Answer:
a) The pressure amplitude at this distance is [tex]\Delta P_{max}=5.8 \times 10^{3} N/m^{2}[/tex]
b) The displacement amplitude is [tex]s_{max}=1.74 \times 10^{-4} m[/tex]
c) The distance at which the sound intensity level drops to 50 dB is [tex]r=89,200Km[/tex]
Explanation:
a) We need to convert to dB to watts, so [tex]P_{dB}=10\times log_{10}(P_{W})[/tex], then [tex]P_{W}=10^{\frac{P_{dB}}{10}= 10^{\frac{100}{10}} = 10^{10} W[/tex].
Then we can find Intensity at 100m, so [tex]I=\frac{P_{pro}}{4 \pi r^{2} }= \frac{10^{10}W}{4 \pi (100m)^{2} }=75.6 \times 10^{3} W/m^{2}[/tex].
And finally, the pressure amplitude at this distance will be [tex]\Delta P_{max}=\sqrt{\rho vI}=\sqrt{(1.29kg/m^{3})(343m/s)(75.6 \times 10^{3} W/m^{2})} =5.8 \times 10^{3} N/m^{2}[/tex]
b) Using [tex]s_{max}=\frac{\Delta P_{max}}{\rho \omega v}[/tex], with [tex]\omega=2 \pi f=2 \pi (12000s^{-1})=\pi 24 \times 10^{3} s^{-1}[/tex], thus,
[tex]s_{max}=\frac{\Delta P_{max}}{\rho \omega v}= \frac{5.8 \times 10^{3} N/m^{2}}{(1.29kg/m^{3}) (\pi 24 \times 10^{3} s^{-1}) (343m/s) }= 1.74 \times 10^{-4} m[/tex]
c) We need to find intensity, so [tex]10 log_{10} (\frac{I}{I_{0}}) = 50[/tex], then,
[tex]I=10^{5} I_{0}=10^{5} \times 10^{-12} W/m^{2}=1.00 \times 10^{-7} W/m^{2}[/tex].
Finally, the distance at which the sound intensity level drops to 50 dB will be
[tex]r=\sqrt{\frac{P_{pro}}{4 \pi I}} =\sqrt{\frac{10^{10}W}{4 \pi (1.00 \times 10^{-7} W/m^{2})}}=89.2 \times 10^{6}m=89,200Km[/tex]
