Example 53 The acoustic pressure at hearing threshold is 20

Example 5-3: The acoustic pressure at hearing threshold is 2.0 10^-5Pa

a)Assuming that the surface of the eardrum is 0.60 cm², what is the corresponding force?

b)We consider that this force is transmitted integrally to the oval window. The surface of the oval window is 0.03 cm². what is the corresponding pressure?

c)The pressure at the eardrum is amplified further as a result of the lever action of the ossicles. Calculate the pressure at the oval window

d)Calculate the amplification realized by the middle ear in dB

Solution

a.

the force is calculated as follows:

    F = PA = 2.0X10^-5 Pa * 0.60 cm² = 2.0X10^-5 Pa * 0.60X10^-4 m2 = 1.2e-9 N

b.

the pressure is calculated as follows:

   P = F/A = 1.2e-9 N / 0.03X10^-4 m² = 4.e-4 Pa

C.

the pressure is,

   dP = 4.e-4 Pa - 2.0X10^-5 Pa = 3.8e-4 Pa

d.

B = (20)log[P/P0] = 20 log[ 4.e-4 / 2.e-5] = 26 dB