Question

In technical normal state air the initial phase of a 310Hz frequency, 0.07Pa sound pressure amplitude free, plane wave traveling in +x direction is 60degree. Determine the sound pressure in the sound field 18minutes later, 175m far from the origin! Let calculate the sound pressure level as well, in the same place!

Answer 1

At normal state the speed of air is given as 343m/s. So wavelength for this frequency is given as = 343m/s/310Hz = 1.10645 m

now wavenumber k = 2/ = 5.67868 m^-1 Now given that at t= 0 ; x = 0 ; = 60^o

w = 2f = 620

So the wave P = P_osin(kx - wt +)

=> P = 0.07sin(kx - wt + 60^o)

So after 18minutes = 18 x 60 s = 1080s we have;

P = 0.07sin(kx - 620 x 1080s + /3)

Now for origin and x = 175m we have

P = 0.07sin(- 620 x 1080s + /3) = 0.06062 Pa

and

P = 0.07 sin(k x 175m - 620 x 1080s + /3) = 0.061356 Pa

Now sound pressure level at both places is given as

SPL|_{x =0} = 20log₁₀(P/P_o); P_o = 2 x 10^-5 Pa

=> SPL|_{x =0} = 20log₁₀(0.06062/2x10^-5) dB = 70 dB

SPL|_x=175m = 20log₁₀(0.061356/2x10^-5) dB = 70 dB