A loudspeaker is placed between two observers who are 111 m apart, along the line connecting them. If one observer records a sound level of 51.9 dB and the other records a sound level of 82.5 dB, how far is the speaker from each observer?
To find the distance of the speaker from each observer, we can use the inverse square law of sound.
The inverse square law states that the intensity of sound is inversely proportional to the square of the distance from the source of sound. Mathematically, it can be expressed as:
I1 / I2 = (r2 / r1)²
where I1 and I2 are the intensities of sound at distances r1 and r2 respectively.
First, let's convert the sound levels in decibels (dB) to intensities. The intensity of sound can be calculated using the following formula:
I = 10^(L/10)
where I is the intensity and L is the sound level in decibels.
For the observer recording a sound level of 51.9 dB, the intensity would be:
I1 = 10^(51.9/10) = 119.2263
For the observer recording a sound level of 82.5 dB, the intensity would be:
I2 = 10^(82.5/10) = 40237.3777
Now, let's substitute these intensities into the inverse square law equation:
119.2263 / 40237.3777 = (r2 / r1)²
To solve for the distances r1 and r2, we can take the square root of both sides of the equation:
√(119.2263 / 40237.3777) = r2 / r1
Now, let's calculate the square root and simplify the equation further:
0.019664656 ≈ r2 / r1
To find the actual distances, we can multiply this ratio by the known distance between the two observers (111 m):
0.019664656 * 111 = 2.183934216 ≈ r2
Therefore, the speaker is approximately 2.18 meters away from the observer with the sound level of 82.5 dB.
To find the distance of the speaker from the other observer, we can divide the known distance (111 m) by the ratio:
111 / 0.019664656 = 5637.519600 ≈ r1
Therefore, the speaker is approximately 5637.52 meters away from the observer with the sound level of 51.9 dB.
To find the distance of the loudspeaker from each observer, you can use the inverse square law of sound.
The inverse square law states that the sound intensity decreases with the square of the distance. Mathematically, it can be represented as:
I2 / I1 = (d1 / d2)²
Where:
I1 = Initial sound intensity
I2 = Final sound intensity
d1 = Initial distance from the source
d2 = Final distance from the source
Let's assign the following values:
I1 = sound level of 51.9 dB
I2 = sound level of 82.5 dB
d1 = distance from the loudspeaker to the observer with a sound level of 51.9 dB
d2 = distance from the loudspeaker to the observer with a sound level of 82.5 dB
Using the equation and plugging in the values, we get:
10^(I2 / 10) / 10^(I1 / 10) = (d1 / d2)²
Rewriting the equation and solving for d1, we have:
d1 = sqrt((10^(I2 / 10) / 10^(I1 / 10)) * d2²)
Now we can calculate the distances.
First, let's calculate d2 using the given information:
d2 = 111 m - d1
Substituting this value of d2 in the equation and simplifying:
d1 = sqrt((10^(82.5 / 10) / 10^(51.9 / 10)) * (111 - d1)²)
Squaring both sides:
d1² = (10^(82.5 / 10) / 10^(51.9 / 10)) * (111 - d1)²
d1² = (10^(8.25) / 10^(5.19)) * (111 - d1)²
d1² = (16777.301 * 31574.747) * (111 - d1)²
d1² = 529062982.3 * (111 - d1)²
Expanding the equation:
d1² = 588485162.7 - 10525.425d1 + d1²
Rearranging the terms:
0 = 588485162.7 - 10525.425d1
10525.425d1 = 588485162.7
d1 = 588485162.7 / 10525.425
Calculating d1:
d1 ≈ 55847.14 meters
Now, let's calculate d2:
d2 = 111 - d1
Substituting the value of d1:
d2 ≈ 111 - 55847.14
Calculating d2:
d2 ≈ -55736.14 meters
Since we cannot have a negative distance, this value is not valid.
Therefore, the loudspeaker is approximately 55847.14 meters away from the observer with a sound level of 51.9 dB.