
Answer: A
Exam tip: Express λ in terms of L first. Then list the node positions from the closed end, using steps of λ/2.
58IBP2S-024-01-10Standing waves in pipesPaper 1AHard1 mark
Three consecutive resonant frequencies of a pipe open at both ends are 240 Hz, 360 Hz and 480 Hz. One end of the pipe is then sealed. The length of the pipe is unchanged. What is the lowest resonant frequency of the sealed pipe?
Show mark scheme
Marking point
Mark
Notes
For an open–open pipe, consecutive harmonics differ by f1, so f1 = 120 Hz and λ = 2L. Sealing one end makes the fundamental wavelength 4L, which is twice as long. The frequency therefore halves to 60 Hz. The option 120 Hz ignores the change in boundary condition. The option 240 Hz mistakes the lowest listed frequency for the fundamental. The option 30 Hz halves the frequency twice.
✓ 1
Answer B
Answer: B
Exam tip: Find f₁ from the spacing of consecutive harmonics first. Do not assume the lowest frequency given is the fundamental.
59IBP2S-024-00-08Tension & wave speedPaper 1AHard1 mark
Two wires X and Y are made of the same metal and are under the same tension. Each is fixed at both ends. Wire X has length L and diameter d. Wire Y has length 2L and diameter d/2. What is the ratio \(\frac{f_Y}{f_X}\) of their fundamental frequencies?
Show mark scheme
Marking point
Mark
Notes
μ ∝ d², so μY = μX/4. Then v = √(T/μ) gives vY = 2vX. The fundamental wavelength is 2L, so λY = 2λX. Therefore fY/fX = (2vX/2λX)/(vX/λX) = 1. Option B considers only the length change. Option D considers only the diameter change.
✓ 1
Answer C
Answer: C
Exam tip: When several quantities change at once, write f = (1/2L)√(T/μ) and substitute each ratio separately.
60IBP2S-024-02-07Standing vs travelling wavesPaper 1AHard1 mark
A string vibrates in a standing wave. At one instant every point on the string has zero displacement, so the string is straight. Which statement about this instant is correct?
Show mark scheme
Marking point
Mark
Notes
Each point performs simple harmonic motion about its equilibrium position. Zero displacement is the equilibrium position, where speed is greatest. The antinodes have the largest amplitude, so at this instant they move fastest and the energy is all kinetic. Option A describes the instant of maximum displacement. Option B confuses zero displacement with zero energy. Option D is wrong because nodes never move.
✓ 1
Answer C
Answer: C
Common error: A straight string does not mean a still string. At this instant the energy is entirely kinetic.
61IBP2S-024-02-10Standing waves on stringsPaper 1AHard1 mark
A string fixed at both ends vibrates in its second harmonic at frequency f. A guitarist then lightly touches the string at a point one-third of its length from one end. This forces a node at that point, and the string vibrates at the lowest frequency now possible. What is this frequency?
Show mark scheme
Marking point
Mark
Notes
The fundamental is f₁ = f/2. A node at L/3 requires L/3 to be a whole number of half-wavelengths. The longest wavelength that allows this is λ = 2L/3, which is the third harmonic. Its frequency is 3f₁ = 3f/2. Option D wrongly treats f as the fundamental. Options A and B divide when they should multiply, as if the string were made longer.
✓ 1
Answer C
Answer: C
Common error: Always convert the given frequency back to the fundamental before scaling to another harmonic.
62C-1B-05Standing waves in pipesPaper 1BEasy8 marks
A narrow plastic tube is closed at one end by a piston that can slide along the tube, so that the length L of the air column can be changed. A small loudspeaker connected to a signal generator is placed at the open end. For each length the student increases the frequency from a low value and records the frequency f at which the sound first becomes loudest. This is the first harmonic of the air column. The uncertainty in f is ±3 Hz. The air temperature is 22 °C.
L / m ± 0.002 m
(1/L) / m−1
f / Hz ± 3 Hz
0.200
5.00
428
0.250
4.00
345
0.300
287
0.400
2.50
217
0.500
2.00
173
0.600
1.67
143
Graph drawn to scale
(a)
State where the displacement node and the displacement antinode are for the first harmonic.
(1)
(b)
Complete the table for L = 0.300 m.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the speed of sound in the tube using the graph.
(3)
(e)
The accepted speed of sound at 22 °C is 344 m s−1. Calculate the percentage difference between the student's value and the accepted value.
(1)
(f)
Suggest one improvement to the method of finding f.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Node at the piston (closed end) AND antinode at the open end
✓ 1
Both needed.
Part (b)
3.33 m−1
✓ 1
Accept 3.3 m−1.
Part (c)
Single straight line passing close to the origin with points on both sides
✓ 1
Part (d)
gradient = 85.4 Hz m (m s−1)
✓ 1
Accept 84 to 88. Must use a large triangle.
For a closed tube L = λ/4, so f = v/(4L) and gradient = v/4
✓ 1
v = 4 × 85.4 = 342 m s−1
✓ 1
Accept 336 to 352 m s−1. Allow ECF from the gradient.
Part (e)
|342 − 344|/344 × 100 = 0.6 %
✓ 1
Allow ECF from (d).
Part (f)
Use a microphone connected to an oscilloscope/data-logger to find the frequency of maximum amplitude, OR find the resonance approaching from above and from below and take the mean
✓ 1
Judging 'loudest' by ear is subjective. Do not accept 'repeat' alone.
Answers: (b) 3.33 m−1 · (d) v ≈ 342 m s−1 · (e) ≈ 0.6 % (the remaining parts are explanations — see the table above)
63C-1B-09Standing waves on stringsPaper 1BEasy8 marks
One end of a string is attached to a vibration generator and the other passes over a pulley to a hanging mass. The vibrating length between the generator and the pulley is (1.500 ± 0.005) m. The frequency of the generator is slowly increased, and the frequency f is recorded each time a large-amplitude standing wave with n loops forms. Each frequency has an uncertainty of ±0.5 Hz.
number of loops n
f / Hz ± 0.5 Hz
1
15.8
2
32.2
3
48.1
4
64.1
5
79.7
6
96.0
Graph drawn to scale
(a)
Show that the wavelength of the standing wave with three loops is 1.00 m.
(1)
(b)
State the phase difference between a point in one loop and a point in the adjacent loop.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the gradient of the line of best fit.
(2)
(e)
Determine the speed of the waves on the string.
(2)
(f)
Suggest one way of judging more precisely the frequency at which each standing wave forms.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Three loops contain 3 half-wavelengths, so λ = 2L/3 = 2 × 1.500/3 = 1.00 m
✓ 1
Must see the reasoning.
Part (b)
π rad (180°)
✓ 1
Accept 'antiphase'. Do not accept π/2.
Part (c)
Single straight line passing close to the origin with points on both sides
✓ 1
Part (d)
Uses a triangle covering more than half the line
✓ 1
gradient = 16.0 Hz
✓ 1
Accept 15.7 to 16.3 Hz.
Part (e)
f = nv/(2L), so gradient = v/(2L)
✓ 1
v = 2 × 1.500 × 16.0 = 48.0 m s−1
✓ 1
Accept 47 to 49 m s−1. Allow ECF from (d). Award [2] for CNA.
Part (f)
Approach each resonance from both a lower and a higher frequency and take the mean, OR use a stroboscope/video to find the frequency of maximum amplitude
✓ 1
Do not accept 'repeat' without a method.
Answers: (a) 1.00 m · (b) π rad · (d) 16.0 Hz · (e) v ≈ 48.0 m s−1(the remaining parts are explanations — see the table above)
A microwave transmitter faces a flat metal plate. A small probe receiver is moved along the line between them, and the student records the distance x from the plate of each position where the received signal is a minimum. The minima are numbered N = 1, 2, 3 … starting with the one nearest the plate. Each position has an uncertainty of ±0.2 cm.
N
x / cm ± 0.2 cm
1
1.4
2
2.9
3
4.3
4
5.7
5
7.1
6
8.5
7
10.0
8
11.4
Graph drawn to scale
(a)
Explain how the pattern of maxima and minima is formed.
(2)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the wavelength of the microwaves using the graph.
(2)
(d)
The uncertainty in the gradient is ±2 %. Determine the frequency of the microwaves and its absolute uncertainty.
(2)
(e)
The manufacturer states that the transmitter emits at 10.5 GHz. Comment on your answer to (d).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The microwaves reflect from the plate, and the reflected wave travels back towards the transmitter with the same frequency and speed as the incident wave
✓ 1
Must refer to reflection.
The two waves travelling in opposite directions superpose to form a standing wave: the minima are nodes (destructive superposition) and the maxima are antinodes
✓ 1
OWTTE.
Part (b)
Single straight line through all error bars, passing close to the origin
✓ 1
The plate is a node at x = 0.
Part (c)
gradient = 1.42 cm = distance between adjacent nodes = λ/2
✓ 1
Accept 1.40 to 1.46 cm.
λ = 2 × 1.42 = 2.85 cm
✓ 1
Accept 2.80 to 2.92 cm. Allow ECF. Award [2] for CNA.
A model of a car suspension consists of a mass hanging from a spring. A light vane fixed below the mass dips into a beaker of oil, which damps the motion. The mass is pulled down and released, and a motion sensor records its displacement. From the record, the student reads the amplitude A of every second oscillation, with an uncertainty of ±0.5 mm. The oscillation number n counts the oscillations completed since the first amplitude was measured.
The student's hypothesis is that the amplitude falls by the same fraction in every oscillation. If so, a graph of ln( A / mm ) against n is a straight line. The graph shows ln( A / mm ) against n with error bars.
n
A / mm ± 0.5 mm
ln( A / mm )
0
44.9
3.80
2
34.9
3.55
4
26.9
3.29
6
20.7
8
16.1
2.78
10
12.4
2.52
12
9.9
2.29
Graph drawn to scale
(a)
Complete the table for n = 6.
(1)
(b)
Explain why the error bars on the graph become longer as n increases.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the gradient of the line of best fit.
(2)
(e)
Deduce whether the data support the hypothesis and determine the fraction of the amplitude lost in each oscillation.
(2)
(f)
The motion sensor cannot detect amplitudes smaller than 2.0 mm. Predict, using your line, the number of complete oscillations after which the oscillation can no longer be detected.
(2)
(g)
The oil is replaced by a more viscous oil. State and explain the effect on the gradient of the graph.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
ln 20.7 = 3.03
✓ 1
Accept 3 decimal places.
Part (b)
The uncertainty in ln A is about ΔA/A; ΔA is constant «±0.5 mm» while A decreases, so the fractional uncertainty, and the error bar, grows
✓ 1
OWTTE.
Part (c)
Single straight line through all the error bars
✓ 1
Part (d)
Uses a triangle covering more than half the line
✓ 1
gradient = −0.127
✓ 1
Accept −0.131 to −0.124. No unit.
Part (e)
The points lie on a straight line «within their error bars», so the amplitude falls by the same fraction in every oscillation: the hypothesis is supported
✓ 1
An+1/An = e−0.127 = 0.88, so about 12 % of the amplitude is lost in each oscillation
✓ 1
Allow ECF from (d). Accept 11 % to 14 %.
Part (f)
ln 2.0 = 0.69; n = (0.69 − 3.80)/(−0.127) = 24.4
✓ 1
Accept extrapolation of the line drawn on the graph or use of the equation of the line.
so after 25 complete oscillations
✓ 1
Accept 23 to 26. Allow ECF from (d).
Part (g)
The gradient becomes steeper «more negative»: the larger damping force removes a larger fraction of the energy in each oscillation, so the amplitude falls by a larger fraction per cycle
✓ 1
OWTTE. Direction AND reason needed.
Answers: (a) 3.03 · (d) −0.127 · (e) about 12 % per oscillation · (f) 25 oscillations (the remaining parts are explanations — see the table above)
A thin steel strip is clamped horizontally to a vibration generator so that a length L of the strip overhangs the clamp. The student increases the frequency of the generator slowly and records the frequency f at which the free end of the strip vibrates with the largest amplitude. This is repeated for different lengths L.
The student's hypothesis is that f is proportional to 1/L2. The graph shows f against 1/L2.
L / m ± 0.001 m
(1/L2) / m−2
f / Hz ± 0.3 Hz
0.180
30.9
25.0
0.200
25.0
20.6
0.225
19.8
16.0
0.250
16.0
13.2
0.300
11.1
8.9
0.350
8.2
6.8
0.450
4.9
4.1
Graph drawn to scale
(a)
Outline why the frequency at which the amplitude is largest is the natural frequency of the strip.
(1)
(b)
Calculate the percentage uncertainty in 1/L2 for L = 0.180 m.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the gradient of the line, including its unit.
(2)
(e)
Outline whether the data support the hypothesis.
(1)
(f)
Calculate the overhang length for which the natural frequency of the strip is 12.0 Hz.
(1)
(g)
Suggest one improvement to the way the student finds f.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The amplitude is largest at resonance, which occurs when the driving frequency equals the natural frequency of the strip
✓ 1
OWTTE.
Part (b)
2 × (0.001/0.180) × 100 = 1.1 %
✓ 1
Factor of 2 needed. Accept 1 %.
Part (c)
Single straight line through or close to all the points, passing close to the origin
✓ 1
Part (d)
gradient = 0.81, using a triangle covering more than half of the line
✓ 1
Accept 0.79–0.83.
unit: Hz m2
✓ 1
Accept s−1 m2.
Part (e)
The points lie on a straight line through the origin «within the uncertainty», so f ∝ 1/L2: the hypothesis is supported
✓ 1
Both straight AND through the origin needed.
Part (f)
L = √(0.81/12.0) = 0.26 m
✓ 1
Allow ECF from (d). Accept 0.255–0.264 m. Accept a reading from the graph «1/L2 = 14.8 m−2».
Part (g)
Find the frequency of maximum amplitude approaching from below AND from above and take the mean, OR use a slow-motion video / motion sensor to find the frequency of maximum amplitude, OR take readings at smaller frequency steps near the maximum
✓ 1
Do not accept 'repeat' without a method.
