#### Questions

Discuss different modes of vibrations in an air column of a pipe open at both the ends.

Draw neat labelled diagrams for modes of vibration of an air column in a pipe when it is open at both ends

Hence derive an expression for fundamental frequency in each case.

#### Solution

**First mode or fundamental mode:**

In this mode of vibration, there is one node at the centre of the pipe and two antinodes, one at each open end as shown in figure (a).

Let,

v = wave velocity in air

n = fundamental frequency

λ = wavelength

L = length of air column

v = nλ ….(1)

Also `L=lambda/2`

`therefore lambda=2L` ...................(2)

From Equation 1 and 2

`v=n2L`

`n=v/(2L)`.......(3)

Equation (3) represents fundamental frequency or lowest frequency of vibration

**ii. Second mode or first overtone:**

In this mode of vibration, there are two nodes and three antinodes as shown in figure (b).

Let,

v = wave velocity in air [As the medium is same, wave

velocity remains same]

n1 = next higher frequency

λ1 = corresponding wavelength

L = length of tube

Velocity of wave is given by,

`v=n_1lambda_1` ..............(4)

Also `L=lambda_1` ..............(5)

From equation (4) and (5),

`v=n_1L`

`n_1=v/L`

`n_1=2xxv/(2L)`.................(6)

From equation (3) and (6)

` n_1 = 2n` .…(7)

Thus, frequency of first overtone (second harmonic) is twice the fundamental frequency

**iii. Third mode or second overtone:**

In this mode of vibration, there are 3 nodes and 4 antinodes as shown in figure (C).

Let,

v = wave velocity in air (As the medium is same, wave velocity remains same)

n2 = next higher frequency

λ2 = corresponding wavelength

L = length of tube

Velocity of wave is given by,

`v=n_2lambda_2 ` ...................(8)

Also `L=(3lambda_2)/2`

`lambda_2=(2L)/3`

From equation (8) and (9),

`v=n_2(2L)/3`

`n_2=(3v)/2L`..............(10)

From equation (3) and (10),

`n_2 = 3n`

Thus, frequency of second overtone is thrice the fundamental frequency.