Question

A rock group is playing in a bar. Sound
emerging from the door spreads uniformly in
all directions. The intensity level of the music
is 66.7 dB at a distance of 5.96 m from the
door.
At what distance is the music just barely
audible to a person with a normal threshold
of hearing? Disregard absorption.
Answer in units of m.

Answer 1

Answer:

5292.64 m

Explanation:

$dB \rightarrow \textrm{sound level}\\ I \rightarrow \textrm{sound intensity}\\ I_0 \rightarrow \textrm{threshold sound intensity}\\ x \rightarrow \textrm{distance of corresponding to threshold intensity of hearing}$

Taking threshold intensity as $1\times 10^{-12} W/m^{2}$ and since it's a constant then sound intensity for 66.7 dB will be

$66.7\;\rm dB = 10\; log_{10}\;\left(\dfrac{I}{I_0} \right)\\ \dfrac{I}{10^{-12}} = 10^{6.67}\\ I = 4.67735\times 10^{-6}\;\rm W/m^2\\ \boxed{I \approx 4.7\times 10^{-6}\;\rm W/m^2}$

Also, since sound intensity is inversely proportional to the square of the distance of the source then the distance can be given by

$\dfrac{I_1}{I_2} = \dfrac{r^2_2}{r^2_1}\\ \dfrac{4.7\times 10^{-6}\;\rm W/m^2}{10^{-12}\;\rm W/m^2} = \dfrac{x^2\;\rm m}{(5.96\;\rm m)^2}\\ x =\sqrt{28012000}\\ \boxed{x \approx 5292.64;\rm m}$

Answer 2

At "5292.64 m" the music just barely audible to a person.

Distance and Sound:

According to the question,

Sound level = 66.7 dB

Threshold intensity = 1 × 10⁻¹² W/m²

Distance = 5.96 m

We know that,

→ 66.7 dB = 10 $log_{10}$ ($\frac{I}{I_0}$)

        $\frac{I}{10^{-12}}$ = $10^{6.67}$

              I = 4.67735 × 10⁻⁶ W/m² or,

                = 4.7 × 10⁻⁶ W/m²

hence,

The distance will be:

→        $\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

By substituting the values, we get

$\frac{4.7\times 10^{-6}}{10^{-12}} = \frac{x^2}{(5.96)^2}$

           x = √28012000

              = 5292.64 m

Thus the answer above is correct.

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