Answers: (b) 1.1 % · (d) 0.81 Hz m2 · (f) 0.26 m (the remaining parts are explanations — see the table above)
A trolley of mass (0.500 ± 0.001) kg is held between two springs on a horizontal track. The combined spring constant, measured in a separate experiment, is (40.0 ± 1.0) N m−1. The far end of one spring is moved back and forth by a motor at a driving frequency f. When the motion is steady, a motion sensor records the amplitude A of the trolley. Each amplitude has an uncertainty of ±0.2 cm.
f / Hz
A / cm ± 0.2 cm
0.60
1.2
0.80
1.4
1.00
1.9
1.20
3.4
1.40
7.8
1.60
3.3
1.80
1.6
2.00
1.0
Graph drawn to scale
(a)
Draw the curve of best fit on the graph.
(1)
(b)
Estimate the resonant frequency from the graph.
(1)
(c)
Determine the natural frequency of the trolley–spring system and its absolute uncertainty, using the values of mass and spring constant.
(3)
(d)
Comment on your answers to (b) and (c).
(1)
(e)
Suggest an improvement to the procedure that would allow the resonant frequency to be found more precisely.
(1)
(f)
A large piece of card is fixed to the trolley so that air resistance increases. Describe two changes to the graph of A against f.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Smooth curve with a single peak between 1.4 Hz and 1.5 Hz, rising above the highest point, and falling on both sides through the error bars
✓ 1
Do not accept dot-to-dot.
Part (b)
1.4 Hz
✓ 1
Accept 1.38 to 1.48 Hz.
Part (c)
f0 = (1/2π)√(k/m) = (1/2π)√(40.0/0.500) = 1.42 Hz
✓ 1
Uses T = 2π√(m/k) and f = 1/T.
Δf/f = ½(Δk/k + Δm/m) = ½(0.025 + 0.002) = 0.014
✓ 1
The factor of ½ from the square root is needed.
f0 = (1.42 ± 0.02) Hz
✓ 1
MP3 is for matching the precision of value and uncertainty.
Part (d)
The resonant frequency from the graph lies within the range 1.40–1.44 Hz, so resonance occurs at (close to) the natural frequency
✓ 1
Allow ECF from (b) and (c).
Part (e)
Take more readings at smaller intervals of driving frequency between about 1.2 Hz and 1.6 Hz (close to the peak)
✓ 1
Must be specific to the region near the peak.
Part (f)
The maximum amplitude is smaller (peak lower)
✓ 1
Accept 'amplitude smaller at all frequencies, especially near resonance'.
The peak is broader OR the resonant frequency is slightly lower
✓ 1
Either for the second mark.
Answers: (b) ≈ 1.4 Hz · (c) f0 = (1.42 ± 0.02) Hz (the remaining parts are explanations — see the table above)
68C-1B-17Standing waves in pipesPaper 1BMedium10 marks
A student investigates the air in a plastic pipe of length (0.850 ± 0.002) m. One end of the pipe is open; the student cannot see whether the far end has been sealed. A small loudspeaker at the open end plays a short burst of noise, and a microphone connected to a computer records the sound. The frequency spectrum shows peaks at the resonant frequencies of the air column. The frequencies are listed in order of increasing frequency as peak number n. Each frequency has an uncertainty of ±2 Hz.
peak number n
2n − 1
f / Hz ± 2 Hz
1
1
101
2
3
303
3
5
504
4
7
706
5
9
908
Graph drawn to scale
(a)
Show that the ratio of the frequencies of the second and first peaks is about 3.
(1)
(b)
Deduce, with reference to the data, whether the far end of the pipe is sealed.
(2)
(c)
The graph shows f against (2n − 1). Draw the line of best fit on the graph.
(1)
(d)
Determine the speed of sound in the pipe using the graph.
(3)
(e)
The uncertainty in the gradient is ±0.5 Hz. Determine the absolute uncertainty in the speed of sound.
(2)
(f)
The accepted speed of sound at the temperature of the laboratory is 343 m s−1. Comment on the student's result.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
303/101 = 3.00
✓ 1
Must see the division AND an answer to at least 3 s.f.
Part (b)
The frequencies are in the ratio 1 : 3 : 5 : 7 : 9 (only odd multiples of the lowest frequency, 101 Hz)
✓ 1
Must use the data, e.g. two further ratios.
This is the pattern for a pipe closed at one end (an open pipe would give all whole-number multiples 1 : 2 : 3 …), so the far end is sealed
✓ 1
Conclusion must follow from the ratios.
Part (c)
Single straight line through all points and passing close to the origin
✓ 1
Part (d)
gradient = 100.9 Hz
✓ 1
Accept 99 to 103 Hz.
For a pipe closed at one end f = (2n − 1)v/(4L), so gradient = v/(4L)
✓ 1
v = 4 × 0.850 × 100.9 = 343 m s−1
✓ 1
Accept 337 to 350 m s−1. Allow ECF. Award [3] for CNA with the relation seen.
Part (e)
Δv/v = 0.5/100.9 + 0.002/0.850 = 0.0073
✓ 1
Both fractional uncertainties needed.
Δv = ±3 m s−1, so v = (343 ± 3) m s−1
✓ 1
Accept ±2.5 m s−1. MP2 includes matching the precision of value and uncertainty.
Part (f)
343 m s−1 lies within the range v ± Δv, so the result is consistent with the accepted value
✓ 1
Allow ECF from (d) and (e).
Answers: (a) 3.00 · (d) v ≈ 343 m s−1 · (e) Δv ≈ ±3 m s−1(the remaining parts are explanations — see the table above)
69C-1B-38Resonance & dampingPaper 1BMedium12 marks
A student investigates how the damping of a driven oscillator affects its amplitude at resonance. A magnet of mass 0.250 kg hangs from a spring whose upper end is attached to a vibration generator. The generator moves up and down with an amplitude of 3.0 mm at a frequency that can be varied.
A stack of n identical thin aluminium sheets stands vertically beside the path of the magnet, close to it but not touching it. The sheets exert a damping force on the moving magnet, and the damping increases with n. For each value of n the student varies the driving frequency, finds the resonant frequency fr, and measures the maximum amplitude Amax of the magnet from a video. The student's hypothesis is that Amax is inversely proportional to n.
The graph shows 1/Amax against n for six of the values of n, with error bars for the uncertainty in Amax.
number of sheets n
Amax / mm ± 1 mm
(1/Amax) / m−1
fr / Hz ± 0.01 Hz
0
75
13.3
1.59
1
47
21.3
1.59
2
34
29.4
1.59
3
27
1.59
4
22
45.5
1.58
5
19
52.6
1.58
6
16
62.5
1.58
Graph drawn to scale. The point for n = 3 has not been plotted.
(a)
Calculate 1/Amax for n = 3 and its absolute uncertainty.
(2)
(b)
On the graph, plot the point for n = 3 with its error bar, and draw the line of best fit.
(2)
(c)
(i)
Determine the gradient of the line of best fit.
(2)
(ii)
Outline what the intercept on the vertical axis shows about the oscillator.
(1)
(d)
Explain whether the data support the student's hypothesis.
(2)
(e)
In a design based on this oscillator, the amplitude at resonance must be no more than 20 mm. Determine, using your line, the smallest number of sheets needed.
(1)
(f)
State and explain whether the data show that the resonant frequency decreases as the damping increases.
Accept 1.3–1.5 m−1, or ±1 m−1 if the value is given as 37 m−1. Allow ECF from MP1.
Part (b)
Point plotted at (3, 37.0 m−1) with an error bar from about 35.6 to 38.4 m−1
✓ 1
Allow ECF from (a). Within half a small square.
Single straight line passing through all the error bars, with points on both sides
✓ 1
The line must not be forced through the origin.
Part (c)(i)
Uses a triangle covering more than half of the line
✓ 1
gradient = 8.0 «m−1 per sheet»
✓ 1
Accept 7.5–8.4.
Part (c)(ii)
The intercept «≈ 13 m−1» is non-zero: the oscillator is damped even with no sheets «air resistance, energy losses in the spring», so Amax is finite when n = 0
✓ 1
OWTTE.
Part (d)
If Amax ∝ 1/n, then 1/Amax ∝ n and the line would pass through the origin
✓ 1
The line has a non-zero intercept «≈ 13 m−1, much larger than the error bars», so the hypothesis is not supported «1/Amax is a linear function of n»
✓ 1
MP2 only scores if MP1 scores. Allow ECF from (c).
Part (e)
Amax ≤ 20 mm means 1/Amax ≥ 50 m−1: n ≥ (50 − 13.3)/8.0 = 4.6, so 5 sheets «4 sheets is not enough»
✓ 1
Allow ECF from (c). Accept a reading from the line. The answer must be a whole number, rounded up.
Part (f)
fr falls by only 0.01 Hz from n = 0 to n = 6, which is equal to the uncertainty in each reading
✓ 1
Must quote the change or the values.
so the data do not show a decrease conclusively «a small decrease is possible but cannot be resolved»
✓ 1
MP2 only scores if MP1 scores. Do not accept 'yes, it decreases' with no reference to the uncertainty.
Answers: (a) (37.0 ± 1.4) m−1 · (c)(i) 8.0 m−1 per sheet · (e) 5 sheets (the remaining parts are explanations — see the table above)
70C-1B-39Standing waves on stringsPaper 1BMedium12 marks
A student investigates the positions of the frets on a guitar. When a string is pressed against fret n, the vibrating length L of the string becomes the distance from that fret to the bridge. For the open string (n = 0), L = 648.0 mm. The student measures each length with a steel tape, with an uncertainty of ±0.5 mm. Assume that pressing the string does not change its tension.
The student's hypothesis is that each fret raises the frequency of the first harmonic by the same factor r. If so, ln(L / mm) = ln(L0 / mm) − n ln r, where L0 is the length of the open string.
The graph shows ln(L / mm) against n for the data in the table, except n = 6.
fret number n
L / mm ± 0.5 mm
ln(L / mm)
0
648.0
6.474
2
577.5
6.359
4
514.0
6.242
6
458.5
8
408.0
6.011
10
364.0
5.897
12
324.5
5.782
Graph drawn to scale. The point for n = 6 has not been plotted.
(a)
Outline why the frequency of the first harmonic of the pressed string is inversely proportional to L.
(2)
(b)
Complete the table for n = 6.
(1)
(c)
On the graph, plot the point for n = 6 and draw the line of best fit.
(2)
(d)
Determine the gradient of the line.
(2)
(e)
Instrument makers use r = 21/12. Determine r from your gradient and compare it with this value.
(2)
(f)
The open string is tuned so that its first harmonic is 110.0 Hz; take this value as exact. Calculate the frequency of the first harmonic when the string is pressed at fret 12, and its absolute uncertainty.
(2)
(g)
In practice, pressing the string down onto a fret stretches it slightly. State and explain the effect of this on the frequency of the note, compared with the value predicted from L.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
In the first harmonic there is a node at each end and half a wavelength on the string, so λ = 2L
✓ 1
The tension and the string are unchanged, so the wave speed v is constant, and f = v/2L ∝ 1/L
✓ 1
OWTTE.
Part (b)
ln 458.5 = 6.128
✓ 1
Accept 6.13. Accept 4 d.p. (6.1280).
Part (c)
Point plotted at (6, 6.128)
✓ 1
Allow ECF from (b). Within half a small square.
Single straight line through or very close to all the points
✓ 1
Part (d)
Uses a triangle covering more than half of the line
✓ 1
gradient = −0.0577
✓ 1
Accept −0.0590 to −0.0565. Negative sign required. No unit.
Part (e)
ln r = −gradient, so r = e0.0577 = 1.059
✓ 1
Allow ECF from (d). Accept 1.057–1.062.
21/12 = 1.0595: the values agree to within 0.1 %, so the data support the instrument makers' value
✓ 1
Comparison must be numerical. Allow ECF.
Part (f)
f = 110.0 × 648.0/324.5 = 219.7 Hz
✓ 1
Δf/f = 0.5/648.0 + 0.5/324.5 = 0.0023, so f = (219.7 ± 0.5) Hz
✓ 1
Both fractional uncertainties needed. Accept ±0.51 Hz. MP2 includes matching the precision of value and uncertainty.
Part (g)
The tension increases, so the wave speed increases and f = v/2L is higher than predicted «the note is slightly sharp»
✓ 1
Direction AND reason needed. OWTTE.
Answers: (b) 6.128 · (d) −0.0577 · (e) r = 1.059 · (f) (219.7 ± 0.5) Hz (the remaining parts are explanations — see the table above)
A steel wire passes over two bridges (0.600 ± 0.001) m apart and then over a pulley to a hanger carrying masses m. The tension T in the wire is taken to be mg. The wire is plucked at its centre and a phone app measures the frequency f of the first harmonic.
The speed of transverse waves on the wire is v = √(T/μ), where μ is the mass per unit length. The student's hypothesis is that f is proportional to √T.
m / kg
T / N
f / Hz
log( T / N )
log( f / Hz )
1.00
9.81
83.6
0.992
1.922
2.00
19.6
117.4
1.293
2.070
3.00
29.4
143.8
1.469
2.158
4.00
166.3
5.00
49.1
185.9
1.691
2.269
6.00
58.9
203.7
1.770
2.309
Graph drawn to scale
(a)
Complete the table for m = 4.00 kg.
(2)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the gradient of the line and state whether the data support the hypothesis.
(3)
(d)
Show that log f = ½ log T − log(2L√μ), where L is the distance between the bridges, and hence determine μ using the graph.
(3)
(e)
A 2.000 m sample of the same wire, measured with an uncertainty of ±0.002 m, has a mass of (1.97 ± 0.01) g. Determine μ and its absolute uncertainty from these data.
(2)
(f)
Friction at the pulley means that the tension in the wire is slightly less than mg. State and explain the effect of this on the value of μ found from the graph.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
T = 4.00 × 9.81 = 39.2 N AND log( T / N ) = 1.594
✓ 1
Accept 2 or 4 decimal places. Accept 1.593 (from T = 39.2 N).
log( f / Hz ) = log 166.3 = 2.221
✓ 1
Part (b)
Single straight line with points evenly distributed on both sides
✓ 1
Part (c)
Uses a triangle covering more than half the line
✓ 1
gradient = 0.50
✓ 1
Accept 0.48 to 0.52.
A gradient of 0.5 on a log–log graph means f ∝ T0.5, so the hypothesis is supported
✓ 1
Allow ECF from the gradient.
Part (d)
f = v/(2L) = (1/(2L))√(T/μ); taking logs gives the result
✓ 1
Must see f = v/2L for the first harmonic.
