## Related questions with answers

The windpipe of a typical whooping crane is about 5.0 ft. long. What is the lowest resonant frequency of this pipe, assuming it is closed at one end? Assume a temperature of $37 ^ { \circ } \mathrm { C }$.

Solutions

VerifiedIn this problem, the windpipe of a whooping crane is typically $L = 5.0~\mathrm{ft}$ long. We calculate the lowest resonant frequency of this pipe, assuming that one end is closed. We assume that the air temeprature is $T = 37~\mathrm{^{\circ}C}$.

The speed of the sound is:

$v = 331 \ \sqrt{1 + \dfrac{T}{273}} = 331 \ \sqrt{1 + \dfrac{37}{273}} = 352.72 \ m/s$

The length of the pipe is:

$L = 5.0 \ ft = 1.524 \ m$

The wavelength is:

$\lambda = 4 \ L = (4) \ (1.524) = 6.096 \ m$

The frequency:

$f = \dfrac{v}{\lambda} = \dfrac{352.72}{6.096} = 58 \ Hz$

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