For a person with normal hearing the faintest sound that can be heard at a frequency of 400hz has a pressure amplitude of about atm. Calculate the corresponding intensity in w/. Take speed of sound in air as 344 m/s and density of air 1.2 kg/.
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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 multiply.gif 10-5 Pa. (At 20°C the bulk modulus for air is 1.42 multiply.gif 105 Pa and v = 344 m/s.)
(a) Calculate the intensity of this sound wave at 20°C.
W/m2
(b) Calculate the sound intensity level of this sound wave at 20°C.
dB
(c) Calculate the displacement amplitude of this sound wave at 20°C.
m
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a)
The sound pressure p (=pressure amplitude of the sound wave) and sound intensity I are related as [1]:
I = p² / Z
where Z is the accoustic impedance of the medium, in which the sound waves propagate.
The accoustic Impedance of air at 20°C is [2]:
Z = 416.9 N∙s/m³ = 416.9 Pa∙s/m
Hence,
I = (6×10⁻⁵Pa)² / 416.9 Pa∙s/m = 8.64×10⁻¹² Pa∙m/s = 8.64×10⁻¹² W/m²
b)
Sound intensity level in dB is defined as [1]:
L = 10∙log₁₀(I/I₀)
with
I₀ = 1.0×10⁻¹² W/m²
Hence;
L = 10∙log₁₀( 8.64×10⁻¹² W/m² / 1.0×10⁻¹² W/m²) = 9.4dB
c)
Displacement is given by [3]:
ξ = p/(Z∙ω) = p/(Z∙2∙π∙f)
where
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