The windpipe of a typical whooping crane is about 5.0 ft long. What is the lowes

Question

The windpipe of a typical whooping crane is about 5.0 ft long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? Assume a temperature of 39

Explanation / Answer

5 ft = 5*0.3048 m = 1.524 m the velocity c of sound at 37° is v = 331 + 0.6*37 = 353.2 m/s The basic wave length is 4 times the length of the pipe = 6.096 m -----> frequency = 353.2 m/s / 6.096 m = 57.939 Hz ----- the next harmonics are odd multiples of the base frequency: fn = (2n + 1)*f1 the first harmonic has frequency f1 (2 + 1)f0 = 3f0 = 173.82 Hz --> wavelength 353.2/173.82 = 2.031 m the second harmonic has frequency f2 = (2*2 + 1)f0 = 289.69 Hz --> wavelength 353.2/289.69 = 1.219 m

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