With a neat labelled diagram, show that all harmonics are present in an air column contained in a pipe open at both the ends. Define end correction.

#### Solution

Stationary waves are produced due to the superposition of two identical simple harmonic progressive waves traveling through the same part of the medium in opposite direction. In the case of pipe open at both ends, the air molecules near the open end are free to vibrate, hence they vibrate with maximum amplitude. Therefore the open end becomes an antinode.

The simplest mode of vibration is the fundamental mode of vibration. (in above fig a ) In this case one node & two antinodes are formed. If λ_{1} be the corresponding wavelength &

`"L be length of the pipe" "L" = lambda_1 / 2`

`lambda_1 = 2 "L"`

If n_{1} be corresponding frequency and v be velocity of sound in air ` v = "n"_1 lambda_1`

`"n"_1 = "v"/lambda_1`

`n_1= "v"/(2"L")......(1) ` This is called 1st harmonic.

The next possible mode of vibration is called the 1st overtone. (in above fig) In this case, two nodes and three antinodes are formed. If λ_{2} and n_{2 }be the corresponding wavelength and frequency,

L = λ_{2}

λ_{2 }= L

But

v n_{2}λ_{2} i.e = λ_{2 }= `v/λ_2` n_{2} = `"v"/"L"`

This is called 2nd harmonic.

The next possible mode of vibration is called 2nd overtone.(in above fig c) In this case three nodes & four antinodes are formed. If λ_{3} and n_{3} be the corresponding wavelength & frequency ,

`L=3lambda_3/2 `

`lambda_3=(2L)/3`

`v=n_3 lambda_3 `

i.e

` n_3=v/lambda_3`

n_{3} = `3"v"/2L` .

This is called 3rd harmonic.

`n_1 : n_2 : n_3 : : 1 : 2 : 3`

Thus in case of pipe open at both end, all harmonics are present.

In this case, the node is forming at the closed-end i.e. at the water surface & antinode is forming at the open end. But antinode is not formed exactly at the open end but slightly outside it since air molecules are free to vibrate. The correction has to be applied is called end correction.