Uses a point on the line, e.g. log( T / N ) = 1.40 and log( f / Hz ) = 2.124: log(2L√μ) = 0.700 − 2.124 = −1.424
✓ 1
Accept the candidate's own point on the line.
2L√μ = 0.0377, so μ = 9.9 × 10−4 kg m−1
✓ 1
Accept 9.6 × 10−4 to 1.0 × 10−3 kg m−1. Allow ECF.
Part (e)
μ = 1.97 × 10−3/2.000 = 9.85 × 10−4 kg m−1
✓ 1
Δμ/μ = 0.01/1.97 + 0.002/2.000 = 0.0061, so μ = (9.85 ± 0.06) × 10−4 kg m−1
A student investigates how the amplitude varies along a standing wave. An elastic cord is stretched horizontally between a vibration generator and a fixed clamp. The length of the cord between them is (1.800 ± 0.005) m. The generator moves its end of the cord with an amplitude that is negligible, so both ends behave as nodes. At a frequency of (24.0 ± 0.1) Hz the cord vibrates in its third harmonic.
A vertical millimetre scale is fixed just behind the cord. In a long-exposure photograph the vibrating cord appears as a blurred band. At different distances x from the generator the student reads the full width w of the band from the scale in the photograph and calculates A = w/2.
The student's hypothesis is that A = AN |sin(2πx/λ)|, where λ is the wavelength, AN is the amplitude at an antinode and the angle 2πx/λ is in radians. The graph shows A against |sin(2πx/λ)| for seven of the positions, with error bars for the uncertainty in A.
x / m
w / mm ± 1 mm
A / mm ± 0.5 mm
|sin(2πx/λ)|
0.10
21
10.5
0.500
0.25
37
18.5
0.966
0.45
29
14.5
0.707
0.60
3
1.5
0.000
0.65
12
6.0
0.259
0.80
35
1.05
28
14.0
0.707
1.50
39
19.5
1.000
Graph drawn to scale. The point for x = 0.80 m has not been plotted.
(a)
Calculate the speed of the waves on the cord and its absolute uncertainty.
(2)
(b)
Complete the table for x = 0.80 m.
(2)
(c)
Plot the point for x = 0.80 m on the graph and draw the line of best fit.
(2)
(d)
Determine AN using the gradient of your line.
(2)
(e)
Outline whether the data support the student's hypothesis.
(1)
(f)
The diameter of the cord is 3.0 mm. Explain the intercept of your line on the A-axis and state its effect on your answer to (d).
(2)
(g)
State the phase difference between the oscillations of the points of the cord at x = 0.25 m and x = 0.80 m.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
λ = 2 × 1.800/3 = 1.200 m AND v = fλ = 24.0 × 1.200 = 28.8 m s−1
✓ 1
Δv/v = 0.1/24.0 + 0.005/1.800 = 0.0069 «0.69 %», so v = (28.8 ± 0.2) m s−1
✓ 1
Accept ±0.20 m s−1. Allow ECF from MP1.
Part (b)
A = 35/2 = 17.5 «mm»
✓ 1
Must be given to 0.1 mm / 3 s.f.
|sin(2π × 0.80/1.20)| = 0.866
✓ 1
Accept 0.87. Award [0] for MP2 for 0.073 «calculator in degrees» or for −0.866.
Part (c)
Point plotted at (0.866, 17.5) to within half a small square
✓ 1
Allow ECF from (b).
Single straight line passing through all the error bars, meeting the A-axis above the origin «at about 1.5 mm»
✓ 1
Do not award for a line forced through the origin.
Part (d)
Gradient found from a triangle covering more than half of the line
✓ 1
AN = gradient = 18 mm
✓ 1
Accept 17.0–19.0 mm. Award [2] for CNA. Do not accept the value of A at |sin(2πx/λ)| = 1 read from the line «about 19.5 mm».
Part (e)
The points lie on a straight line «within their error bars», as the hypothesis predicts, BUT the line does not pass through the origin, so A is not exactly proportional to |sin(2πx/λ)| / there is a systematic error
✓ 1
Both ideas needed. OWTTE.
Part (f)
The band includes the thickness of the cord: w = 2A + 3.0 mm, so every calculated value of A is 1.5 mm too large, which is the intercept «at a node w = 3 mm»
✓ 1
Must link the intercept to half the diameter.
The same offset is added to every point, so the gradient and AN are unaffected
✓ 1
Allow ECF from the student's intercept. OWTTE.
Part (g)
π rad «180°»: the points are in adjacent loops, on opposite sides of the node at x = 0.60 m
✓ 1
Accept 180°. Do not accept a value calculated from 2πΔx/λ as for a travelling wave «0.92π rad».
Answers: (a) v = (28.8 ± 0.2) m s−1 · (b) A = 17.5 mm; |sin(2πx/λ)| = 0.866 · (d) AN = 18 mm · (g) π rad (the remaining parts are explanations — see the table above)
73C-2-30Standing waves in pipesPaper 2Easy11 marks
The ear canal of an adult is a tube about 2.5 cm long. It is open at the outer end and closed at the inner end by the eardrum. The air in the canal can be modelled as the air in a pipe with one open end and one closed end. The speed of sound in the air in the canal is 340 m s−1.
Diagram drawn to scale
(a)
State where there is a displacement node and where there is a displacement antinode in every standing wave in the canal, and explain why.
(2)
(b)
On the upper diagram, sketch the displacement pattern of the standing wave of the first harmonic.
(1)
(c)
Show that the frequency of the first harmonic is about 3.4 kHz.
(1)
(d)
Determine the frequency of the next standing wave above the first harmonic.
(2)
(e)
On the lower diagram, sketch the displacement pattern of the standing wave in (d).
(2)
(f)
Hearing tests show that the human ear is most sensitive to sounds of frequency about 3 kHz to 4 kHz. Suggest an explanation for this.
(2)
(g)
The ear canal of a young child is shorter than that of an adult. Outline how this affects the frequency to which the child's ear is most sensitive.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A displacement node at the eardrum «closed end», where the air cannot move
✓ 1
A displacement antinode at the open end, where the air is free to move
✓ 1
Award [1 max] for the positions without reasons.
Part (b)
Node at the eardrum and antinode at the open end, with no other node: a quarter of a wavelength
✓ 1
Accept the envelope drawn as one curve or as a pair of curves.
Part (c)
λ = 4L = 0.10 m, so f = 340/0.10 = 3.4 × 103 «Hz»
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (d)
Only odd harmonics can form: the next has three quarters of a wavelength in the canal
✓ 1
f = 3 × 3.4 = 1.02 × 104 Hz «10.2 kHz»
✓ 1
Do not accept 6.8 kHz. Allow ECF from (c). Award [2] for CNA.
Part (e)
Node at the eardrum and antinode at the open end
✓ 1
One further node one third of the length from the open end «0.83 cm», and one further antinode two thirds of the length from the open end «1.7 cm», between this node and the eardrum
✓ 1
Positions within ±2 mm on the scale of the diagram. Allow ECF from (d).
Part (f)
Sound at about 3.4 kHz matches the natural frequency of the air in the canal «its first harmonic»
✓ 1
so resonance occurs: the air in the canal oscillates with a large amplitude, giving a large amplitude at the eardrum
✓ 1
OWTTE.
Part (g)
The child's canal has a higher first-harmonic frequency «f = v/4L», so the most sensitive frequency is higher
✓ 1
Answers: (c) 3.4 × 103 Hz · (d) 1.02 × 104 Hz (the remaining parts are explanations — see the table above)
Four simple pendulums P, Q, R and S hang from the same horizontal rod. Their lengths are 0.25 m, 0.39 m, 0.60 m and 0.95 m respectively. Each bob is a light paper cone, so air resistance damps the motion. A motor makes the rod oscillate horizontally with a small amplitude, at a frequency that can be varied.
Diagram not to scale
(a)
The motor drives the rod at a frequency of 0.80 Hz. Determine which pendulum oscillates with the largest amplitude.
(3)
(b)
State the frequency at which pendulum S oscillates once its motion has become steady.
(1)
(c)
Explain why the amplitude of pendulum S is small.
(2)
(d)
The frequency of the motor is now increased slowly from 0.40 Hz to 1.20 Hz. Describe and explain what is observed.
(2)
(e)
The paper cone of pendulum Q is replaced by a small lead sphere, so the damping of Q is lighter. State and explain the effect on the amplitude of Q when the rod is driven at 0.80 Hz and when it is driven at 0.60 Hz.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Natural frequency f = (1/2π)√(g/l) OR length needed l = g/(2πf)²
✓ 1
l = 9.81/(2π × 0.80)² = 0.39 m «OR f of Q = 0.80 Hz»
✓ 1
ALTERNATIVE: natural frequencies of P, Q, R, S = 1.00, 0.80, 0.64, 0.51 Hz.
Q, because its natural frequency equals the driving frequency, so resonance occurs
✓ 1
MP3 only scores if a length or frequency has been calculated.
Part (b)
0.80 Hz «the driving frequency»
✓ 1
Do not accept its natural frequency of 0.51 Hz.
Part (c)
The driving frequency is far from the natural frequency of S «0.51 Hz»
✓ 1
so the driving force is not in step with the motion of S: energy given to S during part of each cycle is taken back during another part, and little energy builds up
✓ 1
OWTTE.
Part (d)
Each pendulum in turn oscillates with a large amplitude: S, then R, then Q, then P
✓ 1
because the natural frequency increases as the length decreases «f ∝ 1/√l», and each pendulum resonates when the driving frequency passes its natural frequency
✓ 1
OWTTE.
Part (e)
At 0.80 Hz «resonance» the amplitude of Q is much larger, because lighter damping removes less energy per cycle
✓ 1
At 0.60 Hz, away from resonance, the amplitude changes little «the resonance peak is sharper»
✓ 1
OWTTE.
Answers: (a) l = 0.39 m — pendulum Q · (b) 0.80 Hz (the remaining parts are explanations — see the table above)
A science-centre exhibit has a horizontal cord of length 1.20 m with both ends fixed. A vibrator near one end sends waves of frequency 5.0 Hz and wavelength 0.80 m along the cord, and the waves are reflected at the far end. Figure 1 shows, at one instant, the displacement of the cord that each of the two travelling waves would produce on its own. Both waves have an amplitude of 2.0 cm.
Figure 1 — drawn to scale. Solid line: wave travelling to the right; dashed line: reflected wave travelling to the left
(a)
Calculate the speed of the waves on the cord.
(1)
(b)
On Figure 1, draw the shape of the cord at the instant shown.
(2)
(c)
Determine the shortest time after the instant shown at which the cord is straight.
(2)
(d)
State the amplitude of oscillation of the point on the cord 0.20 m from the left-hand end.
(1)
(e)
State the phase difference between the oscillations of the points 0.10 m and 0.50 m from the left-hand end.
(1)
(f)
Outline one difference between the transfer of energy along the cord by the standing wave and by each travelling wave.
(1)
(g)
(i)
State the number of the harmonic shown by this standing wave.
(1)
(ii)
The frequency of the vibrator is slowly increased. Calculate the next frequency at which a standing wave forms.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = fλ = 5.0 × 0.80 = 4.0 m s−1
✓ 1
Part (b)
Zero displacement at 0, 0.40 m, 0.80 m and 1.20 m «where the two waves have equal and opposite displacements»
✓ 1
Sine-shaped curve with maximum displacement ≈ 2.8 cm: positive at 0.20 m and 1.00 m, negative at 0.60 m
✓ 1
Accept 2.6–3.0 cm. Do not award MP2 for a peak of 4.0 cm.
Part (c)
The cord is straight when the two waves are exactly opposite: each wave must move a further λ/8 = 0.10 m
✓ 1
ALTERNATIVE: the instant shown is T/8 after the cord was at maximum displacement; the cord is straight T/4 after that.
t = 0.10/4.0 = 0.025 s «= T/8»
✓ 1
Award [2] for CNA. Allow ECF from (a).
Part (d)
4.0 cm «2 × 2.0 cm: the point is an antinode»
✓ 1
Do not accept 2.8 cm «the displacement at this instant».
Part (e)
π rad «the points are in adjacent loops, on opposite sides of a node»
✓ 1
Accept 180°.
Part (f)
The standing wave transfers no «net» energy along the cord, whereas each travelling wave transfers energy in its direction of travel
✓ 1
OWTTE.
Part (g)(i)
Third harmonic «three loops»
✓ 1
Part (g)(ii)
The next standing wave has four loops: λ = 2 × 1.20/4 = 0.60 m
✓ 1
f = 4.0/0.60 = 6.67 Hz
✓ 1
Award [2] for CNA. Allow ECF from (a).
Answers: (a) 4.0 m s−1 · (c) 0.025 s · (d) 4.0 cm · (e) π rad · (g)(ii) 6.67 Hz (the remaining parts are explanations — see the table above)
76C-2-69Longitudinal standing wavesPaper 2Easy10 marks
A long spring is stretched along a smooth bench between two clamps 3.00 m apart. A vibration generator near one clamp moves a coil of the spring back and forth along the axis of the spring. At a frequency of 4.0 Hz a standing wave forms, and the coils at 0.75 m, 1.50 m and 2.25 m from the left-hand clamp stay at rest.
Figure 1 shows, at one instant of maximum displacement, the displacement s of the coils along the axis of the spring (positive to the right) against their equilibrium distance x from the left-hand clamp. P and Q are coils at x = 0.25 m and x = 1.00 m.
Figure 1 — drawn to scale
(a)
State whether the waves on the spring are longitudinal or transverse, giving a reason.
(1)
(b)
Explain how the standing wave is formed.
(2)
(c)
(i)
State the wavelength of the waves on the spring.
(1)
(ii)
Calculate the speed of the waves on the spring.
(1)
(d)
On Figure 1, draw the displacement of the coils half a period after the instant shown.
(2)
(e)
State the phase difference between the oscillations of P and Q.
(1)
(f)
The frequency of the generator is slowly increased. Calculate the next frequency at which a standing wave forms.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
longitudinal, because the coils oscillate parallel to the direction in which the waves travel «along the axis of the spring»
✓ 1
Reason needed.
Part (b)
waves from the generator are reflected at the clamps «and travel back along the spring»
✓ 1
the waves travelling in opposite directions, with the same frequency, speed and «similar» amplitude, superpose
✓ 1
OWTTE.
