The cabin noise level of the new Airbus 380 is 5 decibels lower than that of a 747 Jumbo Jet. By what factor does the noise intensity decrease?
db = 10log(p2/p1) = 5.
10log(p2/p1) = 5,
Divide both sides by 10:
log(p2/p1) = 0.5,
p2/p1 = 10^0.5 = 3.16.
To determine the factor by which the noise intensity decreases, we need to calculate the ratio of the noise intensities between the Airbus 380 and the 747 Jumbo Jet.
The decibel scale is logarithmic, meaning that a decrease or increase in decibels corresponds to a proportional decrease or increase in noise intensity.
The formula for calculating the ratio of noise intensities in decibels is:
Ratio (in decibels) = 10 * log10(Intensity1 / Intensity2)
Given that the cabin noise level of the Airbus 380 is 5 decibels lower than that of the 747 Jumbo Jet, we can substitute the values into the formula:
-5 = 10 * log10(Intensity380 / Intensity747)
Now, let's solve for the ratio of the noise intensities.
Dividing both sides of the equation by 10 gives us:
-0.5 = log10(Intensity380 / Intensity747)
Next, we need to convert the equation from logarithmic form to exponential form. In exponential form, the base 10 is raised to the power of the value on the right-hand side.
10^(-0.5) = Intensity380 / Intensity747
When evaluating 10^(-0.5), we find that it is approximately 0.316.
0.316 = Intensity380 / Intensity747
To find the factor by which the noise intensity decreases, we take the reciprocal of the ratio:
1 / 0.316 ≈ 3.16
Therefore, the noise intensity of the Airbus 380 decreases by a factor of approximately 3.16 compared to the 747 Jumbo Jet.
db = 10log(p2/p1) = -5.
The negative sign means p2 is less than p1 and the ratio is less yhan 1.
10log(p2/p1) = -5,
Divide both sides by 10:
log(p2/p1) = -0.5,
p2/p1 = 10^-0.5 = 0.316.