# physics

posted by
**micole**
.

At a recent rock concert, a dB meter registered 112 dB when placed 1.70 m in front of a loudspeaker on the stage.

(a) What was the power output of the speaker, assuming a hemispherical radiation of the sound and neglecting absorption in the air?

(b) How far away would the intensity level be a somewhat reasonable 85 dB?

I used the equation I = P/(4pi r^2); Rearranging the equation I did: 112/4pi 1.7^2 = 4067.482 W, but this was incorrect. I also converted 112 dB to 11.2 bels, but also wrong. Any suggestions?

(a) Convert 112 dB to the equivalent power per unit area. Then multiply that by the hemisphere area 2 pi R^2, with R = 1.70 m.

(b) Ypur answer weas incorrect because they asked for a distance and you gave them a power number.

Dropping from 112 dB to 85 dB is a 27 dB reduction in power per area. That corresponds to a 10^(27/10) = 501 reduction factor in power per area. Using the inverse square law, that can be achieved with a sqrt(501) = 22.4 increase in distance. That would put the listener 1.70 x 22.4 = 38 meters away