Answer:
4009.21658986 Hz
Explanation:
v = Speed of sound in air = 348 m/s
L = Length of tube = 2.17 cm
Fundamental frequency for a tube where one end is closed and the other end open is given by
[tex]f=\dfrac{v}{4L}\\\Rightarrow f=\dfrac{348}{4\times 2.17\times 10^{-2}}\\\Rightarrow f=4009.21658986\ Hz[/tex]
The fundamental frequency of the canal is 4009.21658986 Hz