Question

If a human ear canal can be thought of as resembling an organ pipe, closed at one end, that resonates at a fundamental frequency of 2,825 Hz, what is the length of the canal? Use normal body temperature 37

Answer 1

Answer 2

2825 Hz = {(331 m/s) * sqrt((37+273)/273)] / (4 * length)
2825 Hz = (352.72) / (4 * length)
0.1249 = (4 * length)
Length = 0.0312142 m
In cm 3.124

Answer 3

v sound = 331.4 + 0.6 Tc = 331.4 + 0.6 * 37 = 353.6 m /s

f1 = v sound / 4L
4L * f1 = v sound ---- 4L = 353.6 / 2825 = 0.12516 --- L = 3.129 cm