Question

For a person with normal hearing, the faintest sound that can be at a frequency of 400 Hz has pressure amplitude of about 6.0 xx 10^(-5) Pa . Calculate the corresponding intensity in W//m^(2) . Take speed of sound in air as 344 m//s and density of air 1.2 kg//m^(3) .

Answer 1

The corresponding intensity is:

• For a person with normal hearing, the faintest sound has

               Frequency, f = 400 Hz

               Pressure Amplitude, Δp = 0.00006 Pa

               Speed of sound in air, v = 344 m/s

               Density of air, ρ= 1.2 $kg/m^3$

• Intensity, I = $\frac{v * dp^2}{2B}$

• But B = ρ$v^2$

• I = $\frac{dp^2}{2 * density * v} = \frac{0.00006^2}{2 * 1.2 * 344} = 4.4 * 10^{-12} W/m^2$

• Hence, corresponding intensity  $= 4.4 * 10^{-12} W/m^2$