Suppose that you and I are having a singing contest, and you produce a sound intensity (measured in W/ m2) about 32 times greater than mine. How many dB louder than me are you singing ?
His is 30dB.
For 79 times greater the answer is 19dB
For 63 times greater the answer is 18dB
How do I calculate for 32 times greater
Let the two intensities be I1 and I2.
10 log(10)I2 - 10log(10)I1
= dB difference
= 10 log(10)[I2/I1]
= 10log(10)32 = 15.1 dB
It doesn't matter what sound intensity power units are used. It is the intensity ratio that matters.
To calculate the decibel (dB) difference for a given ratio, you can use the logarithmic equation:
dB = 10 * log10(ratio)
In this case, you want to find the dB difference when the sound intensity ratio is 32 times greater. You can plug in the ratio into the equation to get the answer:
dB = 10 * log10(32)
Using a calculator, you can calculate the logarithm of 32 to the base 10:
log10(32) ≈ 1.505
Now plug this value back into the equation:
dB ≈ 10 * 1.505
dB ≈ 15.05
Therefore, if your sound intensity is about 32 times greater than mine, you would be singing approximately 15.05 dB louder.