Edexcel International A Level Physics · Unit 2 · Waves and Electricity
Superposition and stationary waves
40 free, high-quality past-paper-style questions on Superposition and stationary waves for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.40 new this month · 138 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U2-2-A-01Unit 2 · ASEasy1 mark
Two sources of waves are described as coherent. Which statement must be true?
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Answer
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Notes
Coherent sources have a constant phase difference, which also means they have the same frequency. Their amplitudes need not be equal, and whether they superpose constructively depends on the path difference to each point.
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Answer A
Answer: A
Exam tip: Define coherence in terms of phase: “a constant phase difference”. “The same frequency” on its own is not enough, because two separate sources at the same frequency can drift in phase.
Waves from two coherent sources that vibrate in phase meet at a point. The path difference at the point is 0.20 m and the wavelength is 0.80 m. What is the phase difference between the two waves at the point?
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Phase difference = 2π × path difference ÷ λ = 2π × 0.20 ÷ 0.80 = π/2 = 1.6 rad (1.57 rad). A quarter of a wavelength of path is a quarter of a cycle.
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Answer C
Answer: C
Exam tip: One whole wavelength of path difference is one whole cycle, 2π rad. Find the fraction of a wavelength first, then multiply by 2π.
Which statement about a stationary wave on a string is correct?
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Between two adjacent nodes every point moves up and down together: they are in phase, though with different amplitudes. Amplitude varies from zero at nodes to a maximum at antinodes, there is no net energy transfer along a stationary wave, and adjacent antinodes are λ/2 apart.
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Answer B
Answer: B
Exam tip: Learn the three contrasts with a progressive wave: amplitude varies with position, phase is the same within a loop, and no energy is transferred along the wave.
A student strikes a tuning fork and holds it so that both prongs just touch the surface of water in a shallow tray. Circular ripples spread out from each prong and overlap.
(a)
State the principle of superposition.
(2)
(b)
Explain why the two prongs act as coherent sources.
(1)
(c)
State what is meant by a wavefront.
(1)
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Part (a)
when two (or more) waves meet at a point
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the resultant displacement is the (vector) sum of the individual displacements
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Part (b)
the prongs vibrate at the same frequency with a constant phase difference between them (because they are parts of the same fork)
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Part (c)
a line (or surface) joining neighbouring points on a wave that are in phase, e.g. along one crest
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Exam tip: In the superposition statement, say “displacement”, not “amplitude”: the displacements add at every instant, and the resultant amplitude follows from that.
One end of a long coiled spring is tied to a table leg. A student shakes the other end from side to side at a steady frequency, and at certain frequencies a stationary wave forms on the spring.
(a)
Explain how the stationary wave is formed.
(3)
(b)
State the distance between adjacent nodes in terms of the wavelength λ.
(1)
(c)
State one difference between the energy transfer in this stationary wave and in a progressive wave on the spring.
(1)
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Part (a)
the wave from the student reflects at the table leg
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the incident and reflected waves have the same frequency and travel in opposite directions
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they superpose: at nodes they are always in antiphase and cancel; at antinodes they are in phase and reinforce
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Part (b)
λ/2
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Part (c)
a stationary wave has no net transfer of energy along the spring, whereas a progressive wave transfers energy in its direction of travel
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Exam tip: “Reflection”, “same frequency”, “opposite directions” and “superpose” are the four ideas an explanation of a stationary wave must contain.
The steel cable of a zip wire has a mass of 18.0 kg and a length of 40.0 m. The tension in the cable is 2.50 kN. A technician strikes one end of the cable, sending a transverse pulse along it.
(a)
Calculate the mass per unit length of the cable.
(1)
(b)
Calculate the speed of the pulse along the cable.
(2)
(c)
Calculate the time for the pulse to travel the length of the cable.
(1)
(d)
Deduce the tension that would halve this time.
(1)
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Part (a)
μ = 18.0 ÷ 40.0 = 0.450 kg m−1
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Part (b)
v = √(T/μ) = √(2500 ÷ 0.450)
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= 74.5 m s−1
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Part (c)
40.0 ÷ 74.5 = 0.537 s (allow error carried forward)
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Part (d)
the speed must double, and v ∝ √T, so the tension must be four times as large: 10.0 kN
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Exam tip: Convert kN to N before using v = √(T/μ). Leaving 2.50 in the formula makes the speed too small by a factor of √1000.
