Solving for Noise Level at a Different Distance

This problem involves calculating the change in sound level (noise level) as the distance from a point source changes. We can use the relationship between sound level, intensity, and distance.

Key Principles

• Sound intensity ($I$) from a point source follows the inverse square law: $I \propto \frac{1}{r^2}$, where $r$ is the distance from the source.
• Sound level ($L$) in decibels (dB) is related to intensity ($I$) by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$, where $I_0$ is a reference intensity.
• The difference in sound levels between two distances ($r_1$ and $r_2$) is given by: $L_1 - L_2 = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.
• Combining these, $L_1 - L_2 = 10 \log_{10} \left( \frac{r_2^2}{r_1^2} \right) = 20 \log_{10} \left( \frac{r_2}{r_1} \right)$.

Step-by-Step Calculation

1. Identify Given Values:
   • Initial noise level, $L_1 = 70$ dB
   • Initial distance, $r_1 = 20$ m
   • Final distance, $r_2 = 80$ m
   • Noise level at final distance, $L_2 = ?$
2. Apply the Formula:

   Use the formula relating sound level difference to distance:

   $L_1 - L_2 = 20 \log_{10} \left( \frac{r_2}{r_1} \right)$

3. Substitute Values:

   Substitute the known values into the formula:

   $70 \text{ dB} - L_2 = 20 \log_{10} \left( \frac{80 \text{ m}}{20 \text{ m}} \right)$

4. Simplify the Ratio:

   Calculate the ratio of the distances:

   $\frac{80}{20} = 4$

   The equation becomes:

   $70 \text{ dB} - L_2 = 20 \log_{10}(4)$

5. Calculate the Logarithm:

   Calculate the value of $\log_{10}(4)$. Note that $\log_{10}(4) \approx 0.602$.

   $70 \text{ dB} - L_2 = 20 \times 0.602$

   $70 \text{ dB} - L_2 \approx 12.04 \text{ dB}$

6. Solve for $L_2$:

   Rearrange the equation to find $L_2$:

   $L_2 = 70 \text{ dB} - 12.04 \text{ dB}$

   $L_2 \approx 57.96 \text{ dB}$

7. Final Answer:

   Rounding the result to the nearest whole number gives $58$ dB. Therefore, the noise level at 80 m is approximately 58 dB.