Decibel Drop of 1 Singer in Choir

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AI Thread Summary
In the discussion about the decibel drop when one singer stops in a choir of ten, participants explore how to calculate the change in sound level. The combined decibel level is initially calculated using the formula for sound intensity, but confusion arises regarding the logarithmic rules for adding decibel levels. It is clarified that the change in decibels can be determined by the difference in intensity from ten to nine singers. The calculation yields a decibel change of approximately 0.457 dB when one singer stops. Overall, the conversation emphasizes understanding sound intensity and the logarithmic nature of decibel calculations.
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Homework Statement


Imagine a choir of 10 singers, all of different frequencies, but each singing at the same decibel level β if heard
separately. Suddenly one gets bashful and stops singing. By how many decibels does the sound of the full choir drop?

Homework Equations



I = I1 +I2 + I3 ...β = 10log10(I/Io)

Io = 1.0 * 10^-2

The Attempt at a Solution



can i just call the combined decibel level of 10 singers x

x = 10log10(10I/(10^-2)

then call the combined decibel level of 9 singers y

y = 10log10(9I/10^-2)

then subtract them, i have a feeling there is not enough information to solve this :(
 
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You don't actually need the intensities. Do you know how to add decibel levels?
 
Mindscrape said:
You don't actually need the intensities. Do you know how to add decibel levels?

no, do i divide them based on a logarithmic rule ??

logx-logy = log(x/y)

or am i way off here

beta change = 10log10(10/9) = .457 according to my calculator
 
Well, let's forget about the general case of adding decibels, and just focus on how the sound intensities will contribute to the total decibel level

dBtot=10log(I1/I0 + I2/I0 + ... + I10/I0)

In this case, all the sound sources are the same. Simplify what I started, use a log rule, and I think you'll be able figure out the rest.
 
vande060 said:
no, do i divide them based on a logarithmic rule ??

logx-logy = log(x/y)

or am i way off here

beta change = 10log10(10/9) = .457 according to my calculator

Let 10n be the intensity of each singer.Then

β= 10*log[10n/10-12] = 10(12 + n)

Similarly write down the equation for 10 β and 9 β and find the difference.
 
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