OCR A Level Physics A: Stationary waves, harmonics and resonance tubes (H556) – 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 cover section 4.4.4 (Stationary waves) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. Use 340 m/s for the speed of sound in air where needed. Revise first with the study guide or the revision notes.

Questions

1. (a) Define a node and an antinode of a stationary wave. [2] (b) State the two conditions that the progressive waves must meet to form a stationary wave on a string. [2]

2. State three differences between a stationary wave and a progressive wave. [3]

3. A microwave transmitter of frequency 10.5 GHz faces a flat metal plate (speed of microwaves 3.00 x 10^8 m/s). A probe receiver is moved along the line between them.

(a) Describe how the probe shows that a stationary wave has formed. [2] (b) Six consecutive maximum readings span 7.1 cm. Calculate the wavelength. [2] (c) Calculate the speed of the microwaves. [2]

4. A string of length 0.90 m is fixed at both ends and vibrates in its 2nd harmonic at 240 Hz.

(a) Describe the pattern, stating the number of nodes and antinodes. [2] (b) State the wavelength. [1] (c) Calculate the speed of the progressive waves on the string. [2] (d) State the frequency of the fundamental mode. [1] (e) For the 4th harmonic, calculate the frequency and the distance between adjacent nodes. [2]

5. The graph of displacement against position along a string at one instant shows a stationary wave. Nodes are at 0, 30 cm and 60 cm. There is an antinode at 15 cm with displacement +4.0 mm, and an antinode at 45 cm with displacement -4.0 mm.

(a) State the wavelength. [1] (b) State the amplitude of each progressive wave that formed the pattern. [1] (c) Half a period later, state the displacements at 15 cm and at 45 cm. [2] (d) State the phase difference between the particles at 15 cm and 45 cm. [1]

6. A tube closed at one end is 0.45 m long.

(a) Calculate the frequency of its fundamental mode. [2] (b) Calculate the next two frequencies at which the air column resonates. [2] (c) Explain why the tube does not resonate at twice the fundamental frequency. [1]

7. A resonance tube experiment uses three tuning forks. For each, the lengths L1 and L2 of the air column at the first two resonances are measured.

Frequency / Hz	L1 / m	L2 / m
320	0.262	0.798
384	0.217	0.664
480	0.173	0.530

(a) Describe how the readings are obtained. [3] (b) Calculate the speed of sound from the 384 Hz readings. [2] (c) Calculate the speed from the other two forks and find the mean of all three. [2] (d) Explain why using L2 - L1 gives a more reliable wavelength than 4 x L1. [2]

8. A string of fixed tension has its fundamental frequency f measured for different lengths L.

L / m	0.30	0.45	0.60	0.75	0.90
f / Hz	120	80	60	48	40

(a) Explain why a graph of f against 1/L should be a straight line through the origin. [2] (b) Two points on the best-fit line are (3.33, 120) and (1.11, 40), with 1/L in m^-1. Calculate the gradient. [2] (c) Determine the speed of the progressive waves on the string. [2]

9. A tube open at both ends is 1.50 m long.

(a) Calculate all the frequencies between 100 Hz and 400 Hz at which the air column resonates. [3] (b) For the 3rd harmonic, state how many nodes and how many antinodes the pattern has. [2] (c) Calculate the distance between adjacent antinodes in the 3rd harmonic. [2]

10. A string is driven at one end and fixed at the other. Give an account of why large steady loops appear on it, and why this happens only at certain driving frequencies. [6]

11. A tube closed at one end resonates at 255 Hz and, as the next resonance above it, at 425 Hz.

(a) Show that these are consecutive odd harmonics and determine the fundamental frequency. [3] (b) Calculate the length of the tube. [2] (c) The closed end is now opened. Calculate the new fundamental frequency. [2]

Answers

1. (a) A node is a point on a stationary wave where the amplitude is zero [1]. An antinode is a point where the amplitude is a maximum [1]. [2] (b) The waves must have the same frequency (and the same wave speed) [1] and travel in opposite directions along the same line [1]. [2] Examiner insight: Define nodes and antinodes by amplitude, not by where the string is still at one instant.

2. Any three of the following, 1 mark each: energy is not carried along a stationary wave, whereas a progressive wave transfers it [1]; amplitude varies with position in a stationary wave but is constant in a progressive wave [1]; phase is the same within a loop of a stationary wave but changes steadily along a progressive wave [1]; or the stationary pattern stays fixed but the progressive profile moves. [3] Examiner insight: Give a point about each wave; “stationary waves have nodes” alone is incomplete.

3. (a) Move the probe along the line between transmitter and plate [1]. The reading rises and falls, with maxima and minima at fixed positions about half a wavelength apart [1]. [2] (b) Six maxima give 5 gaps, each lambda/2 [1]. lambda = 2 x 7.1 / 5 = 2.84 cm [1] [2] (c) c = f lambda = 10.5 x 10^9 x 2.84 x 10^-2 [1] = 2.98 x 10^8 m/s [1] [2] Examiner insight: Show the count of gaps explicitly; it is an easy step to get wrong.

