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.

Answer

3.33

Explanation

In case of vibration in air column ,the antinpdes at the open ends (s) are located at a small distance outside the open end.This small distance is known as end correction.
The value of end correction e=0.3d, where d is the diameter of the tube. In case of a tube open at both ends,the effective length of the air column.
` implies 1'=1+2e` Where e=0.3d
Fundamental frequency in case `f_0=v/(21)`
`implies 320=(320)/(2(1+2e))`
`1+2e=1/2`
`0.48+2(0.3d)=1/2 implies d=1//30m=3.33cm`
Now one end is closed,the effective length of the tube
`1'=1+e`
`f_0=v/(4 1')=(v)/(4(1+e)) implies f_0=(320xx100)/(4(48+0.3xx3.33))=163.27 Hz`