The windpipe of one typical whooping crane is 5.3 feet long. What is the fundamental resonant frequency of the bird's trachea, modeled as a narrow pipe closed at one end? (Assume a temperature of 35°C.)
Hz
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To find the fundamental resonant frequency of a narrow pipe closed at one end, we can use the formula:
f = (v / 2L)
Where:
f is the frequency (in Hz)
v is the speed of sound in air (which depends on temperature)
L is the length of the pipe
In this case, we already have the length of the windpipe (5.3 feet), but we need to determine the speed of sound in air at a temperature of 35°C.
The speed of sound in air can be calculated using the formula:
v = 331.4 + (0.6 * T)
Where:
v is the speed of sound in m/s
T is the temperature in °C
Plugging in the temperature of 35°C into the formula, we get:
v = 331.4 + (0.6 * 35)
v = 331.4 + 21
v = 352.4 m/s
Now we can calculate the fundamental resonant frequency:
f = (352.4 / (2 * 5.3))
Simplifying:
f = 352.4 / 10.6
f = 33.28 Hz
Therefore, the fundamental resonant frequency of the whooping crane's trachea would be approximately 33.28 Hz.