Node and antinode spacing
Adjacent nodes are λ/2 apart, and so are adjacent antinodes. A node and the next antinode are λ/4 apart. Here λ is the wavelength of the component progressive waves, so λ = 2 × (node-to-node distance).
N–N = A–A = λ/2 · N–A = λ/4 · v = fλ = f × 2(N–N)What this actually means
This one relation carries most of the numerical marks in stationary waves. Measure a node-to-node distance, double it to get λ, then use v = fλ.
Measure across several loops and divide, not just one gap. If a detector passes through 20 antinodes over 1.90 m, that is 10 wavelengths, so λ = 0.190 m. Averaging over many gaps kills the reading error.
The factor of two exists because the two component waves are moving in opposite directions, so their relative phase changes twice as fast with position as it would for one wave alone.
Adjacent-node distance and loop length are the same thing. One loop of the envelope is half a wavelength, which is why a fundamental on a string of length L has λ = 2L.
The wavelength you get is that of the progressive waves. Saying the stationary wave has wavelength equal to the node spacing is wrong and is punished.
Taking the distance between adjacent nodes as a whole wavelength, which halves every speed and frequency you calculate.
Prove it — watch it be true
- Select the string system with mode n = 4 and switch on the measurement brackets.
- Read the N–N bracket and confirm it equals half the wavelength shown for the component waves.
- Read the N–A bracket and confirm it is exactly half of the N–N value.