Q 10 c (iii) (2 Viewers)

Ivy Mike · Oct 24, 2011

I keep hearing that people got ln(2)/0.1 for this question... starting to panic a little bit.

This is what I did. I'd appreciate it if anybody 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

someth1ng · Oct 24, 2011

I did something similar along those lines but ended with ln2 - I did your method later on but I'm not sure what the "correct" method is.

Tasha270494 · Oct 24, 2011

yeah. see I did that too and everyone's saying different stuff.
I guess we interpreted it as "by how many TIMES is it greater when doubled"
rather than the question that asked "by HOW MUCH will sound have increased when doubled"

.. I dont know.

Ivy Mike · Oct 24, 2011

Tasha270494 said:
yeah. see I did that too and everyone's saying different stuff.
I guess we interpreted it as "by how many TIMES is it greater when doubled"
rather than the question that asked "by HOW MUCH will sound have increased when doubled"

.. I dont know.

Yeah I see what you're trying to say...

When I take L2-L1 I get ln(2)/0.1 so hopefully they'll both be right answers.

IMO the wording for a couple of the questions was a bit vague.

mathemalia · Oct 24, 2011

Ivy Mike said:
I keep hearing that people got ln(2)/0.1 for this question... starting to panic a little bit.

This is what I did. I'd appreciate it if anybody 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

$I = 10^-12 * e^{0.1L}$
$L =\frac{ln\frac{I}{10^{-12}}}{0.1}$

If you double I, it becomes 2I:

$L1 = \frac{ln\frac{2I}{10^{12}}}{0.1}$

$L1 = \frac{ln2 + lnI - ln10^{12}}{0.1}$
$L1 = \frac{ln2}{0.1} +\frac{ ln\frac{I}{10^{-12}}}{0.1}$

friend177 · Oct 24, 2011

Ivy Mike said:
I keep hearing that people got ln(2)/0.1 for this question... starting to panic a little bit.

This is what I did. I'd appreciate it if anybody 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

Q10 A iii

timinator1993 · Oct 24, 2011

Yea, definitley poor wording, they should take both answers :/

CecilyMare · Oct 24, 2011

Yay. I got 10ln2

Ivy Mike · Oct 24, 2011

CecilyMare said:
Yay. I got 10ln2

Post your working?

michaeljennings · Oct 24, 2011

Ivy Mike said:
Post your working?

Generally, it would be the form
$I=Ae^{kL}$
Where A=initial intensity, so we're given that $A=10^{-12}$
You want to find how L changes, when the initial intensity doubles.
That is,
$2(10^{-12})= 10^{-12}e^{0.1L}$
This then becomes
$2= e^{0.1L}$
$ln2= lne^{0.1L}$
Therefore, when the original intensity doubles, loudness becomes:
$L=\frac{ln2}{0.1}$
$=10ln2=6.93$

CecilyMare · Oct 24, 2011

Therefore, the loudness has increased by 10ln2 decibels

:3

michaeljennings one is better.

elvishtears · Oct 24, 2011

yeah I ended up getting around 7% increased loudness, Just doubled the value in ii and showed the ratio

Neologism · Oct 24, 2011

Ivy Mike said:
I keep hearing that people got ln(2)/0.1 for this question... starting to panic a little bit.

This is what I did. I'd appreciate it if anybody 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

I did the exact same as you! I got 1.025... and then I said that the loudness increased by 2.5% :/ Am I wrong? pretty sure I am

barbernator · Oct 24, 2011

Neologism said:
I did the exact same as you! I got 1.025... and then I said that the loudness increased by 2.5% :/ Am I wrong? pretty sure I am

I got that as well, but then i realised after the exam that the increase for each time the intensity doubles is not linear. so it doesn't increase by a fixed percentage, rather a fixed amount of decibels, so the answer is 10ln2 not 2.5% increase

perpetuity · Oct 24, 2011

it's way to easy u idiots

Tasha270494 · Oct 25, 2011

barbernator said:
I got that as well, but then i realised after the exam that the increase for each time the intensity doubles is not linear. so it doesn't increase by a fixed percentage, rather a fixed amount of decibels, so the answer is 10ln2 not 2.5% increase

yeah, I think your right..

L $L=\frac{ln(10^{12})}{0.1}=276.31..$
Double Intensity;
$L=\frac{ln(10^{12}\times 2)}{0.1}=283.24..$
Quadruple Intensity;
$L=\frac{ln(10^{12}\times 4)}{0.1}=290.173..$

If you look at it, difference = constant of 10ln2
Percentage is slightly different between them..
(and I used exact values on my calculator)

Q 10 c (iii) (2 Viewers)

Ivy Mike

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someth1ng

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Tasha270494

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Ivy Mike

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mathemalia

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