204k views
4 votes
The human ear canal is about 2.8 cm long. If it is regarded as a tube that is open at one end and closed at the eardrum, what is the fundamental frequency around which we would expect hearing to be most sensitive?[3.1 kHz]

User Garry Wong
by
5.2k points

2 Answers

6 votes

Final answer:

The human ear canal resonates like a tube closed at one end, with fundamental frequencies that the ear is most sensitive to within the 2000 to 5000 Hz range, depending on the length of the canal.

Step-by-step explanation:

The human ear canal acts as a resonator similar to a tube that is closed at one end and open at the other. Given that the human ear canal ranges in length from 1.80 cm to 2.60 cm and using the body temperature of 37°C to calculate the speed of sound within the ear, which affects the resonant frequencies, we can determine the range of the fundamental frequencies for the average population. Considering that the fundamental frequency of a tube closed at one end is determined by the formula f = v / (4L), where f is the frequency, v is the speed of sound, and L is the length of the tube, one can calculate the range of fundamental resonant frequencies that the ear canal would be most sensitive to, which typically lies within the 2000 to 5000 Hz range. This sensitivity range correlates to the range of frequencies at which the human ear is most adept at picking up sound intensity, as depicted in the intensity versus frequency graph of the human ear.

User Uzluisf
by
4.9k points
4 votes

Answer:

f = 3.1 kHz

Step-by-step explanation:

given,

length of human canal =2.8 cm = 0.028 m

speed of sound = 343 m/s

fundamental frequency = ?

The fundamental frequency of a tube with one open end and one closed end is,


f = (v)/(4L)


f = (343)/(4* 0.028)


f = (343)/(0.112)

f = 3062.5 Hz

f = 3.1 kHz

hence, the fundamental frequency is equal to f = 3.1 kHz

User Rav
by
6.1k points