help pl0x (1 Viewer)

[jason2kool]

http://oi49.tinypic.com/anjvao.jpg

Question is from 2u hsc 2011, but the solutions that they gave confused me,
can i get some help please?
Thanks in advance

 

[Ph03nix]

What part exactly are you confused with?

 

[jason2kool]

part 2 and 3

 

[qwerty44]

http://www.boardofstudies.nsw.edu.a...pdf_doc/mathematics-hsc-sample-answers-11.pdf

 

[Ph03nix]

For Part 2

If Intensity is greater than 8.1 x 10^-9, ear damage occurs so that means the greatest intensity at which no ear damage occurs is 8.1 x 10^-9. So to find the maximum loudness of a sound so that no ear damage occurs, you have to sub in 8.1 x 10^-9 for I in the equation and solve for L.

8.1 x 10^-9 = 10^-12 x e^0.1L
8.1 x 10^3 = e^0.1L
ln(8.1 x 10^3) = 0.1L
ln(8.1 x 10^3)/0.1 = L

Part 3
You have to find the difference in loudness when intensity is doubled. So

I = 10^-12 x e^0.1L
For Intensity to have doubled, loudness must have changed so
Let L = L* for 2I
2I = 10^=12 x e^0.1(L*)

Also,
2I = 2 x I
= 2 x 10^-12 x e^0.1L

So,

2 x 10^-12 x e^0.1L = 10^-12 x e^0.1(L*)
2e^0.1L = e^0.1L*
ln (2 x e^0.1L) = 0.1L*
ln 2 + 0.1L = 0.1L*
ln 2 = 0.1L* - 0.1L
L* - L = ln2/0.1



Sorry for horrible format ^

 

[Leffife]

(ii)
8.1 x 10^(-9) = 10^(-12) x e^(0.1L)
e^(0.1L) = (8.1 x 10^(-9)) / 10^(-12)
e^(0.1L) = 8100
0.1L loge = loge 8100
L = loge8100/0.1
L = 89.996 ...... etc
L = 90 (nearest whole number)

Conclusion: maximum loudness is 90db.

 

[Leffife]

I think there are more than one option for part (iii) but anyways i'll use my favourite method.

I = Ioe^(0.1L)

Let I = 2Io
Thus, 2Io = Ioe^(0.1L)
e^(0.1L) = 2
0.1L = loge2
L = 10loge2
L = 6.931..... etc
L = 7 (nearest whole number)

Conclusion: There was an increase of 7db.

 

[Leffife]

Please also check if I did any errors, since I usually do silly mistakes somewhere.