- #1
Dr. S
- 13
- 0
Thank you for the help!
Last edited:
Sound Intensity Level and Hearing Problem
- Thread starter Dr. S
- Start date
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Homework Help Overview
The discussion revolves around sound intensity levels and the application of the inverse square law in calculating sound intensity at different distances from a source. Participants are exploring the relationship between sound intensity and distance, particularly in the context of a hearing problem scenario.
Discussion Character
- Exploratory, Assumption checking
Approaches and Questions Raised
- Some participants suggest using the inverse square law to relate sound intensity at different distances, while others seek clarification on how to apply this concept in the given context.
Discussion Status
The discussion is active, with participants sharing their thoughts on the use of the inverse square law. There is a focus on understanding how to apply this principle rather than reaching a definitive conclusion.
Contextual Notes
Participants are navigating the specifics of sound intensity calculations and the assumptions underlying the use of the inverse square law in this scenario.
- #2
Dr. S
- 13
- 0
Could really use some input!
- #3
Curious3141
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Dr. S said:
It's correct, but I would've just used the inverse square law to work out the sound intensity at 2m from the one at 3m instead of calculating the power of the source.
- #4
Dr. S
- 13
- 0
Curious3141 said:
Inverse square law? How I would I go about using that here?
Thank you, by the way. :)
- #5
Curious3141
Homework Helper
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Dr. S said:
[itex]\frac{I_{2m}}{I_{3m}} = \frac{3^2}{2^2}[/itex].
And you're welcome.
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