Two speakers are both emitting sounds, the first with an intensity level of 1 x 10^{-11} \frac{W}{m^2} and the second with an intensity of 1 x 10^{-8} \frac{W}{m^2}.
What is the db level of the first?
\beta =
What is the difference in the decibel (db) level of the two sounds?
\Delta \beta =
db1 = 10*Log(I/Io)
db1 = 10*Log(10^(-11)/10^(-12) =
10*Log(10^(-11)*10^12) = 10*Log(10) =
10*1 = 10.
db2 = 10*Log(10^(-8)/10^(-12) =
10*Log(10^(-8)*10^12) = 10*Log(10^4) =
10 * 4 = 40.
40 - 10 = 30db Difference.
To find the dB (decibel) level of a sound, you can use the formula:
β = 10 log10(I/I0)
Where β is the dB level, I is the intensity of the sound, and I0 is the reference intensity.
For the first speaker, the intensity level is 1 x 10^(-11) W/m^2. To find the dB level, we need to find the ratio of the intensity to the reference intensity. The reference intensity is generally taken to be the threshold of hearing, which is 1 x 10^(-12) W/m^2.
So for the first speaker:
β1 = 10 log10((1 x 10^(-11))/(1 x 10^(-12)))
Simplifying the above expression gives:
β1 = 10 log10(10)
Since log10(10) = 1, the dB level of the first speaker is:
β1 = 10 x 1
β1 = 10 dB
Now let's calculate the difference in the decibel level of the two sounds.
To find the difference in dB level, use the formula:
Δβ = β2 - β1
Where Δβ is the difference in dB levels, β2 is the dB level of the second speaker, and β1 is the dB level of the first speaker.
For the second speaker, the intensity level is 1 x 10^(-8) W/m^2. Following the same steps as before, we can calculate the dB level of the second speaker.
β2 = 10 log10((1 x 10^(-8))/(1 x 10^(-12)))
Simplifying the expression gives:
β2 = 10 log10(10,000)
Since log10(10,000) = 4, the dB level of the second speaker is:
β2 = 10 x 4
β2 = 40 dB
Now we can find the difference in dB level:
Δβ = 40 dB - 10 dB
Therefore, the difference in the decibel level of the two sounds is:
Δβ = 30 dB