Respuesta :
The question is incomplete. Here is the complete question.
To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity when the intensity as measured in W/m² changes by a multiplicative factor. The number of decibels increase by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is
[tex]\beta=10log(\frac{I}{I_{0}} )dB[/tex],
where [tex]I_{0}[/tex] is a reference intensity. for sound waves, [tex]I_{0}[/tex] is taken to be [tex]10^{-12} W/m^{2}[/tex]. Note that log refers to the logarithm to the base 10.
Part A: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity, i.e. [tex]I=10I_{0}[/tex]? Express the sound intensity numerically to the nearest integer.
Part B: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity, i.e. [tex]I=100I_{0}[/tex]? Express the sound intensity numerically to the nearest integer.
Part C: Calculate the change in decibels ([tex]\Delta \beta_{2},\Delta \beta_{4}[/tex] and [tex]\Delta \beta_{8}[/tex]) corresponding to f = 2, f = 4 and f = 8. Give your answer, separated by commas, to the nearest integer -- this will give an accuracy of 20%, which is good enough for sound.
Answer and Explanation: Using the formula for sound intensity level:
A) [tex]I=10I_{0}[/tex]
[tex]\beta=10log(\frac{10I_{0}}{I_{0}} )[/tex]
[tex]\beta=10log(10 )[/tex]
β = 10
The sound Intensity level with intensity 10x is 10dB.
B) [tex]I=100I_{0}[/tex]
[tex]\beta=10log(\frac{100I_{0}}{I_{0}} )[/tex]
[tex]\beta=10log(100)[/tex]
β = 20
With intensity 100x, level is 20dB.
C) To calculate the change, take the f to be the factor of increase:
For [tex]\Delta \beta_{2}[/tex]:
[tex]I=2I_{0}[/tex]
[tex]\beta=10log(\frac{2I_{0}}{I_{0}} )[/tex]
[tex]\beta=10log(2)[/tex]
β = 3
For [tex]\Delta \beta_{4}[/tex]:
[tex]I=4I_{0}[/tex]
[tex]\beta=10log(\frac{4I_{0}}{I_{0}} )[/tex]
[tex]\beta=10log(4)[/tex]
β = 6
For [tex]\Delta \beta_{8}[/tex]:
[tex]I=8I_{0}[/tex]
[tex]\beta=10log(\frac{8I_{0}}{I_{0}} )[/tex]
β = 9
Change is
[tex]\Delta \beta_{2},\Delta \beta_{4}[/tex], [tex]\Delta \beta_{8}[/tex] = 3,6,9 dB