4. (a) One node at each end and one in the middle (3 nodes) [1], with 2 antinodes, so two loops [1]. [2] (b) Two loops fit the length, so lambda = L = 0.90 m [1] [1] (c) v = f lambda = 240 x 0.90 [1] = 216 m/s [1] [2] (d) The 2nd harmonic is twice the fundamental, so f1 = 120 Hz [1] [1] (e) f4 = 4 x 120 = 480 Hz [1]. lambda = 2L/4 = 0.45 m, so node spacing = lambda/2 = 0.225 m [1] [2] Examiner insight: The nth harmonic on a fixed string has n loops.

5. (a) Adjacent nodes are 30 cm apart, so lambda = 60 cm [1] [1] (b) The antinode amplitude is the sum of the two, so each is 4.0 / 2 = 2.0 mm [1] [1] (c) After half a period the pattern is inverted: -4.0 mm at 15 cm [1] and +4.0 mm at 45 cm [1]. [2] (d) They are in adjacent loops, so they are in antiphase: 180 degrees (pi rad) [1] [1] Examiner insight: For phase statements, refer to the loops on the graph.

6. (a) Closed tube, so L = lambda/4 and lambda = 4 x 0.45 = 1.80 m [1]. f1 = v / lambda = 340 / 1.80 = 189 Hz [1] [2] (b) The next resonances are the 3rd and 5th harmonics: 3 x 189 = 567 Hz [1] and 5 x 189 = 944 Hz [1]. [2] (c) Twice the fundamental would need an antinode at the closed end or a node at the open end, which does not fit the boundary conditions [1]. [1] Examiner insight: Link the missing harmonic to the end conditions.

7. (a) Hold a sounding fork of known frequency over a tube standing in water and raise or lower the tube to change the air column [1]. Record the lengths at the first two loud sounds, L1 and L2 [1]. Calculate lambda = 2 (L2 - L1) and then v = f lambda [1]. [3] (b) L2 - L1 = 0.664 - 0.217 = 0.447 m, so lambda = 0.894 m [1]. v = 384 x 0.894 = 343 m/s [1] [2] (c) 320 Hz: lambda = 2 x 0.536 = 1.072 m, v = 343 m/s. 480 Hz: lambda = 2 x 0.357 = 0.714 m, v = 343 m/s (342.7) [1]. The mean of the three values is 343 m/s [1]. [2] (d) The antinode forms slightly above the tube mouth, so L1 is a little less than lambda/4 and 4 x L1 is too small [1]. The same offset is in both L1 and L2, so it cancels in the difference [1]. [2] Examiner insight: Name the source of error (the antinode position) and say why the difference cancels it.

8. (a) For the fundamental, lambda = 2L and v = f lambda, so f = v / 2L [1]. The speed is constant at fixed tension, so f is proportional to 1/L, giving a straight line that passes through the origin, with gradient v / 2 [1]. [2] (b) Gradient = (120 - 40) / (3.33 - 1.11) [1] = 36 m/s [1] [2] (c) The gradient equals v / 2 [1], so v = 2 x 36 = 72 m/s [1] [2] Examiner insight: Use points far apart on the line and say what the gradient represents.

9. (a) For an open tube, f = n v / 2L = n x 340 / 3.00 = 113.3n Hz [1]. n = 1, 2, 3 give 113 Hz, 227 Hz and 340 Hz [1]. n = 4 gives 453 Hz, which is above 400 Hz, so there are only three [1]. [3] (b) Both ends are antinodes, so the 3rd harmonic has 4 antinodes [1] and 3 nodes [1]. [2] (c) lambda = 2L/3 = 1.00 m [1]. Adjacent antinodes are lambda/2 = 0.50 m apart [1]. [2] Examiner insight: Show how the upper limit was applied by testing the next harmonic.

10. A progressive wave travels along the string and reflects from the fixed end [1]. The reflected wave has the same frequency and speed as the incident wave but travels in the opposite direction [1]. The two waves superpose and the displacements add [1]. At nodes the waves are always in antiphase, so they cancel and the displacement is always zero [1]. At antinodes they are always in phase, so they reinforce to give the maximum amplitude [1]. A steady pattern needs a node at both ends, so a whole number of half-wavelengths must fit; this happens only at the harmonic frequencies (resonance) [1]. [6] Examiner insight: A six-mark explanation needs a chain: reflection, superposition, cancelling and adding, then the frequency condition.

11. (a) Resonances of a closed tube are f1, 3 f1, 5 f1 and so on, so consecutive resonances differ by 2 f1 [1]. 425 - 255 = 170 Hz, so f1 = 85 Hz [1]. Then 255 / 85 = 3 and 425 / 85 = 5, so they are the 3rd and 5th harmonics [1]. [3] (b) Closed tube: L = lambda/4 = v / 4 f1 [1] = 340 / (4 x 85) = 1.00 m [1] [2] (c) Open at both ends, f1 = v / 2L [1] = 340 / 2.00 = 170 Hz [1] [2] Examiner insight: For “show that”, state the spacing is 2 f1 and test the ratios.

Where marks are usually lost

• Using lambda instead of lambda/2 for the node spacing, or counting maxima rather than gaps.
• Giving even harmonics for a tube closed at one end.
• Taking the second resonance of a resonance tube to be the 2nd harmonic.
• Explaining stationary-wave formation without mentioning reflection and superposition.

Next steps

• Read the revision notes and the study guide.
• Return to the OCR A Level Physics hub and tick items off the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Section 4.4.4, Stationary waves.