Jeulin 222 044 Mode D'emploi page 19

Periodic Phenomena
Kundt's tube with analog outputs
Ref : 22204
The simplest state corresponds to the existence of pressure antinodes at the loudspeaker
end and pressure nodes at the open end. Schematically, we have the following distribution:
Open end
The distance between two nodes is λ / 2. If L is the length of the tube, it must satisfy the
relation L = λ / 2.
As we have the relation: λ = v/N, we observe that the length of the tube must be:
For example, for L = 50 cm and the value of v = 346 m.s
frequency of the sound emitted by the loudspeaker is:
However, other modes of vibration are possible. The simplest, described below,
corresponds to the following standing state:
This time, the new wave length λ' must be such that L = 2. λ' / 2 or: L = λ'
which gives a frequency of:
With the proposed tube: N' = 692 Hz, which is easy to obtain.
In general, the length of the tube must represent an integer k times half the wave length of
the sound emitted:
Which corresponds to a sound frequency of:
Note: These successive remarkable states are obtained by fixing the frequency of the LFG,
When a system of standing waves is attained, it manifests itself by an increase in the sound
intensity coming from the pipe (when it is open), audible to the ear.
ENGLISH
•
Case when the pipe is open
N
N
which powers the loudspeaker.
V N
V
'/
/2
L = λ / 4 = v / 4N, or N = v / 2L
N = 346 / (2 x 0.5) = 346 Hz
N' = 2 x N
L = k λ / 2
N' = k N = k. v / 2L
18
V
Loudspeaker
N
'/
-1
(speed of sound at 25 °C), the