The method and the results are on the stationary waves lab. Do the analysis yourself first, then check the answers below.
Quick answer: A stationary wave forms when two identical waves travel in opposite directions and superpose, giving fixed nodes and antinodes. These answers go with the Physics 11-12 lab on stationary waves: the graph, the calculations, the uncertainties and the conclusion explained, for all 8 analysis questions.
Model answers use the sample table. Your own readings will differ slightly, but L plotted against 1/f should be a straight line with a small negative intercept equal to minus the end correction.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB C.4
At the water surface there is a node (air cannot move) and near the open end there is an antinode, a distance e above the tube. The shortest pattern joining a node to an antinode is a quarter of a wavelength. L + e = λ/4.
Question 2 Line equation
Since λ = v/f, L + e = v/(4f). So L = (v/4)(1/f) − e, a straight line when L (y-axis) is plotted against 1/f (x-axis), with gradient v/4 and intercept −e.
Question 3 Slope of the line
1/f for 256 Hz = 3.906 × 10⁻³ s; for 512 Hz = 1.953 × 10⁻³ s. Gradient = (0.158 − 0.326) ÷ (0.001953 − 0.003906) = 86.02 m/s (metres per second, since L is in m and 1/f in s).
Question 4 Speed of sound
v = 4 × gradient = 4 × 86.02 = 344 m/s.
Question 5 End correction
Intercept = L − gradient × (1/f) using the 256 Hz point = 0.326 − 86.02 × 0.003906 = -0.01 m, so e = 1 cm. The tube radius is 2.0 cm, so 0.6r = 1.2 cm is of the same order.
Question 6 Percentage difference
|344 − 343| ÷ 343 × 100 = 0.31 %.
Question 7 Second resonance
L₂ − L₁ = λ/2, so λ = 2 × (0.493 − 0.158) = 0.67 m. v = fλ = 512 × 0.67 = 343 m/s. The end correction cancels in the difference.
Question 8 Uncertainty and improvement
3 mm ÷ 158 mm × 100 = 1.9 %. To improve, find the quietest positions either side of the maximum and take the midpoint, or use a loudspeaker and signal generator to give a steady tone.
