Question

The sound intensity level 5.0 m from a large power saw is 100 dB. At what distance will the sound be a more tolerable 80 dB?

Short answer

The sound will be more tolerable at$50m$.

Step-by-step solution

Given information

The sound intensity level from a large power saw is $100dB$.

The distance is$5.0m$

The distance at which sound is more tolerable

The intensity of the wave is given by $I=\frac{P}{a}=\frac{P}{4{\mathrm{\pi r}}^2}$.

Here, $P=I_1{a}_1$and $P=I_2{a}_2$.

Now,

$$\begin{gathered} r_2=\sqrt{\frac{a_2}{4\mathrm{\pi }}} \\ =\sqrt{\frac{{\displaystyle \frac{P}{I_2}}}{4\mathrm{\pi }}} \\ =\sqrt{\frac{{\displaystyle \frac{I_1{a}_1}{I_2}}}{4\mathrm{\pi }}} \\ =\sqrt{\frac{{\displaystyle \frac{I_1}{I_2}}\left(4{\mathrm{\pi r}}_1^2\right)}{4\mathrm{\pi }}} \\ =r_1\sqrt{\frac{I_1}{I_2}} \end{gathered}$$

Substitution

Substitute given values into the obtained equation

$$\begin{gathered} r_2=5.0\sqrt{100} \\ =5\times 10 \\ =50m \end{gathered}$$

Therefore, the sound will be more tolerable at$50m$.