Consider the human outer ear auditory canal as a tube open at one end and as a 2.5-cm-long resonator with rigid walls. What is the resonant frequency of the auditory canal? The speed of sound is 343 m/s.
A)34.3 Hz
B)3430 Hz
C)6860 Hz
D)13720 Hz

Question

Quizplus · Accepted Answer

The resonant frequency of a tube open at one end is given by $f = \frac{v}{4L}$, where $v$ is the speed of sound and $L$ is the length of the tube. Substituting $v = 343 \, \text{m/s}$ and $L = 0.025 \, \text{m}$ (since 2.5 cm = 0.025 m), we get $f = \frac{343}{4 \times 0.025} = 3430 \, \text{Hz}$.

Consider the Human Outer Ear Auditory Canal as a Tube