Intensity/Decibel question (1 Viewer)

[o_0]

I think this was in question 10? part a?

what did you guys get for the last part, where it asked for something like when intensity is doubled, how much is loudness increased?

 

[Tavin]

I can't do latex... forgive me. But I believe the question was I = 10^-12 x e^0.1L, yes?

So doubled intensity would've been 2 = e^0.1L

0.1L x lne = ln 2

L = (ln2/0.1)

Which came out to be 10ln2 or 6.93... Which was roughly 7 decibels. So every 7 decibels increased in loudness, the intensity doubled.

 

[someth1ng]

I just got ln2.

You need to do the L of the double I over L of the normal I.

 

[friend177]

I can't do latex... forgive me. But I believe the question was I = 10^-12 x e^0.1L, yes?

So doubled intensity would've been 2 = e^0.1L

0.1L x lne = ln 2

L = (ln2/0.1)

Which came out to be 10ln2 or 6.93... Which was roughly 7 decibels. So every 7 decibels increased in loudness, the intensity doubled.

Yeah something close to 7.

 

[Tavin]

I just got ln2.

I could be wrong.

 

[o_0]

are you sure thats right???

 

[friend177]

I could be wrong.

You'll get a mark if you show stuff do get that

 

[Tavin]

You'll get a mark if you show stuff do get that

Could you explain where I went wrong?

 

[friend177]

Could you explain where I went wrong?

Basically what Tavin did

 

[jamesfirst]

I just made 2 = e^0.1 L


In 2 / 0.1 = L


L = what ever the answer is.

 

[mathemalia]

are you sure thats right???

If you double I, it becomes 2I:

 

[Tavin]

Basically what Tavin did

...I AM TAVIN. What did I do wrong?

 

[o_0]

,.. i don't understand your working out..


i let L2=10ln(2I*10^12), and L1=10ln(I*10^12)... got a percentage increase of about 7.7 percent?

 

[friend177]

...I AM TAVIN. What did I do wrong?

Oh lol I thought you were the other dude, dont know lol I know my answer was close to 7

 

[damo_g]

I did this slightly different but got the same answer. Now that i see the way Tavin did it i don't know why that didn't occur to me.

I doubled the intensity from part (ii) and worked out the loudness. Then i took this new number and subtracted the loudness from part (ii) and got like 6.9...
So i got the loudness increased by 6.9... if the intensity doubled. My answer worked out to be the same amount as ln2/0.1.

would i still get full marks?

 

[someth1ng]

Weird.

 

[MrBrightside]

Yeah I got something close to 7, think it was like 6.93... I just doubled the previous answer from ii and used it in iii. Should be correct.

 

[damo_g]

Yeah I got something close to 7, think it was like 6.93... I just doubled the previous answer from ii and used it in iii. Should be correct.

Ye that's exactly what i did and got the 6.93 bit.

 

[Tavin]

I did this slightly different but got the same answer. Now that i see the way Tavin did it i don't know why that didn't occur to me.

I doubled the intensity from part (ii) and worked out the loudness. Then i took this new number and subtracted the loudness from part (ii) and got like 6.9...
So i got the loudness increased by 6.9... if the intensity doubled. My answer worked out to be the same amount as ln2/0.1.

would i still get full marks?

Presuming I'm right, (which I may not be) then yes, that's fine. I merely did it from the intensity at 0 decibels to 10ln2 decibels, and you did the same, simply at a different point.

I COULD BE WRONG THOUGH. Several people have suggested it, just haven't explained.

 

[someth1ng]

the 6.9... answer is 10ln2 - I just got ln2...1 mark down the drain but I have decent working for the rest I suppose.