Answer:
* in a tube with both ends open
λ = 2L / n n = 1, 2, 3,…
* tube with one end open and the other closed.
λ = 4L / n n = (2n ’+ 1 )
Explanation:
When a sound wave enters a pipe, a resonance process occurs, whereby only some wavelengths can occur.
* in a tube with both ends open
in this case there are maximums at each end, so if the length of the had is L
λ = 2L 1st harmonic
λ = 2 L / 2 2nd harmonic
λ = 2L / 3
λ = 2L / n n = 1, 2, 3,…
* In the case of a tube with one end open and the other closed.
At the open end there is a belly and at the closed end a node
λ = 4L 1st harmonic
λ = 4L / 3 3rd harmonic
λ = 4L / 5
λ = 4L / n n = 1, 3, 5,… odd
n = (2n ’+ 1 ) where n’ are all integers
