Solving the Sound Level Problem: An Introduction to Environmental Engineering

[ChloeYip]

Introduction to environmental engineering

1. Homework Statement

A sound level reading of 132 dB was taken near a construction site where some
chippers were being used. When only one chipper was working, the sound level
reading was 122 dB. Assume that all the chippers are located at the same point.
Estimate the number of chippers in operation when the reading of 132 dB was
obtained.

Homework Equations

(The attached photo)

The Attempt at a Solution

I have no idea how to do it...

Can anyone help me?
Thanks

 

[gneill]

Your attachment spells out how to sum sound sources given their dB levels. Just imagine that instead of having different L1, L2, L3, etc., you have n copies of one sound level L1. You'll need to exercise a bit of your algebra of logarithms and exponents knowledge.

 

[ChloeYip]

I have found a website: http://www.jiskha.com/display.cgi?id=1303792267 of similar question...

Just imagine that instead of having different L1, L2, L3, etc., you have n copies of one sound level L1.

I only understand the formula in the website (ie 10*Log(10)I1/I2) is derived by "10logI1-10logI2=10*Log(10)I1/I2"
However, I don't understand why it list 10*Log(10)I1/I2 as the formula... why I1/I2 ? but not (n-1)/n? (where n is number of chippers)
Thanks

 

[gneill]

I have found a website: http://www.jiskha.com/display.cgi?id=1303792267 of similar question...

I only understand the formula in the website (ie 10*Log(10)I1/I2) is derived by "10logI1-10logI2=10*Log(10)I1/I2"
However, I don't understand why it list 10*Log(10)I1/I2 as the formula... why I1/I2 ? but not (n-1)/n? (where n is number of chippers)
Thanks

Sound levels (in dB) don't sum algebraically because they are based on logarithms. When you sum logs you are effectively multiplying the underlying quantities, not adding them. When you take the difference between logs you are effectively dividing one by the other.

Sound intensity is a linear quantity (Watts per square meter) that can be summed algebraically. That is, if there are sound sources with Intensities $I_1, I_2, ...,I_n$ then the intensities sum: $I_{sum} = I_1 + I_2 +,...,+ I_n$.

So when you want to sum up the contributions of independent sound sources which are specified in terms of dB, you need to sum their intensities, not their levels, then convert the resulting intensity sum back to the level in dB.

Now you should know that sound level L in dB is defined in terms of a reference power level $I_o = 10^{-12}W/m^2$, so that by definition $L = 10 log(\frac{I}{I_o})$ . Thus:

$\frac{L}{10} = log(\frac{I}{I_o})$
$10^{\frac{L}{10}} = \frac{I}{I_o}$

So you can sum the quantities $10^{\frac{L_i}{10}}$ to obtain the net $\frac{I}{I_o}$. That is what the formula in your attachment is showing:
$$L_{com} = 10 log \left[ \sum_{i = 1}^n \left(10^{\frac{L_i}{10}} \right) \right]$$
The summation sign is summing the individual $\frac{I_i}{I_o}$ quantities and the result is converted to the net sound level in dB by taking the log and multiplying by 10.

 

[ChloeYip]

L=10log(I/Io)

It is the definition? I haven't seen it before... Thanks for telling me...
The attached photo are all things my instructor told us about the topic...

Why is intensity ratio I1/I2 is the number of chippers? I still don't understand... Isn't "10*Log(10)I1/I2" meaning the difference of conditions of sound levels of with and without 1 chippers?

Thanks

 

[gneill]

Why is intensity ratio I1/I2 is the number of chippers? I still don't understand... Isn't "10*Log(10)I1/I2" meaning the difference of conditions of sound levels of with and without 1 chippers?

That expression will give you the difference in sound level (in dB) between the two scenarios where I1 and I2 are intensities (power).

The intensity ratio is the ratio of power associated with the two scenarios: n sources versus 1 source in this case. But sound levels are not intensities. Sound levels are measured in dB, which are logarithm based. The ratio of the intensities will give you how many sources of intensity $I_2$ make up a total intensity of $I_1$, since power sums algebraically (P = P1 + P2 + P3 + ...).

So, you are looking for the ratio $n = I_1/I_2$ to determine how many chippers of intensity $I_2$ will raise the dB level from 122 to 132 dB. Write out the math using the definition of dB:

$L_1 - L_2 = 10 log \left(\frac{I_1}{I_o} \right) - 10 log \left(\frac{I_2}{I_o} \right)$
$\Delta dB = 10 \left[ log \left(\frac{I_1}{I_o} \right) - log \left(\frac{I_2}{I_o} \right) \right]$
$\Delta db = 10 log\left( \frac{I_1}{I_2} \right)$

You know the $\Delta dB$ from the given information. Solve for the intensity ratio.

 

[ChloeYip]

n sources versus 1 source in this case

I still don't understand... When L1, it is of n chippers; when L2, it is of n-1 chippers... when they minus each other, is it give level of 1 chipper?

According to the given formula(though I still don't understand), I solve it in this way... Am I correct?

To be honest, I feel like I don't even understand what's sound power and intensity... Are they both of symbol "L"? Are their relationship just sound intensity=10log(sound power)?
I am totally frustrated...
Thanks

 

[gneill]

I still don't understand... When L1, it is of n chippers; when L2, it is of n-1 chippers... when they minus each other, is it give level of 1 chipper?

According to the original problem statement:
L1 = 132 dB is for n chippers
L2 = 122 dB is for 1 chipper

To be honest, I feel like I don't even understand what's sound power and intensity... Are they both of symbol "L"? Are their relationship just sound intensity=10log(sound power)?
I am totally frustrated...

Power is Watts, sound Intensity (symbol $I$) is Watts/m², and sound level (symbol $L$) given in units of $dB$ is $10 log(I/I_o)$, where $I_o$ is a fixed reference intensity of 10^-12 W/m².

Think of decibels as a way to compare a quantity (in this case a sound intensity) to a fixed reference value using a logarithmic scale. So first a sound intensity $I$ is divided by the reference value that it is being compared to, $I_o$. That leaves a unitless number that represents how many times larger $I$ is than the reference value. Then that ratio is then converted to dB units by the operation of taking its logarithm base 10 and multiplying by 10.

In this problem we are given two sound levels in dB. Call them $L_1 = 132~dB$ and $L_2 = 122~dB$. But in order to answer the question you need to know the ratio $I_1/I_2$. From the definition of sound level in dB we can write:

$L_1 = 10 log(\frac{I_1}{I_o})~~~~~~$and$~~~~~~~L_2 = 10 log(\frac{I_2}{I_o})$

where $I_1$ and $I_2$ are the sound intensities in Watts/m² that correspond to those to sound levels. Now you could extract the two intensities from the above two expressions, then form the ratio of intensities to solve the problem. Or you could use the property of dB that says that the difference in sound levels is the dB scaled ratio of their intensities for which I offered a proof in my last post. Here it is again:

Let $ΔdB = L_1 - L_2$. Then from the definitions of $L_1$ and $L_2$ we can write:

$ΔdB = 10 log(\frac{I_1}{I_o}) - 10 log(\frac{I_2}{I_o})$

which simplifies to

$ΔdB = 10 log(\frac{I_1}{I_2})$

So the ratio $I_1/I_2$ can be easily obtained from the difference in sound levels.