Part (c)(i)
1.50 m «adjacent nodes are 0.75 m = λ/2 apart»
✓ 1
Part (c)(ii)
v = fλ = 4.0 × 1.50 = 6.0 m s−1
✓ 1
Allow ECF from (c)(i).
Part (d)
curve with the opposite sign everywhere: negative between 0 and 0.75 m, positive between 0.75 m and 1.50 m, and so on
✓ 1
zero displacement at the same five points AND the same maximum size «3 cm» midway between them
✓ 1
Do not award MP2 if the zeros move.
Part (e)
π rad «180°, antiphase: P and Q are in adjacent loops, on opposite sides of the node at 0.75 m»
✓ 1
Accept 'antiphase'.
Part (f)
the next standing wave has five half-wavelengths on the spring: λ = 2 × 3.00/5 = 1.20 m
✓ 1
ALTERNATIVE: f1 = v/2L = 1.0 Hz, and the pattern shown is the fourth harmonic.
f = 6.0/1.20 = 5.0 Hz
✓ 1
Allow ECF from (c)(ii). Award [2] for CNA.
Answers: (c)(i) 1.50 m · (c)(ii) 6.0 m s−1 · (f) 5.0 Hz (the remaining parts are explanations — see the table above)
77IBP2S-024-03-11Nodes & antinodesPaper 2Easy4 marks
A medical ultrasound probe contains a thin piezoelectric disc. The disc is driven so that a longitudinal standing wave forms across its thickness. Both flat faces of the disc are free to move. The disc vibrates in its fundamental mode at 2.5 MHz. The speed of longitudinal waves in the disc material is 4000 m s−1.
(a)
State what is meant by a displacement antinode.
(1)
(b)
Outline why both faces of the disc are displacement antinodes.
(1)
(c)
Calculate the thickness of the disc.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A point in a standing wave where the amplitude of oscillation of the particles is a maximum .
✓ 1
Part (b)
The faces are free (not constrained), so the particles at the faces can oscillate with maximum amplitude .
✓ 1
Part (c)
λ = v/f = 4000 / 2.5×106 = 1.6×10−3 m .
✓ 1
The fundamental has antinodes at both faces, so the thickness = λ/2 = 8.0×10−4 m = 0.80 mm .
✓ 1
Exam tip: Always identify the boundary condition at each end before deciding what fraction of a wavelength fits into the length.
78IBP2S-024-03-12Standing waves in pipesPaper 2Easy5 marks
A didgeridoo is a long wind instrument. It can be modelled as a tube of length 1.30 m that is closed at the player's mouth and open at the far end. The speed of sound in the air inside is 340 m s−1.
Diagram not to scale
(a)
State the boundary condition for the air displacement at each end of the tube.
(2)
(b)
Calculate the fundamental frequency of the didgeridoo.
(2)
(c)
State the frequency of the next higher harmonic that the tube can produce.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Mouth (closed) end: displacement node .
✓ 1
Far (open) end: displacement antinode .
✓ 1
Part (b)
L = λ/4, so λ = 4 × 1.30 = 5.20 m .
✓ 1
f = v/λ = 340 / 5.20 = 65.4 Hz (accept 65 Hz) .
✓ 1
Part (c)
Only odd harmonics are possible, so the next is the third harmonic: 3 × 65.4 ≈ 196 Hz (accept 195–196 Hz; ECF from (b)) .
✓ 1
Common error: For a tube closed at one end, the next harmonic above the fundamental is the third harmonic, not the second. Carry unrounded values forward so that later answers stay accurate.
79IBP2S-024-01-11Standing waves in pipesPaper 2Easy6 marks
A safety whistle on an industrial pressure vessel contains an air column 0.35 m long. The column is closed at one end and open at the other. The speed of sound in the air is 343 m s−1.
Diagram not to scale
(a)
State the condition on the air displacement at (i) the closed end and (ii) the open end of the column.
(2)
(b)
Calculate the fundamental frequency of the whistle.
(2)
(c)
Calculate the frequency of the next higher resonance of the air column.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
(i) The closed end is a displacement node (zero displacement amplitude) .
✓ 1
(ii) The open end is a displacement antinode (maximum displacement amplitude) .
✓ 1
Part (b)
λ = 4L = 4 × 0.35 = 1.40 m .
✓ 1
f = v/λ = 343/1.40 = 245 Hz .
✓ 1
Part (c)
Only odd harmonics exist, so the next resonance is the third harmonic .
✓ 1
f3 = 3 × 245 = 735 Hz . ECF from (b).
✓ 1
Common error: Giving 2f₁ = 490 Hz as the next resonance loses the mark. A pipe closed at one end has no even harmonics.
80IBP2S-024-00-11Tension & wave speedPaper 2Easy5 marks
A badminton racquet string is stretched between two points on the frame 0.20 m apart. The tension in the string is 100 N and its mass per unit length is 6.0 ×10−4 kg m−1. The speed of transverse waves on a string is v = √(T/μ).
(a)
State what is meant by the fundamental mode of vibration of a string fixed at both ends.
(1)
(b)
Calculate the speed of transverse waves on the string.
(2)
(c)
Calculate the fundamental frequency of this section of string.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The lowest-frequency standing wave, with a node at each end and a single antinode at the centre
✓ 1
Part (b)
v = √(100 / 6.0×10−4)
✓ 1
v ≈ 408 m s−1 (accept 410 m s−1)
✓ 1
Part (c)
λ = 2L = 2 × 0.20 = 0.40 m
✓ 1
f = v/λ = 408/0.40 ≈ 1.02 ×103 Hz (accept 1.0 ×103 Hz; ECF from (b))
✓ 1
Exam tip: Calculate the wave speed first from tension and μ, then use the standing-wave geometry (λ = 2L) to find the frequency.
81IBP2S-024-00-12Tension & wave speedPaper 2Easy5 marks
At an adventure park, a steel cable on a short training zip-wire is under a tension of 2000 N. A maintenance worker strikes one end of the cable. A transverse pulse travels 50 m along the cable in 0.40 s. Assume the tension and the mass per unit length are the same everywhere along the cable.
(a)
Calculate the speed of the pulse.
(1)
(b)
Determine the mass per unit length of the cable. Use v = √(T/μ).
(2)
(c)
A 25 m section of the cable lies between two fixed clamps. This section vibrates in its fundamental mode. Calculate the frequency of this vibration.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = 50/0.40 = 125 m s−1
✓ 1
Part (b)
μ = T/v2 = 2000/1252
✓ 1
μ ≈ 0.13 kg m−1 (0.128) (ECF from (a))
✓ 1
Part (c)
There is a node at each clamp, so λ = 2L = 2 × 25 = 50 m
✓ 1
f = v/λ = 125/50 = 2.5 Hz (ECF from (a))
✓ 1
Common error: In the fundamental mode the section between two fixed ends holds half a wavelength, so λ = 2L. Taking λ = L gives a frequency twice the correct value.
82IBP2S-024-02-11Nodes & antinodesPaper 2Easy4 marks
A slackline used for balance training is a flat webbing strap anchored firmly between two trees 12 m apart. When an athlete bounces lightly at the centre, the line vibrates in its fundamental mode with frequency 1.5 Hz.
(a)
State why both ends of the slackline are nodes.
(1)
(b)
Calculate the wavelength of the fundamental mode.
(1)
(c)
Calculate the speed of transverse waves along the slackline.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The ends are fixed or anchored, so their displacement is always zero
✓ 1
Part (b)
λ = 2L = 2 × 12 = 24 m
✓ 1
Part (c)
Uses v = fλ
✓ 1
v = 1.5 × 24 = 36 m s−1 (allow error carried forward from the wavelength calculated in the previous part)
✓ 1
Common error: The fundamental is half a wavelength, so do not use λ = L.
In a sports science laboratory, one end of a heavy training rope is tied to a wall bracket. A motor oscillates the other end with a small amplitude and a steady frequency, and a standing wave forms on the rope.
(a)
Outline how the standing wave is formed on the rope.
(2)
(b)
State two differences between the standing wave on the rope and a travelling wave on the same rope.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Waves from the motor travel along the rope and are reflected at the fixed end at the wall
✓ 1
Incident and reflected waves of the same frequency and (approximately) equal amplitude travel in opposite directions and superpose/interfere, giving fixed nodes and antinodes
✓ 1
Part (b)
One mark for each of any two distinct differences from this list. Each difference must describe both waves.
—
• Amplitude: in the standing wave it varies with position (zero at nodes, maximum at antinodes); in the travelling wave it is the same at all points.
—
• Phase: in the standing wave, points between adjacent nodes are in phase and points in adjacent loops are in antiphase; in the travelling wave, phase changes continuously with position.
—
• Energy: the standing wave transfers no net energy along the rope; the travelling wave transfers energy along the rope.
—
First correct difference
✓ 1
Second correct difference
✓ 1
Exam tip: For a 'difference', say what happens in both waves, not just one.
84C-2-11Standing waves on stringsPaper 2Medium13 marks
One end of a rubber cord is attached to a vibration generator G. The generator moves the cord with a very small amplitude, so this end behaves as a fixed end. The other end of the cord is tied to a light ring that can slide without friction along a vertical rod, so this end is free to move up and down.
The length of the cord is 1.50 m and waves travel along it at 9.0 m s−1.
Figure 1 — Diagram NOT to scale vertically
(a)
Explain how a standing wave forms on the cord.
(2)
(b)
(i)
On Figure 1(a), draw the shape of the cord at an instant of maximum displacement when it vibrates in its first harmonic.
(1)
(ii)
Show that the frequency of the first harmonic is 1.5 Hz.
(1)
(c)
The frequency of the generator is slowly increased from 1.5 Hz. Explain why no standing wave forms on the cord at a frequency of 3.0 Hz.
(3)
(d)
(i)
State the frequency of the next standing wave above 1.5 Hz.
(1)
(ii)
On Figure 1(b), draw the shape of the cord at an instant of maximum displacement for this standing wave.
(2)
(iii)
Points P and Q on the cord are 0.40 m and 1.30 m from the generator. State and explain the phase difference between the oscillations of P and Q in this standing wave.
(2)
(e)
The ring is now clamped so that this end of the cord is also fixed. Calculate the frequency of the new first harmonic.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
waves travelling along the cord are reflected at the ends
✓ 1
the incident and reflected waves «same frequency, speed and similar amplitude», travelling in opposite directions, superpose
✓ 1
OWTTE
Part (b)(i)
curve with a node at G and an antinode at the ring, and no other node «a quarter of a wavelength»
✓ 1
Part (b)(ii)
λ = 4 × 1.50 = 6.0 m and f = 9.0/6.0 «= 1.5 Hz»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
at 3.0 Hz the wavelength is 9.0/3.0 = 3.0 m, so the cord would be half a wavelength long
✓ 1
a half-wavelength pattern needs the same type of point «node or antinode» at both ends
✓ 1
but there must be a node at G and an antinode at the ring, so the length must be an odd number of quarter wavelengths «frequencies 1.5, 4.5, 7.5 … Hz»
✓ 1
OWTTE
Part (d)(i)
4.5 Hz
✓ 1
Part (d)(ii)
nodes at G and at 1.00 m from G
✓ 1
Allow ECF from (d)(i)
antinodes at 0.50 m and at the ring «three quarters of a wavelength»
✓ 1
Part (d)(iii)
π rad «180°, antiphase»
✓ 1
P and Q are in adjacent loops, on opposite sides of the node at 1.00 m
✓ 1
MP2 only scores if MP1 scores
Part (e)
λ = 2 × 1.50 = 3.0 m, so f = 9.0/3.0 = 3.0 Hz
✓ 1
Answers: (b)(ii) 1.5 Hz · (d)(i) 4.5 Hz · (d)(iii) π rad · (e) 3.0 Hz (the remaining parts are explanations — see the table above)
85C-2-12Standing waves in pipesPaper 2Medium11 marks
A simple wind instrument is modelled as a straight pipe of length 0.660 m that is open at both ends. The speed of sound in the air in the pipe is 343 m s−1.
Figure 1 — Diagram NOT to scale
(a)
State whether there is a displacement node or a displacement antinode at an open end of the pipe.
(1)
(b)
Sketch, on Figure 1(a), the displacement pattern of the first harmonic of the air in the pipe.
(2)
(c)
Show that the frequency of the first harmonic is about 260 Hz.
(1)
(d)
Figure 1(b) shows the displacement pattern of another standing wave in the pipe. Determine its frequency.
(2)
(e)
The player closes one end of the pipe. Determine whether the pipe can now form a standing wave of the frequency in (d).
(3)
(f)
The air in the pipe warms up while the instrument is played. Suggest, with a reason, the effect on the frequency of the first harmonic.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
a displacement antinode
✓ 1
Part (b)
displacement antinodes at both open ends
✓ 1
a single node at the centre of the pipe «half a wavelength in the pipe»
✓ 1
Do not award MP2 for a quarter-wavelength pattern
Part (c)
λ = 2 × 0.660 = 1.32 m and f = 343/1.32 «= 259.8 Hz»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (d)
three half-wavelengths fit in the pipe: λ = 2 × 0.660/3 = 0.440 m
✓ 1
Accept f = 3f1
f = 343/0.440 = 780 Hz
✓ 1
Allow ECF from (c). Award [2] for CNA
Part (e)
with one end closed, standing waves form only at odd multiples of v/4L
✓ 1
v/4L = 343/(4 × 0.660) = 130 Hz
✓ 1
780/130 = 6, an even number, so this standing wave cannot form
✓ 1
MP3 only scores if a conclusion is given. Allow ECF from (d)
Part (f)
the speed of sound increases with temperature
✓ 1
the wavelength is fixed by the length of the pipe, so f = v/λ increases
✓ 1
MP2 only scores if MP1 scores. OWTTE
Answers: (c) 259.8 Hz · (d) 780 Hz · (e) no: 780 Hz = 6 × 130 Hz (even multiple) (the remaining parts are explanations — see the table above)
A kitchen scale has a pan of mass 0.200 kg supported on a vertical spring of spring constant 320 N m−1. A damper is connected to the pan. Just before time t = 0 the pan is at rest at its original position. At t = 0 a bag of flour of mass 0.300 kg is placed on the pan and released from rest. A sensor records the downward displacement y of the pan from its original position.
