In the case of an air column vibrating in a pipe closed at one end, there are longitudinal stationary waves of sound formed in it.At the closed end, the air molecules are not free to vibrate and hence over there a displacement node can only be formed.
Vibration is the least.
Conversely, at the open end the molecules are free to vibrate. Hence over there, always a displacement antinode is formed.
Vibration is the maximum.
We know that the distance between a node and an adjacent antinode is always λ/4.
In the first mode, the length of the air column L=λ1/4.
Frequency f=V/λ
Hence frequency becomes f1=V4L as shown in figure (a)
In the 2nd mode, the length of the air column L=3λ2/4.
Hence frequency becomes f2=3V4L as shown in figure (b). ∴f2=3f1
In the 3rd mode, the length of the air column L=5λ3/4.
Hence frequency becomes f3=5V4L as shown in figure (c). ∴f3=5f1
Similarly f4=7f1
This shows that only odd multiples of the first fundamental frequency are present in such a case.
- Harmonics means the multiples of the fundamental frequency (may or may not) present in a case of a stationary wave.
- Overtones means the frequencies higher than the fundamental one really present.
You can thus see that only odd harmonics are present as overtones in the case of an air column vibrating in a pipe closed at one end.
______________________________________________________________________________________________
λ1=81173
λ2=81170
f1−f2=10 smaller λ means bigger f. Hence not f2−f1
V=?
We know that f=Vλ
f1−f2=10
∴Vλ1−Vλ2=10
∴V×(λ−11−λ−12)=10
V×(173−17081)=10
V=81×103=270
velocity of sound in air =270m/s