Question : Find the ratio of the length of a closed pipe to that of an open pipe in order that the second overtone of the former is in unison with the fourth overtone of the latter.
Doubt by Niyati
Solution :
We know,.
For a closed Organ Pipe
𝜈n = (2n-1)v/4L
Frequency for second overtone i.e. n=3
𝜈3=5v/4L — (1)
𝜈3=5v/4L — (1)
For an open Organ Pipe
𝜈n = nv/2L'
Frequency for fourth overtone i.e. n=5
𝜈5=5v/2L' — (2)
𝜈5=5v/2L' — (2)
Where 𝜈n is the frequency of nth harmonic, v is the velocity of the sound in the air and L is the length of the organ pipe.
According to Question
For unison
𝜈3 = 𝜈5
For unison
𝜈3 = 𝜈5
5v/4L = 5v/2L'
5v/5v = 4L/2L'
1 = 2L/L'
1/2=L/L'
L/L'=1/2
L:L=1:2
Hence, required ratio for the length of two organ pipes is 1:2
5v/5v = 4L/2L'
1 = 2L/L'
1/2=L/L'
L/L'=1/2
L:L=1:2
Hence, required ratio for the length of two organ pipes is 1:2