Question

The auditory canal behaves like a resonant tube to aid in hearing. One end terminated at the eardrum, while the other opens to the outside. Typically, the canal is about 2.4 cm long. At which frequency would it resonate in its first harmonic?
4.2 kHz
3.6 kHz
2.9 kHz
5.7 kHz

Answer 1

Answer:

3.6 kHz

Explanation:

The pipes behave like a closed pipe . The end open is the end of the air canal outside the ear and the closed end is the eardrum.

The first harmonic will be as seen in the figure attached.

The length of the first harmonic will be λ/4.

λ/4=2.4 cm

λ=2.4 * 4=9.6 cm 0.096 m

Speed of Sound- 344 m/s(in air)

velocity(v) * Time Period(T) = Wavelength (λ)

Also, Time Period(T)= \frac{\textrm{1}}{\textrm{Frequency(f)}}

\frac{\textrm{Velocity}}{\textrm{Wavelength}}=\frac{\textrm{1}}{\textrm{Time Period}} =Frequency

Plugging in the values into the equation,

Frequency = $\frac{344}{0.096}$ Hz

                  = 3583.3 Hz≈3600 Hz= 3.6 kHz

Frequency= 3.6 kHz

Answer 2

Answer:

3.6 kHz

Explanation:

The auditory canal is a closed pipe because it has one closed end, the end terminated at the eardrum.

The length of the first harmonic of a closed pipe is given as;

L = λ/4 --------------------  (i)

where L = Length

and λ = wavelength

2.4 = λ/4

λ = 2.4 x 4 = 9.6 cm

Also, v = fλ  ------------------ (ii)

where v = speed of sound in air = 344 m/s

f = frequency of wave in Hertz

f = v/λ ------------------ (iii)

convert 9.6 cm to m = 0.096 m

substitute for λ and v in (iii)

$f = \frac{344}{0.096} = 3583.33$

3583.33 Hz = 3600 Hz = 3.6 kHz