YAPhysicsT: Sound Waves

[Ricemarine]

Having a hard time trying to figure out this question since Power is not known...

-One loud speaker is put between two people 110m apart
-One records a level of 60 dB, the other 80 dB

-How far is the speaker for each person?

So from what I can tell, person(1) has a distance of R(for radius), the other has a distance of 110-R.

So from the equations I have to utilize.
(I = P/A or I = P/4piR^2)
and
Beta = 10dB * log(I/Io) in which Io = 10^-12 w/m^2...

Plugging in 60 dB for beta, I get [ log I = 60dB + log(10^-12 w/m^2) ] / 10dB
in which I = 63095 w/m^2 (not sure)...

But then now I'm wondering what would I do about Power, since the Power is not actually known....

Thanks...

 

[BrownTown]

You don't even need to know what the absolute values are, 20dB more is 100 times as powerful, and since the power drops off as the inverse square of the distance you simply know that one person is 10 times closer than the other so one person is 10m away and the other is 100m away. At least that seems correct after a few seconds of looking at it, I've never had any classes on such things, but jsut knowing the units.

 

[silverpig]

Originally posted by: BrownTown
You don't even need to know what the absolute values are, 20dB more is 100 times as powerful, and since the power drops off as the inverse square of the distance you simply know that one person is 10 times closer than the other so one person is 10m away and the other is 100m away. At least that seems correct after a few seconds of looking at it, I've never had any classes on such things, but jsut knowing the units.

Not only that, but the numbers work out really easily.

 

[BrownTown]

Originally posted by: silverpig

Originally posted by: BrownTown
You don't even need to know what the absolute values are, 20dB more is 100 times as powerful, and since the power drops off as the inverse square of the distance you simply know that one person is 10 times closer than the other so one person is 10m away and the other is 100m away. At least that seems correct after a few seconds of looking at it, I've never had any classes on such things, but jsut knowing the units.

Not only that, but the numbers work out really easily.

Yeah, thats why I posted it since I wasn't really 100% sure, but then again I've had teachers do that on purpose and have it actually be the wrong answer to try and trick you.

 

[Tiamat]

Can you assume inverse square law, or is there a reverberant field to muck this problem up?

n/m I didnt see the 110m part which helps simplify this problem greatly to just inverse square law for all intents and purposes.

 

[Ricemarine]

Originally posted by: BrownTown

Originally posted by: silverpig

Originally posted by: BrownTown
You don't even need to know what the absolute values are, 20dB more is 100 times as powerful, and since the power drops off as the inverse square of the distance you simply know that one person is 10 times closer than the other so one person is 10m away and the other is 100m away. At least that seems correct after a few seconds of looking at it, I've never had any classes on such things, but jsut knowing the units.

Not only that, but the numbers work out really easily.

Yeah, thats why I posted it since I wasn't really 100% sure, but then again I've had teachers do that on purpose and have it actually be the wrong answer to try and trick you.

So ok, I understand that part now, since I remember my teacher discussed about this... But to be sure, I would like to double check if you guys agree with this work (to avoid massive knockoffs for not showing work, or for using the wrong equations to solve)...

so for the decibel sound level, I can plug in

Beta = 10 * logx (what should I do about the I/Io?)
---which translates to---
(80dB - 60dB) = 10 * logx
in which it would become
2 dB = logx
100 = x (since logx inversed = 10^x)

Wait......

 

[Ricemarine]

Wait, I'm confused by something then... if 20 dB is 100 times as powerful, does't that mean that Person(1) is 100 times closer than Person(2)? 😕

 

[dighn]

inverse square law: p = k/(r^2), 1/10 reduction in r causes p to increase by 100x

look at it this way: p1/p2 = 100 = (k/r1^2) / (k/r2^2) = (r2/r1)^2 => 10 = r2/r1

 

[Ricemarine]

Originally posted by: dighn
inverse square law: p = k/(r^2), 1/10 reduction in r causes p to increase by 100x

look at it this way: p1/p2 = 100 = (k/r1^2) / (k/r2^2) = (r2/r1)^2 => 10 = r2/r1

Is there a proof I could possibly use or a workaround due to the fact my professor hasn't discussed the inverse square law? The inverse square law isn't even in my book.

 

[dighn]

Originally posted by: Ricemarine

Originally posted by: dighn
inverse square law: p = k/(r^2), 1/10 reduction in r causes p to increase by 100x

look at it this way: p1/p2 = 100 = (k/r1^2) / (k/r2^2) = (r2/r1)^2 => 10 = r2/r1

Is there a proof I could possibly use or a workaround due to the fact my professor hasn't discussed the inverse square law? The inverse square law isn't even in my book.

I = P/4piR^2

that IS an inverse square equation, I = k/r^2 where k = P/4pi. you don't need to know P because when you do I1/I2 the P becomes irrelevant