Question

An air column which has a fixed length 0.78 m is closed at one end and open at the other end. It produces a 2nd resonant frequency of 335 Hz. Calculate the speed of sound in air.

Answer 1

A closed at one end tube will have approximate resonances of:

$$f = \dfrac{nv}{4L}$$

where $n$ is an odd number (order of resonant frequency), $v$ is the speed of sound, and $L = 0.78m$ is the length of the tube. Since $f = 335Hz$, and it is the 2nd resonant frequency the corresponding order will be $n=3$. Thus, obtain the expression for the speed of sound:

$$v = \dfrac{4fL}{n}\\ v = \dfrac{4\cdot 335\cdot 0.78}{3} = 348.4m/s$$

Answer. 348.4 m/s.