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sound and auditory mechanics

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impact loud speaker upon particle distribution in the air

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sound amplitude depends on variation of sound pressure modulation of atmospheric pressure = 100.000 Pascal

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p eff = (½ √2) p max hearing threshold = 0.00002 Pa SPL (dB) = 20×log (p/ p ref )

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reference 2.10 -5 Pa p ref, 0 dB SPL SPL = 20×log (p/p ref )

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when you double sound pressure, sound intensity increases with ?

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when you double sound pressure, sound intensity increases with 6 dB 20 log p 1 /p 2 = 20 log 2 = 20×0.3 = 6

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Question When two students talk non-synchroneously with a sound intensity of 60 dB SPL each, what do they produce together ?

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Answer When two students talk non-synchroneously with a sound intensity of 60 dB SPL each, they produce together 2 times more energy = 63 dB total when you double sound pressure, sound intensity increases with 6 dB 20 log p 1 /p 2 = 20 log 2 = 20×0.3 = 6 but energy increases with 3 dB 10 log e 1 /e 2 = 10 log 2 = 10×0.3 = 3

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Question When two students talk synchroneously with a sound intensity of 60 dB SPL each, what do they produce together ?

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Answer When two students talk synchroneously with a sound intensity of 60 dB SPL each, they produce together 66 dB total

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Question What is the total sound intensity in a room with - A radio 70 dB SPL -Two students speaking asynchronous, each 60 dBSPL - one plane flying over with 80 dB SPL perceived in the room

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Question What is the total sound intensity in a room with -A radio 70 dB SPL -Two students speaking asynchronous, each 60 dBSPL - one plane with 80 dB SPL First transfer into energies, then sum and transfer in dB again. Result = 80.1 dB

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sound intensity decreases with r²

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Normal hearing threshold 1000Hz 0 dB SPL Falling leaves 10 dB SPL Whispering 20 dB SPL Very soft talking in a room 40 dB SPL Normal speact 1at 1 m 60 dB SPL Loud conservation with shouting 80 dB SPL Pneumatic hammer 100 dB SPL Disco 110 dB SPL Very loud sound speaker 120 dB SPL Starting airplane at 20 m. 130 à 140 dB SPL Pain threshold 130 à 140 dB SPL

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resonance and impedance

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auditory canal = open pipe (1, 3, 5 enz.) x ¼ 27 mm = ¼

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(1, 3, 5 enz.) x ¼ 27 mm = ¼ = 108 mm f = 3100 Hz gehoorgang = open orgelpijp resonantiegebied = 2000- 5000 Hz

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transition air - liquid acoustic impedance Z = p / u (in Rayleigh like Ohm) p: pressure needed u: velocity impedance endolympfe 56000 impedance air = 410 factor 135: 97% reflection: therefore ossicles resonance including ossicular chain: 1000 Hz

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impedance 1.resistence: frequency independent transfer sound energy in heat 2.stifness= elasticity that decreases with frequency 3.inertia increases with frequency compliance (in ml) = 1 / impedance

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Hefboomwerking Middenoor 17x 1.3x 2x

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17x1.3x2=44.2 pressure gain 10 log(44.2 2 ) = 33 dB theoretical gain measured: ≤ 30 dB

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function inner ear - mechanical-electricial transition by the inner hair cells - frequency analysis by macromechanics of the basilar membrane - increasing sensitivity by micro-mechanics by the outer hair cells

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Cochlear model Ovale venster Ronde venster Helicotrema (verbinding Scala vestibuli en Scala tympani) http://www.iurc.montp.inserm.fr/cric/audition/start.htm

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- mechanical-electricial transition by the inner hair cells tip link – Hudspeth spring model

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- increasing sensitivity by micro-mechanics by the outer hair cells: the cochlear amplifier

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- deflecion towards kinociulim decreases receptor potential results in: mechanical deformation of the cortical lattice leading to a shortening in cell body length and increase in diameter moving the basilary membrane further away from the kinocilium

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Mechanics outer hair cells + membrane: second resonator and cochlear amplifier Animation : http://cc.usu.edu/~dgsinex/courses/SHS311_notes/2-ear/corti.htmhttp://cc.usu.edu/~dgsinex/courses/SHS311_notes/2-ear/corti.htm Efferent innervation: function selective hearing

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Micro mechanics adds energy to the tranverse wave

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Hearing. How hearing works First, the ear gathers the compressional waves. Next, the ear amplifies the waves. In the ear, the amplified waves are converted.

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