Question

At a rock concert, a dB meter registered 130 dB when placed 2.2 m in front of a loudspeaker on the stage.

a. What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air?

b. How far away would the sound level be a somewhat reasonable 95 dB ? answer with units

Answer 1

Initial distance from the loudspeaker = R₁ = 2.2 m

Sound intensity level of the loudspeaker at a distance of 2.2 m = ₁ = 130 dB

Sound intensity of the loudspeaker at a distance of 2.2 m = I₁

I₁ = 10 W/m²

Power output of the speaker = P

P = I₁(4R₁²)

P = 4I₁R₁²

P = 4(10)(2.2)²

P = 608.21 W

New distance from the loudspeaker so that sound level is 95 dB = R₂

Sound intensity level of the loudspeaker at the new distance = ₂ = 95 dB

Sound intensity of the loudspeaker at the new distance = I₂

I₂ = 3.162 x 10^-3 W/m²

P = I₂(4R₂²)

P = 4I₂R₂²

608.21 = 4(3.162x10^-3)R₂²

R₂ = 123.72 m

a) Power output of the speaker = 608.21 W

b) Distance from the loudspeaker so that the sound level is 95 dB = 123.72 m