The faintest sound the human ear can detect at a frequency of 1000 Hz correspond to an intensity of about `1.00xx10^-12(W)/(m^2)`, which is called threshold of hearing. The loudest sounds the ear can tolerate at this frequency correspond to an intensity of about `1.00(W)/(m^2)`, the threshold of pain. Detemine the pressure amplitude and displacement amplitude associated with these two limits. Take speed of sound`=342(m)/(s)` and density of air`=1.20(kg)/(m^3)`

Explanation

Think about the quietest environment you have ever experienced. It is likely that the intensity of sound in even this quietest environment is higher than the threshold of hearing. Because we are given intesities and asked to calculate pressure and displacement amplitudes, this problem requires the concepts discussed in this section.
To find the pressure amplitude at the threshold of hearing,
`trinangleP_(max)=sqrt(2rhovI)`
`=sqrt(2(1.20(kg)/(m^3))(342(m)/(s))(1.00xx10^(-12)(W)/(m^2))`
`=2.87xx10^-5(W)/(m^2)`
Calculate the corresponding displacement amplitude using
`S_(max)=(triangleP_(max))/(rhovomega)=(2.87xx10^-5(N)/(m^2))/((1.20(kg)/(m^3))(342(m)/(s))(2pixx1000Hz))`
`=1.11xx10^(-11)m`
In a similar manner, one finds that the loudest sounds the human ear can tolerate correspond to a pressure amplitude of `28.7(N)/(m^2)` and a displacement amplitude equal to `1.11xx10^-5m`.
Because atmospheric pressure is about `10^5(N)/(m^2)`, the result for the pressure amplitude tells us that the ear is sensitive to pressure fluctuations as small as 3 parts in `10^(10)`. The displacement amplitude is also a remarkably small number. If we compare this result for `S_(max)` to the size of an atom (about `10^(-10)`m), we see that the ear is an extremely sensitive detector of sound waves.