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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: Burnttoast on October 02, 2008, 05:07:03 pm

Title: Sound Question Help! :O
Post by: Burnttoast on October 02, 2008, 05:07:03 pm: Heya.
Um.. i'm just wondering if anyone could please help explain to me these questions..
They are to do with sound...
1) A new material is claimed to 'absorb 90% of the sound energy that falls upon it'. How many dB is this equivalent to?
answer:. checkpoints (yes.. these questions are from checkpoints) says that if 90% absorbed, the intensity is reduced by a factor of 10. Where does this factor 10 come from?! :O .. and then it says L=10log10 ..
even if it is a factor of 10.. don't' you still need the Io thing.. 1x10^-12 ... hmmm
and
2) Another company advertises that its sound insulation material 'reduces sound levels by a massive 20dB'. What percentage of incident energy is absorbed if their claim is correct?
checkpoints says... that L=-20 => Incident=10^-2..... Intensity is reduced by a factor of 100.
i really.. don't understand these factor things.
if anyone could explain that would be awesome :)
thanksss =D
Title: Re: Sound Question Help! :O
Post by: Collin Li on October 02, 2008, 05:15:30 pm: The factor of 10 is because the sound intensity has gone from 100% to 10%.

100% is 10 times bigger than 10%.

You want to measure a change in the sound intensity level, so you only need to know the factor change (10).

Someone may be able to explain the idea of "factors" better, but essentially it's got to do with logarithm laws.

$\log a + \log b = \log ab$

The addition of decibels (logarithms) are a measure of sound intensity factors.
Title: Re: Sound Question Help! :O
Post by: Burnttoast on October 02, 2008, 05:33:38 pm: so.. you take away the 90% from 100% to see how much.. intensity is left..
which gives 10%..
and then to find out the factor.. you do 100/10 ?
it kinda makes sense. thanks =D
Title: Re: Sound Question Help! :O
Post by: Mao on October 02, 2008, 05:53:05 pm: the formula you want is $\Delta L (\mbox{Db})=10\log_{10} \left( \frac{I_2}{I_1}\right)$ , with a reference sound level (threshold of hearing of 1kHz sound) $1\times 10^{-12}\; \mbox{W m^{-1}}=0\; \mbox{dB}$

decibels more or less allow comparison between two different sound intensities with a logarithmic scale.

in the first case, a sound is reduced to 10% of its original intensity, it has decreased by a factor of 10. the new sound level would change by $10\log_{10} \frac{1}{10}=-10$

in the second case, a reduction of 20dB: $-20 = 10 \log_{10} \frac{1}{k}\implies \frac{1}{k}=10^{-2}=0.01$
so k in this case is 100, and the transmitted sound is one hundredth of the original sound intensity. hence, it has been reduced by a factor of 100.
Title: Re: Sound Question Help! :O
Post by: Burnttoast on October 02, 2008, 06:08:11 pm: since.. is been reduced by 10%..
10% =1/10
so it becomes 10log (1/10)

so if it had been reduced by 20%
20%=2/10=1/5
so it becomes 10log (1/5)..
sorry >.<

thanks heaps though =D