Question

The human ear canal is about 2.6 cm long and can be regarded as a tube open at one end and closed at the eardrum. What is the fundamental frequency around which we would expect hearing to be most sensitive

Answer 1

Answer:

f = v/λ

 = v/4*L

 = 343 / 4 (0.026)

= 3120 Hz  

Answer 2

Answer:

3.298 K Hz

Explanation:

The Human ear canal can be attributed to a tube which is open at the end and closed at the other end. As shown in the figure attached the length of the canal is 2.6 cm representing L in the diagram.

To obtain our fundamental frequency the expression below is used;

f = v/λ ........................1

The length of the human ear canal is given as 2.6 cm and the speed of sound if not given is assumed to be 343 m/s

where f is the fundamental frequency;

v is the speed of sound = 343 m/s;

λ is the wavelength.

Calculating for the wavelength for the first harmonics using equation 2;

λ =  4 L.....................2

L is the length of the ear canal = 2.6 cm = $\frac{2.6}{1000}$ = 0.026 m

λ = 4 x 0.026 m

λ = 0.104 m

Now lets find the fundamental frequency using equation 1

f = v/λ ........................1

$f = \frac{343 m/s}{0.104 m}$

f = 3298 Hz

f = 3298 Hz/1000

f = 3.298 K Hz

Therefore the fundamental frequency around which would expect the hearing to be more sensitive would be 3.298 K Hz