Question

In: Physics

At a rock concert, a dB meter registered 130 dB when placed 2.2 m in front...

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

Solutions

Expert Solution

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


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