The questions are on the stationary waves test. Take it under real conditions before reading the mark scheme.
Quick answer: A stationary wave forms when two identical waves travel in opposite directions and superpose, giving fixed nodes and antinodes. This mark scheme gives the model answers and the marks for the 8 questions of the Physics 11-12 test on stationary waves, from violin string to end correction idea, with the equivalent answers to accept.
Model answers with a marking guide. Award method marks for correct use of the node and antinode conditions even when the final value is wrong, and check each answer carries the right unit.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB C.4
Question 1 Violin string (3 marks)
- λ = 2L = 1.8 m.
- f₁ = v/2L = 180 ÷ 1.8 = 100 Hz.
- 3f₁ = 300 Hz, with 3 antinodes.
Marking: 1 mark for each part.
Question 2 Closed pipe resonances (3 marks)
- f₁ = v/4L = 343 ÷ 1.60 = 214.4 Hz.
- 3f₁ = 643.1 Hz and 5f₁ = 1072 Hz.
Marking: (a) 1; (b) 1 for each frequency.
Question 3 Reading the loops (3 marks)
- The 4th harmonic (4 loops).
- 4 × λ/2 = 1.20 m, so λ = 0.60 m.
- v = fλ = 200 × 0.60 = 120 m/s.
Marking: 1 mark for each part.
Question 4 Two kinds of wave (3 marks)
(1) Energy is transferred along a progressive wave but not along a stationary wave. (2) All points on a progressive wave have the same amplitude; a stationary wave has nodes of zero amplitude and antinodes of maximum amplitude. (3) Phase changes steadily along a progressive wave, but points between adjacent nodes oscillate in phase and neighbouring loops are in antiphase. One mark for each valid difference.
Marking: 1 mark for each of three correct differences.
Question 5 Odd harmonics (2 marks)
The closed end must be a node and the open end an antinode. A length of L = (2n − 1)λ/4 with n = 1, 2, 3… fits these conditions, so only f₁, 3f₁ and 5f₁ can occur. Even multiples would put a node at both ends or an antinode at both ends.
Marking: 1 for node/antinode at the ends; 1 for odd quarter wavelengths.
Question 6 Doubling the tension (2 marks)
f ∝ √T, so f = 220 × √2 = 311 Hz.
Marking: 1 for f ∝ √T; 1 for the answer.
Question 7 Fretting a string (2 marks)
For constant v, f ∝ 1/L. f = 110 × 0.64 ÷ 0.48 = 147 Hz.
Marking: 1 for f ∝ 1/L; 1 for the answer.
Question 8 End correction idea (2 marks)
The antinode at an open end forms slightly outside the pipe (about 0.6r), so the effective length is longer than the physical length. If this is ignored, the length used is too small, so the calculated frequency is too high.
Marking: 1 for antinode beyond the end; 1 for frequency too high.
