A Level Physics 8.1 Stationary Waves MCQs – 30 Practice Questions with Answers (Cambridge 9702)
Free practice MCQs for Cambridge International AS & A Level Physics (9702), Section 8.1 Stationary Waves. Test yourself on how stationary waves form, stationary waves on strings and in air columns, then open the Show Answer panels to check the worked solutions.
Formula recap: adjacent nodes are λ/2 apart, a node and the next antinode are λ/4 apart, and f = v/λ. For a string fixed at both ends the fundamental has L = λ/2; for a closed pipe L = λ/4 and for an open pipe L = λ/2. See the full A Level Physics Formula Sheet for every equation.
Formation, Nodes & Antinodes
Q1. A stationary wave is set up on a stretched string. How is the stationary wave formed?
- (A) Two progressive waves of the same frequency travel in the same direction and superpose
- (B) Two progressive waves of the same frequency and amplitude travel in opposite directions and superpose
- (C) A single progressive wave travels along the string and is absorbed at the far end
- (D) Two progressive waves of very different frequencies travel in opposite directions and superpose
Show Answer
Answer: B – A stationary wave forms when two progressive waves of the same frequency and (usually similar) amplitude travel in opposite directions through the same region and superpose, often when a wave is reflected back along itself at a fixed end.
Q2. What is a node of a stationary wave?
- (A) A point that remains at zero displacement at all times
- (B) A point where the displacement is always at its maximum value
- (C) A point that moves along the wave with speed v = fλ
- (D) A point where the two superposing waves are always in phase
Show Answer
Answer: A – At a node the two waves always cancel, so the displacement stays permanently zero. The largest displacement occurs at an antinode.
Q3. Which statement about an antinode is correct?
- (A) The particles there do not oscillate
- (B) The particles there oscillate with the smallest amplitude on the wave
- (C) The particles there oscillate with the maximum amplitude
- (D) The particles there are permanently displaced from equilibrium
Show Answer
Answer: C – At an antinode the two waves always reinforce, giving oscillation with the maximum amplitude, equal to the sum of the amplitudes of the two superposing waves.
Q4. On a stationary wave, adjacent nodes are 0.40 m apart. What is the wavelength of the waves that formed it?
- (A) 0.20 m
- (B) 0.40 m
- (C) 1.60 m
- (D) 0.80 m
Show Answer
Answer: D – Adjacent nodes (and adjacent antinodes) are separated by λ/2, so λ = 2 × 0.40 = 0.80 m.
Q5. A stationary wave has wavelength 1.2 m. What is the distance between a node and the nearest antinode?
- (A) 0.60 m
- (B) 0.30 m
- (C) 1.20 m
- (D) 0.15 m
Show Answer
Answer: B – A node and the adjacent antinode are separated by λ/4 = 1.2 / 4 = 0.30 m.
Q6. In a stationary wave, how do the phases of oscillation of particles lying between two consecutive nodes compare?
- (A) They are all in phase with each other
- (B) They are all in antiphase with each other
- (C) The phase increases steadily from one node to the next
- (D) Each particle is exactly a quarter of a cycle behind its neighbour
Show Answer
Answer: A – All particles between two consecutive nodes move in phase (they reach maximum displacement together), whereas particles on opposite sides of a node are in antiphase.
Q7. Two particles in a stationary wave are on opposite sides of a node. What is their phase difference?
- (A) 0
- (B) 90° (π/2 rad)
- (C) 180° (π rad)
- (D) 360° (2π rad)
Show Answer
Answer: C – Particles either side of a node oscillate in antiphase: when one is at maximum positive displacement the other is at maximum negative displacement, a phase difference of 180° or π rad.
Q8. Which statement about energy transfer in a stationary wave is correct?
- (A) Energy is transferred in the direction of the faster wave
- (B) Energy is transferred from each node to the next antinode
- (C) Energy is transferred rapidly along the wave in both directions
- (D) There is no net transfer of energy along the wave
Show Answer
Answer: D – The two superposing waves carry equal energy in opposite directions, so there is no net energy transfer along a stationary wave – a key difference from a progressive wave.
Q9. How does the amplitude of oscillation vary along a stationary wave, compared with a progressive wave?
- (A) In both waves every particle oscillates with the same amplitude
- (B) In a stationary wave the amplitude varies with position, from zero at a node to a maximum at an antinode; in a progressive wave all particles have the same amplitude
- (C) In a stationary wave all particles have the same amplitude; in a progressive wave the amplitude varies with position
- (D) In a stationary wave the amplitude increases steadily from one end to the other
Show Answer
Answer: B – The amplitude in a stationary wave depends on position (zero at nodes, maximum at antinodes). In a progressive wave every particle oscillates with the same amplitude (ignoring damping).
