Question

At a rock concert, a dB meter registered 125 dB when placed 2.9 m in front of a loudspeaker on the stage. The intensity of the reference level required to determine the sound level is 1.0×10−12W/m2.

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 86 dB ?

Answer 1

Initial distance from the loudspeaker = R₁ = 2.9 m

Sound level of the loudspeaker at a distance of 2.9 m from it = ₁ = 125 dB

Sound intensity of the loudspeaker at a distance of 2.9 from it = I₁

I₁ = 3.162 W/m²

Power output of the speaker = P

P = I₁(4R₁²)

P = 4(3.162)(2.9)²

P = 334.2 W

New distance from the loud speaker = R₂

Sound level at this new distance = ₂ = 86 dB

Sound intensity at this new distance = I₂

I₂ = 3.981 x 10^-4 W/m²

P = I₂(4R₂²)

334.2 = 4(3.981x10^-4)R₂²

R₂ = 258.5 m

A) Power output of the speaker = 334.2 W

B) Distance at which the sound level would be 86 dB = 258.5 m