Easy question, cant get the right result but have all fomulars

  • #1
mimi.janson
80
0
easy question, can't get the right result but have all fomulars!

Homework Statement



hi everyone i have a question i need to solve, and i aready found out what i need to do, but i seem to be getting a too big result and i don't get why,so i would appreciate it if someone would explain what i was doing wrong.

it is explained that the sound intensity (I) is defined as the sound power (P) per unit A (4*∏*r2)

Something is placed in point A and the sound it sends 10 meters to point B the volume of sound is 110 deciBell. so i had to find the sound power (P)

Homework Equations


I have this following formula i can use

I=P/(4*∏*r2)




The Attempt at a Solution



so i thought i would be able to solve it by isolating P and tried this way

110=P/(4*∏*102)

110=P/1256,63

110*1256,63=(P/1256,63)*1256,63

138230,07 Watt=P


But it seems like it is too big of a result ?
what do you guys think ? and thanks for everyone who would try to help me out
 

Answers and Replies

  • #2
Muphrid
834
2


The 110 dB volume is not a sound intensity. You should have some formula to convert from volume to intensity (which is measured in watts per square meter). This formula should involve logarithms as well as some reference sound intensity level (probably called [itex]I_0[/itex]).

Edit: in other words, a volume in decibels means that you have something 110 dB louder than some baseline intensity, and without knowing that baseline, you can't make sense of the equation at all.
 
  • #3
mimi.janson
80
0


The 110 dB volume is not a sound intensity. You should have some formula to convert from volume to intensity (which is measured in watts per square meter). This formula should involve logarithms as well as some reference sound intensity level (probably called [itex]I_0[/itex]).

Edit: in other words, a volume in decibels means that you have something 110 dB louder than some baseline intensity, and without knowing that baseline, you can't make sense of the equation at all.


Ok so i did convert it and the 110 dB became 0,1 I is that right ? and then i tried to put it in the formula from before I=P/(4*∏*r2)

0,1=P/(4*∏*102)
125 watt=P

did i do it right this time ?
 
  • #4
Muphrid
834
2


If your reference intensity is [itex]I_0 = 10^{-12}[/itex] watts per square meter, sure. That does seem to be standard.
 
  • #5
mimi.janson
80
0


If your reference intensity is [itex]I_0 = 10^{-12}[/itex] watts per square meter, sure. That does seem to be standard.

Yes that is the reference intensity i need to use, since it is the given one in my notes.

But if it is asked to determine the function that gives the dB volume as the function of the meters. would I be able to use I=P/(4*∏*r2) again and that would be f(x)=P/(4*∏*x2) ? but that would leave me with both P and x undefined. Do you have any idea on how this is supposed to be done?

Because the want the sound volume in dB as a function of the distance, but i will always have two undefined if i have to make these informations to a function which are the distance and sound volume so how can i possibly make a function with so many undefined?
 
  • #6
Muphrid
834
2


No, now that you've found the power P, that is the P you use. You say [itex]I = I(r)[/itex] by the formula you have, and the volume [itex]L(r) = 10 \log \frac{I(r)}{I_0}[/itex].
 
  • #7
mimi.janson
80
0


No, now that you've found the power P, that is the P you use. You say [itex]I = I(r)[/itex] by the formula you have, and the volume [itex]L(r) = 10 \log \frac{I(r)}{I_0}[/itex].

I get it now i already found out what i must do. thank you
 
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