Why choose /100 over x100?Why Divide by 100 when solving decibels?

[silversurf]

I'm having difficulty understanding the solution to this question. "A source creates a sound that is observed at 200 Watts per meter squared. If the intensity level of the observed sound decreases by 20 decibels what is the decrease in observed intensity?

The way I solved it was I saw 20 decibels as a decrease in intensity of 100 so I subtracted 100 from 200 to get 100 Watts per meter squared but the solution divides 200 by 100 and then subtracts this from the original number: 200 - 2 = 198. Why do they solve it like this and not the way I did?

 

[test1234]

I'm having difficulty understanding the solution to this question. "A source creates a sound that is observed at 200 Watts per meter squared. If the intensity level of the observed sound decreases by 20 decibels what is the decrease in observed intensity?

The way I solved it was I saw 20 decibels as a decrease in intensity of 100 so I subtracted 100 from 200 to get 100 Watts per meter squared but the solution divides 200 by 100 and then subtracts this from the original number: 200 - 2 = 198. Why do they solve it like this and not the way I did?

Hi silversuf, 20dB is a decrease in intensity of 100 times. Therefore, the intensity level of the observed sound is 1/100 times that of the original sound.
$20(dB)=-10~log_{10} \frac{I_{new}}{I_{original}}$
$\frac{I_{new}}{I_{original}}=10^{-2}$
$I_{new}=10^{-2} I_{original}=10^{-2} \times 200 W/m^2=2W/m^2$
Hence the decrease in observed intensity $= (200-2) W/m^2 =198W/m^2$

 

[controlswhiz]

Good day,

The sound level in decibels is

$β = 10log_{10}(\frac{I}{I_0})$

Where beta is measured n decibels, $I$ the intensity in question and $I_0$ is the reference intensity. Since you have a decrease of 20 decibels in your reference intensity (the source), you may re-write the above equation as

$20 = 10log_{10}(I_{0}) - 10log_{10}(I)$

multiplying by -1 in both sides of the equation and using the laws of logarithms,

$-20 = 10log_{10}(\frac{I}{I_0})$

the -20 can be thought of as $-β$ in the above expression, indicating that this is a decrease in sound level with respect to the reference intensity. For solving the difference in the observed intensity, you will have to solve for $I$ and subtract this number from the reference intensity.

P.S. I think this post should be in the Homework & Coursework Questions area.

Cheers.

 

[silversurf]

Thank you both so much!

 

[PeterO]

I'm having difficulty understanding the solution to this question. "A source creates a sound that is observed at 200 Watts per meter squared. If the intensity level of the observed sound decreases by 20 decibels what is the decrease in observed intensity?

The way I solved it was I saw 20 decibels as a decrease in intensity of 100 so I subtracted 100 from 200 to get 100 Watts per meter squared but the solution divides 200 by 100 and then subtracts this from the original number: 200 - 2 = 198. Why do they solve it like this and not the way I did?

To get a feel for a question like this I rely on a couple of reference examples.

#1
Halving or doubling the intensity means a 3 dB change (halving -3dB, doubling +3dB)
This is supposedly the smallest change the average human can notice - meaning that if the sound level went from 90dB to 91dB not many people would notice apparently

#2
Changing the intensity by a factor of 10 means a change of 10dB - I just like that both figures are 10, the only time the factor change in Intensity = the dB change in Sound level.

To get a 20dB change, you just have to have a 10dB change, then a second 10dB change.

The first means the intensity is down by a factor of 10 (from 200 to 20) and the second means another factor of 10 (from 20 to 2)

SO the intensity has changed from 200 to 2 ; a change of 198.

Perhaps more simply: dB changes are additions and subtractions, intensity changes a factor changes.

so 20dB change was +20dB or -20dB, The intensity change of 100 was x100 or /100.