Questions: What is the sound intensity level of a rock concert with intensity of 0.7 W / m^2?

Solution Steps

We need to find the sound intensity level (in decibels) of a rock concert given its intensity, \( I = 0.7 \, \mathrm{W/m^2} \).

The sound intensity level \( L \) in decibels (dB) is given by the formula: \[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \] where \( I_0 \) is the reference intensity, typically \( I_0 = 1 \times 10^{-12} \, \mathrm{W/m^2} \).

Substitute \( I = 0.7 \, \mathrm{W/m^2} \) and \( I_0 = 1 \times 10^{-12} \, \mathrm{W/m^2} \) into the formula: \[ L = 10 \log_{10} \left( \frac{0.7}{1 \times 10^{-12}} \right) \]

Calculate the ratio inside the logarithm: \[ \frac{0.7}{1 \times 10^{-12}} = 0.7 \times 10^{12} \]

Now, compute the logarithm: \[ \log_{10} (0.7 \times 10^{12}) = \log_{10} (0.7) + \log_{10} (10^{12}) \] \[ \log_{10} (0.7) \approx -0.1549 \quad \text{and} \quad \log_{10} (10^{12}) = 12 \] \[ \log_{10} (0.7 \times 10^{12}) = -0.1549 + 12 = 11.8451 \]

Multiply by 10 to get the sound intensity level: \[ L = 10 \times 11.8451 = 118.451 \]

Final Answer