OxfordAQA International AS and A-level Physics: Superposition and stationary waves (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow section 3.5.5 (Principle of superposition of waves and formation of stationary waves) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4. It is shared International AS and A-level content. Where a question needs the speed of light or sound, or g, the value is given or stated in the answer.

Read the study guide first if you need the explanations, and keep the revision notes open for the formulae. See also the course hub, the printable checklist and the free diagnostics.

Questions

1. (a) State what is meant by a node on a stationary wave. [1] (b) State two conditions needed for two progressive waves to form a stationary wave. [2]

2. Two waves of equal amplitude and frequency travel in opposite directions along a string. At one instant the waves coincide exactly. Explain, using the displacements of the two waves, why the resultant displacement is zero at every point one quarter of a period later. [3]

3. A string 1.20 m long, with both ends fixed, vibrates in its third harmonic when driven at 150 Hz. (a) Find the wavelength of the waves on the string. [1] (b) Find the speed at which waves travel along the string. [1] (c) State how many nodes there are, counting both ends. [1] (d) Find the frequency of the first harmonic. [1]

4. A string of length 0.48 m has tension 72 N and mass per unit length 1.5 × 10⁻³ kg m⁻¹. (a) Calculate the frequency of the first harmonic. [3] (b) The tension is halved with the length unchanged. Calculate the new first harmonic frequency. [1]

5. A microwave transmitter faces a metal plate. A small receiving probe slides along the line joining them. The first signal minimum recorded is at 6.0 cm from the plate and the sixth is at 13.5 cm. (a) Calculate the wavelength of the microwaves. [2] (b) Calculate the frequency, using c = 3.00 × 10⁸ m s⁻¹. [1] (c) Suggest why the signal at a minimum is not exactly zero. [1]

6. A loudspeaker emits sound of frequency 680 Hz towards a flat board. Take the speed of sound as 340 m s⁻¹. (a) Calculate the wavelength and the distance between adjacent positions of minimum loudness. [2] (b) A microphone is at a position of minimum loudness. State the distance it must move along the line to reach the next position of maximum loudness. [1] (c) Explain why the loudness is low at the minimum. [1]

7. Compare a stationary wave with a progressive wave in terms of amplitude, phase and energy transfer. [3]

8. A student measures the first harmonic frequency f of a string for different vibrating lengths l, with the tension fixed.

l / m	0.30	0.40	0.50	0.70	0.90
f / Hz	140	105	84.0	60.0	46.7

(a) Use the first-harmonic equation to explain why plotting f against 1/l should give a straight line through the origin. [2] (b) Find the gradient of that graph from the first and last data points. [2] (c) The string has mass per unit length 2.1 × 10⁻³ kg m⁻¹. Use your gradient to calculate the tension. [2]

9. Describe an investigation of how the first harmonic frequency of a stretched string depends on its tension. Your account should cover the apparatus, the measurements, control of variables, and how the results would be analysed. [7]

10. A 1.50 m length of wire has a mass of 3.6 g. A section of length 0.80 m of this wire is stretched between two fixed supports and vibrates in its third harmonic at 345 Hz. (a) Calculate the frequency of the first harmonic and the mass per unit length of the wire. [2] (b) Calculate the wavelength, the wave speed and the tension. [4] (c) The tension is changed so that the first harmonic is 110 Hz. Calculate the new tension. [2]

11. A stationary wave is formed by two progressive waves, each of amplitude 2.5 mm. (a) State the amplitude of the stationary wave at an antinode. [1] (b) At a point a distance λ/8 from a node, the amplitude is 2 × 2.5 × sin 45° mm. Calculate it. [2] (c) State the phase relationship between oscillations at two points on opposite sides of a node. [1] (d) State the phase relationship between two points in the same loop. [1] (e) Explain why no energy is carried along the medium by a stationary wave. [1]

Answers

1. (a) A point on the wave where the amplitude is zero (the displacement is always zero). [1] (b) The waves must have the same frequency. [1] They must travel in opposite directions (and overlap in the same region). [1] [3] Examiner insight: “Opposite directions” and “same frequency” are two separate points, so both must appear and neither can be implied by the other.

2. In a quarter of a period each wave moves a quarter of a wavelength, one to the left and one to the right. [1] So one wave has moved half a wavelength relative to the other, and the waves are in antiphase at every point. [1] Their amplitudes are equal, so the displacements are equal and opposite and the sum is zero everywhere. [1] [3] Examiner insight: An explanation that says only “they cancel” lacks the reasoning; show how far each wave moves and why that puts them in antiphase.

3. (a) λ = 2l/3 = 2.40/3 = 0.80 m [1] (b) v = fλ = 150 × 0.80 = 120 m s⁻¹ [1] (c) Third harmonic has three loops, so 4 nodes [1] (d) f₁ = 150/3 = 50 Hz [1] [4] Examiner insight: Counting loops and counting nodes give different numbers; write the pattern down first so that you do not confuse them.

4. (a) T/μ = 72 / 1.5 × 10⁻³ = 48 000 m² s⁻² [1] v = √48 000 = 219 m s⁻¹ [1] f₁ = v/2l = 219 / 0.96 = 228 Hz [1] (b) f₁ scales with √T: 228 / √2 = 161 Hz [1] [4] Examiner insight: Keep an unrounded value for v until the end, and give the answer to three significant figures, matching the data.

