Question

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 frequrncy of resonance for the tube.

Answer 1

Correct Answer - A::C
Tube open at both ends :
(a) `v = (v)/(2(l + 0.6D))` :, `320 = (320)/(2(0.48 + 0.6 xx D))`
…(ii) `0.48 + 0.6D = 0.5` rArr `0.6D = 0.02`
rArr `D = (0.02)/(60) xx 100 cm = 3.33cm`
Tube closed at both ends : .... (iii)`v = (v)/(4(l + 0.3D)) = `320 = (320)/(4(0.48 + 0.3 xx 0.033))`
`~~ 163 Hz`