The eardrum is displaced surprisingly little when a sound wave strikes it. Some

Question

The eardrum is displaced surprisingly little when a sound wave strikes it. Some idea of the displacement of the eardrum by a sound wave can be obtained by using Equation 7.1: I = 2 pi^2 and rho v^2 A^2 The displacement amplitude, A, of the air molecules gives us a lower limit to the displacement of the eardrum. At the threshold of hearing at 3.8 kHz, the sound intensity is about 10^-12 W/m^2. The threshold of pain begins at an intensity of 1.0 W/m^2 f. What is the amplitude of eardrum vibrations at these intensities? The density of air is 1.3kg/m^3. 10^-3nm (at threshold of hearing) (This value is ~ 3/100 the diameter of a hydrogen atom!) 10^3nm (at threshold of pain)

Explanation / Answer

when intensity is 10^(-12) W/m^2

10^(-12)=2*pi^2*343*1.3*(3.8*10^3)^2*A^2

==>A=sqrt(10^(-12)/(2*pi^2*343*1.3*(3.8*10^3)^2)

=2.805*10^(-12) m

=2.805*10^(-3) nm

part b:

at threshold of pain, intensity=1 W/m^2

1=2*pi^2*343*1.3*(3.8*10^3)^2*A^2

==>A=sqrt(1/(2*pi^2*343*1.3*(3.8*10^3)^2))

=2.805*10^(-6) m

=2.805*10^3*10^(-9) m

=2.805*10^3 nm

so in both the cases, the box need to be filled with 2.805.

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