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)
[tex]I=\frac{P^2}{2pv}[/tex]
where
P=pressure
p=density of air
v=velocity
[tex]I=\frac{6*10^-5^2}{2*1.2*344} = 4.36*10^-12[/tex] 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
