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 = 0.0195 kg / m ) sustains a tension of 332 N because the wire is stretched between two poles that are 13.05 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?
Explanation / Answer
The tension in the string is T = 332 N The linear density of the string is = 0.0195 kg/m The length of the string (distance between the poles) L = 13.05 m The frequency of the fundamental vibration of the wave in a string is f = 1/2L (T/) Where L is the legth of vibrating segment T is the applied tension is the linear density f = 1/2(13.05 m) (332 N / 0.0195 kg/m) f = 4.999 Hz f ˜ 5 Hz The lowest possible requency a human ear can hear is f' = 20 Hz The number of the harmonic which produces the required frequency is n = 20 Hz/5 Hz n = 4 The second harmonic of the string will produce the moaning sound in the string The tension in the string is T = 332 N The linear density of the string is = 0.0195 kg/m The length of the string (distance between the poles) L = 13.05 m The frequency of the fundamental vibration of the wave in a string is f = 1/2L (T/) Where L is the legth of vibrating segment T is the applied tension is the linear density f = 1/2(13.05 m) (332 N / 0.0195 kg/m) f = 4.999 Hz f ˜ 5 Hz The lowest possible requency a human ear can hear is f' = 20 Hz The number of the harmonic which produces the required frequency is n = 20 Hz/5 Hz n = 4 The second harmonic of the string will produce the moaning sound in the string The second harmonic of the string will produce the moaning sound in the string