Question

At a rock concert, a dB meter registered 130 dB when placed 2.5 m in front of a loudspeaker on 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 85 dB?

Answer 1

Answer:

785.398 W

444.5698 m

Explanation:

I = Intensity of sound

r = Distance

The intensity of sound is given by

$\beta=10log\frac{I}{I_0}\\\Rightarrow 130=10log\frac{I}{10^{-12}}\\\Rightarrow 13=log\frac{I}{10^{-12}}\\\Rightarrow 10^{13}=\frac{I}{10^{-12}}\\\Rightarrow I=10^{-12}\times 10^{13}\\\Rightarrow I=10\ W/m^2$

Power

$P=IA\\\Rightarrow P=I4\pi r^2\\\Rightarrow P=10\times 4\pi\times 2.5^2\\\Rightarrow P=785.398\ W$

The power output of the speaker is 785.398 W

If $\beta=85\ db$

$\beta=10log\frac{I}{I_0}\\\Rightarrow 85=10log\frac{I}{10^{-12}}\\\Rightarrow 8.5=log\frac{I}{10^{-12}}\\\Rightarrow 10^{8.5}=\frac{I}{10^{-12}}\\\Rightarrow I=10^{-12}\times 10^{8.5}\\\Rightarrow I=10^{-3.5}\ W/m^2$

$P=IA\\\Rightarrow P=I4\pi r^2\\\Rightarrow r=\sqrt{\frac{P}{I4\pi}}\\\Rightarrow r=\sqrt{\frac{785.398}{10^{-3.5}\times 4\pi}}\\\Rightarrow r=444.5698\ m$

The distance would be 444.5698 m