Intensity/Decibel question (1 Viewer)

o_0 · Oct 24, 2011

oh foot nuts!

i just realised i got all the working out BUT, i concluded that "therefore, when intensity is doubled, loudness increases by x1.077"

IE L2/L1 = 1.077
i realised that 1.077 x 90 = 6.93,

so wrong or right?

Ivy Mike · Oct 24, 2011

This is what I did. I'd appreciate it if somebody could tell me if I did something wrong.

Normal Intensity:

$1 = 10^{-12} \cdot e^{0.1L_{1}}$
$10^{12}=e^{0.1L_{1}}$
$ln(10^{12})/0.1 = L_{1}$

Double Intensity:

$2 = 10^{-12}\cdot e^{0.1L_{2}}$
$2\times 10^{12}= e^{0.1L_{2}}$
$ln(2 \times 10^{12})= 0.1L_{2}$
$L_{2}= ln(2\times 10^{12})/0.1$

$L_{2}/L_{1} = ln(2\times 10^{12})/ln(10^{12})$
$= 1.025085833$ times louder

Tavin · Oct 24, 2011

o_0 said:
oh foot nuts!

i just realised i got all the working out BUT, i concluded that "therefore, when intensity is doubled, loudness increases by x1.077"

IE L2/L1 = 1.077
i realised that 1.077 x 90 = 6.93,

so wrong or right?

With that working, you should get it, but the wording might cost you mark.

Because 10x1.077 =/= about 17.
Only works for 90.

o_0 · Oct 24, 2011

Ivy Mike said:
This is what I did. I'd appreciate it if somebody could tell me if I did something wrong.

Normal Intensity:

$1 = 10^{-12} \cdot e^{0.1L_{1}}$
$10^{12}=e^{0.1L_{1}}$
$ln(10^{12})/0.1 = L_{1}$

Double Intensity:

$2 = 10^{-12}\cdot e^{0.1L_{2}}$
$2\times 10^{12}= e^{0.1L_{2}}$
$ln(2 \times 10^{12})= 0.1L_{2}$
$L_{2}= ln(2\times 10^{12})/0.1$

$L_{2}/L_{1} = ln(2\times 10^{12})/ln(10^{12})$
$= 1.025085833$ times louder

heya yeah.. my brother just explained it to me,

the percentage increase is always changing, so you were meant to calculate L2 - L1 = 6.93 for all values of I

FUCK!!! hahaha i also calculated ratio instead of difference, but at least 1 out of 2 marks hopefully

SnowTau · Oct 24, 2011

I got ln1024 or 10ln2, however you wanna write it.

CecilyMare · Oct 24, 2011

there is already a thread on it.

can someone make a thread on the last q of q10 pls

OldMathsGuy · Oct 24, 2011

Assume an intensity of "c". Then use an intensity of "2c" and compare the two. My latex skills are non-existent but with god use of log rules, you are left with a difference of 10 x ln(2) = 6.93.
I think the reason why this was a q10 question is because you had to algebraically show this worked in general rather than through a specific set of values.

Best Regards
OldMathsGuy

umz93 · Oct 24, 2011

I stated that i wanted intensity to equal 1

barbernator · Oct 24, 2011

umz93 said:
I stated that i wanted intensity to equal 1

That works, i evaluated it as a bloody percentage instead of a difference

OmmU · Oct 24, 2011

jamesfirst said:
I just made 2 = e^0.1 L

In 2 / 0.1 = L

L = what ever the answer is.

barbernator said:
That works, i evaluated it as a bloody percentage instead of a difference

I think I did the above, then evaluated as a % increase.

o_0 · Oct 24, 2011

OmmU said:
I think I did the above, then evaluated as a % increase.

it doesnt work...

the percentage increase isn't linear

Neologism · Oct 24, 2011

DAMNNN I got this answer, I wrote it down but did other stuff after it, idk why just freaked a little. Will I still get the marks, I had the answer right there?

aheyhey · Oct 26, 2011

Yes I second 10ln2.
I left it exact cos they didn't ask for an approximation ... at least from what I can recall.

MrBrightside · Oct 26, 2011

aheyhey said:
Yes I second 10ln2.
I left it exact cos they didn't ask for an approximation ... at least from what I can recall.

Yeah I got that as well.

Intensity/Decibel question (1 Viewer)

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