Question

At a rock concert, a dB meter registered 131 dB when placed 2.6 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

Answer:

Explanation:

A) 131 dB = 10*log(I / 1e-12W/m²)

where I is the intensity at 2.6 m away.

13.1 = log(I / 1e-12W/m²

1.25e13= I / 1e-12W/m²

I = 1.25 x10^1W/m²

power = intensity * area

P = I * A = 12.5W/m² * 4π(2.6m)² =1061 W ◄

B) 86 dB = 10*log(I / 1e-12W/m²)

8.6 = log(I / 1e-12W/m²)

3.98e8 = I / 1e-12W/m²

I = 3.98e-4 W/m²

area A = P / I = 1061W / 3.98e-4W/m² = 2.66e6 m²

A = 4πr²

2.66e6 m² = 4πr²

r = 14.5m ◄