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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 mm2 and the area of stirrup is about 3.2 mm2. 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 3×10−2 Pa from its normal equilibrium pressure value which is wqual to 105 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 1.42×105 Pa. Bulk modulus of fluid is 2.18×109 Pa.
Find the pressure amplitude of given sound wave in the fluid of inner ear.
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 mm2 and the area of stirrup is about 3.2 mm2. 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 3×10−2 Pa from its normal equilibrium pressure value which is wqual to 105 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 1.42×105 Pa. Bulk modulus of fluid is 2.18×109 Pa.
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 1.42×105 Pa. Bulk modulus of fluid is 2.18×109 Pa.
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Solution
The correct option is D 0.4 Pa
Given : Area of stirrup As=3.2mm2=3.2×10−6 m2 Area of eardrum A=43mm2=43×10−6m2
Pressure amplitude at eardrum Po=3×10−2 Pa
Force exerted by sound wave in air on eardrum
F=PoA=3×10−2×43×10−6=129×10−8N
According to question , force exerted on eardrum is equal to force exerted by ossicles on the inner ear.
Let pressure amplitude in the fluid of inner ear be p′o
∴ p′o=FAs=129×10−83.2×10−6=0.4 Pa
Given : Area of stirrup As=3.2mm2=3.2×10−6 m2
Area of eardrum A=43mm2=43×10−6m2
Pressure amplitude at eardrum Po=3×10−2 Pa
Force exerted by sound wave in air on eardrum
F=PoA=3×10−2×43×10−6=129×10−8N
According to question , force exerted on eardrum is equal to force exerted by ossicles on the inner ear.
Let pressure amplitude in the fluid of inner ear be p′o
∴ p′o=FAs=129×10−83.2×10−6=0.4 Pa
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