A 60 man choir sings with the sound intensity of 70 dB in a 12m distance from the choir.
a) What is the average sound intensity in this distance?
b) What is the sound intensity in a 20 m distance ?
Help please :)
a) To find the average sound intensity in a given distance, we need to calculate the sound intensity level (SIL) by using the formula:
SIL = 10 * log10(I/I0)
Where:
- SIL is the sound intensity level in decibels (dB)
- I is the sound intensity in watts per square meter (W/m^2)
- I0 is the reference sound intensity (I0 = 10^(-12) W/m^2)
Given:
- Sound intensity level of the choir, SIL1 = 70 dB
- Distance from the choir, d1 = 12 m
First, we need to convert the sound intensity level from decibels to watts per square meter using the inverse formula:
I1 = I0 * 10^(SIL1/10)
Substituting the given values:
I1 = (10^(-12) W/m^2) * 10^(70/10)
= (10^(-12) W/m^2) * 10^7
= 10^(-12 + 7) W/m^2
= 10^(-5) W/m^2
Next, to find the average sound intensity (I_avg), we use the inverse square law which states that the sound intensity decreases proportionally to the square of the distance from the source:
I_avg = I1 / (d1^2)
Substituting the given values:
I_avg = 10^(-5) W/m^2 / (12^2)
= 10^(-5) W/m^2 / 144
≈ 6.94 * 10^(-8) W/m^2
Therefore, the average sound intensity in the 12 m distance from the choir is approximately 6.94 * 10^(-8) W/m^2.
b) To find the sound intensity in a 20 m distance (I2), we can use the inverse square law formula again:
I2 = I_avg * (d2^2)
Substituting the given values:
I2 = 6.94 * 10^(-8) W/m^2 * (20^2)
= 6.94 * 10^(-8) W/m^2 * 400
= 2.776 * 10^(-5) W/m^2
Therefore, the sound intensity in a 20 m distance from the choir is approximately 2.776 * 10^(-5) W/m^2.
To answer these questions, we need to understand the inverse square law for sound intensity. According to this law, the sound intensity decreases inversely as the square of the distance increases. In other words, if you double the distance from the sound source, the sound intensity will decrease by a factor of four.
Now let's solve the questions:
a) What is the average sound intensity in this distance (12m)?
The average sound intensity can be calculated by dividing the total sound intensity by the number of sound sources. In this case, the total sound intensity is given as 70 dB for a 60-man choir. However, since we're looking for the average sound intensity at a specific distance, we need to take into account the inverse square law.
To do this, we'll use the formula:
Average Sound Intensity = Total Sound Intensity / (Number of Sources * Distance^2)
In this case, the number of sources is 60 and the distance is 12m:
Average Sound Intensity = 70 dB / (60 * (12m)^2)
Simplifying the equation:
Average Sound Intensity = 70 dB / (60 * 144m^2)
Average Sound Intensity ≈ 70 dB / 103680m^2
Average Sound Intensity ≈ 0.000675 dB/m^2
b) What is the sound intensity in a 20m distance?
To calculate the sound intensity at a new distance, we can use the inverse square law. Let's assume the sound intensity at a distance of 12m is 70 dB. Now we want to calculate the sound intensity at a distance of 20m:
Sound Intensity at 20m = Sound Intensity at 12m * (Distance at 12m^2 / Distance at 20m^2)
Using the given values:
Sound Intensity at 20m = 70 dB * (12m^2 / 20m^2)
Sound Intensity at 20m = 70 dB * (144m^2 / 400m^2)
Sound Intensity at 20m ≈ 25.2 dB
So, the sound intensity at a 20m distance is approximately 25.2 dB.