When a sound wave enters the ear, it sets the eardrum into oscillation, which in turn causes oscillation of 3 tiny bones in the middle ear called ossicles. This oscillation is finally transmitted to the fluid filled in inner portion of the ear termed as inner ear, the motion of the fluid disturbs hair cells within the inner ear which transmit nerve impulses to the brain with the information that a sound is present. The theree bones present in the middle ear are named as hammer, anvil and stirrup. Out of these the stirrup is the smallest one and this only connects the middle ear to inner ear as shown in the figure below. The area of stirrup and its extent of connection with the inner ear limits the sensitivity of the human ear consider a person's ear whose moving part of the eardrum has an area of about `50mm^2` and the area of stirrup is about `5mm^2`. The mass of ossicles is negligible. As a result, force exerted by sound wave in air on eardum and ossicles is same as the force exerted by ossicles on the inner ear. Consider a sound wave having maximum pressure fluctuation of `4xx10^-2Pa` from its normal equilibrium pressure value which is equal to `10^5Pa`. Frequency of sound wave in air is `332(m)/(s)`. Velocity of sound wave in fluid (present in inner ear) is `1500(m)/(s)`. Bulk modulus of air is `1.42xx10^5Pa`. Bulk modulus of fluid is `2.18xx10^9Pa`.
Q. Find the pressure amplitude of given sound wave in the fluid of inner ear.

Options

A. `0.03Pa`
B. `0.04Pa`
C. `0.3Pa`
D. `0.4Pa`

Answer

D

Explanation

The force exerted on inner ear is same as that of the force exerted on eardrum, due to negligible mass of ossicles.
`P_(max)=(F_(max))/("area of stirrup")`
`(P_("max on eardrum")xxA_("eardrum"))/(A_("stirrup"))`
Pressure amplitude is given by
`triangleP_0=P_(max)-P_("normal value")`
`=((P_0+(triangleP_0)_(on "eardrum"))xxA_("eardrum"))/(A_("stirrup"))-(P_0xxA_("eardrum"))/(A_("stirrup"))`
`=((triangleP_0)_("on eardrum")xxA_("eardrum"))/(A_("stirrup"))`
`=(4xx10^-2xx50xx10^-6)/(5xx10^-6)=0.4Pa`