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Sometimes, when the wind blows across a long wire, a low-frequency "moaning" sound is produced. The 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 = 0.0180 kg / m ) sustains a tension of 350 N because the wire is stretched between two poles that are 17.43 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?

Respuesta :

Answer:

N = 5 harmonics

Explanation:

As we know that frequency of the sound is given as

[tex]f = \frac{N}{2L}\sqrt{\frac{T}{\mu}}[/tex]

now we have

[tex]T = 350 N[/tex]

[tex]\mu = 0.0180 kg/m[/tex]

L = 17.43 m

now we have

[tex]f = \frac{N}{2(17.43)}\sqrt{\frac{350}{0.0180}}[/tex]

[tex]f = 4 N[/tex]

if the lowest audible frequency is f = 20 Hz

so number of harmonics is given as

[tex]20 = 4 N[/tex]

N = 5 harmonics

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