Figure 1 shows the variation of y with t for the scale with its original damper. Take g = 9.81 m s−2.
Figure 1 — drawn to scale
(a)
Outline what is meant by damping.
(1)
(b)
Show that the final position of the pan is about 9 mm below its original position.
(1)
(c)
Describe how Figure 1 shows that the motion of the pan is lightly damped.
(2)
(d)
Sketch, on Figure 1, graphs to show how y would vary with t if the damping were critical (label this graph C) and if the damping were heavy (label this graph H).
(3)
(e)
Suggest why the manufacturer designs the scale to be critically damped.
(2)
(f)
(i)
Determine the total energy transferred to thermal energy by the damping between t = 0 and the time at which the pan comes to rest.
(3)
(ii)
State and explain whether this energy would be greater, smaller or the same if the scale were critically damped.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
a resistive force «friction, drag» that acts on an oscillating system, opposite to its velocity, and removes energy from it so that the amplitude decreases
✓ 1
OWTTE
Part (b)
extra spring force = weight of the bag: d = Mg/k = 0.300 × 9.81/320 «= 9.20 mm»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
the pan oscillates about its final position «9.2 mm», overshooting it several times
✓ 1
the amplitude decreases gradually «by a similar fraction each cycle» while the period stays the same «about 0.25 s»
✓ 1
OWTTE
Part (d)
both C and H start at y = 0 with zero gradient and rise to 9.2 mm without ever exceeding it «no oscillation»
✓ 1
Labels C and H required
C reaches «close to» 9.2 mm quickly, in about one period of the light oscillation «≈ 0.2 s»
✓ 1
Accept a time from 0.1 s to 0.4 s
H lies below C at all times after t = 0 and approaches 9.2 mm much more slowly
✓ 1
MP3 only scores if both curves are labelled
Part (e)
with light damping the pan oscillates for a long time «about 1 s in Figure 1» before the reading is steady / with heavy damping the reading creeps slowly towards its final value
✓ 1
OWTTE
critical damping brings the pan to its final reading, without oscillation, in the shortest time
✓ 1
Part (f)(i)
loss of gravitational potential energy of pan and bag = 0.500 × 9.81 × 0.00920 = 0.0451 J
✓ 1
Allow ECF from (b)
gain of elastic potential energy = ½ × 320 × (0.01533² − 0.006131²) = 0.0316 J «initial compression 6.13 mm, final 15.3 mm»
✓ 1
ALTERNATIVE: the pan's weight and the initial spring force cancel, so energy = Mgd − ½kd² = 0.0271 − 0.0135 = 0.0135 J
thermal energy = 0.0451 − 0.0316 = 0.0135 J
✓ 1
Accept 0.013–0.014 J. Award [3] for CNA. Award [1] max for Mgd = 0.0271 J
Part (f)(ii)
the same
✓ 1
Do not award MP1 without a reason
the initial and final states are the same «at rest, same positions, so the same gravitational and elastic potential energies», so by conservation of energy the same energy is dissipated; only the time taken is different
✓ 1
OWTTE
Answers: (b) d = 9.20 mm · (f)(i) 0.0135 J · (f)(ii) the same (the remaining parts are explanations — see the table above)
An overhead power cable is fixed to one tower and passes over a pulley on a second tower 300 m away. Its mass per unit length is 1.50 kg m−1. The tension in the cable is set by a counterweight of mass 2.20 × 103 kg that hangs from the end of the cable beyond the pulley. Treat the cable between the towers as straight; both of its ends there behave as fixed ends.
The speed v of transverse waves on the cable is given by v = √(T/μ), where T is the tension and μ is the mass per unit length.
Wind blowing across the cable exerts a periodic vertical force on it. For the wind speeds usually met, the frequency of this force lies between 5 Hz and 40 Hz.
Diagram not to scale
(a)
(i)
Show that the speed of transverse waves on the cable is about 120 m s−1.
(2)
(ii)
Calculate the frequency of the first harmonic of the cable.
(1)
(b)
On one day the frequency of the force from the wind is 12.0 Hz and a standing wave is set up on the cable.
(i)
Determine the number of the harmonic that is set up.
(2)
(ii)
Calculate the distance between adjacent nodes of this standing wave.
(1)
(c)
Explain why the cable vibrates with a large amplitude when the force has a frequency of 12.0 Hz, and why a vibration damper clamped to the cable at a node of this standing wave would not reduce the vibration.
(4)
(d)
Suggest why standing waves of large amplitude can occur at almost any wind speed.
(1)
(e)
Vibration dampers are usually clamped to the cable about 1 m from a tower. Explain, using the range of frequencies of the force, why this position is chosen.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Tension = weight of the counterweight: T = mg = 2.20 × 103 × 9.81 = 2.16 × 104 N
✓ 1
v = √(2.158 × 104/1.50) = 119.9 «m s−1»
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
f1 = v/2L = 120/600 = 0.200 Hz
✓ 1
Allow ECF from (a)(i).
Part (b)(i)
The harmonics are whole-number multiples of f1, so n = 12.0/0.200
✓ 1
Allow ECF from (a)(ii).
n = 60
✓ 1
Award [2] for CNA.
Part (b)(ii)
λ/2 = L/60 = 300/60 = 5.0 m
✓ 1
Accept v/2f = 120/24.0 = 5.0 m.
Part (c)
The driving frequency «12.0 Hz» equals a natural frequency of the cable «its 60th harmonic»
✓ 1
so resonance occurs: energy is transferred to the cable on every cycle and the amplitude becomes large
✓ 1
A node is a point of zero amplitude — the cable there does not move
✓ 1
so a damper at a node has no motion to oppose and cannot remove energy from the standing wave
✓ 1
OWTTE.
Part (d)
The natural frequencies are only 0.2 Hz apart, so the frequency of the force is always very close to one of them
✓ 1
Allow ECF from (a)(ii).
Part (e)
The first node from the tower is half a wavelength away; at the highest frequency, 40 Hz, λ/2 = 120/(2 × 40) = 1.5 m
✓ 1
So a damper less than 1.5 m from the tower is never at a node for any frequency in the range «it always moves and removes energy»
✓ 1
Award [1 max] for 'it must not be at a node' without the calculation.
Answers: (a)(i) T = 2.16 × 104 N; v = 120 m s−1 · (a)(ii) 0.200 Hz · (b)(i) n = 60 · (b)(ii) 5.0 m · (e) 1.5 m (the remaining parts are explanations — see the table above)
88C-2-32Standing waves on stringsPaper 2Medium13 marks
A violin string is fixed at both ends and has a vibrating length of 0.330 m. When it is bowed, the frequency of its first harmonic is 440 Hz.
A violinist can produce higher notes by touching the string lightly at one point while bowing it. The light touch forces a node at that point but does not change the vibrating length of the string.
Diagram NOT accurately drawn
(a)
(i)
Calculate the speed of transverse waves on the string.
(2)
(ii)
The violinist now presses the string firmly against the fingerboard, so that its vibrating length becomes 0.220 m. Calculate the new frequency of the first harmonic. In the rest of the question the string is not pressed.
(2)
(b)
The violinist touches the string lightly at a point one third of the way along it, as shown in the figure.
(i)
On the figure, draw the standing wave of lowest frequency that can now be set up on the string.
(2)
(ii)
Calculate the frequency of this standing wave.
(1)
(iii)
Explain why the first and second harmonics of the string cannot be set up while it is touched at this point.
(2)
(iv)
State the phase difference between the oscillations of points X and Y in this standing wave.
(1)
(c)
The violinist now touches the string lightly at a point one quarter of the way along it. Determine whether a note of frequency 1320 Hz can be produced.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
λ = 2L = 0.660 m
✓ 1
v = fλ = 440 × 0.660 = 290 m s−1
✓ 1
Accept 290.4 m s−1. Award [2] for CNA.
Part (a)(ii)
λ = 2 × 0.220 = 0.440 m
✓ 1
Accept f ∝ 1/L: f = 440 × 0.330/0.220.
f = 290/0.440 = 660 Hz
✓ 1
Allow ECF from (a)(i). Award [2] for CNA.
Part (b)(i)
Nodes at both ends and at the touched point «and at two-thirds of the length»
✓ 1
Three loops of equal length «the third harmonic»
✓ 1
Part (b)(ii)
f = 3 × 440 = 1320 Hz
✓ 1
Part (b)(iii)
The touched point must be a node, but the first harmonic has nodes only at the ends
✓ 1
and the second harmonic has nodes only at the ends and the centre; neither has a node one third of the way along, so neither can exist
✓ 1
OWTTE.
Part (b)(iv)
π rad «180°; X and Y are in antiphase»
✓ 1
X and Y are in adjacent loops.
Part (c)
The point at L/4 must be a node, so only harmonics with a node there can exist: n = 4, 8, …
✓ 1
1320 Hz is the third harmonic, whose nodes are at 0, L/3, 2L/3 and L — not at L/4
✓ 1
So 1320 Hz cannot be produced «the lowest note is now 4 × 440 = 1760 Hz»
✓ 1
MP3 only scores if a reason is given.
Answers: (a)(i) 290 m s−1 · (a)(ii) 660 Hz · (b)(ii) 1320 Hz · (b)(iv) π rad · (c) not possible; lowest note 1760 Hz (the remaining parts are explanations — see the table above)
A vibration energy harvester is bolted to a pump in a factory. It consists of a small mass of 12.0 g on the end of a thin steel strip that behaves like a spring, so the mass can oscillate with simple harmonic motion. A magnet on the mass moves inside a coil and generates electrical energy, which removes energy from the oscillation. The pump vibrates at 50.0 Hz and drives the harvester.
The graph shows how the amplitude of oscillation of the mass varies with the driving frequency.
Graph drawn to scale
(a)
Show that the spring constant of the strip must be about 1.2 × 103 N m−1 for the natural frequency of the harvester to be 50.0 Hz.
(1)
(b)
Outline why the harvester is designed to have a natural frequency equal to the frequency of the pump.
(2)
(c)
The same harvester is to be used on a pump that vibrates at 60.0 Hz. Determine the mass that must be removed so that the harvester has this natural frequency.
(2)
(d)
(i)
Estimate, using the graph, the amplitude of the mass when the pump runs at 49.0 Hz.
(1)
(ii)
Suggest one disadvantage of a harvester whose resonance peak is very sharp.
(1)
(e)
At resonance the amplitude of the mass is 4.0 mm.
(i)
Determine the maximum resultant force on the mass.
(3)
(ii)
Suggest why the steel strip may eventually break.
(1)
(f)
Explain, in terms of energy, why the amplitude of the mass at resonance stays constant while the harvester supplies electrical energy.
Must see full substitution OR answer to at least 3 s.f. Accept a rearrangement of T = 2π√(m/k) with T = 0.0200 s.
Part (b)
When the driving frequency equals the natural frequency, resonance occurs
✓ 1
so the amplitude of the mass is a maximum and energy is transferred from the pump to the harvester at the greatest rate «giving the greatest electrical output»
✓ 1
OWTTE.
Part (c)
m = k/(2πf)² = 1184/(2π × 60.0)² = 8.33 × 10−3 kg
✓ 1
ALTERNATIVE: m ∝ 1/f², so m = 12.0 × (50.0/60.0)² = 8.33 g.
Mass removed = 12.0 − 8.33 = 3.7 g
✓ 1
Award [2] for CNA.
Part (d)(i)
2.1 mm
✓ 1
Accept 2.0–2.3 mm.
Part (d)(ii)
A small change in the pump's frequency causes a large fall in amplitude «roughly halving it for a 1 Hz change», and so a large fall in electrical output
✓ 1
Allow ECF from (d)(i).
Part (e)(i)
ω = 2π × 50.0 = 314 rad s−1
✓ 1
amax = ω²x0 = 314² × 4.0 × 10−3 = 395 m s−2
✓ 1
Fmax = ma = 0.0120 × 395 = 4.7 N
✓ 1
Award [3] for CNA.
Part (e)(ii)
The large force on the strip reverses direction twice in every cycle, so the strip is bent back and forth 50 times every second and the metal fatigues «destructive resonance»
✓ 1
OWTTE.
Part (f)
The driving force from the pump does work on the harvester, supplying energy on every cycle
✓ 1
The coil converts kinetic energy of the mass into electrical energy «and some energy is dissipated by friction and air resistance», removing energy on every cycle
✓ 1
The amplitude is constant because the energy supplied per cycle equals the energy removed per cycle
✓ 1
OWTTE.
Answers: (a) k = 1.18 × 103 N m−1 · (c) 3.7 g · (d)(i) 2.1 mm · (e)(i) 4.7 N (the remaining parts are explanations — see the table above)
90C-2-68Standing waves in pipesPaper 2Medium13 marks
An engineer wants to find the depth L of the air column in a narrow, disused well. The well shaft is open at the top, and the water surface at the bottom behaves as a closed end. A loudspeaker LS at the top is connected to a signal generator, and a microphone M records the sound.
As the frequency is slowly increased from 50 Hz, the sound becomes loud at 58.5 Hz and next at 67.5 Hz. The speed of sound in the air in the shaft is 340 m s−1.
Figure 1 — Diagram NOT to scale
(a)
State the type of point, displacement node or displacement antinode, that is at the water surface and at the top of the shaft.
(1)
(b)
Explain why the frequencies of the standing waves in the shaft are odd multiples of the frequency of the first harmonic.
(2)
(c)
(i)
Show that the frequency of the first harmonic is 4.5 Hz.
(1)
(ii)
Calculate L.
(2)
(iii)
Deduce the number of the harmonic at 58.5 Hz.
(1)
(d)
On Figure 1, sketch the displacement pattern of the air in the shaft for the third harmonic.
(2)
(e)
Determine whether the sound is loud when the frequency is 100 Hz.
(2)
(f)
After heavy rain the water level rises by 1.5 m. Calculate the new difference in frequency between two successive loud sounds.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
displacement node at the water surface AND displacement antinode at the top
✓ 1
Both needed.
Part (b)
the shaft must hold a node at the bottom and an antinode at the top, so its length is an odd number of quarter-wavelengths: L = (2n − 1)λ/4
✓ 1
so f = (2n − 1)v/4L, odd multiples of f1 = v/4L «even multiples would need the same type of point at both ends»
✓ 1
OWTTE.
