At a rock concert, a dB meter registered 134 dB when placed 2.2 m in front of a

Question

At a rock concert, a dB meter registered 134 dB when placed 2.2 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? Express your answer to two significant figures and include the appropriate units. B. How far away would the sound level be 84 dB ? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

a)

power = intensity x area

P = IA

But intensity level beta = 10 dB*log(I/Io)

I = 1 x 10^-12 W/m^2

134 = 10dB*log(I/10^-12)

13.4 = log(I/10^-12)

I/10^-12 = = 10^(13.4)

I/10^-12 = 2.511 x 10^13

I = 25.11 W/m^2

P = IA = I*4pi*r^2

P = 25.11 x W/m^2 x 4 x pi x 2.2^2

P = 1527.22 W

B)

for beta = 84 dB

84 dB = 10 dB*log(I/Io)

I/Io = 10^(8.4)

I/Io = 2.512 x 10^8

I = 2.512 x 10^-4 W/m^2

Surface area = power output / intensity

Surface area = 1527.22 / (2.512 x 10^-4)

Surface area = 6079972.33

r = sqrt(6079972.33 / (4*pi))

r = 695.578 m

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