The diagram shows wavefronts spreading from two coherent sources S1 and S2 that vibrate in phase with equal amplitude. Solid lines are crests and dashed lines are troughs. At which point is the amplitude of the resultant oscillation a minimum?
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S lies on a crest (solid line) from one source and a trough (dashed line) from the other. The waves arrive in antiphase (path difference 1.5λ), so they cancel: a minimum at S. At P and R crest meets crest, and at Q trough meets trough, so all three are maxima.
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Answer D
Answer: D
Common error: Thinking trough + trough gives a minimum. Two troughs meeting give a deeper trough: the waves are in phase there, so it is a maximum.
The diagram shows a stationary wave on a string fixed at both ends, with a scale beneath it. What is the wavelength of the progressive waves that form the stationary wave?
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There are four loops in 1.20 m, so each loop (node to node) is 0.30 m. One loop is half a wavelength, so λ = 2 × 0.30 = 0.60 m.
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Answer B
Answer: B
Exam tip: Count the loops, divide the length by the number of loops, then double: each loop is λ/2.
Two triangular pulses travel towards each other along a long rope. Figure 1 shows the rope at time t = 0. Each pulse moves at 0.50 m s−1 and keeps its shape. Point P is 0.90 m from the left-hand end of the rope.
(a)
Calculate the time at which the centres of both pulses reach P.
(1)
(b)
State the displacement of P at this time.
(1)
(c)
Determine the displacement of P at t = 0.80 s.
(2)
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Part (a)
each centre is 0.50 m from P (0.40 m and 1.40 m from the end), so t = 0.50 ÷ 0.50 = 1.0 s
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Part (b)
(+3.0) + (−2.0) = +1.0 cm
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Part (c)
the centres are now at 0.80 m and 1.00 m, each 0.10 m from P, so pulse 1 gives +1.5 cm and pulse 2 gives −1.0 cm at P
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resultant displacement = +0.5 cm
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Exam tip: Superposition works point by point: find each pulse’s displacement at the point separately, keeping its sign, then add them.
Figure 1 shows the displacement pattern of a stationary sound wave in a tube that is open at both ends. A scale is placed beneath the tube. The speed of sound is 340 m s−1. Ignore end corrections.
(a)
Explain why there is a displacement antinode at each open end of the tube.
(1)
(b)
Use Figure 1 to determine the wavelength of this stationary wave.
(1)
(c)
Calculate the frequency of the sound.
(1)
(d)
Calculate the fundamental frequency of the tube.
(1)
(e)
One end of the tube is now closed. Calculate the new fundamental frequency.
(1)
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Part (a)
the air at an open end is free to move, so it oscillates with the largest amplitude
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Part (b)
nodes at 0.15 m and 0.45 m are λ/2 apart, so λ = 2 × 0.30 = 0.60 m
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Part (c)
f = 340 ÷ 0.60 = 567 Hz
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Part (d)
fundamental: λ = 2L = 1.20 m, f = 340 ÷ 1.20 = 283 Hz
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Part (e)
λ = 4L = 2.40 m, f = 340 ÷ 2.40 = 142 Hz
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Common error: Assuming the tube length is one wavelength (or half) without looking at the pattern. Count node-to-node spacings (λ/2) and node-to-antinode spacings (λ/4).
A stationary wave is set up on a stretched rope by a motor vibrating at 150 Hz. Adjacent nodes are 0.24 m apart. What is the speed of the waves on the rope?
Figure: stationary wave on the rope; N = node (not to scale)
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Adjacent nodes are λ/2 apart, so λ = 2 × 0.24 = 0.48 m. v = fλ = 150 × 0.48 = 72 m s−1.
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Answer C
Answer: C
Common error: Taking the node spacing as the wavelength. One loop of a stationary wave is half a wavelength, so double the node spacing.
A drainpipe of length 0.85 m is blocked at its lower end and open at its upper end. Wind blowing across the top makes the air inside resonate at its lowest possible frequency. The speed of sound is 340 m s−1. Ignore end corrections. What is this frequency?
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Closed at one end: node at the blocked end, antinode at the open end, so the fundamental has L = λ/4. λ = 4 × 0.85 = 3.4 m and f = 340 ÷ 3.4 = 100 Hz.
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Answer A
Answer: A
Model sketch
Figure: fundamental mode in the pipe, displacement envelope (not to scale)
Exam tip: Mark N at a closed end and A at an open end. The simplest pattern from N to A is a quarter of a wavelength.
A nylon washing line has a mass per unit length of 3.0 × 10−3 kg m−1 and is under a tension of 48 N. What is the speed of transverse waves along the line?
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v = √(T/μ) = √(48 ÷ 3.0 × 10−3) = √(1.6 × 104) = 1.3 × 102 m s−1 (126 m s−1).
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Answer C
Answer: C
Common error: Forgetting the square root. Check the units: N ÷ kg m−1 = m2 s−2, which is a speed squared.
In an investigation of a vibrating string, a student keeps the vibrating length and the string the same and changes the tension T. For each tension she records the fundamental frequency f. Which graph should be a straight line through the origin?
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f = (1/2L)√(T/μ), so f2 = T/(4L2μ). With L and μ constant, f2 against T is a straight line through the origin with gradient 1/(4L2μ).
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Answer B
Answer: B
Exam tip: Square both sides to remove a square root, then match the result to y = mx. Plotting f against √T would also work.
Two loudspeakers at the edge of a sports field are connected to the same amplifier and emit sound in phase at a frequency of 500 Hz. The speed of sound is 340 m s−1. A spectator stands at point R, which is 25.00 m from one loudspeaker and 27.04 m from the other.
Figure: two loudspeakers emitting in phase and the spectator at R (not to scale)
(a)
Calculate the wavelength of the sound.
(1)
(b)
Show that the spectator hears a loud sound at R.
(2)
(c)
The spectator walks to a point S where the path difference is 2.38 m. Deduce what she hears at S.
(1)
(d)
Suggest why the sound at S is not completely silent.
(1)
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Notes
Part (a)
λ = 340 ÷ 500 = 0.68 m
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Part (b)
path difference = 27.04 − 25.00 = 2.04 m
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= 3.0λ, a whole number of wavelengths, so the waves arrive in phase and superpose constructively
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Part (c)
2.38 ÷ 0.68 = 3.5λ, so the waves arrive in antiphase: a quiet sound / a minimum
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Part (d)
the waves have different amplitudes at S because S is at different distances from the two loudspeakers / sound also reaches S by reflection from nearby surfaces, so the cancellation is incomplete
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Exam tip: Always write the path difference as a number of wavelengths, then say “in phase” or “in antiphase” before naming a maximum or minimum.
A heavy rope of length 1.50 m is fixed at both ends and driven by a motor. At a frequency of 84 Hz a stationary wave with three loops is formed. Point X is 0.25 m from one end of the rope and point Y is 0.75 m from the same end.
(a)
Calculate the wavelength of the waves on the rope.
(1)
(b)
Calculate the speed of the waves on the rope.
(1)
(c)
Calculate the fundamental frequency of the rope.
(1)
(d)
Compare the amplitudes and the phases of the oscillations of X and Y, and state the displacement of Y at the instant X has its maximum positive displacement.
(3)
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Part (a)
three loops = 3λ/2 = 1.50 m, so λ = 1.00 m
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Part (b)
v = fλ = 84 × 1.00 = 84 m s−1
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Part (c)
84 ÷ 3 = 28 Hz
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Part (d)
X and Y are both at antinodes (0.25 m and 0.75 m are midway between nodes at 0, 0.50 m and 1.00 m), so they have the same (maximum) amplitude
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they are in adjacent loops, so they oscillate in antiphase / phase difference π rad
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Y has its maximum negative displacement (equal in size to that of X)
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Model sketch
Figure: three-loop stationary wave at the instant X has its maximum positive displacement (solid) and half a cycle later (dashed)
Exam tip: Mark the nodes first (every λ/2 from a fixed end), then decide which loop each point is in. Same loop: in phase. Adjacent loops: antiphase.
One tube of a set of pan pipes is closed at its lower end and open at its upper end. Its length is 0.180 m. The speed of sound in air is 343 m s−1. Ignore end corrections.
(a)
Explain why there is a displacement node at the closed end and a displacement antinode at the open end.
(2)
(b)
Calculate the fundamental frequency of the tube.
(2)
(c)
Calculate the next two frequencies at which the tube resonates.
(1)
(d)
Explain why the tube cannot resonate at 953 Hz.
