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.

Options

A. `5.29 cm`
B. `3.33 cm`
C. `4.78 cm`
D. `4.29 cm`

Answer

B

Explanation

When an air column in a tube vibrates , the antinodes at the open end(s) are located at a small distance outside the open end. This small distance is called as end correction .
Approximate end correction ` = 0.3 d`
when `d` is the diameter of the tube .
In case of a tube open at both ends , the effective length of the tube that should be taken in calculation will now be `l`.
`rArr l' = l + 2 e` where `e = 0.3 d`
`v = ( c )/( 2 l)`
`rArr 320 = (320)/( 2( l + 2 e))`
`l + 2 e = 0.5`
`0.48 + 2 (0.3 d) = 0.5`
`rArr d = 1//30 m = 3.33 cm`