If the intensity is increased by a factor 20,by how many decibel is the sound level increased
10 log base ten of power ratio
or in this case
10 log 20
about 13 dB
If by "intensity" the question means amplitude rather than power then it is 20 log or about 26 dB
To determine the increase in decibel (dB) when the intensity is increased by a factor of 20, we can use the formula:
dB = 10 * log10(I2 / I1)
Where:
- dB is the change in sound level
- I2 is the new intensity
- I1 is the initial intensity
Since the intensity is increased by a factor of 20, I2 = 20 * I1.
Let's substitute the values into the formula and calculate the change in decibel:
dB = 10 * log10((20 * I1) / I1)
= 10 * log10(20)
= 10 * 1.301
= 13.01 dB
Therefore, when the intensity is increased by a factor of 20, the sound level is increased by 13.01 decibels.
To determine how many decibels the sound level is increased when the intensity is increased by a factor of 20, we can use the formula for calculating the change in decibels (dB). The formula is:
dB = 10 * log10(I2/I1)
Where I1 is the initial intensity and I2 is the final intensity. In this case, since the intensity is increased by a factor of 20, I2 = 20 * I1.
Substituting this value into the formula, we get:
dB = 10 * log10(20 * I1/I1)
Simplifying further:
dB = 10 * log10(20)
Now, we can calculate the value inside the log10 function, which is 20.
dB = 10 * log10(20)
Using a calculator or a mathematical software, we find that log10(20) is approximately 1.30.
Thus:
dB ≈ 10 * 1.30
Calculating the value, we get:
dB ≈ 13
Therefore, when the intensity is increased by a factor of 20, the sound level is increased by approximately 13 decibels.