Q10. A stationary wave is formed by two waves each of amplitude 3.0 mm. What is the amplitude of oscillation at an antinode?
- (A) 6.0 mm
- (B) 3.0 mm
- (C) 1.5 mm
- (D) zero
Show Answer
Answer: A – At an antinode the displacements of the two waves always add, so the amplitude is 3.0 + 3.0 = 6.0 mm, twice the amplitude of each progressive wave.
Stationary Waves on Strings
Q11. A string of length 0.80 m is fixed at both ends and vibrates in its fundamental mode. What is the wavelength of the stationary wave?
- (A) 0.40 m
- (B) 0.80 m
- (C) 1.60 m
- (D) 3.20 m
Show Answer
Answer: C – Fixed ends are nodes. In the fundamental mode the string is one half-wavelength long, L = λ/2, so λ = 2L = 2 × 0.80 = 1.60 m.
Q12. In a Melde-type experiment, a string of length 1.2 m vibrates in its fundamental mode. The speed of transverse waves on the string is 60 m s-1. What is the frequency of vibration?
- (A) 25 Hz
- (B) 50 Hz
- (C) 72 Hz
- (D) 144 Hz
Show Answer
Answer: A – Fundamental wavelength λ = 2L = 2 × 1.2 = 2.4 m. Then f = v/λ = 60 / 2.4 = 25 Hz.
Q13. A string fixed at both ends vibrates in its second harmonic. How many nodes are there on the string, including the two ends?
- (A) 2
- (B) 4
- (C) 5
- (D) 3
Show Answer
Answer: D – In the second harmonic the string contains one full wavelength (two loops), with nodes at both fixed ends and one node in the middle: 3 nodes in total.
Q14. A string fixed at both ends vibrates in its third harmonic, so that L = 3λ/2. How many antinodes are there?
- (A) 2
- (B) 3
- (C) 4
- (D) 6
Show Answer
Answer: B – The third harmonic has three loops, each with one antinode at its centre, so there are 3 antinodes (and 4 nodes, including the two ends).
Q15. The fundamental frequency of a stretched string is 40 Hz. What is the frequency of its third harmonic?
- (A) 13 Hz
- (B) 80 Hz
- (C) 120 Hz
- (D) 160 Hz
Show Answer
Answer: C – For a string fixed at both ends the harmonic frequencies are whole-number multiples of the fundamental: third harmonic = 3 × 40 = 120 Hz.
Q16. Waves travel at 120 m s-1 on a string of length 0.50 m fixed at both ends. The string vibrates in its second harmonic. What is the frequency?
- (A) 60 Hz
- (B) 120 Hz
- (C) 480 Hz
- (D) 240 Hz
Show Answer
Answer: D – In the second harmonic L = λ, so λ = 0.50 m. Then f = v/λ = 120 / 0.50 = 240 Hz (twice the fundamental of 120 Hz).
Q17. The length of a vibrating string is doubled while the tension and the wave speed on the string stay the same. What happens to the fundamental frequency?
- (A) It halves
- (B) It stays the same
- (C) It doubles
- (D) It increases by a factor of four
Show Answer
Answer: A – The fundamental wavelength is λ = 2L, so f = v/2L: frequency is inversely proportional to length. Doubling L halves the fundamental frequency.
Q18. The tension in a string is increased to four times its original value, with the length unchanged. The wave speed is proportional to the square root of the tension. What happens to the fundamental frequency?
- (A) It stays the same
- (B) It doubles
- (C) It increases by a factor of four
- (D) It halves
Show Answer
Answer: B – Speed v is proportional to √T, so four times the tension gives √4 = 2 times the speed. Since f = v/2L with L fixed, the fundamental frequency doubles.
Q19. A string fixed at both ends vibrates in its fourth harmonic. How many antinodes are produced?
- (A) 2
- (B) 3
- (C) 4
- (D) 5
Show Answer
Answer: C – The nth harmonic on a string fixed at both ends has n loops, each containing one antinode. The fourth harmonic therefore has 4 antinodes (and 5 nodes).
Q20. A string of length 0.90 m fixed at both ends vibrates with three loops (segments). Waves travel on the string at 180 m s-1. What is the frequency of vibration?
- (A) 100 Hz
- (B) 200 Hz
- (C) 600 Hz
- (D) 300 Hz
Show Answer
Answer: D – Each loop is λ/2 long, so λ/2 = 0.90 / 3 = 0.30 m and λ = 0.60 m. Then f = v/λ = 180 / 0.60 = 300 Hz.
Stationary Waves in Air Columns
Q21. A stationary wave is set up in a pipe that is closed at one end. Which pairing of boundary conditions is correct?
