Questions: A rock band is holding a concert in the local park and using three 1000-Watt speakers (totaling 3000 Watts) to project the sound. You visit the concert and consider two seats. Seat A is an average distance of 20 meters from the speakers. Seat B is an average distance of 50 meters from the speakers. How does the decibel rating of the two seats compare when the speakers are blasting at 100% efficiency? The sound at Seat A is 2 dB higher than the sound at Seat B. The sound at Seat A is 2.5 dB higher than the sound at Seat B. The sound at Seat A is 3 dB higher than the sound at Seat B. The sound at Seat A is 8 dB higher than the sound at Seat B.

Solution Steps

• The sound intensity \( I \) at a distance \( r \) from a point source is given by \( I = \frac{P}{4 \pi r^2} \), where \( P \) is the power of the source.
• For Seat A (\( r_A = 20 \) meters): \[ I_A = \frac{3000}{4 \pi (20)^2} \]
• For Seat B (\( r_B = 50 \) meters): \[ I_B = \frac{3000}{4 \pi (50)^2} \]

• The ratio of the intensities \( \frac{I_A}{I_B} \) is: \[ \frac{I_A}{I_B} = \frac{\frac{3000}{4 \pi (20)^2}}{\frac{3000}{4 \pi (50)^2}} = \left( \frac{50}{20} \right)^2 = \left( \frac{5}{2} \right)^2 = \frac{25}{4} \]

• The difference in decibels \( \Delta L \) between two intensities is given by: \[ \Delta L = 10 \log_{10} \left( \frac{I_A}{I_B} \right) \]
• Substituting the ratio: \[ \Delta L = 10 \log_{10} \left( \frac{25}{4} \right) = 10 \log_{10} (6.25) \]
• Using the logarithm property: \[ \log_{10} (6.25) \approx 0.796 \]
• Therefore: \[ \Delta L \approx 10 \times 0.796 = 7.96 \approx 8 \text{ dB} \]

• The sound at Seat A is 8 dB higher than the sound at Seat B.

Final Answer