Answer to Question #170797 in Physics for dhm

Question

A 4.28 m deep well acts as a closed pipe. When wind blows across the top , what is the third harmonic (f3) that it creates?

(speed of sound =343 m/s)

Answer 1

Let's first find the wavelength of the wave. For the third harmonic in the closed pipe there is the following length-wavelength relationship:

$$\lambda=\dfrac{4}{3}L=\dfrac{4}{3}\cdot4.28\ m=5.706\ m.$$

Finally, we can find the third harmonic frequency from the wave speed formula:

$$v=f\lambda,$$ $$f=\dfrac{v}{\lambda}=\dfrac{343\ \dfrac{m}{s}}{5.706\ m}=60.1\ Hz.$$