Answer:
Part a)
[tex]I = 3.16 \times 10^{-5} W/m^2[/tex]
Part b)
[tex]P_o = 0.162 Pa[/tex]
Explanation:
Part a)
Level of sound = 75 dB
now we know that
[tex]L = 10 Log\frac{I}{I_0}[/tex]
here we know that
[tex]I_0 = 10^{-12} W/m^2[/tex]
now we have
[tex]75 = 10 Log(\frac{I}{10^{-12}})[/tex]
[tex]I = 3.16 \times 10^{-5} W/m^2[/tex]
Part b)
Intensity of sound wave is given as
[tex]I = \frac{1}{2}\rho A^2\omega^2 c[/tex]
here we know that
[tex]A = \frac{P_o}{Bk}[/tex]
so we have
[tex]I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c[/tex]
[tex]I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}[/tex]
now we know
[tex]\rho = 1.2 kg/m^3[/tex]
[tex]c = 340 m/s[/tex]
[tex]B = 1.4 \times 10^5 Pa[/tex]
now we have
[tex]3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}[/tex]
[tex]P_o = 0.162 Pa[/tex]
