The human ear canal is, on average, about 2.5 cm (25 mm) long, though some sources mention a small range around this average (approximately 2.2 to 3.1 cm depending on individuals and measurement methods). It functions as a resonant cavity that is closed at one end (the eardrum) and open at the other (the outer ear). This design influences hearing sensitivity by selectively amplifying certain sound frequencies.
Given the speed of sound as 340 m/s and the ear canal length as approximately 2.5 cm (0.025 m), the ear canal acts like a tube that is closed at one end and open at the other, so its fundamental resonant frequency can be calculated. The fundamental frequency f1f_1f1 for a tube open at one end and closed at the other is given by:
f1=v4Lf_1=\frac{v}{4L}f1=4Lv
where:
- vvv is the speed of sound (340 m/s)
- LLL is the length of the tube (0.025 m)
Plugging in the values:
f1=3404×0.025=3400.1=3400 Hzf_1=\frac{340}{4\times 0.025}=\frac{340}{0.1}=3400\text{ Hz}f1=4×0.025340=0.1340=3400 Hz
This means the ear canal naturally amplifies sounds around 3400 Hz, which corresponds to the frequencies where human hearing is very sensitive. Higher odd harmonics (3rd, 5th, etc.) occur at frequencies:
fn=(2n−1)×f1,n=1,2,3,…f_n=(2n-1)\times f_1,\quad n=1,2,3,\ldots fn=(2n−1)×f1,n=1,2,3,…
So the next resonant frequencies would be 3 × 3400 = 10200 Hz, 5 × 3400 = 17000 Hz, etc., within the range of hearing frequencies. Thus, the ear canal length of about 2.5 cm plays a significant role in our sensitivity to frequencies around 3.4 kHz by acting as a resonant cavity.
