Question

An open-pipe resonator has a length of 2.39 m. Calculate the frequency of its third harmonic if the speed of sound is 343 m/s.

Answer 1

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, $l=\dfrac{3\lambda}{2}$

So,

$\dfrac{3v}{2f}=2.39$

f is frequency in its third harmonics

$f=\dfrac{3v}{2\times 2.39}\\\\f=\dfrac{3\times 343}{2\times 2.39}\\\\f=215.27\ Hz$

So, the frequency of its third harmonic is 215.27 Hz.