Question

Q1.If the sound 10 meters from the stage at a rock concert is at the 120dB what is the intensity level 100 meters away

Answer 1

QUESTION:

Q1. If the sound $r_1 = 10$ meters from the stage at a rock concert is at the $L_1 = 120\,\mathrm{dB}$ what is the intensity level $L_2$ $r_2 = 100$ meters away

SOLUTION:

Intensity level 10 meters from the stage is

$$L_1 = 10 \lg \frac{I_1}{I_0}$$ $$\frac{L_1}{10} = \lg \frac{I_1}{I_0}$$ $$10^{\frac{I_1}{10}} = \frac{I_1}{I_0}$$ $$I_1 = I_0 10^{\frac{I_1}{10}}$$

On the other hand sound intensity is sound power $P_{\mathrm{acc}}$ per unit area $A$:

$$I_1 = \frac{P_{\mathrm{acc}}}{4\pi r_1^2}$$ $$P_{\mathrm{acc}} = 4\pi r_1^2 I_1$$ $$P_{\mathrm{acc}} = 4\pi r_1^2 \cdot I_0 10^{\frac{I_1}{10}}$$

And

$$I_2 = \frac{P_{\mathrm{acc}}}{4\pi r_2^2}$$ $$I_2 = \frac{4\pi r_1^2 \cdot I_0 10^{\frac{I_2}{10}}}{4\pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2 \cdot I_0 10^{\frac{I_2}{10}}$$

Hence intensity level $100\,\mathrm{m}$ away is

$$L_2 = 10 \lg \frac{I_2}{I_0} = 10 \lg \left(\frac{\left(\frac{r_1}{r_2}\right)^2 \cdot I_0 10^{\frac{I_2}{10}}}{I_0}\right) = 10 \lg \left((\frac{r_1}{r_2})^2 10^{\frac{I_2}{10}}\right) = 10 (\lg 10^{\frac{I_2}{10}} + \lg \left(\frac{r_1}{r_2}\right)^2) = 10 \left(\frac{L_1}{10} - 2 \lg \frac{r_2}{r_1}\right) = L_1 - 20 \lg \frac{r_2}{r_1}$$ $$L_2 = 98.4\,\mathrm{dB}$$

ANSWER:

$$L_2 = 98.4\,\mathrm{dB}$$