Jeulin 222 044 Mode D'emploi page 20

Periodic Phenomena
Kundt's tube with analog outputs
Ref : 22204
The simplest state corresponds to the existence of pressure antinodes at the closed end
and pressure nodes at the loudspeaker. Schematically, we have the following distribution:
The distance between the nodes and antinodes is λ/4. If L is the length of the tube, it must
satisfy the relation L = λ /4. As the speed of sound v in air is linked to the wave length and
the frequency N of the wave emitted by the loudspeaker by λ = v/N, we observe that the
length of the pipe must be:
For example, for L = 50 cm and a value of v = 346 m.s
the frequency of the sound emitted by the loudspeaker is:
Note: The 5 cm diameter loudspeaker is unable to generate a sound of such low frequency
However, other modes of vibration are possible. The following one corresponds to the
standing state:
Open end
This time, the new wave length λ' is such that L = λ'/2 + λ'/4 = 3. λ'/4, which gives a
frequency of:
Or, the numerical value is:
In general, the length of the tube must represent an odd number (2k + 1) of one fourth the
wavelength of the sound emitted:
Which corresponds to a sound frequency of:
N = (2k + 1) v / 4L (v = speed of sound in air)
ENGLISH
•
Case when the pipe is closed
V
thus making it impossible to perform the experiment.
L
L
/4
L = λ / 4 = v / 4N, or N = v / 4L
N = 346 / (4 x 0.5) = 173 Hz
N' = v / λ' = 3.v / 4L = 3.N
N' = 3.N = 3 x 173 = 519 Hz
L = (2k + 1). λ / 4
19
N
Loudspeaker
-2 (speed of sound in air at 22°C),
Loudspeaker