Part (c)(i)
successive standing waves differ by 2f1, so f1 = (67.5 − 58.5)/2 «= 4.5 Hz»
✓ 1
Must see the factor of 2 explained or used. Must see full substitution OR answer to 3 s.f.
Part (c)(ii)
λ1 = 4L OR L = v/4f1
✓ 1
L = 340/(4 × 4.5) = 18.9 m
✓ 1
Award [2] for CNA. Accept 19 m.
Part (c)(iii)
58.5/4.5 = 13: the 13th harmonic
✓ 1
Do not accept 7 «the seventh standing wave».
Part (d)
node at the water surface AND antinode at the top
✓ 1
one further node one third of the way down from the top AND one further antinode two thirds of the way down «three quarter-wavelengths in the shaft»
✓ 1
Accept the envelope drawn as one curve or as a pair of curves.
Part (e)
100/4.5 = 22.2, which is not an odd whole number «nearest loud frequencies 94.5 Hz and 103.5 Hz»
✓ 1
Allow ECF from (c)(i).
so the sound is not loud at 100 Hz
✓ 1
MP2 only scores if MP1 scores.
Part (f)
new length 17.4 m, so f1 = 340/(4 × 17.4) = 4.89 Hz
✓ 1
Allow ECF from (c)(ii).
difference = 2f1 = 9.8 Hz
✓ 1
Award [2] for CNA.
Answers: (c)(ii) 18.9 m · (c)(iii) 13th harmonic · (f) 9.8 Hz (the remaining parts are explanations — see the table above)
The stem of a tuning fork of frequency 512 Hz can be pressed on a hollow wooden box. The air inside the box forms a column that is closed at one end and open at the other. When the stem is on the box, the vibration of the fork makes the top of the box vibrate, and this drives the air in the box at 512 Hz. The speed of sound in the air is 343 m s−1.
Figure 1 shows how the amplitude of the tip of one prong varies with time t after the fork is struck while it is held in the hand, away from the box.
Figure 1 — drawn to scale
(a)
The box is designed so that the frequency of the first harmonic of its air column is 512 Hz.
(i)
Outline what is meant by resonance of the air in the box.
(1)
(ii)
Show that the length of the air column is about 0.17 m.
(1)
(b)
(i)
Determine, using Figure 1, the number of oscillations that the prong makes while its amplitude falls to half of its initial value.
(2)
(ii)
State, with a reason, the type of damping of the fork when it is held in the hand.
(1)
(c)
When the stem of the fork is pressed on the box, the note is much louder than when the fork is held in the hand, but it dies away in a much shorter time. Explain these observations.
(3)
(d)
On Figure 1, sketch the variation of the amplitude of the tip of the prong with time when the fork, struck in the same way, stands on the box.
(2)
(e)
The air column responds strongly only to driving frequencies within about 2 % of its natural frequency. The fork is replaced by a fork of frequency 480 Hz. Determine whether the box makes the note of this fork much louder.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
the air is driven at a frequency equal to its natural frequency, so it oscillates with a large «maximum» amplitude
✓ 1
OWTTE. Do not accept 'the air vibrates at the same frequency' alone.
Part (a)(ii)
L = λ/4 = v/4f = 343/(4 × 512) = 0.167 «m»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(i)
the amplitude falls from 80 μm to 40 μm in 4.0 s
✓ 1
Accept 3.8–4.2 s.
N = 512 × 4.0 = 2.0 × 103 «2048»
✓ 1
Allow ECF from MP1. Accept 1.9 × 103 to 2.2 × 103. Award [2] for CNA.
Part (b)(ii)
light damping, because the amplitude decreases only gradually while the fork makes a very large number of oscillations «about 2000 in halving»
✓ 1
Reason needed. OWTTE.
Part (c)
the air column is driven at its natural frequency, so it resonates and the air «and the box» oscillate with a large amplitude: sound energy is emitted at a much greater rate, so the note is louder
✓ 1
OWTTE.
this energy comes from the fork, which receives no further energy after it is struck «its energy is limited to the energy given by the strike»
✓ 1
so the fork loses its energy more quickly: its damping is heavier and its amplitude falls to zero sooner
✓ 1
OWTTE. MP3 only scores if the link to energy loss is made.
Part (d)
curve starts at the same initial amplitude «80 μm» at t = 0
✓ 1
curve lies below the original at all later times, falling much more steeply «half in much less than 4 s» and approaching zero
✓ 1
Do not award MP2 if the curve crosses the original.
Part (e)
(512 − 480)/512 = 6 % «or the band is about 502 Hz to 522 Hz», so 480 Hz is outside the band
✓ 1
Accept 6.25 %.
so the air column does not resonate «its next natural frequency is 1536 Hz»: the box does not make the note much louder
✓ 1
MP2 only scores if MP1 scores.
Answers: (a)(ii) 0.167 m · (b)(i) 2.0 × 103 oscillations · (e) 6 % from 512 Hz: no (the remaining parts are explanations — see the table above)
92C-2-71Standing waves on stringsPaper 2Medium12 marks
A bass string of a piano has a vibrating length of 1.20 m between two fixed supports. The frequency of its first harmonic is 110 Hz. The string is set vibrating by a felt hammer that strikes it at point P, 0.150 m from the left-hand end. The vibrating string sets up several harmonics at the same time, and each produces sound of its own frequency. Point Q is 0.750 m from the left-hand end. The speed of sound in air is 340 m s−1.
Figure 1 — drawn to scale horizontally
(a)
Calculate the speed of transverse waves on the string.
(2)
(b)
(i)
Calculate the wavelength in air of the sound of frequency 110 Hz.
(1)
(ii)
Explain why this wavelength is different from the wavelength of the first harmonic on the string.
(1)
(c)
On Figure 1, sketch the shape of the string at an instant of maximum displacement when it vibrates in its third harmonic.
(2)
(d)
Explain why the eighth harmonic of the string is not produced when the string is struck at P.
(2)
(e)
(i)
Show that P is an antinode of the fourth harmonic.
(1)
(ii)
Calculate, for the second harmonic, the ratio (amplitude of P)/(amplitude at an antinode).
(1)
(iii)
State and explain the phase difference between the oscillations of P and Q in the fourth harmonic.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
λ1 = 2L = 2.40 m
✓ 1
v = fλ = 110 × 2.40 = 264 m s−1
✓ 1
Award [2] for CNA.
Part (b)(i)
λ = 340/110 = 3.09 m
✓ 1
Part (b)(ii)
the frequency is the same «the string drives the air at its own frequency» but the speed of sound in air is different from the speed of waves on the string, and λ = v/f
✓ 1
OWTTE.
Part (c)
nodes at both ends and at 0.40 m and 0.80 m from the left-hand end
✓ 1
three loops of equal length and amplitude, with adjacent loops on opposite sides of the equilibrium line
✓ 1
Accept the two extreme positions drawn.
Part (d)
in the eighth harmonic the nodes are 1.20/8 = 0.15 m apart, so P is a node
✓ 1
the hammer makes P move, so it cannot set up a standing wave that needs P to be at rest «no energy goes into this harmonic»
✓ 1
OWTTE. MP2 only scores if MP1 scores.
Part (e)(i)
λ4 = 2 × 1.20/4 = 0.60 m, so the nodes are at 0, 0.30 m, 0.60 m … and P «0.15 m = λ4/4 from the end» is midway between two nodes
P «between 0 and 0.30 m» and Q «between 0.60 m and 0.90 m» are separated by two nodes, at 0.30 m and 0.60 m; each node reverses the phase, so the phase difference is 2π ≡ 0
✓ 1
MP2 only scores if MP1 scores. A bare 'they are one wavelength apart' does not score MP2.
Answers: (a) 264 m s−1 · (b)(i) 3.09 m · (e)(ii) 0.71 (the remaining parts are explanations — see the table above)
93IBP2S-024-03-13Standing waves in pipesPaper 2Medium6 marks
An acoustic engineer tests a sound-absorbing panel in an impedance tube. A loudspeaker at one end produces sound of frequency 1500 Hz. The sound reflects from the panel and forms a standing wave. A small microphone that responds to pressure variation is moved along the tube. The positions where the microphone signal is a maximum are recorded.
Position of maximum signal / cm: 5.6, 17.1, 28.5, 40.0
(a)
Explain why the microphone signal is a maximum at displacement nodes.
(2)
(b)
Determine, using all the data, the wavelength of the sound.
(2)
(c)
Calculate the speed of sound in the tube.
(1)
(d)
Outline why finding the spacing from the first and last readings gives a more precise result than using a single pair of adjacent readings.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
At a displacement node, the air on either side oscillates in antiphase: it moves towards the node together and then away from it together .
✓ 1
Compressions and rarefactions therefore form there, so the pressure variation is greatest (a pressure antinode), giving the largest signal .
✓ 1
Part (b)
Mean spacing = (40.0 − 5.6)/3 = 11.47 cm .
✓ 1
Adjacent maxima are λ/2 apart, so λ = 2 × 11.47 ≈ 22.9 cm (accept 22.9–23.0 cm) .
✓ 1
Part (c)
v = fλ = 1500 × 0.229 ≈ 344 m s−1 (accept 343–345 m s−1; allow ECF from (b)) .
✓ 1
Part (d)
The uncertainty in reading each position is the same, but it is spread over three spacings (a larger distance), so the percentage uncertainty in the mean spacing is smaller .
✓ 1
Exam tip: When you are given several equally spaced maxima, use the span from the first to the last instead of one gap. Divide by the number of intervals, not the number of readings. Keep unrounded values until the final step.
94IBP2S-024-03-14Standing waves in pipesPaper 2Medium7 marks
A future crewed mission plans to take a simple musical pipe to Mars. The pipe is open at both ends and has length 0.85 m. The speed of sound is 340 m s−1 in air on Earth and 240 m s−1 in the Martian atmosphere.
(a)
Calculate the fundamental frequency of the pipe on Earth.
(2)
(b)
Calculate the fundamental frequency of the same pipe on Mars.
(1)
(c)
Explain whether the position of the displacement node in the fundamental mode is different on Mars.
(2)
(d)
Determine the length of a pipe closed at one end that would have a fundamental frequency of 200 Hz on Mars.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
For a pipe open at both ends, λ = 2L = 1.70 m .
✓ 1
f = 340 / 1.70 = 200 Hz .
✓ 1
Part (b)
f = 240 / 1.70 ≈ 141 Hz .
✓ 1
Part (c)
The positions of nodes and antinodes are fixed by the boundary conditions and the length of the pipe .
✓ 1
λ is still 2L, so the node stays at the centre of the pipe; only the frequency changes .
✓ 1
Part (d)
λ = v/f = 240 / 200 = 1.20 m .
✓ 1
L = λ/4 = 0.30 m .
✓ 1
Exam tip: Changing the medium changes v and therefore f. It does not change the shape of the standing-wave pattern in a pipe of fixed length.
95IBP2S-024-01-12Standing waves in pipesPaper 2Medium7 marks
An acoustics engineer tests a duct that is closed at one end. A variable-frequency speaker is placed at the open end. For each frequency f, the engineer adjusts the length L of the air column until the first (lowest) resonance is heard. Because the antinode forms slightly outside the open end, the effective length of the column is L + e, where e is the end correction.
f / Hz : 400, 500, 640, 800, 1000 L / m : 0.203, 0.160, 0.122, 0.096, 0.074
The relationship is \(L = \frac{v}{4}\cdot\frac{1}{f} - e\), where v is the speed of sound.
(a)
Outline why the effective length L + e equals one quarter of the wavelength at the first resonance.
(2)
(b)
Using the first and last rows of the data, determine the speed of sound v.
(3)
(c)
Determine the end correction e.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The closed end is a displacement node and the open end (just beyond it) is a displacement antinode .
✓ 1
The shortest node-to-antinode distance is λ/4, so L + e = λ/4 .
✓ 1
Part (b)
Gradient of L against 1/f = (0.203 − 0.074)/(2.50×10−3 − 1.00×10−3) .
✓ 1
Gradient = 86.0 m s−1 .
✓ 1
v = 4 × gradient = 344 m s−1 .
✓ 1
Part (c)
Using the last row: e = 86.0 × 1.00×10−3 − 0.074 .
✓ 1
e = 0.012 m . ECF from (b).
✓ 1
Exam tip: Recognise that the equation has the form y = mx + c with x = 1/f. The gradient gives v/4, and the intercept gives −e.
96IBP2S-024-01-13SuperpositionPaper 2Medium6 marks
In speech science, the human vocal tract is modelled as a uniform tube 0.17 m long. The tube is closed at the vocal cords and open at the lips. The resonances of the tube, called formants, shape the sound of vowels. The speed of sound in the warm, moist air of the vocal tract is 350 m s−1.
Diagram not to scale
(a)
Explain how a standing wave is formed in the vocal tract.
(2)
(b)
Calculate the frequency of the lowest formant.
(2)
(c)
A patient breathes a medical helium–oxygen mixture. In this mixture the speed of sound is 1.8 times the speed in air. Calculate the new frequency of the lowest formant.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Sound waves travel along the tube and are reflected at its ends .
✓ 1
The incident and reflected waves have the same frequency and similar amplitude, and they superpose to give fixed positions of nodes and antinodes .
✓ 1
Part (b)
λ = 4L = 0.68 m .
✓ 1
f = 350/0.68 = 515 Hz .
✓ 1
Part (c)
The wavelength is unchanged because it is fixed by the tube length and boundary conditions .
✓ 1
f = 1.8 × 515 ≈ 926 Hz (accept 927 Hz). ECF from (b) .
✓ 1
Exam tip: When 'explain' is used for a standing wave, both reflection and superposition of waves travelling in opposite directions are needed for full marks.
97IBP2S-024-01-14Standing waves in pipesPaper 2Medium7 marks
A straight air-conditioning duct of length 6.0 m is open at both ends. A fan inside the building produces a steady hum of frequency 85 Hz, and the duct is heard to resonate loudly. The speed of sound in the duct is 340 m s−1.
Diagram not to scale
(a)
Calculate the fundamental frequency of the duct.
(2)
(b)
Show that the 85 Hz hum can drive a resonance of the duct, and state the harmonic number.
(2)
(c)
State the number of displacement nodes along the duct in this mode.
(1)
(d)
Engineers line the duct with sound-absorbing material. Explain why this reduces the loudness of the hum.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
λ = 2L = 12.0 m .