- (A) Antinode at the closed end, node at the open end
- (B) Node at the closed end, antinode at the open end
- (C) Nodes at both ends
- (D) Antinodes at both ends
Show Answer
Answer: B – Air at the closed end cannot move, so it is a displacement node; air at the open end is free to move with maximum amplitude, so it is (approximately) an antinode.
Q22. A closed pipe of length 0.25 m resonates in its fundamental mode. What is the wavelength of the sound?
- (A) 0.25 m
- (B) 0.50 m
- (C) 2.0 m
- (D) 1.0 m
Show Answer
Answer: D – A closed pipe fits a quarter-wavelength in its fundamental mode, L = λ/4, so λ = 4L = 4 × 0.25 = 1.0 m.
Q23. A pipe closed at one end has length 0.85 m. The speed of sound in air is 340 m s-1. What is the fundamental frequency of the pipe?
- (A) 100 Hz
- (B) 200 Hz
- (C) 400 Hz
- (D) 850 Hz
Show Answer
Answer: A – For the closed pipe, λ = 4L = 4 × 0.85 = 3.4 m. Then f = v/λ = 340 / 3.4 = 100 Hz.
Q24. A closed pipe has a fundamental frequency of 100 Hz. Which frequency is the next higher frequency at which the pipe can resonate?
- (A) 150 Hz
- (B) 200 Hz
- (C) 300 Hz
- (D) 500 Hz
Show Answer
Answer: C – A closed pipe supports only odd harmonics: f, 3f, 5f, … The next resonance after 100 Hz is 3 × 100 = 300 Hz. Even harmonics such as 200 Hz are missing.
Q25. A pipe open at both ends has length 0.50 m. The speed of sound in air is 340 m s-1. What is its fundamental frequency?
- (A) 170 Hz
- (B) 680 Hz
- (C) 85 Hz
- (D) 340 Hz
Show Answer
Answer: D – An open pipe has antinodes at both ends, so the fundamental fits half a wavelength: λ = 2L = 1.0 m. Then f = v/λ = 340 / 1.0 = 340 Hz.
Q26. An open pipe and a closed pipe have the same fundamental frequency of 200 Hz. Which statement about their harmonics is correct?
- (A) Both pipes produce all harmonics: 200, 400, 600 Hz, …
- (B) The open pipe produces all harmonics, but the closed pipe produces only 200, 600, 1000 Hz, …
- (C) The closed pipe produces all harmonics, but the open pipe produces only the odd ones
- (D) Neither pipe can produce a harmonic above the fundamental
Show Answer
Answer: B – An open pipe supports all harmonics (f, 2f, 3f, …), while a closed pipe, with a node at one end and an antinode at the other, can only fit an odd number of quarter-wavelengths, giving odd harmonics only: 200, 600, 1000 Hz, …
Q27. A closed pipe has a fundamental frequency of 150 Hz. An open pipe of exactly the same length is used instead. What is the fundamental frequency of the open pipe?
- (A) 300 Hz
- (B) 150 Hz
- (C) 75 Hz
- (D) 450 Hz
Show Answer
Answer: A – For the same length L, the closed-pipe fundamental has λ = 4L, while the open-pipe fundamental has λ = 2L – half the wavelength, so twice the frequency: 2 × 150 = 300 Hz.
Q28. In a resonance-tube experiment with a closed pipe, the measured resonant length is always slightly greater than λ/4. Why is this end correction needed?
- (A) Because the node forms slightly inside the closed end
- (B) Because sound travels faster just outside the tube
- (C) Because the antinode forms slightly beyond the open end of the tube
- (D) Because the tuning fork loses energy to the air
Show Answer
Answer: C – The air just outside the open end also vibrates, so the effective antinode lies a small distance (the end correction, about 0.6 times the tube radius) beyond the physical end of the tube. The true quarter-wavelength is therefore a little longer than the measured tube length.
Q29. An open pipe has length 0.34 m. The speed of sound in air is 340 m s-1. What is the fundamental frequency of the pipe?
- (A) 250 Hz
- (B) 500 Hz
- (C) 1000 Hz
- (D) 125 Hz
Show Answer
Answer: B – For an open pipe, λ = 2L = 2 × 0.34 = 0.68 m. Then f = v/λ = 340 / 0.68 = 500 Hz.
Q30. A pipe closed at one end has a fundamental frequency of 85 Hz. Which of the following frequencies will NOT cause this pipe to resonate?
- (A) 85 Hz
- (B) 255 Hz
- (C) 425 Hz
- (D) 170 Hz
Show Answer
Answer: D – A closed pipe resonates only at odd multiples of the fundamental: 85 (1f), 255 (3f) and 425 Hz (5f) are allowed, but 170 Hz = 2f is an even harmonic and does not resonate.
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