Help! Assignment due tonight ! Sound And Waves Question

  • #1

Homework Statement


The sound intensity level 15.0 m from a point source is 64.0 dB. At what distance will it be 53.0 dB?


Homework Equations



Beta = 10log(I/Io)

I2=(r1/r2)^2 I1

The Attempt at a Solution



Found the intensity of 64dB @ 15m I1= 2.51*10^-6

Found the intensity of 53.0dB I2= 1.99 * 10 ^ -7

I used the second forumula ^ to solve for r2...

I got r2 = 10.5m.... not sure what to do from here or if I am on the right path......... frustrated
 

Answers and Replies

  • #2
berkeman
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Homework Statement


The sound intensity level 15.0 m from a point source is 64.0 dB. At what distance will it be 53.0 dB?


Homework Equations



Beta = 10log(I/Io)

I2=(r1/r2)^2 I1

The Attempt at a Solution



Found the intensity of 64dB @ 15m I1= 2.51*10^-6

Found the intensity of 53.0dB I2= 1.99 * 10 ^ -7

I used the second forumula ^ to solve for r2...

I got r2 = 10.5m.... not sure what to do from here or if I am on the right path......... frustrated
The new distance will have to be greater, since the 2nd sound power level is lower, right?

Try thinking of it this way... What is the difference between the two power levels in dB?

And for that difference in dB, what does the ratio of the two Intensities need to be? And what does that imply for the ratio of the radii?
 
  • #3
gneill
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Looks like something went wrong in your solving for r2 in the last step. Take a close look at that.
 
  • #4
The new distance will have to be greater, since the 2nd sound power level is lower, right?

Try thinking of it this way... What is the difference between the two power levels in dB?

And for that difference in dB, what does the ratio of the two Intensities need to be? And what does that imply for the ratio of the radii?
The ratio of the two intensities is... (2.51*10^-6/1.99*10^-7) = 12.6, so the intensities decreased by a factor of 12.6 ...

so what does that mean for distance..
 
  • #5
berkeman
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The ratio of the two intensities is... (2.51*10^-6/1.99*10^-7) = 12.6, so the intensities decreased by a factor of 12.6 ...

so what does that mean for distance..
You had it in one of your equations. Intensity falls as a function of the radius _______
 
  • #6
You had it in one of your equations. Intensity falls as a function of the radius _______
(r1/r2)^2
 
  • #8
There you go!
ive tried this a i cant get it .. i dont know what im missing, assignments due in 2 hrs and I aiming for a 100.
 
  • #9
collinsmark
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ive tried this a i cant get it .. i dont know what im missing, assignments due in 2 hrs and I aiming for a 100.
Show us what you have (using your I2=(r1/r2)2 I1 equation [as mentioned within the last few or so posts]). :smile:
 
  • #10
Show us what you have (using your I2=(r1/r2)2 I1 equation [as mentioned within the last few or so posts]). :smile:
(1.99*10^-7)=(15/x)^2 * 2.51 * 10^-6

7.93*10^-14=(15/x)^2


6.29*10^-27=15/x

r2(x)= 2.38 *10^27.......large number something is wrong
 
  • #11
berkeman
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(1.99*10^-7)=(15/x)^2 * 2.51 * 10^-6

7.93*10^-14=(15/x)^2


6.29*10^-27=15/x

r2(x)= 2.38 *10^27.......large number something is wrong
You said that the ratio is 12.6, right? so 12.6 = the ratio of the radii squared. Take the square root to get the ratio, and that tells you how much farther away the second radius is...
 
  • #12
collinsmark
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(1.99*10^-7)=(15/x)^2 * 2.51 * 10^-6

7.93*10^-14=(15/x)^2
Something went wrong with exponents.
10a/10b = 10a-b, not 10a+b. :wink:

[Edit: Or just use berkeman's advice in the previous post.]
 
  • #13
You said that the ratio is 12.6, right? so 12.6 = the ratio of the radii squared. Take the square root to get the ratio, and that tells you how much farther away the second radius is...
square root =3.55+15=18.6..
 
  • #14
collinsmark
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square root =3.55+15=18.6..
Don't panic! :tongue2:

But you shouldn't be adding there. Go back to your equation. You know that

I1/I2 = (x/(15.0 m))2

And you've already determined that I1/I2 = 12.6.

So x is .... :smile:
 
  • #15
Don't panic! :tongue2:

But you shouldn't be adding there. Go back to your equation. You know that

I1/I2 = (x/(15.0 m))2

And you've already determined that I1/I2 = 12.6.

So x is .... :smile:
53.2 m! i got it after, that was the right answer...appreciate the help! once i get frustrated with a problem i shut down..ill have to make sure i understand whats going on in this for future exams !
 
  • #16
Also, in the above equation with the workings I had (15/x) not (x/15)^2, so i could have figured it out using this method and the other method correct???
 
  • #17
collinsmark
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53.2 m! i got it after, that was the right answer...appreciate the help! once i get frustrated with a problem i shut down..ill have to make sure i understand whats going on in this for future exams !
Great! :smile:

Now that you've got the answer, I want to show you something that you should know (definitely for future exams).

[tex] 10 \ \log \left( \frac{a}{b}\right)^2 = 20 \ \log \left( \frac{a}{b}\right) [/tex]

Keeping the above in mind, here is a much easier way to solve the original problem that we've been working on. The problem statement says,
The sound intensity level 15.0 m from a point source is 64.0 dB. At what distance will it be 53.0 dB?
64 dB - 53 dB = 11 dB.

And you know the power flux decreases as a function of r2 (this has been discussed in previous posts in this thread). So,
11 dB = 20 log (x/(15 m))​
Solve for x. :wink: (yes, it's really that simple. Practice stuff like this for awhile, and you'll get the hang of it.)
 
  • #18
collinsmark
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Also, in the above equation with the workings I had (15/x) not (x/15)^2, so i could have figured it out using this method and the other method correct???
We inverted everything, remember?.

12.58925 = (x/(15.0 m))2

1/12.58925 = ((15 m)/x)2

And where does this 12.58925 come from?

10 log (12.58925) = 11 dB!
 
  • #19
Makes so much sense... this problem was really not that hard. :redface:
 
  • #20
berkeman
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Beauty, collinsmark.
 

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