Question

30 seconds of exposure to 115 dB sound can damage your hearing, but a much quieter 94 dB may begin to cause damage after 1 hour of continuous exposure. You are going to an outdoor concert, and you'll be standing near a speaker that emits 50 W of acoustic power as a spherical wave. What minimum distance should you be from the speaker to keep the sound intensity level below 94 dB?

Answer 1

Answer:

39.8 m ≈ 40 m

Explanation:

power (P) = 50 W

sound intensity level ($p$) = 94 dB

the distance (r) can be gotten from the equation I = $\frac{power}{4nr^{2} }$ (take not that π is shown as $n$)

making r the subject of the formula we have r = $\sqrt{\frac{power}{4nI} }$   (take not that π is shown as $n$)

But to apply this equation we need to get the value of the intensity (I)

• we can get the intensity (I) from the formula sound intensity level ($p$) = 10 log₁₀$(\frac{I}{I'})$

• rearranging the above formula we have intensity (I) = $I' x 10^{\frac{p}{10} }$

• I' = reference intensity = 1 x$10^{-12} W/m^{2}$

• now substituting all required values into the formula for intensity (I)

• I = $1 x 10^{-12} x 10^{\frac{94}{10} }$ = 0.00251 $W/m^{2}$

now that we have the value of the intensity (I)  we can substitute it into the formula for the distance (r)

distance (r) = $\sqrt{\frac{power}{4nI} }$

r = $\sqrt{\frac{50}{4x3.142x0.00251} }$ = 39.8 m ≈ 40 m