At a rock concert, the sound intensity 1.0 in front of the bank of loudspeakers is 0.10 . A fan is 30 from the loudspeakers. Her eardrums have a diameter of 8.4 .How much energy is transferred to each eardrum in one second?
2 answers
The answer is in joules but I have no idea how to get there using the sound intensity formulas :(
The sound travels in all directions .. so it's power is spread out over the surface of a sphere.
The radius of this sphere increases as it travels, distributing the power (P) over a greater and greater area (A) that reduces it's intensity (I) in W/m²
Intensity, I = P/A .. for a given source power the intensity I ∝ 1/A ∝ 1/d²
I ∝ 1/d² ..
I1 / I2 = (d2)² / (d1)²
Intensity at 30m .. I2 = I1 x (d1)² / (d2)² ..
I2 = 0.10W/m² x (1.0m)² / (30.0m)² .. .. I2 = 1.10^-4 (J/s)/m² .. W/m²
Energy collected by ear drum each sec .. E = I2 x Ear drum area
E = 1.10^-4 (J/s)/m² x (π.[4.20^-3m]²) .. .. .. ►E = 6.10^-9 J
The radius of this sphere increases as it travels, distributing the power (P) over a greater and greater area (A) that reduces it's intensity (I) in W/m²
Intensity, I = P/A .. for a given source power the intensity I ∝ 1/A ∝ 1/d²
I ∝ 1/d² ..
I1 / I2 = (d2)² / (d1)²
Intensity at 30m .. I2 = I1 x (d1)² / (d2)² ..
I2 = 0.10W/m² x (1.0m)² / (30.0m)² .. .. I2 = 1.10^-4 (J/s)/m² .. W/m²
Energy collected by ear drum each sec .. E = I2 x Ear drum area
E = 1.10^-4 (J/s)/m² x (π.[4.20^-3m]²) .. .. .. ►E = 6.10^-9 J
1 answer