(1)
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Part (a)
at the closed end the air molecules cannot move (along the tube), so the displacement is always zero: a node
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at the open end the air can move freely, so the oscillation has its largest amplitude: an antinode
953 Hz is twice the fundamental, which would need a whole number of half wavelengths in the tube (the same kind of point at both ends); with a node at one end and an antinode at the other only odd multiples of λ/4 fit
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Model sketch
Figure: fundamental mode in the closed tube, displacement envelope (not to scale)
Common error: Listing 2f1 as a resonance of a tube closed at one end. Only odd harmonics are possible: f1, 3f1, 5f1 …
A student is given a length of string, a vibration generator, a signal generator, a pulley clamped to the bench and a set of slotted masses.
(a)
Describe how she could investigate how the fundamental frequency of the string depends on its vibrating length, with the tension and the string kept the same. Your answer should include how she would analyse her results.
(6)
(b)
State one safety precaution for this investigation.
(1)
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Part (a)
Indicative content:
—
string attached to the vibration generator, passed over the pulley with a fixed hanging mass so the tension (mg) stays constant
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vibrating length L measured from the vibration generator to the top of the pulley with a metre rule
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frequency increased slowly from a low value until the string vibrates in one loop with maximum amplitude; frequency read from the signal generator
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repeat the frequency reading (approaching from above and below) and calculate a mean
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move the vibration generator to change L, and record at least five different lengths
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plot f against 1/L: a straight line through the origin shows f ∝ 1/L
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Part (b)
keep feet away from under the hanging masses / place a cushioned box beneath them / wear eye protection in case the string snaps
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Exam tip: In a method answer, name the independent, dependent and control variables, say how each is measured and with what, and finish with the graph you would plot and the shape that confirms the relationship.
An ultrasound transducer emits sound of frequency 40.0 kHz towards a flat metal plate, which reflects it straight back. A small microphone moved along the line between the transducer and the plate detects a series of positions of minimum signal. A student measures the distance across 10 spacings between adjacent minima as 42.5 mm.
(a)
Explain why positions of minimum signal occur.
(2)
(b)
Explain why the student measures across 10 spacings rather than one.
(1)
(c)
Calculate the speed of the ultrasound in air.
(2)
(d)
The uncertainty in the distance of 42.5 mm is ± 0.5 mm. Calculate the percentage uncertainty in this distance.
(1)
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Answer
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Part (a)
the incident and reflected waves have the same frequency and travel in opposite directions, forming a stationary wave
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at the minima (nodes) the two waves always arrive in antiphase and cancel
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Part (b)
the measured distance is larger, so the percentage uncertainty in the measurement (and in the spacing) is smaller
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Part (c)
spacing = 4.25 mm = λ/2, so λ = 8.50 mm
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v = fλ = 40.0 × 103 × 8.50 × 10−3 = 340 m s−1
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Part (d)
0.5 ÷ 42.5 × 100 = 1.2%
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Exam tip: Minima of a stationary wave are nodes, half a wavelength apart. Divide by the number of spacings, not the number of minima, before doubling.
Two loudspeakers vibrate in phase. The graph shows how the displacement at point X caused by each loudspeaker varies with time. X is further from loudspeaker 2 than from loudspeaker 1, and the path difference is less than one wavelength. The speed of sound is 340 m s−1. What is the path difference at X?
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From the graph the period is 4.0 ms and the wave from loudspeaker 2 lags by 1.0 ms, a quarter of a period, so the phase difference is π/2. λ = 340 × 4.0 × 10−3 = 1.36 m, so the path difference = λ/4 = 0.34 m (equivalently 340 × 1.0 × 10−3).
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Answer A
Answer: A
Exam tip: A time lag Δt between two waves at a point is a fraction Δt/T of a cycle. The same fraction of a wavelength is the path difference.
The diagram shows the displacement pattern of a stationary sound wave in a tube closed at one end. The speed of sound is 340 m s−1. Ignore end corrections. What is the frequency of the sound?
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From the closed end: node → antinode → node → antinode, which is three quarter-wavelengths. So 0.51 m = 3λ/4, λ = 0.68 m and f = 340 ÷ 0.68 = 500 Hz.
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Answer C
Answer: C
Exam tip: Count the quarter-wavelengths between a node and the next antinode. A tube closed at one end always holds an odd number of them.
A student measures the speed v of transverse waves on a cord for several values of the tension T and plots the graph shown. What is the mass per unit length of the cord?
