A pipe of length 6.8 m is closed at one end and sustains a standing wave at its second overtone. determine the distance between a node and an adjacent antinode.

For a pipe closed at one end, the second overtone is the 3rd natural frequency as shown in the diagram.

For the fundamental frequency,

λ = 4L

where

λ = wavelength

L = the length of the tube

For the second overtone (3rd frequency)

λ = (4/3) L

Because L = 6.8 m, therefore

λ = (4/3) * (6.8 m) = 9.0667 m

The distance between a node and an adjacent antinode is λ/2, which is

9.0667/2 = 4.533 m

Answer: 4.5 m