Question

What would be the sound level produced by a chorus of 33 singers

Answer 1

Since dB (decibels) is a logarithm unit, we need to understand the formula for the sound level before calculating the new sound level for 33 singers:

`dB=10\log\mleft((I)/(I_0)\mright)`

Where I and I0 are the relative powers of the sound.

If we will have 33 singers, that means the value of I will be 33 times higher, so we have:

`\begin{gathered} dB1=10\log\mleft((I)/(I_0)\mright)=73.2\\ \\ dB2=10\log\mleft(33(I)/(I_0)\mright)=10\left(\log\mleft((I)/(I_0)\mright)+\log\mleft(33\mright)\right?\\ \\ =10\log\mleft((I)/(I_0)\mright)+10\log\mleft(33\mright)\\ \\ =73.2+10\cdot1.52\\ \\ =73.2+15.2\\ \\ =88.4\text{ dB} \end{gathered}`

Therefore the new sound level is 88.4 dB.