To solve this problem, we can use the inverse square law for sound intensity. According to this law, the sound intensity decreases as the distance from the source increases. It can be expressed as:
I1/I2 = (r2/r1)^2
Where:
- I1 represents the sound intensity at the first observer,
- I2 represents the sound intensity at the second observer,
- r1 represents the distance between the loudspeaker and the first observer, and
- r2 represents the distance between the loudspeaker and the second observer.
Using the formula for sound level in decibels (dB), which is given by:
L = 10 log10(I/I0)
Where:
- L represents the sound level in decibels,
- I represents the sound intensity, and
- I0 represents the reference sound intensity (usually 10^(-12) W/m^2).
Let's begin by calculating the sound intensity in watts/m^2 at each observer, using the formula:
I = I0 * 10^(L/10)
First, let's calculate the sound intensity at the first observer:
I1 = (10^(-12)) * 10^(51.8/10)
I1 = (10^(-12)) * 10^5.18
I1 = 10^(-7.18)
I1 ≈ 7.943 × 10^(-8) W/m^2
Next, let's calculate the sound intensity at the second observer:
I2 = (10^(-12)) * 10^(75.8/10)
I2 = (10^(-12)) * 10^7.58
I2 = 10^(-4.42)
I2 ≈ 2.638 × 10^(-5) W/m^2
Now, we can use the inverse square law to find the ratio of distances:
(I1/I2) = (r2/r1)^2
(7.943 × 10^(-8))/(2.638 × 10^(-5)) = (r2/r1)^2
Dividing both sides by (2.638 × 10^(-5)):
(7.943 × 10^(-8))/(2.638 × 10^(-5)) ≈ (r2/r1)^2
Using a calculator, we find that the ratio of distances is approximately 0.003.
Taking the square root of both sides:
√(0.003) ≈ √(r2/r1)^2
0.0548 ≈ r2/r1
Multiplying both sides by the distance r1:
0.0548 * r1 ≈ r2
Now, we know that the two observers are initially 113m apart, so:
r1 + r2 = 113
Substituting the value of r2:
0.0548 * r1 + r1 = 113
Combining like terms:
1.0548 * r1 = 113
Dividing both sides by 1.0548:
r1 ≈ 107.09
Now we can find the value of r2 by subtracting r1 from the initial distance of 113m:
r2 ≈ 113 - 107.09
r2 ≈ 5.91
Therefore, the loudspeaker is approximately 107.09m away from the first observer and 5.91m away from the second observer.