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Sound Intensity/Power Question. Help.

  • #1

Homework Statement


A point source emits 32.0 W of sound isotropically. A small microphone intercepts the sound in an area of 0.560 cm2, 205 m from the source. What is the sound intensity at that distance?

= 6.06×10-5 W/m^2

What is the power intercepted by the microphone?



Homework Equations





The Attempt at a Solution



Got the answer for part A, no im stuck on part B............erg.

I figured, well i found the intesnity at 250m so;

I=P/A .. therefore,

P=I*A
= (6.06*10^-5)(0.000056m^2)
= 3.00 * 10^-9 W..... but this was wrong. No idea what im doing wrong here.

These are the only questions i can't get on a long assignment... frustated.
 

Answers and Replies

  • #2
collinsmark
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P=I*A
= (6.06*10^-5)(0.000056m^2)
= 3.00 * 10^-9 W..... but this was wrong. No idea what im doing wrong here.
Check your precision/significant figures. :wink:
 
  • #3
Check your precision/significant figures. :wink:
When I use less then that it says use more sig figs !.... its an online assignment so it tells me when im wrong or right. I've calculated these questions a million times a million different times but no luck..
 
  • #4
collinsmark
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When I use less then that it says use more sig figs !.... its an online assignment so it tells me when im wrong or right. I've calculated these questions a million times a million different times but no luck..
I don't necessarily mean the number of the significant figures. :smile: Rather, I mean that there is some sort of big rounding/truncation error in your calculation.

The answer isn't three point zero zero...., It's three point something something... :wink:
 
  • #5
I don't necessarily mean the number of the significant figures. :smile: Rather, I mean that there is some sort of big rounding/truncation error in your calculation.

The answer isn't three point zero zero...., It's three point something something... :wink:
When i do the calculations on my calulator it just says.... 0.000000003, and nothing follows trails after the 3..

When I calculated part A) i didnt take any sig figs and multiplied it out by 0.0000560m
 
  • #6
collinsmark
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When i do the calculations on my calulator it just says.... 0.000000003, and nothing follows trails after the 3..

When I calculated part A) i didnt take any sig figs and multiplied it out by 0.0000560m
Check to see if your calculator has a setting to display things in scientific notation.

If not, you'll have to do things manually.

Try multiplying
(6.06 x 10-5)(5.6 x 10-5) = 6.06 x 5.6 x 10-5-5
 
  • #7
Check to see if your calculator has a setting to display things in scientific notation.

If not, you'll have to do things manually.

Try multiplying
(6.06 x 10-5)(5.6 x 10-5) = 6.06 x 5.6 x 10-5-5
Got it.. thanks a lot i would have never figured this one out... calculator never had scientific notation.
 
  • #8
collinsmark
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Got it.. thanks a lot i would have never figured this one out... calculator never had scientific notation.
It's great to know you figured it out. :smile:

But if you're dealing with sound intensity problems now, be prepared to deal with this sort of issue a lot! Sound intensities can vary over a huge dynamic range, and your going to need be able to represent very small numbers (such as 0.000000000001) to very large numbers.

If you haven't learned it yet, you are probably about to be introduced to the concept of decibels. Decibels were created and used for this very reason (although when it was invented, people didn't have calculators, so there's more to it.) Also, decibels let you perform calculations using addition instead of multiplication (that's the other reason). Decibels are still used today all the time, and not just for sound intensity, but also for radio frequency wave intensity (cell phones, wireless routers, Blutooth devices, etc), and other things.

But you might wish to consider a new calculator that has the capability of scientific notation, and one capable of exponential and logarithmic functions (decibels are all about exponentials and logarithms). :wink: (Or alternately, just get used to manually doing the things that your calculator can't do.)
 
  • #9
It's great to know you figured it out. :smile:

But if you're dealing with sound intensity problems now, be prepared to deal with this sort of issue a lot! Sound intensities can vary over a huge dynamic range, and your going to need be able to represent very small numbers (such as 0.000000000001) to very large numbers.

If you haven't learned it yet, you are probably about to be introduced to the concept of decibels. Decibels were created and used for this very reason (although when it was invented, people didn't have calculators, so there's more to it.) Also, decibels let you perform calculations using addition instead of multiplication (that's the other reason). Decibels are still used today all the time, and not just for sound intensity, but also for radio frequency wave intensity (cell phones, wireless routers, Blutooth devices, etc), and other things.

But you might wish to consider a new calculator that has the capability of scientific notation, and one capable of exponential and logarithmic functions (decibels are all about exponentials and logarithms). :wink: (Or alternately, just get used to manually doing the things that your calculator can't do.)
Thank you!, appreciate the information!.. could you check out my other questions if you have sometime! i got 2 hrs until assignment is due and i need some help....
 

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