Demonstrating stationary waves in an air column
A loudspeaker of fixed frequency is held over a tube whose effective air column length is varied by a plunger or by raising the water level. A loud sound is heard whenever resonance occurs, first at L₁ ≈ λ/4, next at L₂ ≈ 3λ/4, so L₂ − L₁ = λ/2 and the speed of sound is v = fλ = 2f(L₂ − L₁). A node forms at the closed end and an antinode at the open end.
v = fλ = 2f (L₂ − L₁)What this actually means
Two ways to run it and both are examined. Either fix the length and sweep the frequency, or fix the frequency and slide the plunger. The resonance-tube version fixes f and varies L, which is the safer one to describe.
Use the difference of the two resonance lengths, never a single length, because that difference is exactly λ/2 and cancels the end correction automatically.
The loud sound is the observable. Say the amplitude of vibration of the air column becomes large at resonance, so a much louder sound is heard, and the sound dies away again on either side.
Removing the plunger turns it into a both-ends-open pipe with antinodes at each end, which changes the harmonic series to all integers. Questions do sometimes switch mid-way.
The physical reason for the closed-end node is that the air at a rigid surface cannot be displaced along the tube. The open-end antinode exists because air there is free to move.
Using the first resonance length alone as λ/4 rather than taking the difference of two resonance lengths, which leaves the end correction in your answer.
Prove it — watch it be true
- Select the closed pipe with mode n = 1 and confirm the pattern is a quarter of a wavelength with a node at the closed end.
- Step to the next available mode and confirm it is three quarters of a wavelength.
- Read the N–N bracket and confirm the extra length added between the two modes is λ/2, which is the quantity the resonance tube measures.