A train sounds its horn as it approaches an intersection. The horn can just be heard at a level of 42 dB by an observer 10 km away.
(a) What is the average power generated by the horn?
W
(b) What intensity level of the horn's sound is observed by someone waiting at an intersection 53 m from the train? (Treat the horn as a point source and neglect any absorption of sound by the air.)
2dB
To solve this problem, we will use the formulas related to sound power, intensity, and the decibel scale.
(a) Let's find the power generated by the horn.
Step 1: Convert the sound level from decibels (dB) to intensity using the formula:
\(I = I_0 \times 10^{\left(\frac{\text{dB}}{10}\right)}\)
Where:
- \(I\) is the sound intensity in watts per meter squared (W/m^2)
- \(I_0\) is the reference intensity, which is \(10^{-12}\) W/m^2 for sound in air.
Given that the sound level is 42 dB, we can calculate the sound intensity:
\(I = (10^{-12} \, \text{W/m^2}) \times 10^{\left(\frac{42}{10}\right)}\)
\(I \approx 1.2 \times 10^{-6} \, \text{W/m^2}\)
Step 2: Calculate the average power generated by the horn using the formula:
\(P = I \times A\)
Where:
- \(P\) is the average power in watts (W)
- \(A\) is the area over which the sound is spreading.
As the sound spreads spherically from the horn, the area can be calculated using the formula:
\(A = 4 \pi r^2\)
Given that the observer is 10 km away from the horn (which is 10,000 m), we can calculate the power:
\(P = (1.2 \times 10^{-6} \, \text{W/m^2}) \times (4 \pi (10,000)^2)\)
\(P \approx 1.52 \times 10^8 \, \text{W}\)
So, the average power generated by the horn is approximately 152 million watts (W).
(b) We want to find the intensity level observed by someone waiting at an intersection 53 m from the train.
First, let's find the sound intensity at that distance using the inverse square law. The inverse square law states that the intensity of sound decreases inversely with the square of the distance from the source.
\(I_2 = \left(\frac{r_1}{r_2}\right)^2 \times I_1\)
Where:
- \(I_1\) is the initial sound intensity at a distance \(r_1\) from the source
- \(I_2\) is the sound intensity at a distance \(r_2\) from the source
Given that the initial sound intensity is \(I \approx 1.2 \times 10^{-6} \, \text{W/m^2}\) (from part a) and the distance from the source to the intersection is 10 km + 53 m = 10,053 m:
\(I_2 = \left(\frac{10,000^2}{10,053^2}\right) \times 1.2 \times 10^{-6}\)
\(I_2 \approx 1.198 \times 10^{-6} \, \text{W/m^2}\)
Next, we can convert the sound intensity back to the decibel (dB) scale using the formula:
\(\text{dB} = 10 \times \log_{10}\left(\frac{I}{I_0}\right)\)
Given the sound intensity \(I_2\), we can calculate the sound level observed at the intersection:
\(\text{dB} = 10 \times \log_{10}\left(\frac{1.198 \times 10^{-6}}{10^{-12}}\right)\)
\(\text{dB} \approx 112.75 \, \text{dB}\)
So, the intensity level of the horn's sound observed by someone waiting at the intersection 53 m from the train is approximately 112.75 dB (which can be rounded to 113 dB).