Sound intensity is inversely proportional to the square of the distance from the sound source; that is
I= k/ (r)^(2),
where I is the intensity, r is the distance from the sound source, and k is a constant.
Suppose that you are sitting a distance R from the TV, where its sound intensity is I (base 1). Now you move to a seat five times as far from the TV, a distance 5R away, where the sound intensity is I(base 2).
what is the relationship between I(base 1) & I (base 2)?
What is the relationship between the decibel levels associated with I(base 1) & I (base 2)
-Round the decibel level for I(base 2) to the nearest integer.
The decibel level for I(base 2) is about __?__dB lower than I (base 1)
I = k/r^2
Now replace r with 5r, and you have
I' = k/(5r)^2 = k/25r^2 = (1/25)(k/r^2)
If r is replaced with nr, then I gets scaled down by a factor of 1/n^2
To determine the relationship between I(base 1) and I(base 2), we can use the equation for sound intensity mentioned: I = k/(r)^2.
Let's start by considering the two distances from the TV:
- For the first case (I(base 1)), you are sitting a distance R away from the TV.
- For the second case (I(base 2)), you have moved to a seat five times as far, a distance 5R away.
Substituting these values into the equation, we get:
I(base 1) = k/(R)^2
I(base 2) = k/(5R)^2
To find the relationship between I(base 1) and I(base 2), we can compare the two expressions:
I(base 1) / I(base 2) = (k/(R)^2) / (k/(5R)^2)
= (k/(R)^2) * ((5R)^2/k)
= (k * 25R^2) / (k * R^2)
= (25R^2) / (R^2)
= 25
Therefore, we can conclude that I(base 1) is 25 times greater than I(base 2).
Now let's move on to the relationship between the decibel levels associated with I(base 1) and I(base 2).
The decibel level formula for sound intensity is: L = 10 * log10(I/I0), where L is the decibel level, I is the sound intensity, and I0 is a reference intensity.
Assuming I(base 1) is the reference intensity (I0), we can use the formula to find the decibel level for I(base 1):
L(base 1) = 10 * log10(I(base 1)/I(base 1))
= 10 * log10(1)
= 10 * 0
= 0 dB
The decibel level for I(base 1) is 0 dB.
For I(base 2), we need to calculate the decibel level using the same formula. Since we want it rounded to the nearest integer, we'll use the closest whole number.
L(base 2) = 10 * log10(I(base 2)/I(base 1))
Substituting the values of I(base 1) and I(base 2) into the equation, we get:
L(base 2) = 10 * log10((k/(5R)^2)/(k/(R)^2))
= 10 * log10((k/(5R)^2) * ((R)^2/k))
= 10 * log10((1/25) * (R^2/R^2))
= 10 * log10(1/25)
= 10 * log10(0.04)
Using a calculator, we find log10(0.04) ≈ -1.39794.
L(base 2) ≈ 10 * (-1.39794)
L(base 2) ≈ -13.9794 dB
Lastly, if we round -13.9794 dB to the nearest whole number, the decibel level for I(base 2) is about -14 dB lower than I(base 1).
Therefore, the answer to your question is: The decibel level for I(base 2) is about -14 dB lower than I(base 1).