ANSWER:100dB
from the sound intensity level,the sound intensity is calculated as:
[tex]\beta[/tex]₁=[tex]\beta[/tex]=(10dB)㏒₁₀(l₁/l₀)
inserting numbers:
120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²
Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which
can be used to solve for l₁ and get
l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²
since the sound intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in
ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:
l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²
since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:
[tex]\beta[/tex]₂=[tex]\beta[/tex]=(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)
[tex]\beta[/tex]₂=(10dB)㏒₁₀(2.04×10¹⁰)
[tex]\beta[/tex] =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB
