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A cable with a linear density of \mu=0.109~\text{kg/m}μ=0.109 kg/m is hung from telephone poles. The tension in the cable is 572 N. The distance between poles is 19.9 meters. The wind blows across the line, causing the cable resonate. A standing waves pattern is produced that has 4.5 wavelengths between the two poles. Assuming room temperature air, what is the frequency of the hum?

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whitneytr12

Answer:

[tex] f=16.46 Hz[/tex]

Explanation:

The equation of the speed of a mechanical wave in terms of the tension and linear density, of the cable in our case, is given by:

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

Where:

  • T is the tension of the cable (T = 572 N)
  • μ is the linear density of the cable (μ = 0.109 kg/m)

And we know that v = λ*f

  • λ is the wavelength
  • f is the frequency

Because a standing waves pattern is produced that has 4.5 wavelengths between the two poles and the distance between poles is 19.9 meters, the value of the wavelength is: λ = 19.9/4.5 = 4.4 m.

Therefore, the frequency will be:

[tex]\lambda f=\sqrt{\frac{T}{\mu}}[/tex]

[tex]f=\frac{1}{\lambda}\sqrt{\frac{T}{\mu}}[/tex]

[tex]f=\frac{1}{4.4}\sqrt{\frac{572}{0.109}}[/tex]

[tex] f=16.46 Hz[/tex]  

I hope it helps you!

       

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