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To review the concepts that play roles in this problem, consult Multiple-Concept Example 4. Sometimes, when the wind blows across a long wire, a low-frequency "moaning" sound is produced. This sound arises because a standing wave is set up on the wire, like a standing wave on a guitar string. Assume that a wire (linear density = 0.0163 kg/m) sustains a tension of 47.8 N because the wire is stretched between two poles that are 8.12 m apart. The lowest frequency that an average, healthy human ear can detect is 20.0 Hz. What is the lowest harmonic number n that could be responsible for the "moaning" sound?

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To solve this problem we will start from the definition of the string length as a function of the harmonic number and the wavelength. In turn, we will use the velocity expression on a string defined as the square root between the tension and the linear density. Finally, we will apply the wavelength equivalence as a function of speed and frequency, an expression that will help us find the harmonic. Let's start by defining the relationship of the [tex]n^{th}[/tex]harmonic as

[tex]L= \frac{n\lambda}{2}[/tex]

Here,

n = Number of harmonic

[tex]\lambda[/tex]= Wavelength

At the same time the velocity is defined as,

[tex]v = \sqrt{\frac{T}{\rho}}[/tex]

Here

T= Tension

[tex]\rho =[/tex]  Density

Replacing with our values we have that the velocity is

[tex]v = \sqrt{\frac{47.8}{0.0163}}[/tex]

[tex]v = 54.15 m/s[/tex]

Through the expression of the wavelength we have to

[tex]\lambda = \frac{f}{v}[/tex]

Replacing at the first equation we have,

[tex]L= \frac{n\lambda}{2}[/tex]

[tex]L = \frac{n(\frac{f}{v})}{2}[/tex]

Rearranging to find n,

[tex]n = \frac{2fL}{v}[/tex]

The value of the frequency was given at the statement as 20.0 Hz, then

[tex]n = \frac{2(20)(8.12)}{54.15}[/tex]

[tex]n = 5.99 \approx 6[/tex]

Therefore the smallest integer n is 6.

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