How we can hear the sound and task of the ear.

We consider the ear as presser transducer because the ear can convert the energy of sound waves into electrical energy. You can learn here how does the ear work with sounds waves. The ear is divided into three part as the following.

  1. Outer ear
  2. Middle ear
  3. Inner ear

The cross-section of the ear is shown in this fig. The outer ear consists of an external canal. It is open to the atmosphere at one end and terminates at the eardrum at the other end. The auditory canal is 25.0 mm long. The area of the eardrum is 80 mm2. The auditory canal is equivalent to an organ pipe that one end is closed. The human ear is most sensitive to sounds of frequencies around 3000 Hz. The minimum intensity of sound which the ear can detect is 10-12 Wm-2. The sound intensity level of 160 dB may rupture the eardrum. The intensity (I) of a sound wave, expressed in terms of its pressure amplitude (Pm), is given by the following equation. \Huge I=\frac{P_{m}^{2}}{2v\rho}

I = the intensity of  sound wave,  ρ = the density of air, v =the speed of sound

The important parts of the middle ear are three small linked bones. They are the hammer, the anvil, and the stirrup. We’ve named like this because of their respective shapes. These tree bones function is equivalent to a lever system. Its one arm, hammer, is coupled to the eardrum. Its other arm, stirrup, is coupled to the oval window (area=4 mm2 ) of the inner ear. A schematic representation of the lever, piston, and action of the middle ear are shown in this fig. The inner ear consists of a small spiral. The spiral-shaped tube called cochlea, filled with a fluid. In this figure, you can see the cochlea is straightened form.

The cochlea is divided lengthwise into three canals. These three canals are separated from each other by membranes.

As the pressure wave passes along the first canal it causes transverse displacements of the basilar membrane. The basilar membrane separates the second canal from the third canal. It has been found that the basilar membrane is made of thousands of parallel fibers which run across it.

The fibers of the basilar membrane towards the base of the cochlea are short and stuff. They vibrate very rapidly. And they sensitive to high notes. The fibers of the basilar membrane towards the apexes of the cochlea are long and more flexible. That so they vibrate more slowly. And also they are sensitive to low notes. This is how the inner eare resolves frequencies. This is the task of the inner ear that resolving the frequencies from the source.

Calculate the fundamental resonant frequency of the ear.

We consider the auditory canal to be an organ pipe closed at one end. Lets we assume the speed of sound in air is
330 ms-1 .

v =  330 ms-1  ,
λ = 4×(length of auditory canal)= 4×2.5×10-2 m,
f = v/λ
Therefore f = 3300 Hz

Note – This value depends on the other value of variables. Therefore fundamental resonant frequency is not constant. It will be around the 3300 Hz. Some readings above are not constant they will be changed according to the man or women. example – the length of the auditory canal, oval window area. However, you can get an approximate value.

The pressure variation of the standing wave is maximum at the eardrum when the auditory canal resonates because ( A displacement node at a closed end) corresponds to a pressure antinode.

Determine the pressure amplitude of the sound waves with the minimum intensity of sound.

The density of air is 1.25 kg m-3 , we can find it using the following equation.

 \Huge P_m=\sqrt{2v\rho I}
after applying the values for v, ρ and I. then Pm =2.75×10-5 Pa. The force acting on the eardrum Fe=Pm.A = (2.75×10-5  )×(80×10-6 ) = 2.2×10-9  N. Considering the lever action, the force generated on the oval window Fo=2Fe=4.4×10-9  N

  • The sound intensity level that ruptures the eardrum is 160dB.
  • The sound intensity that ruptures the eardrum is  104  Wm-2.

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