At a rock concert, a dB meter registered 133 dBwhen placed 2.4 m in front of a l

Question

At a rock concert, a dB meter registered 133 dBwhen placed 2.4 m in front of a loudspeaker on the stage.

Part A

What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air?

Part B

How far away would the sound level be a somewhat reasonable 94 dB ?

At a rock concert, a dB meter registered 128 dBwhen placed 2.9 m in front of a loudspeaker on the stage.

Part A

What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air?

Part B

How far away would the sound level be a somewhat reasonable 88 dB ?

Explanation / Answer

here,

part A)

If the dB are sound-intensity dB (dB-SIL), then the reference level is I0 = 10^-12 W/m^2

133 = 10*log(I/I0) = 10*[log(I) + 12]

13.3 = log(I) + 12

log(I) = 1.3, I = 19.95 W/m^2

This is covering an area of 4*pi*(2.4^2),

so the total power is 19.95*4*pi*(2.4^2) W.

Power , P = 3613.66 W

Part B)

94 dB = 10*[log(I) + 12]

log(I) = - 2.6

I = 2.5 * 10^-3 W/m^2

The distance is such that the area covered by 3613.66 W yields 10^-3 W/m^2:

3613/(4*pi*r^2) = 2.5 * 10^-3

r = 214.56 m

the distance of the sound level is 214. 56 m

-------------------------

part A)

If the dB are sound-intensity dB (dB-SIL), then the reference level is I0 = 10^-12 W/m^2

128 = 10*log(I/I0) = 10*[log(I) + 12]

12.8 = log(I) + 12

log(I) = 0.8, I = 6.31 W/m^2

This is covering an area of 4*pi*(2.9^2),

so the total power is 6.31*4*pi*(2.9^2) W.

Power , P = 666.52 W

Part B)

88 dB = 10*[log(I) + 12]

log(I) = - 3.2

I = 0.63 * 10^-3 W/m^2

The distance is such that the area covered by 666.52 W yields 10^-3 W/m^2:

666.52/(4*pi*r^2) = 0.63 * 10^-3

r = 290.23 m

the distance of the sound level is 290.23 m

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