At a recent rock concert, a dB meter registered 112 dB when placed 1.70 m in front of a loudspeaker on the stage.
(a) What was the power output of the speaker, assuming a hemispherical radiation of the sound and neglecting absorption in the air?
(b) How far away would the intensity level be a somewhat reasonable 85 dB?
I used the equation I = P/(4pi r^2); Rearranging the equation I did: 112/4pi 1.7^2 = 4067.482 W, but this was incorrect. I also converted 112 dB to 11.2 bels, but also wrong. Any suggestions?
(a) Convert 112 dB to the equivalent power per unit area. Then multiply that by the hemisphere area 2 pi R^2, with R = 1.70 m.
(b) Ypur answer weas incorrect because they asked for a distance and you gave them a power number.
Dropping from 112 dB to 85 dB is a 27 dB reduction in power per area. That corresponds to a 10^(27/10) = 501 reduction factor in power per area. Using the inverse square law, that can be achieved with a sqrt(501) = 22.4 increase in distance. That would put the listener 1.70 x 22.4 = 38 meters away
To get the correct answer for part (a), you need to convert the sound level in decibels (dB) to the equivalent power per unit area. Here's how you can do it:
1. Start by converting the decibel (dB) value to the equivalent power ratio using the equation:
Power ratio = 10^(dB/10)
In this case, the power ratio would be: Power ratio = 10^(112/10)
2. Next, multiply the power ratio by the power per unit area at a distance of 1.70 m, assuming hemispherical radiation of sound. The power per unit area at a distance R from a point source is given by:
Power per unit area = Power output / (4πR^2)
So, for this problem, the power per unit area at a distance of 1.70 m is:
Power per unit area = Power output / (4π(1.70^2))
Substitute the power ratio and calculate the power output.
For part (b), you need to find the distance at which the intensity level drops to a reasonable 85 dB. Here's how you can approach it:
1. Convert the dB value to the power ratio using the same formula as before:
Power ratio = 10^(dB/10)
In this case, the power ratio would be: Power ratio = 10^(85/10)
2. Using the inverse square law, the power per unit area decreases with the square of the distance. So, to achieve a 27 dB reduction in power per area, we need a distance where the power per unit area is reduced by a factor of 501.
Power ratio reduction factor = 10^(27/10)
Distance increase factor = sqrt(Power ratio reduction factor)
3. Multiply the original distance (1.70 m) by the distance increase factor to find the new distance at which the intensity level would be 85 dB.
New distance = Original distance × Distance increase factor
Substitute the values and calculate the new distance.
Remember to keep track of the units and use appropriate values for π (pi) in your calculations.
(a) To determine the power output of the speaker, we need to convert the dB reading to the equivalent power per unit area.
1. Start by converting 112 dB to the equivalent power per unit area in watts/m² using the formula:
Power per area (W/m²) = 10^(dB/10)
Substituting the value dB = 112, we get:
Power per area (W/m²) = 10^(112/10)
Calculating this, we find:
Power per area (W/m²) = 158489.3192 W/m²
2. Next, multiply the power per area by the surface area of the hemisphere, which is 2πR^2, where R is the distance from the loudspeaker. In this case, R = 1.70 m.
Power output of the speaker (W) = Power per area (W/m²) x Surface area of hemisphere (2πR^2)
Substituting the values, we have:
Power output of the speaker (W) = 158489.3192 W/m² x 2π(1.70)^2
Calculating this, we find:
Power output of the speaker (W) ≈ 541894.9748 W (rounded to the nearest watt)
Therefore, the power output of the speaker is approximately 541,895 watts.
(b) To find the distance at which the intensity level would be a somewhat reasonable 85 dB, we need to determine the increase in distance required to reduce the power per area by 27 dB.
Using the inverse square law, we know that the power per area decreases inverse proportionally to the square of the distance.
1. Convert the reduction in power per area from 112 dB to 85 dB. This corresponds to a reduction of 27 dB:
Reduction factor = 10^(27/10) ≈ 501
2. Determine the increase in distance needed to achieve this reduction factor. As the distance increases, the power per area decreases.
Increase in distance = sqrt(Reduction factor) ≈ sqrt(501) ≈ 22.4
3. Finally, calculate the new distance by multiplying the increase in distance by the original distance of 1.70 m:
New distance = 1.70 m x 22.4 ≈ 38 meters
Therefore, the listener would need to be approximately 38 meters away to achieve a somewhat reasonable 85 dB intensity level.