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v2 = T/μ, so the gradient is 1/μ. Gradient = 4.0 × 103 ÷ 10 = 400 N−1 m2 s−2, giving μ = 1 ÷ 400 = 2.5 × 10−3 kg m−1.
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Answer A
Answer: A
Common error: Ignoring the power of ten in an axis label. “v2 / 103 m2 s−2” means a reading of 4.0 is 4.0 × 103 m2 s−2.
In an investigation of a vibrating string, the tension is kept at 20 N and the vibrating length L is varied. The graph shows the fundamental frequency f plotted against 1/L, with a best-fit line through the origin. What is the mass per unit length of the string?
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Answer
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Notes
f = v/2L, so the gradient of f against 1/L is v/2. Gradient = 150 ÷ 3.0 = 50 m s−1, so v = 100 m s−1. Then μ = T/v2 = 20 ÷ 1002 = 2.0 × 10−3 kg m−1.
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Answer B
Answer: B
Exam tip: Write the equation in the form y = mx before using a gradient: here f = (v/2) × (1/L), so the gradient is v/2, not v.
A string of length 1.40 m and mass per unit length 2.5 × 10−3 kg m−1 is fixed at both ends and driven at a frequency of 240 Hz. Figure 1 shows the string at time 0 (solid line) and half a period later (dashed line).
(a)
Use Figure 1 to determine the wavelength of the waves on the string.
(1)
(b)
Calculate the speed of the waves on the string.
(1)
(c)
Calculate the tension in the string.
(2)
(d)
The tension is unchanged. Calculate the frequency at which the string vibrates in three loops.
Exam tip: On a displacement–distance graph of a stationary wave, the nodes are where every curve crosses the axis. Measure across several loops and divide, for a more precise node spacing.
Two progressive waves of the same amplitude and frequency travel in opposite directions along a stretched cord and superpose to form a stationary wave. Figure 1 shows the displacement caused by each wave at time t = 0: wave A travels to the right and wave B travels to the left.
(a)
Use Figure 1 to determine the resultant displacement of the cord at t = 0 at (i) x = 0.20 m and (ii) x = 0.30 m.
(2)
(b)
The point at x = 0.30 m is a node. Explain why, and state the position of the next node along the cord.
(2)
(c)
Determine the amplitude of oscillation of the cord at x = 0.10 m.
(2)
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Answer
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Notes
Part (a)
(i) 2.0 + 0 = 2.0 mm
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(ii) about +1.4 + (−1.4) = 0
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Part (b)
at x = 0.30 m the oscillations caused by the two waves are always in antiphase with equal amplitudes, so they cancel at every instant
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nodes are λ/2 = 0.40 m apart, so the next node is at 0.70 m
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Part (c)
x = 0.10 m is λ/4 = 0.20 m from the node at 0.30 m, so it is an antinode, where the two waves are always in phase
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amplitude = 2.0 + 2.0 = 4.0 mm
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Common error: Taking the resultant displacement at one instant as the amplitude. At x = 0.10 m the displacement at t = 0 is only about 2.8 mm, yet the point is an antinode with amplitude 4.0 mm.
Noise-cancelling headphones contain a microphone and a small loudspeaker. The loudspeaker produces a cancelling wave that should be an exactly inverted copy of the noise reaching the ear. Figure 1 shows the displacement–time graphs, at the ear, of a steady noise and of the cancelling wave produced by a circuit that responds slightly too slowly.
(a)
State the conditions on the phase difference and the amplitudes of the two waves for complete cancellation.
(1)
(b)
Use Figure 1 to determine the frequency of the noise.
(1)
(c)
Use Figure 1 to determine how late the cancelling wave is, and calculate the phase error this causes, in radians.
(2)
(d)
The circuit delay stays the same for all frequencies. Explain why the headphones cancel high-frequency noise less well than low-frequency noise.
(2)
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Answer
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Notes
Part (a)
phase difference of π rad (antiphase) and equal amplitudes
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Part (b)
period = 2.0 ms, so f = 1 ÷ 2.0 × 10−3 = 5.0 × 102 Hz (500 Hz)
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Part (c)
a trough of the noise is at 1.50 ms but the matching crest of the cancelling wave is at 1.75 ms, so it is 0.25 ms late
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phase error = 2π × 0.25 ÷ 2.0 = 0.79 rad (π/4)
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Part (d)
a higher frequency has a shorter period, so the same delay is a larger fraction of a cycle and the phase error is larger
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the waves are further from antiphase, so they cancel less and the resultant amplitude is larger
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Exam tip: A time delay Δt between two waves of period T is a phase difference of 2πΔt/T. The same delay matters more when the period is shorter.
