Question

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 calls within the inner ear which transmit nerve impulses to the brain with information that a sound is present. The three 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 eat whose moving part of the eardrum has an area of about 43 mm${}^2$ and the area of stirrup is about 3.2 mm${}^2$. The mass of ossicles is negligible. As a result, force exerted by sound wave in air on eardrum and ossicles is same as the force exerted by ossicles on the inner ear. Consider a sound wave having maximum pressure fluctuation of $3\times {10}^{-2}$ Pa from its normal equilibrium pressure value which is wqual to ${10}^5$ Pa. Frequency of sound wave is 1200 Hz.
Data: Velocity 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 $1.42\times {10}^5$ Pa. Bulk modulus of fluid is $2.18\times {10}^9$ Pa.

Find the displacement amplitude of given sound wave in the fluid of inner ear.

• A. 4.4$\times {10}^{11}$m
• B. 8 $\times {10}^{11}$m
• C. 3.65$\times {10}^{-11}$m
• D. 8.1$\times {10}^{-12}$m

Answer 1

Answer: (A), (D)

The correct options are
A 4.4$\times {10}^{11}$m
D 8.1$\times {10}^{-12}$m

The displacement amplitude in the fluid of inner ear is given by:

$A=\frac{P_{ma{x}_{innterear}}}{B_{fluid}{k}_{innerear}}$

where, $k_{innerear}$ is the compressibility of inner ear $=\frac{\omega }{v_{innerear}}=\frac{2\pi \left(1000\right)}{1500}=4.2rad/m$

$P_{ma{x}_{innerear}}=\frac{P_{ma{x}_{air}}{S}_{eardrum}}{S_{stirrup}}$$=\frac{(3\times {10}^{-2})\times 43}{3.2}=0.4Pa$

Hence, amplitude $A=\frac{0.4}{2.18\times {10}^9\times 4.2}$$=4.4\times {10}^{-11}m$