physics help?

Question

The windpipe of a typical whooping crane is about 5.3 feet long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? (Assume a temperature of 32°C.)

Answer 1

v = 331.3√(1 + 32/273.15) m/s
1m = 100cm(1 in/2.54 cm)(1ft/12 in)
1m = (25/7.62) ft.
v = (8282.5/7.62)√(1 + 32/273.15) fps
λ/2 = 5.3 ft.
λ = 10.6 ft.
f = v/λ
f = (8282.5/80.772)√(1 + 32/273.15) Hz
f ≈ 108.3819 Hz ≈ 108 Hz