A tube of a certain diameter and of length `48cm` is open at both ends. Its fundamental frequency is found to be `320 Hz`. The velocity of sound in air is `320 m//sec`. Estimate the diameter of the tube.
One end of the tube is now closed. Calculate the lowest frequency of resonance for the tube.

Explanation

The displacement curves of longitudinal waves in a tube open at both ends is shown in figure-6.74(a) and (b).

Let r be the radius ofthe tube. We know that antinodes occur slightly outside the tube at a distance 0.6 r from the tube end.
The distance between two antinodes is given by
`(lambda)/(2) = 48 + 2 xx 0.6 r`
We have `lambda = (v)/(n) = (32000)/(320) = 100 cm`
`50 = 48 + 1.2 r`
`r = (2)/(1.2)`
= 1.67 cm
The daimeter of the tube is
`D = 2r = 3.33 cm`
When one end is closed, then
`(lambda)/(4) = 48 + 0.6r`
= 48 + 0.6 + xx 1.67
= 49
`lambda = 4 xx 49`
= 196 cm
`n = (v)/(lambda) = (32000)/(196)` = 163.3 Hz