Answer:
f = 215.27 Hz
Explanation:
We have,
An open-pipe resonator has a length of 2.39 m.
It is required to find the frequency of its third harmonic if the speed of sound is 343 m/s.
For third harmonic, its length is equal to, [tex]l=\dfrac{3\lambda}{2}[/tex]
So,
[tex]\dfrac{3v}{2f}=2.39[/tex]
f is frequency in its third harmonics
[tex]f=\dfrac{3v}{2\times 2.39}\\\\f=\dfrac{3\times 343}{2\times 2.39}\\\\f=215.27\ Hz[/tex]
So, the frequency of its third harmonic is 215.27 Hz.
