Intensity/Decibel question (1 Viewer)

o_0

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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?
 
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Ivy Mike

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

Normal Intensity:





Double Intensity:







times louder
 

Tavin

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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

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

Normal Intensity:





Double Intensity:







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

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

OldMathsGuy

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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
 

Neologism

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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

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

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