5. (a) Five gaps between six minima: 13.5 − 6.0 = 7.5 cm, so λ/2 = 7.5/5 = 1.5 cm [1]; λ = 3.0 cm [1] (b) f = c/λ = 3.00 × 10⁸ / 0.030 = 1.0 × 10¹⁰ Hz [1] (c) The reflected wave is weaker than the incident wave, so cancellation is not complete. [1] [4] Examiner insight: Counting six minima as six gaps is the standard slip here, and it gives the wrong wavelength.

6. (a) λ = v/f = 340 / 680 = 0.50 m [1]; adjacent minima are λ/2 = 0.25 m apart [1] (b) λ/4 = 0.125 m [1] (c) The incident and reflected waves arrive in antiphase there, so they cancel (almost completely). [1] [4] Examiner insight: Say “in antiphase” and name both waves; “the sound waves cancel” alone is too vague.

7. Amplitude: constant along a progressive wave, but varies from zero at nodes to a maximum at antinodes in a stationary wave. [1] Phase: changes steadily along a progressive wave; in a stationary wave points in one loop are in phase and points in neighbouring loops are in antiphase. [1] Energy: a progressive wave transfers energy along the medium; a stationary wave does not. [1] [3] Examiner insight: A comparison needs both waves described for the same feature; stating only the stationary-wave property earns less.

8. (a) f = (1/2l)√(T/μ), so with T and μ fixed, f is directly proportional to 1/l. [1] A graph of f against 1/l is therefore a straight line through the origin. [1] (b) 1/l values are 3.33 and 1.11 m⁻¹. Gradient = (140 − 46.7) / (3.33 − 1.11) [1] = 42 m s⁻¹ [1] (c) Gradient = v/2, so v = 84 m s⁻¹ [1]; T = μv² = 2.1 × 10⁻³ × 84² = 15 N (14.8 N) [1] [6] Examiner insight: Pick points widely separated on the line, show both subtractions, and say what the gradient represents before you use it.

9. Apparatus: string over a bridge and then a pulley with a hanger carrying slotted masses, driven by a vibration generator connected to a signal generator. [1] Tension T = mg, found from the hanging mass. [1] Keep the vibrating length l between the bridge and pulley fixed, and use the same string so that μ is fixed. [1] For each tension, increase the frequency slowly until a single large loop forms and adjust to find the frequency of maximum amplitude, which is f₁. [1] Change the mass in equal steps and repeat each measurement, taking the mean. [1] Plot f² against T. A straight line through the origin confirms f² ∝ T. [1] The gradient is 1/(4l²μ), which gives μ, or can be compared with a measured μ. [1] [7] Examiner insight: “Control variables” earns nothing unless you name them (length and mass per unit length) and say how they are held fixed.

10. (a) f₁ = 345/3 = 115 Hz [1]; μ = 3.6 × 10⁻³ / 1.50 = 2.4 × 10⁻³ kg m⁻¹ [1] (b) λ = 2l/3 = 1.60/3 = 0.533 m [1]; v = fλ = 345 × 0.5333 = 184 m s⁻¹ [1]; T = μv² [1] = 2.4 × 10⁻³ × 184² = 81 N (81.3 N) [1] (c) T = μ(2lf₁)² [1] = 2.4 × 10⁻³ × (1.60 × 110)² = 74 N (74.3 N) [1] [8] Examiner insight: The mass of 1.50 m of wire gives μ, not the 0.80 m that vibrates; show the conversion from grams to kilograms.

11. (a) 2 × 2.5 = 5.0 mm [1] (b) 2 × 2.5 × sin 45° = 5.0 × 0.7071 = 3.5 mm [1] (3.54 mm); the answer is less than the antinode amplitude [1] (c) The oscillations are in antiphase (a phase difference of π rad). [1] (d) The oscillations are in phase. [1] (e) The two waves carry equal energy in opposite directions, so there is no net energy transfer along the medium. [1] [6] Examiner insight: Give the answer to (b) with a sensible number of significant figures and comment on whether it lies between zero and the antinode amplitude.

Where marks are usually lost

• Writing that adjacent nodes are one wavelength apart instead of half a wavelength.
• Counting minima instead of gaps when finding a wavelength from microwave or sound readings.
• Explaining formation of a stationary wave without saying that the two waves have the same frequency and opposite directions.
• Saying “cancel” without the phrase “in antiphase” or a reference to equal and opposite displacements.
• Forgetting to convert grams to kilograms when finding mass per unit length.
• Using the whole string length when the question states a shorter vibrating length.
• In an experiment, naming the variable to change but not the quantities to keep constant.
• Plotting f against T and expecting a straight line, instead of f² against T.
• Using “fundamental” or “overtone” when the question asks for the first or other harmonic.

Next steps

• Condense the topic with the revision notes.
• Re-read the explanations in the study guide.
• Use the course hub and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), specification Version 4.4, for exams from May/June 2020 onwards, published by OxfordAQA. These questions cover section 3.5.5.