Question

The intesity of a wave is I=2π2ρvf2A2, where ρ is the density of the medium, v is the speed of wave, f is the frequency, and A is the amplitude. The density of air is 1.29 kg/m3, the speed of sound in air is 343 m/s, the intensity of the reference level required to determine the sound level is 1.0×10−12W/m2.

What would be the sound level (in dB ) of a sound wave in air that corresponds to a displacement amplitude of vibrating air molecules of 0.13 mm at 480 Hz ?

Answer 1

The given intensity of wave in the given question is  I=2π2ρvf2A2, and this looks a bit weird equation. I would thus interpert it as

I=0.5ρ[(2*pi*f)^2](A^2)v

From the given

density of air rho= 1.29 kg/m3

the speed of sound in air v= 343 m/s

f= 480Hz

A= 0.13mm

Putting the values in the equation of Intensity we get I=33.973 W/m^2

Now the equation for finding the sound level is

β=10*log(I/I_0) ...... And the value of I_0= 1.0×10−12W/m2. (Given). And  I=33.973 W/m^2 (obtained).

Hence we get  sound level as 135 dB..