Question #170797

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)


1
Expert's answer
2021-03-12T07:15:07-0500

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


λ=43L=434.28 m=5.706 m.\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=fλ,v=f\lambda,f=vλ=343 ms5.706 m=60.1 Hz.f=\dfrac{v}{\lambda}=\dfrac{343\ \dfrac{m}{s}}{5.706\ m}=60.1\ Hz.

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