Question
The sound level produced by one singer is
71.4 dB.
What would be the sound level produced
by a chorus of 21 such singers (all singing at
the same intensity at approximately the same
distance as the original singer)?
Answer in units of dB.
71.4 dB.
What would be the sound level produced
by a chorus of 21 such singers (all singing at
the same intensity at approximately the same
distance as the original singer)?
Answer in units of dB.
Answers
GPT-4o mini
Sound levels in decibels (dB) are measured on a logarithmic scale. This means that when you combine sound levels, you can't simply add the dB values together. Instead, you need to convert the levels back to intensity, combine them, and then convert back to dB.
The sound intensity \( I \) in relation to the sound level \( L \) in dB is given by the formula:
\[
L = 10 \log_{10}\left(\frac{I}{I_0}\right)
\]
where \( I_0 \) is the reference intensity, typically \( 10^{-12} \, \text{W/m}^2 \).
To find the new sound level produced by 21 singers, follow these steps:
1. **Convert the single singer's level to intensity.**
Given:
\[
L = 71.4 \, \text{dB}
\]
We can find the intensity \( I \):
\[
I = I_0 \cdot 10^{L/10}
\]
\[
I = 10^{-12} \cdot 10^{71.4/10} = 10^{-12} \cdot 10^{7.14} = 10^{-12 + 7.14} = 10^{-4.86} \, \text{W/m}^2
\]
2. **Multiply the intensity by the number of singers.**
If one singer produces intensity \( I \), then 21 singers produce:
\[
I_{total} = 21 \cdot I = 21 \cdot 10^{-4.86} \, \text{W/m}^2
\]
3. **Convert the total intensity back to dB.**
Now we find the sound level \( L_{total} \):
\[
L_{total} = 10 \log_{10}\left(\frac{I_{total}}{I_0}\right)
\]
Since \( I_{total} = 21 \cdot 10^{-4.86} \), we can calculate:
\[
L_{total} = 10 \log_{10}\left(\frac{21 \cdot 10^{-4.86}}{10^{-12}}\right)
\]
\[
L_{total} = 10 \log_{10}(21) + 10 \log_{10}(10^{-4.86}) + 10 \log_{10}(10^{12})
\]
\[
L_{total} = 10 \log_{10}(21) - 48.6 + 120
\]
First, we calculate \( 10 \log_{10}(21) \):
\[
\log_{10}(21) \approx 1.322
\]
Thus,
\[
10 \log_{10}(21) \approx 13.22
\]
Plugging in this value:
\[
L_{total} \approx 13.22 - 48.6 + 120
\]
\[
L_{total} \approx 84.62 \, \text{dB}
\]
Thus, the sound level produced by a chorus of 21 singers would be approximately **84.6 dB**.
The sound intensity \( I \) in relation to the sound level \( L \) in dB is given by the formula:
\[
L = 10 \log_{10}\left(\frac{I}{I_0}\right)
\]
where \( I_0 \) is the reference intensity, typically \( 10^{-12} \, \text{W/m}^2 \).
To find the new sound level produced by 21 singers, follow these steps:
1. **Convert the single singer's level to intensity.**
Given:
\[
L = 71.4 \, \text{dB}
\]
We can find the intensity \( I \):
\[
I = I_0 \cdot 10^{L/10}
\]
\[
I = 10^{-12} \cdot 10^{71.4/10} = 10^{-12} \cdot 10^{7.14} = 10^{-12 + 7.14} = 10^{-4.86} \, \text{W/m}^2
\]
2. **Multiply the intensity by the number of singers.**
If one singer produces intensity \( I \), then 21 singers produce:
\[
I_{total} = 21 \cdot I = 21 \cdot 10^{-4.86} \, \text{W/m}^2
\]
3. **Convert the total intensity back to dB.**
Now we find the sound level \( L_{total} \):
\[
L_{total} = 10 \log_{10}\left(\frac{I_{total}}{I_0}\right)
\]
Since \( I_{total} = 21 \cdot 10^{-4.86} \), we can calculate:
\[
L_{total} = 10 \log_{10}\left(\frac{21 \cdot 10^{-4.86}}{10^{-12}}\right)
\]
\[
L_{total} = 10 \log_{10}(21) + 10 \log_{10}(10^{-4.86}) + 10 \log_{10}(10^{12})
\]
\[
L_{total} = 10 \log_{10}(21) - 48.6 + 120
\]
First, we calculate \( 10 \log_{10}(21) \):
\[
\log_{10}(21) \approx 1.322
\]
Thus,
\[
10 \log_{10}(21) \approx 13.22
\]
Plugging in this value:
\[
L_{total} \approx 13.22 - 48.6 + 120
\]
\[
L_{total} \approx 84.62 \, \text{dB}
\]
Thus, the sound level produced by a chorus of 21 singers would be approximately **84.6 dB**.