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 dist

Question

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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Thông Đạt 6 months 2021-07-30T00:18:47+00:00 1 Answers 7 views 0

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    2021-07-30T00:19:48+00:00

    Answer:

     f=16.46 Hz

    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:

    v=\sqrt{\frac{T}{\mu}}

    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:

    \lambda f=\sqrt{\frac{T}{\mu}}

    f=\frac{1}{\lambda}\sqrt{\frac{T}{\mu}}

    f=\frac{1}{4.4}\sqrt{\frac{572}{0.109}}

     f=16.46 Hz  

    I hope it helps you!

           

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