The psychological sensation of loudness B is measured in decibels and is defined with reference to the intensity, , of a barely audible sound by the following equation. In this equation, I is the intensity of the sound in question.
B=10log 1/1o
If the sound of a jet engine during takeoff is 140 decibels and the sound of a rock concert is 115 decibels, how does the intensity of the sound of the jet engine compare to the intensity of the sound of the rock concert?
I got that the intensity of the concert would be louder...is this right?
check your typing
your equation contains no l
besides that, what is 1o ?
im sorry the 1 is l and the zero is the lower exponet
You mean I think:
Db = 10 log (I/Io)
no, wrong. The higher the Db the louder
140 = 10 log (Ijet/Io)
115 = 10 log (Iband/Io)
14-11.5 = log (Ijet/IIband)
10^2.5 = Ijet/Iband = 316
the jet intensity (power) is 316 times the rock concert
Thanks i see what i did wrong
is it 21?
To compare the intensity of the sound of the jet engine to the intensity of the sound of the rock concert, we need to use the equation you provided:
B = 10log(I/1o)
Given that the sound of the jet engine is 140 decibels and the sound of the rock concert is 115 decibels, we can substitute these values into the equation:
For the jet engine:
B1 = 140
For the rock concert:
B2 = 115
Now, we can rearrange the equation to solve for the ratio of intensities, I1/I2:
B1 = 10log(I1/1o)
140 = 10log(I1/1o)
B2 = 10log(I2/1o)
115 = 10log(I2/1o)
Next, we can divide the equations to find the ratio:
(10log(I1/1o))/(10log(I2/1o)) = 140/115
Simplifying the left side:
log(I1/1o)/log(I2/1o) = 140/115
Using logarithm properties, we can rewrite the equation:
log(I1/1o) = log(I2/1o) * 140/115
Now, we can solve for the ratio of intensities:
I1/1o = (I2/1o)^(140/115)
Since the intensities are proportional to the square of the sound intensities, we can square both sides:
(I1/1o)^2 = [(I2/1o)^(140/115)]^2
Now, we can simplify and compare the intensities:
(I1/1o)^2 = (I2/1o)^(280/115)
This means that the intensity of the jet engine (I1) is equal to the intensity of the rock concert (I2) raised to the power of (280/115).
Based on our calculation, the intensity of the jet engine is greater than the intensity of the rock concert. Therefore, the jet engine is louder than the rock concert.