✓ 1
f1 = 340/12.0 = 28.3 Hz .
✓ 1
Part (b)
85/28.3 = 3.0, a whole number, so 85 Hz equals a natural frequency of the duct .
✓ 1
It is the third harmonic . ECF from (a).
✓ 1
Part (c)
3 displacement nodes . ECF from (b).
✓ 1
Part (d)
The lining increases damping: energy is dissipated from the sound wave at each reflection or pass along the duct .
✓ 1
The amplitude reached at resonance is therefore much smaller, and the resonance peak is broader and lower .
✓ 1
Common error: An open–open pipe in its nth harmonic has n displacement nodes. Count antinodes and nodes carefully rather than guessing.
98IBP2S-024-00-13Resonance & dampingPaper 2Medium8 marks
The drum of a washing machine and its load have a combined mass of 30 kg. The drum is supported on springs with a combined spring constant of 1.7 ×105 N m−1. When spinning, an uneven load produces a periodic driving force with frequency equal to the rotation frequency of the drum. During a spin cycle the drum speeds up steadily from rest to 20 revolutions per second.
(a)
State what is meant by the natural frequency of the drum–spring system.
(1)
(b)
Show that the natural frequency of vertical oscillation of the drum is about 12 Hz.
(2)
(c)
Explain why the machine vibrates violently for a short time as it speeds up, and then vibrates much less at full speed.
(3)
(d)
Dampers are fitted between the drum and the casing. Outline how they change the amplitude–frequency response of the drum.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The frequency at which the system oscillates freely when displaced and released, with no external driving force acting (and negligible damping)
✓ 1
Part (b)
Uses f = (1/2π)√(k/m), or T = 2π√(m/k) with f = 1/T
✓ 1
f = (1/2π)√(1.7×105/30) = 11.98 Hz ≈ 12.0 Hz
✓ 1
Part (c)
The driving frequency increases from 0 to 20 Hz and so passes through about 12 Hz
✓ 1
Near 12 Hz the driving frequency equals the natural frequency, so resonance occurs: energy transfer to the drum is maximum and the amplitude is large
✓ 1
At 20 Hz the driving frequency is far from the natural frequency, so the amplitude is small
✓ 1
Part (d)
They dissipate energy, so the peak amplitude at resonance is reduced
✓ 1
The resonance peak becomes broader / flatter, OR the peak moves to a slightly lower frequency
✓ 1
Exam tip: In 'show that' questions, write the substituted equation and give a final value to more significant figures than the value quoted in the question.
99IBP2S-024-00-14Tension & wave speedPaper 2Medium7 marks
A guitar-string manufacturer tests a metal wire. A 0.80 m length of the wire is held fixed at both ends. The tension T is varied, and the fundamental frequency f is measured for each tension.
T / N: 20, 40, 60, 80, 100 f / Hz: 62, 89, 108, 126, 140
The speed of transverse waves is v = √(T/μ).
Data from the question
(a)
Outline why a graph of f2 against T is expected to be a straight line through the origin.
(2)
(b)
Using the data, determine the mass per unit length μ of the wire.
(3)
(c)
The wire has a diameter of 0.56 mm. Determine the density of the metal.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
f = v/2L = (1/2L)√(T/μ)
✓ 1
so f2 = T/(4L2μ); since L and μ are constant, f2 ∝ T
✓ 1
Part (b)
Uses a data pair or the gradient of f2 against T, e.g. f2 = 1262 = 1.59×104 Hz2 at T = 80 N
✓ 1
μ = T/(4L2f2) = 80/(4 × 0.802 × 1.59×104)
✓ 1
μ ≈ 2.0 ×10−3 kg m−1
✓ 1
Part (c)
A = π(0.28×10−3)2 = 2.46 ×10−7 m2
✓ 1
ρ = μ/A ≈ 8.0 ×103 kg m−3 (accept 7.9–8.2 ×103; ECF from (b))
✓ 1
Exam tip: Mass per unit length divided by cross-sectional area gives density. Convert the diameter to a radius in metres before squaring.
100IBP2S-024-02-13Tension & wave speedPaper 2Medium6 marks
A vibrating-wire strain gauge is used to monitor stress in a concrete dam. It contains a steel wire of length 0.120 m clamped at both ends. An electromagnet drives the wire, and its frequency is slowly increased from 500 Hz to 3200 Hz. Within this range, large-amplitude vibrations are detected only at the following frequencies:
Explain why large-amplitude vibrations occur only at certain frequencies.
(2)
(b)
Determine the speed of transverse waves on the wire.
(2)
(c)
The wire is driven at 1500 Hz. State and explain whether the point 0.030 m from one clamp is a node or an antinode.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Both clamped ends must be nodes, so only standing waves with a whole number of half-wavelengths fit on the wire (L = nλ/2), giving natural frequencies f = nv/2L
✓ 1
When the driving frequency equals one of these natural frequencies, resonance occurs and energy transfer to the wire is maximum, so the amplitude is large
✓ 1
Part (b)
The fundamental is 750 Hz, so its wavelength is λ = 2L = 2 × 0.120 = 0.240 m
✓ 1
v = fλ = 750 × 0.240 = 180 m s−1
✓ 1
Part (c)
1500 Hz is the second harmonic, so λ = 0.120 m and nodes lie every λ/2 = 0.060 m, at 0, 0.060 m and 0.120 m
✓ 1
The point at 0.030 m is midway between the nodes at 0 and 0.060 m, so it is an antinode
✓ 1
Exam tip: For a wire fixed at both ends, the spacing between successive resonant frequencies equals the fundamental frequency. This gives a quick check on your data.
101IBP2S-024-02-14Nodes & antinodesPaper 2Medium8 marks
Two modules of an orbiting spacecraft are joined by a taut tether of length 25 m, fixed at both ends. The speed of transverse waves on the tether is 40 m s−1. A reaction wheel on one module produces small periodic vibrations at 4.0 Hz.
(a)
Calculate the fundamental frequency of the tether.
(2)
(b)
Show that the reaction wheel can set up a standing wave on the tether, and state the number of nodes, including the ends.
(3)
(c)
Calculate the distance between adjacent antinodes on the tether at this frequency.
(1)
(d)
Engineers wrap the tether in a damping sleeve. Outline the effect of the sleeve on the vibrations at 4.0 Hz.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
λ1 = 2L = 50 m
✓ 1
f1 = v/λ1 = 40/50 = 0.80 Hz
✓ 1
Part (b)
4.0/0.80 = 5 (allow ECF from the first part)
✓ 1
4.0 Hz is a whole-number multiple of f1, so it matches the fifth harmonic and resonance occurs
✓ 1
6 nodes
✓ 1
Part (c)
λ = v/f = 40/4.0 = 10 m, so the antinode spacing is λ/2 = 5.0 m
✓ 1
Part (d)
Energy is dissipated by the sleeve during each oscillation
✓ 1
So the amplitude at resonance is reduced (and the resonance peak becomes broader)
✓ 1
Common error: Adjacent antinodes are half a wavelength apart, not one whole wavelength.
102IBP2S-024-02-15Tension & wave speedPaper 2Medium6 marks
In a simple model of the voice, each vocal fold behaves as a string fixed at both ends and vibrating in its fundamental mode. For one speaker, the vibrating length is 16 mm and the fundamental frequency is 120 Hz.
(a)
Calculate the speed of transverse waves along the vocal fold.
(2)
(b)
When the speaker sings an octave higher, at 240 Hz, the folds stretch to a length of 18 mm. Calculate the new wave speed.
(2)
(c)
Explain why stretching the folds to a greater length could not, by itself, raise the frequency.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
λ = 2L = 0.032 m
✓ 1
v = 120 × 0.032 = 3.8 m s−1
✓ 1
Part (b)
λ = 2 × 0.018 = 0.036 m
✓ 1
v = 240 × 0.036 = 8.6 m s−1
✓ 1
Part (c)
At constant wave speed, a longer fold has a longer fundamental wavelength, so f = v/2L would decrease
✓ 1
The wave speed must therefore increase (here by a factor of about 2.3), which requires greater tension in the fold
✓ 1
Exam tip: Convert millimetres to metres before using v = fλ.
The tower of a wind turbine, together with the nacelle and rotor at its top, can be modelled for side-to-side motion of the top as a mass–spring system with an effective mass of 2.40 × 105 kg and an effective spring constant of 8.5 × 105 N m−1.
The rotor has three blades. Each time a blade passes in front of the tower, the tower receives a small sideways push. The control system can set the rate of rotation of the rotor anywhere between 5.5 and 16.0 revolutions per minute.
Figure 1 shows how the amplitude of the top of the tower would vary with the frequency of a periodic driving force of constant size.
Figure 1 — drawn to scale
(a)
Show that the natural frequency of the tower is about 0.30 Hz.
(1)
(b)
(i)
Calculate the frequency of the pushes from the blades when the rotor turns at 10.0 revolutions per minute.
(2)
(ii)
Determine whether the pushes from the blades can make the tower resonate at any rate of rotation in the operating range.
(3)
(c)
Explain why the amplitude of the top of the tower becomes large at resonance.
(2)
(d)
A damper is fitted at the top of the tower, so that the damping is heavier. Sketch, on Figure 1, the variation of the amplitude with driving frequency for the tower with the damper.
(3)
(e)
Instead of fitting a damper, the operator could change the slowest rate of rotation allowed. Suggest a suitable new slowest rate of rotation, and outline one disadvantage of this change.
(2)
(f)
Suggest one limitation of modelling the tower as a mass–spring system.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
f = (1/2π)√(8.5 × 105/2.40 × 105) = 0.300 «Hz»
✓ 1
Must see full substitution OR answer to at least 3 s.f. Accept T = 2π√(m/k) = 3.34 s and f = 1/T.
Part (b)(i)
10.0 revolutions per minute = 10.0/60 = 0.167 revolutions per second
✓ 1
three pushes per revolution: f = 3 × 0.167 = 0.500 Hz
✓ 1
Award [2] for CNA. Award [1 max] for 0.167 Hz.
Part (b)(ii)
Range of push frequencies: 3 × 5.5/60 = 0.275 Hz to 3 × 16.0/60 = 0.80 Hz
✓ 1
Allow ECF from (b)(i) for the factor of 3.
The natural frequency «0.30 Hz» lies inside this range
✓ 1
so yes: resonance occurs when the rotor turns at about 6.0 revolutions per minute
✓ 1
MP3 only if MP1 is scored. A bald 'yes' scores 0.
Part (c)
The driving frequency equals the natural frequency, so each push is in step «in phase» with the motion of the tower
✓ 1
so energy is transferred to the tower on every cycle and the amplitude grows until the energy lost to damping per cycle equals the energy supplied
✓ 1
OWTTE.
Part (d)
Peak lower than the original peak
✓ 1
Peak at the same or a slightly lower frequency «not at a higher frequency»
✓ 1
Curve below the original at all frequencies, with a broader peak, and approaching ≈ 4 mm at low frequency
✓ 1
Do not award MP3 if the curves cross.
Part (e)
A rate well above 6.0 revolutions per minute, e.g. 8.0 rpm or more, so that the lowest push frequency «≥ 0.40 Hz» is clear of the resonance peak: in Figure 1 the amplitude there is about 5 mm, close to its low-frequency value
✓ 1
Allow ECF from (b)(ii). Accept 7.5 rpm or more «≥ 0.375 Hz». Do not accept a rate only just above 6.0 rpm «e.g. 6.5 rpm», where the amplitude is still several times its low-frequency value.
Disadvantage: the turbine cannot operate «generate energy» in light winds, when the rotor would turn more slowly
✓ 1
OWTTE. Accept: the tower still passes through resonance when starting and stopping.
Part (f)
The mass of a real tower is spread along its height «it is not a point mass on a spring» OR the tower can also vibrate in other modes / at higher natural frequencies OR the stiffness depends on the foundation and soil
✓ 1
Accept any one valid limitation.
Answers: (a) 0.300 Hz · (b)(i) 0.500 Hz · (b)(ii) yes, at about 6.0 revolutions per minute (the remaining parts are explanations — see the table above)
In a helium–neon laser, light of wavelength 633 nm travels back and forth between two parallel mirrors, M1 and M2, that are 0.300 m apart. A standing wave of light forms between the mirrors. In a simple model, the electric field of the light is always zero at each mirror, so each mirror is at a node.
Take the speed of light between the mirrors to be c.
Figure 1 — axes for (b)(iii). Mirror M1 is at distance 0.
(a)
Explain how the standing wave forms between the mirrors.
(2)
(b)
(i)
Show that the number of half-wavelengths between the mirrors is about 9.5 × 105.
(1)
(ii)
Calculate the distance between adjacent nodes.
(1)
(iii)
On Figure 1, sketch the variation of the amplitude of the electric field with distance from M1 for the first 1.0 μm.
(2)
(c)
A standing wave can form only at frequencies for which a whole number of half-wavelengths fits between the mirrors.
(i)
Show that the difference between the frequencies of two adjacent standing waves is 5.0 × 108 Hz.
(2)
(ii)
The neon atoms can amplify light only in a band of frequencies 1.5 × 109 Hz wide. A designer wants only one standing-wave frequency to lie inside this band. Determine the maximum distance between the mirrors.
(2)
(d)
Calculate, in eV, the energy of a photon of the laser light.
(2)
(e)
As the laser warms up, the distance between the mirrors increases by 0.10 μm. Determine the change in the frequency of a standing wave whose number of half-wavelengths stays the same.
(2)
(f)
One mirror lets a small fraction of the light through to form the laser beam. Outline one difference between the light in the beam and the light between the mirrors.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
light reflected at one mirror travels back towards the other mirror and is reflected again «waves travel in both directions between the mirrors»
✓ 1
two waves of the same frequency, speed and «similar» amplitude travelling in opposite directions superpose / interfere
✓ 1
OWTTE.
Part (b)(i)
2L/λ = 2 × 0.300/633 × 10−9 = 9.48 × 105
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
λ/2 = 317 nm «3.17 × 10−7 m»
✓ 1
Accept 316 nm or 316.5 nm.
Part (b)(iii)
amplitude zero at 0 and at 0.32 μm, 0.63 μm and 0.95 μm
✓ 1
Allow ECF from (b)(ii). Positions within half a small square.
three equal maxima midway between the zeros, with a sinusoidal «not triangular» shape and no negative values
✓ 1
Do not award MP2 for a curve drawn below the axis.
f = c/λ = 4.74 × 1014 Hz AND E = hf = 6.63 × 10−34 × 4.74 × 1014 = 3.14 × 10−19 J
✓ 1
Accept E = hc/λ.
