Question

For a person with normal hearing, the faintest sound that can be heard at a frequency of 400 Hz has a pressure amplitude of about 6.0 10 Pa. Calculate the (a) intensity; (b) sound intensity level; (c) displacement amplitude of this sound wave at 20C

Answer 1

Answer:

a)4.36*10^-12W/m^2

b)L=6.39dB

c)ξ=5.73×10⁻¹¹m

Explanation:

Take speed of sound in air as 344m/s and density of air 1.2kg/m3 .

a)

$I=\frac{P^2}{2pv}$

where

P=pressure

p=density of air

v=velocity

$I=\frac{6*10^-5^2}{2*1.2*344} = 4.36*10^-12$ W/m^2

b)

Sound intensity level in dB is defined as:  

L = 10∙log₁₀(I/I₀)  

with  

I₀ = 1.0×10⁻¹² W/m²  

Hence;  

L = 10∙log₁₀( 4.36x10^-12 W/m² / 1.0×10⁻¹² W/m²) = 6.39dB

c)

Displacement is given by :  

ξ = p/(Z∙ω) = p/(Z∙2∙π∙f)  

where  

Z = 416.9 N∙s/m³ = 416.9 Pa∙s/m

f frequency and ω angular frequency of the sound wave.  

So the amplitude of this sound wave is:  

ξ = 6×10⁻⁵Pa / (416.9 Pa∙s/m ∙ 2∙π∙ 400s⁻¹) = 5.73×10⁻¹¹m