A company with loud machinery needs to cut its sound intensity to 44% of its original level. By how many decibels would the loudness be reduced? Use the formula L=101log I/I_0 decibels formula. Round to the nearest hundredth.
To find the reduction in decibels, we need to calculate the initial and final loudness levels using the given formula.
Given:
L = 101log(I/I_0)
Let's assume the initial sound intensity as I_0.
We are asked to reduce the sound intensity to 44% of its original level. Therefore, the final sound intensity becomes 0.44*I_0.
Using the given formula, we can say:
L = 101log(I/I_0)
Final loudness, L1 = 101log(0.44*I_0/I_0)
= 101log(0.44)
To calculate L1, we can use logarithmic properties;
log(a*b) = log(a) + log(b)
Therefore;
L1 = 101(log(0.44))
= 101(log(4.4) - log(10))
= 101(0.643 - 1)
= 101(-0.357)
Now, let's calculate the initial loudness, L0, using the original sound intensity, I_0.
L0 = 101log(I_0/I_0)
= 101log(1)
= 101(0)
= 0
So, the initial loudness, L0, is 0 decibels.
Therefore, to find the reduction in decibels, we subtract L1 from L0:
Reduction in decibels = L0 - L1
= 0 - 101(-0.357)
Using a calculator we find:
Reduction in decibels ≈ 36.06
Rounded to the nearest hundredth, the reduction in decibels would be approximately 36.06 decibels.