Question

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?

Answer 1

N = 5 harmonics

Step-by-step explanation:

As we know that frequency of the sound is given as

`f = (N)/(2L)\sqrt{(T)/(\mu)}`

now we have

`T = 350 N`

`\mu = 0.0180 kg/m`

L = 17.43 m

now we have

`f = (N)/(2(17.43))\sqrt{(350)/(0.0180)}`

`f = 4 N`

if the lowest audible frequency is f = 20 Hz

so number of harmonics is given as

`20 = 4 N`

N = 5 harmonics