A wire is fixed at A and C and passes over a bridge at B. The tension is the same on both sides of the bridge, and the mass per unit length of the wire is 1.2 × 10−3 kg m−1. A vibrator makes the wire vibrate at 160 Hz, and stationary waves form in both sections as shown in Figure 1.
(a)
Use Figure 1 to determine the wavelength of the waves on the wire.
(1)
(b)
Calculate the speed of the waves on the wire.
(1)
(c)
Calculate the tension in the wire.
(2)
(d)
The frequency is increased. Determine the next frequency at which stationary waves form in both sections, with nodes at A, B and C.
(2)
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Answer
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Notes
Part (a)
AB is one loop, λ/2 = 0.30 m, so λ = 0.60 m (BC: two loops in 0.60 m agrees)
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Part (b)
v = 160 × 0.60 = 96 m s−1
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Part (c)
T = μv2
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= 1.2 × 10−3 × 962 = 11 N (11.1 N)
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Part (d)
a whole number of half-wavelengths must fit into both 0.30 m and 0.60 m; the next possibility is λ/2 = 0.15 m, so λ = 0.30 m
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f = 96 ÷ 0.30 = 320 Hz
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Exam tip: A bridge or clamp holds the wire still, so it is always a node. A section can resonate only when a whole number of half-wavelengths fits into it.
A string on a laboratory sonometer has a fundamental frequency of 120 Hz when its tension is 40 N and its vibrating length is 0.60 m. The tension is increased to 90 N and a bridge reduces the vibrating length to 0.45 m. What is the new fundamental frequency?
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Answer
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Notes
f = (1/2L)√(T/μ), so f ∝ √T/L for the same string. Tension factor √(90/40) = 1.5; length factor 0.60/0.45 = 1.33. New f = 120 × 1.5 × 1.33 = 240 Hz.
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Answer D
Answer: D
Exam tip: In a ratio question write the proportionality first (f ∝ √T/L), then apply each factor separately. A shorter string has a higher frequency.
Two loudspeakers are connected to one signal generator set to 850 Hz, but the leads to one loudspeaker are reversed so that the two loudspeakers vibrate in antiphase. A microphone is 3.40 m from one loudspeaker and 2.80 m from the other. The speed of sound is 340 m s−1. Which statement describes the signal at the microphone?
Figure: loudspeakers driven in antiphase by one signal generator, and the microphone (not to scale)
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Answer
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λ = 340 ÷ 850 = 0.40 m. Path difference = 3.40 − 2.80 = 0.60 m = 1.5λ, which adds a phase difference of π. The loudspeakers already differ by π, so the total is 2π: the waves arrive in phase, giving a maximum.
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Answer D
Answer: D
Exam tip: Add the two sources of phase difference: the phase difference between the sources and 2π × path difference/λ. A total of 2π, 4π … is a maximum.
A stationary wave of wavelength 0.60 m is set up on a string. Point P is 0.10 m from a node N, and point Q is 0.35 m from N on the same side. What is the phase difference between the oscillations of P and Q?
Figure: stationary wave of wavelength 0.60 m, showing node N and points P and Q (not to scale vertically)
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Notes
Nodes are λ/2 = 0.30 m apart, so the next node after N is at 0.30 m. P (0.10 m) is in the first loop and Q (0.35 m) is in the neighbouring loop. Points in adjacent loops oscillate in antiphase: phase difference = π = 3.1 rad.
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Answer D
Answer: D
Common error: Using 2π × separation/λ, which is only true for a progressive wave. On a stationary wave, find which loop each point is in: same loop gives 0, adjacent loops give π.
A student investigates how the fundamental frequency f of a stretched string depends on the tension T. The vibrating length is fixed at 0.800 m and the tension is provided by masses hanging over a pulley. g = 9.81 N kg−1. Results (hanging mass — f): 0.100 kg — 28 Hz; 0.200 kg — 39 Hz; 0.300 kg — 48 Hz; 0.400 kg — 55 Hz; 0.500 kg — 62 Hz.
(a)
Using v = √(T/μ), show that f2 = T/(4L2μ), and explain why a graph of f2 against T should be a straight line through the origin.
(2)
(b)
Calculate the tension and the value of f2 for the 0.300 kg mass.
