   Chapter 14, Problem 55P

Chapter
Section
Textbook Problem

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°C.

To determine
The lowest resonant frequency of the pipe by assuming the pipe is closed at one end.

Explanation

Given Info: The speed of sound in air is 331 m/s and the temperature of air in Celsius is 37°C .

Formula to calculate the speed of the sound in windpipe is given by the relation.

v=vs(T273K)

• v is the speed of the sound in windpipe,
• vs is the speed of the sound in air and.
• T is the temperature.

Substitute 331m/s for vs and 37°C for T to find v.

v=(331m/s)((37+273K)273K)=352.7m/s

In the resonance mode, the wavelength of the sound waves in a pipe closed at one end is,

λ1=4L

• L is the length of the windpipe and,
• λ1 is the fundamental wavelength.

Substitute 5

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