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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]

Respuesta :

Answer:

f = 3.1 kHz

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,

[tex]f = \dfrac{v}{4L}[/tex]

[tex]f = \dfrac{343}{4\times 0.028}[/tex]

[tex]f = \dfrac{343}{0.112}[/tex]

       f = 3062.5 Hz

       f = 3.1 kHz

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

okpalawalter8

The fundamental frequency is mathematically given as

f = 3.1 kHz

What is the fundamental frequency around which we would expect the hearing to be most sensitive?

Question Parameters:

The human ear canal is about 2.8 cm long

Generally, the equation for the fundamental frequency   is mathematically given as

f = v/4L

Therefore

[tex]f = \frac{343}{4*0.028}[/tex]

f = 3062.5 Hz

In conclusion, The frequency is

f = 3.1 kHz

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