Q 10 c (iii) (1 Viewer)

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) (1 Viewer)

Ivy Mike

Member

someth1ng

Retired Nov '14

Tasha270494

Member

Ivy Mike

Member

mathemalia

Member

friend177

Member

timinator1993

Member

CecilyMare

Member

Ivy Mike

Member

michaeljennings

Active Member

CecilyMare

Member

elvishtears

Member

Neologism

Member

barbernator

Active Member

perpetuity

Banned

Tasha270494

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)