(1)
(c)
The gradient of the best-fit line of the graph of f2 against T is 780 Hz2 N−1. Calculate the mass per unit length of the string.
(3)
(d)
Suggest how the student could make sure that each recorded frequency is the fundamental frequency at resonance.
(1)
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Answer
Mark
Notes
Part (a)
fundamental: λ = 2L, so f = v/2L = (1/2L)√(T/μ); squaring gives f2 = T/(4L2μ)
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L and μ are constant, so this has the form y = mx with gradient 1/(4L2μ)
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Part (b)
T = 0.300 × 9.81 = 2.94 N and f2 = 482 = 2.30 × 103 Hz2 (2304 Hz2)
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Part (c)
gradient = 1/(4L2μ), so μ = 1/(4L2 × gradient)
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= 1 ÷ (4 × 0.8002 × 780)
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= 5.0 × 10−4 kg m−1
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Part (d)
check that the string vibrates in a single loop, and adjust the frequency slowly (from below and from above) to find where the amplitude of the antinode is largest, taking the mean
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Model sketch
Figure: f² against T with the best-fit line through the origin (gradient 780 Hz² N⁻¹)
Exam tip: When a gradient is given, write what the gradient equals in symbols before substituting. Here gradient = 1/(4L2μ), so μ follows in one line.
The A string of an acoustic guitar has a vibrating length of 0.650 m and a fundamental frequency of 110 Hz. Its mass per unit length is 6.2 × 10−3 kg m−1. The speed of sound in air is 340 m s−1.
(a)
Calculate the speed of the transverse waves on the string.
(1)
(b)
Calculate the tension in the string.
(2)
(c)
The player presses the string against a fret, reducing the vibrating length to 0.488 m. Calculate the new fundamental frequency.
(2)
(d)
Calculate the wavelength in air of the sound produced by the open string, and explain why it differs from the wavelength of the waves on the string.
(2)
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Answer
Mark
Notes
Part (a)
λ = 2 × 0.650 = 1.30 m; v = 110 × 1.30 = 143 m s−1
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Part (b)
T = μv2
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= 6.2 × 10−3 × 1432 = 130 N (127 N)
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Part (c)
tension and μ are unchanged, so the wave speed is unchanged and f ∝ 1/L
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f = 110 × 0.650 ÷ 0.488 = 147 Hz
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Part (d)
λ = 340 ÷ 110 = 3.09 m
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the frequency is the same in air as on the string, but the wave speed in air is greater, so the wavelength is longer (λ = v/f)
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Exam tip: Frequency is set by the source and stays the same when a wave passes into a new medium; speed and wavelength change together.
The graph shows a string carrying a stationary wave at an instant when it has its maximum displacement. The dashed line shows the string half a period later. Points K, L, M and N are 8 cm, 24 cm, 48 cm and 68 cm from the left-hand end. Which two points oscillate in antiphase with each other and with the same amplitude?
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Answer
Mark
Notes
Nodes are at 0, 40 cm and 80 cm. K and L are in the first loop, M and N in the second, so only pairs from different loops are in antiphase. K is 8 cm from the node at 0 and M is 8 cm from the node at 40 cm, so they have equal amplitudes: K and M. L (24 cm, 16 cm from a node) has a larger amplitude than M.
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Answer D
Answer: D
Exam tip: On a stationary wave, the loop decides the phase (same loop: in phase; adjacent loops: antiphase) and the distance from the nearest node decides the amplitude.
A tuning fork of frequency 480 Hz is held above a tube containing water. The water level is lowered from the top, and resonance is heard first with an air column of 17.0 cm and next with an air column of 52.0 cm, as shown. The antinode at the open end is a small, unknown distance above the top of the tube, the same in both cases. What is the speed of sound in the air?
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Answer
Mark
Notes
Between the first and second resonances the air column gains half a wavelength, and the unknown end distance cancels: λ/2 = 52.0 − 17.0 = 35.0 cm. λ = 0.700 m and v = 480 × 0.700 = 336 m s−1.
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Answer C
Answer: C
Exam tip: Using the difference between two resonance lengths removes the end correction, which is why this method is more accurate than using the first length alone.
A string of vibrating length 0.75 m is fixed at both ends. Its mass per unit length is 4.0 × 10−3 kg m−1. The graph shows the frequencies at which the string resonates, plotted against the harmonic number n. What is the tension in the string?
