Consider a closed end pipe PQ of length L by blowing air from the right side of the pipe the longitudinal standing waves may be formed due to the superposition of two identical longitudinal waves (incident and reflected waves) travelling in the opposite directions with the same speed.
As the displacement of the air particle at the closed end would always be zero (minimum), there would always be a nodal point at the closed end. The displacement of the air particle at open end would always be maximum, so there would always be an antinode at the open end.
The simplest mode in which air column in a closed organ pipe can be set into vibrations is shown in fig (a).
In this case, L = λ1/4
or λ1 = 4L
Hence, the frequency, v1 the of the fundamental mode of vibrations will be,
v1 = v/λ1 = v/4L ...(i)
The first overtone or third harmonic of the vibrations of the closed pipe is shown in the fig (b) and can be produced if air is blown **.
In this case, L = 3λ2/2
or, λ2 = 4L/3
Therefore, the frequency of the first overtone of the closed end pipe is given by
v2 = v/λ2
= 3v/4L = 3v2 ...(ii)
Fig. (c) represents the second overtone or fifth harmonic of the vibrations of the closed end pipe.
In this case, L = 5λ3/4
or λ3 = 4L/5
Hence, the frequency of the second overtone of the closed end pipe is given by:
v3 = v/λ3
= 5v/4L = 5v1 ...(iii)
From equations (i) and (iii), we have observed that v1 : v2 : v3 = 1 : 3 : 5.
Thus, in closed end pipe only odd harmonics are present.