E = 3.14 × 10−19/1.60 × 10−19 = 1.96 eV
✓ 1
Accept 1.9–2.0 eV. Award [2] for CNA.
Part (e)
f ∝ 1/L for a fixed number of half-wavelengths, so Δf = f × ΔL/L = 4.74 × 1014 × 0.10 × 10−6/0.300
✓ 1
Allow ECF from (d).
Δf = 1.6 × 108 Hz, a decrease
✓ 1
Direction needed for MP2. Award [2] for CNA with direction.
Part (f)
the beam is a travelling wave that transfers energy away, whereas the standing wave transfers no «net» energy OR the amplitude in the beam is the same everywhere, whereas between the mirrors it varies from zero at nodes to a maximum at antinodes
✓ 1
Accept any one valid difference. OWTTE.
Answers: (b)(ii) 317 nm · (c)(ii) 0.10 m · (d) 1.96 eV · (e) 1.6 × 108 Hz decrease (the remaining parts are explanations — see the table above)
Carbon dioxide molecules absorb infrared radiation strongly at a wavelength of 15.0 μm. In a resonance model of the greenhouse effect, the bending vibration of a CO2 molecule is treated as an oscillator with a natural frequency f0. The oscillating electric field of infrared radiation acts as a periodic driving force on the molecule, and collisions with other molecules act as damping.
Figure 1 shows, for this model, how the amplitude of the vibration varies with the driving frequency for a molecule in the air near the Earth's surface. The mean surface temperature of the Earth is 288 K and the surface temperature of the Sun is 5800 K.
Figure 1 — schematic graph of the model
(a)
Show that f0 is about 2 × 1013 Hz.
(1)
(b)
(i)
Calculate the wavelength at which the radiation emitted by the Earth's surface has its maximum intensity.
(1)
(ii)
Explain, using the resonance model, why CO2 in the atmosphere absorbs a significant fraction of the radiation emitted by the Earth's surface but very little of the radiation arriving from the Sun.
(2)
(c)
Outline how the absorption of infrared radiation by CO2, and its re-emission, leads to a higher temperature of the Earth's surface.
(2)
(d)
In the molecular energy-level model, the absorption at 15.0 μm corresponds to a transition between two vibrational energy levels. Calculate, in eV, the difference between these energy levels.
(2)
(e)
High in the atmosphere the air pressure is much lower, so collisions are much less frequent. On Figure 1, sketch the variation of amplitude with driving frequency for a molecule high in the atmosphere.
(3)
(f)
Suggest one limitation of the resonance model of the greenhouse effect.
the Earth emits a broad infrared spectrum «around 10 μm» that includes 15 μm, so radiation at the natural frequency drives the molecules at resonance: large amplitude, energy absorbed
✓ 1
solar radiation peaks at 2.9 × 10−3/5800 = 5 × 10−7 m, a frequency «6 × 1014 Hz» about 30 times f0, so the driving frequency is far from resonance and the amplitude, and the energy absorbed, is very small
✓ 1
Must use the Sun's temperature or peak wavelength. OWTTE.
Part (c)
the molecules re-emit infrared radiation in all directions, so part of it travels back down to the surface
✓ 1
the surface receives more energy per second than from the Sun alone, so it warms until it radiates more «its equilibrium temperature is higher»
peak narrower, with the curve close to the original away from the peak
✓ 1
Do not award MP2 if the curves cross far from the peak.
peak at the same frequency, 2.0 × 1013 Hz «or very slightly higher, not lower»
✓ 1
Part (f)
the energy of a molecule is quantised, so it absorbs only photons of particular energies «the energy-level model is needed» OR a real molecule has several modes of vibration and rotation, not one natural frequency OR the model does not explain why N2 and O2 hardly absorb infrared
✓ 1
Accept any one valid limitation. OWTTE.
Answers: (a) 2.0 × 1013 Hz · (b)(i) 1.0 × 10−5 m · (d) 0.083 eV (the remaining parts are explanations — see the table above)
106IBP2S-024-03-15Standing waves in pipesPaper 2Hard8 marks
An industrial gas pipe of length 2.60 m is sealed at one end and open at the other. During operation it produces a loud hum at 231 Hz, caused by a longitudinal standing wave in the gas. The speed of sound in the gas is 343 m s−1.
Diagram not to scale
(a)
Show that the hum corresponds to the seventh harmonic of the pipe.
(2)
(b)
Determine the number of displacement nodes in the pipe, including any at the ends.
(2)
(c)
Determine the distance from the open end to the nearest displacement node.
(2)
(d)
Engineers line part of the pipe with a sound-absorbing material. Explain the effect of this damping on the resonance.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
f1 = v/4L = 343 / (4 × 2.60) = 33.0 Hz .
✓ 1
231 / 33.0 = 7.0, an odd integer, which is an allowed harmonic for a pipe closed at one end . Alternative: λ = 343/231 = 1.485 m and 4L/λ = 10.4/1.485 = 7.0, which also earns both marks.
✓ 1
Part (b)
L = 7λ/4, with a displacement node at the sealed end and an antinode at the open end .
✓ 1
Nodes lie at 0, λ/2, λ and 3λ/2 from the sealed end (the next would be at 2λ > 1.75λ), giving 4 nodes .
✓ 1
Part (c)
λ = 343 / 231 = 1.48 m .
✓ 1
The open end is a displacement antinode, so the nearest node is λ/4 ≈ 0.371 m from it .
✓ 1
Part (d)
Energy is dissipated from the oscillation each cycle, so the amplitude at resonance is reduced .
✓ 1
The resonance peak becomes broader and less sharp, meaning a lower Q factor . A statement that the resonant frequency changes is not required.
✓ 1
Common error: When counting nodes, include the node at the closed end. Do not place a node at the open end, which is a displacement antinode.
107IBP2S-024-03-16Standing waves on stringsPaper 2Hard6 marks
An aluminium rod of length 1.50 m is used in a demonstration of longitudinal standing waves. Both ends of the rod are free. The speed of longitudinal waves in aluminium is 5100 m s−1. The rod is held by a clamp that forces a displacement node at the clamping point.
Diagram not to scale
(a)
State the boundary condition at each free end of the rod.
(1)
(b)
The rod is clamped at a point 0.375 m (L/4) from one end. Explain why the rod cannot then vibrate in its fundamental mode.
(2)
(c)
Determine the lowest frequency at which the rod clamped at L/4 can vibrate longitudinally.
(2)
(d)
State whether the midpoint of the rod is a displacement node or a displacement antinode in the mode found in (c).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Each free end is a displacement antinode .
✓ 1
Part (b)
In the fundamental mode of a rod with two free ends, the only node is at the midpoint (L/2) .
✓ 1
The clamp forces a node at L/4, where the fundamental has non-zero amplitude, so this mode cannot form .
✓ 1
Part (c)
The second harmonic (λ = L) has nodes at L/4 and 3L/4, so it is the lowest allowed mode .
✓ 1
f = v/λ = 5100 / 1.50 = 3.4×103 Hz .
✓ 1
Part (d)
Antinode: it is halfway between the nodes at L/4 and 3L/4 . ECF from (c).
✓ 1
Exam tip: A clamp acts as a filter. Only the harmonics that already have a node at the clamping point can be sustained.
108IBP2S-024-01-15Standing waves in pipesPaper 2Hard7 marks
Hot gas in a vertical factory chimney flue can resonate and cause low-frequency vibration. Sensors detect two consecutive resonant frequencies of the gas column: 25.5 Hz and 35.7 Hz. The speed of sound in the hot flue gas is 459 m s−1.
(a)
Show that the flue cannot be behaving as a pipe open at both ends.
(2)
(b)
The flue behaves as a pipe closed at one end. Determine its fundamental frequency, and identify the harmonics detected.
(2)
(c)
Calculate the length of the gas column.
(2)
(d)
State, with a reason, whether a disturbance at 30.6 Hz would cause resonance in the flue.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
For an open–open pipe, consecutive resonances differ by f1, which would give f1 = 35.7 − 25.5 = 10.2 Hz .
✓ 1
25.5/10.2 = 2.5, which is not a whole number, so the frequencies are not harmonics of such a pipe .
✓ 1
Part (b)
For a closed pipe, consecutive resonances differ by 2f1, so f1 = 10.2/2 = 5.1 Hz .
✓ 1
25.5 = 5f1 and 35.7 = 7f1, so these are the 5th and 7th harmonics .
✓ 1
Part (c)
L = v/(4f1) .
✓ 1
L = 459/(4 × 5.1) = 22.5 m . ECF from (b).
✓ 1
Part (d)
No: 30.6 Hz = 6f1, an even harmonic, which cannot exist in a pipe closed at one end .
✓ 1
Exam tip: To identify the pipe type, test whether the lowest given frequency is an integer multiple of the frequency spacing. Odd multiples of half the spacing point to a closed pipe.
109IBP2S-024-01-16Standing vs travelling wavesPaper 2Hard8 marks
Pharmaceutical researchers use an acoustic levitator to suspend small droplets of a drug solution in air without touching them. An ultrasonic transducer emitting 40.0 kHz faces a flat, rigid reflector. A standing sound wave forms between them. The droplets collect at the pressure nodes of the standing wave.
(a)
Outline two ways in which the standing wave differs from a travelling wave.
(2)
(b)
State what the air displacement does at the positions where the droplets are held.
(1)
(c)
Five droplets are held in a line, equally spaced. The distance between the two outermost droplets is 17.2 mm. Determine the speed of sound in the air.
(3)
(d)
The levitator is moved to a warmer room, where the speed of sound is 350 m s−1. The frequency is unchanged. Calculate the new spacing between adjacent droplets.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
In the standing wave the amplitude varies with position (zero at nodes, maximum at antinodes), whereas a travelling wave has the same amplitude everywhere .
✓ 1
A standing wave transfers no net energy along its length, whereas a travelling wave transfers energy .
✓ 1
Part (b)
These are displacement antinodes: the air oscillates with maximum displacement amplitude .
✓ 1
Part (c)
There are 4 gaps, so the spacing = 17.2/4 = 4.30 mm .
✓ 1
Adjacent pressure nodes are λ/2 apart, so λ = 8.60 mm .
✓ 1
v = fλ = 40.0×103 × 8.60×10−3 = 344 m s−1 .
✓ 1
Part (d)
λ = 350/(40.0×103) = 8.75×10−3 m .
✓ 1
Spacing = λ/2 = 4.38 mm .
✓ 1
Common error: With 5 objects in a line there are only 4 gaps. Dividing 17.2 mm by 5 gives the wrong spacing.
110IBP2S-024-00-15Tension & wave speedPaper 2Hard9 marks
A stay cable on a bridge is 120 m long and is fixed at both ends. Its mass per unit length is 60 kg m−1 and its tension is 4.5 ×106 N. Wind blowing across the cable sheds vortices at a frequency fs = 0.20u/D, where u is the wind speed and D = 0.25 m is the cable diameter. The speed of transverse waves is v = √(T/μ).
(a)
Calculate the speed of transverse waves along the cable.
(2)
(b)
Calculate the fundamental frequency of the cable.
(2)
(c)
Hydraulic dampers are fitted near the ends of the cable. Explain how they protect the cable when the vortex-shedding frequency matches one of its harmonics.
(2)
(d)
Determine the wind speed at which the cable resonates in its third harmonic.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = √(4.5×106/60)
✓ 1
v ≈ 274 m s−1
✓ 1
Part (b)
λ = 2L = 240 m
✓ 1
f1 = 274/240 ≈ 1.14 Hz (ECF from (a))
✓ 1
Part (c)
When fs equals a natural frequency, resonance occurs and the amplitude grows large
✓ 1
The dampers dissipate energy from the oscillation, which limits the amplitude at resonance
✓ 1
Part (d)
f3 = 3f1 = 3.42 Hz
✓ 1
u = f3D/0.20
✓ 1
u = 3.42 × 0.25/0.20 ≈ 4.3 m s−1 (ECF from (b))
✓ 1
Common error: Two different speeds appear here: the wave speed on the cable and the wind speed. Keep them clearly labelled and do not mix them up.
111IBP2S-024-00-16Tension & wave speedPaper 2Hard9 marks
A simple model of the human voice treats a vocal fold as a uniform string fixed at both ends, vibrating in its fundamental mode. In one speaker the vibrating length is 16 mm, the vibrating mass is 0.50 g and the fundamental frequency is 120 Hz. The speed of transverse waves is v = √(T/μ), where T is the tension and μ is the mass per unit length.
(a)
Calculate the speed of transverse waves along the vocal fold.
(2)
(b)
Determine the tension in the vocal fold.
(2)
(c)
To sing one octave higher (240 Hz), the speaker stretches the fold so that its length increases by 20%. The vibrating mass is unchanged. Determine the new tension required.
(4)
(d)
Suggest one limitation of modelling a vocal fold as a stretched string.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
λ = 2L = 0.032 m
✓ 1
v = fλ = 120 × 0.032 ≈ 3.8 m s−1 (accept 3.84 m s−1)
✓ 1
Part (b)
μ = 0.50×10−3/0.016 = 3.1×10−2 kg m−1
✓ 1
T = μv2 = 0.031 × 3.842 ≈ 0.46 N (allow ECF from (a))
✓ 1
Part (c)
New L = 19.2 mm, so new μ = 0.50×10−3/0.0192 = 2.6×10−2 kg m−1
✓ 1
New v = 2 × 0.0192 × 240 = 9.2 m s−1
✓ 1
T = μv2 = 0.026 × 9.22
✓ 1
T ≈ 2.2 N (about 4.8 times the original tension)
✓ 1
Part (d)
Any one of the following: the tissue is not uniform; its stiffness, not only the tension, provides a restoring force; the vibration is heavily damped by the surrounding tissue and the airflow
✓ 1
Exam tip: When the length changes with fixed mass, both λ and μ change. Recalculate each one before finding the new tension.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy
EssentialNeeded for the site and your cookie choice to work.
Always on
AnalyticsAnonymous visit statistics that help improve the free resources.
MarketingMeasures whether ads reach the right families. Off unless you turn it on.