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Answer
Mark
Notes
From the graph the fundamental (n = 1) is 60 Hz. For the fundamental λ = 2L = 1.50 m, so v = 60 × 1.50 = 90 m s−1. T = μv2 = 4.0 × 10−3 × 902 = 32 N (32.4 N).
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Answer C
Answer: C
Exam tip: The gradient of a graph of f against n for a string fixed at both ends equals the fundamental frequency, because fn = n × v/2L.
Two small ultrasonic transmitters S1 and S2 are connected to the same signal generator, so they emit waves in phase with a frequency of 42.5 kHz. Figure 1 is a plan view of the transmitters and a receiver at point X, drawn on a grid of 1.0 cm squares. The speed of sound in air is 340 m s−1.
(a)
Calculate the wavelength of the ultrasound.
(1)
(b)
Use Figure 1 to determine the distances S1X and S2X, and hence the path difference at X.
(2)
(c)
Deduce whether the receiver detects a maximum or a minimum of intensity at X.
(2)
(d)
The frequency of the signal generator is increased slowly from 42.5 kHz. Calculate the frequency at which X first becomes a position of maximum intensity.
(2)
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Answer
Mark
Notes
Part (a)
λ = 340 ÷ (42.5 × 103) = 8.0 × 10−3 m = 0.80 cm
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Part (b)
S1X = 8.0 cm
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S2X = √(6.02 + 8.02) = 10.0 cm, so the path difference = 2.0 cm
the waves arrive in antiphase and superpose destructively, so X is a minimum
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Part (d)
the next maximum needs a path difference of 3λ, so λ = 2.0 ÷ 3 = 0.667 cm
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f = 340 ÷ (6.67 × 10−3) = 51.0 kHz
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Exam tip: Always convert a path difference into a number of wavelengths. A whole number means in phase (maximum); a whole number plus a half means antiphase (minimum).
A student investigates how the fundamental frequency f of a stretched string depends on its mass per unit length μ. She uses five strings of different μ, each with a vibrating length L of 0.600 m and the same tension T, provided by a hanging mass. Figure 1 shows her graph of f against μ−1/2 with a best-fit line.
(a)
Use v = √(T/μ) to show that f = (1/2L)√(T/μ), and explain why the graph should be a straight line through the origin.
(2)
(b)
Determine the gradient of the line.
(2)
(c)
Calculate the tension in the strings.
(2)
(d)
Suggest how the student could find the value of μ for each string.
(1)
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Answer
Mark
Notes
Part (a)
for the fundamental λ = 2L, so f = v/λ = (1/2L)√(T/μ)
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f = (√T/2L) × μ−1/2 with T and L constant, which has the form y = mx
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Part (b)
large triangle on the line, e.g. from (0, 0) to (30, 150)
T = 36 N (allow 35 to 37 N; allow error carried forward)
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Part (d)
measure the mass of a long measured length of the same string on a balance and divide the mass by the length
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Exam tip: Draw the gradient triangle across at least half the line, with corners where the line crosses gridline intersections, and show the readings you used.
Points P and Q are on a string that carries a stationary wave. Figure 1 shows how the displacement of each point varies with time. P is at an antinode. The progressive waves that form the stationary wave travel at 120 m s−1.
(a)
Use Figure 1 to determine the frequency of the stationary wave.
(1)
(b)
State the phase difference between the oscillations of P and Q, and what it shows about their positions on the string.
(2)
(c)
Calculate the wavelength of the progressive waves.
(1)
(d)
The amplitude at a distance d from a node is proportional to sin(2πd/λ). Calculate the shortest possible distance between P and Q.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
period = 5.0 ms, so f = 1 ÷ 5.0 × 10−3 = 2.0 × 102 Hz (200 Hz)
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Part (b)
π rad (antiphase)
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P and Q are in adjacent loops, on opposite sides of a node (or in loops an odd number apart)
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Part (c)
λ = 120 ÷ 200 = 0.60 m
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Part (d)
Q has half the antinode amplitude, so sin(2πd/λ) = 0.5 and d = λ/12 = 0.050 m from a node
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P, at an antinode, is λ/4 = 0.15 m from the same node
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shortest distance PQ = 0.15 + 0.050 = 0.20 m
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Exam tip: On a stationary wave, the phase tells you which loop a point is in, and the amplitude tells you how far it is from the nearest node.
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