Ph siolo forA lied Biomedical En ineerin ll 1. The figure below shows a schematic drawing of the surface area mismatch between the tympanic membrane and the oval window. in put fro m ear canal output to inner ear tympani: \"osscicles\" oval window membrane The pressure p1 acts over the surface of the tympanic membrane with surface area A1 resulting in the force F1. The cylinder represents the chain of ossicles, and can in this case be considered a fossils-so system. meaning no energy is lost in the transduction. F2 is the force acting on the oval window with surface area 24;, which results in pressure p; being transmitted to the inner ear. is} In part A of Module 'r'. we said that there was a gain in sound pressure produced from various mechanisms in the external and middle ears. The figure above describes one component that contributes to that gain. Derive the equation for p: in terms of incoming pressure to the tympanic membrane p1 and the ratio of surface areas of the tympanic membrane A1 and the oval window .42. [23 pts} {b} Sound intensity was defined in part a of Module ? as power per unit area. and can be expressed in terms of pressure and a quantity called charactenstri: acoustic impedance. represented as Z. Z in air is approximately 400 N $me This quantity is an analog to resistance in Dhm's Law. and describes the resistance of sound travel through a substance. They are related by: 2 rr = Pf who]? i' is in \"bits m2 p is in 17's {1] Z is in .\\' s rt:3 The system in the figure produces an increase of sound intensity arriving at the inner ear of 25 dB. Recall that: rise) = 1r: - iagmii\" {2) If the surface area of the tympanic membrane A1 is 55 mmz, find the surface area of the oval window A2. {so pts}