Intensity and Power of isotropic sound

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SUMMARY

The discussion focuses on calculating the sound intensity and power intercepted by a microphone from an isotropic sound source emitting at 3000 Hz with a power of 34 watts. The microphone, with an area of 0.74 cm², is positioned 158 meters away from the source. The correct approach involves using the formula for intensity, I = power/area, while considering the spherical surface area at the distance of 158 meters to determine the effective area for the microphone. The user seeks clarification on the calculations and the relationship between the microphone's area and the total surface area of the sphere at that distance.

PREREQUISITES
  • Understanding of sound intensity and power calculations
  • Familiarity with the concept of isotropic sound sources
  • Knowledge of spherical geometry and surface area calculations
  • Basic principles of acoustics and sound wave propagation
NEXT STEPS
  • Calculate the surface area of a sphere with a radius of 158 meters
  • Determine the fraction of the sphere's surface area represented by the microphone's area
  • Learn about the relationship between sound intensity and distance from a point source
  • Explore the implications of frequency on sound intensity and power distribution
USEFUL FOR

Students in physics, acoustics researchers, audio engineers, and anyone involved in sound measurement and analysis will benefit from this discussion.

turandorf
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Homework Statement


An electronic point source emits sound isotropically at a frequency of 3000 Hz and a power of 34 watts. A small microphone has an area of 0.74 cm2 and is located 158 meters from the point source.
a) What is the sound intensity at the microphone ?

b) What is the power intercepted by the microphone?


Homework Equations


I=(power)/(area)
Avg power=(1/2)(mu)(v)(w^2)A^2 (not sure what all these mean)


The Attempt at a Solution


I tried Intensity=(power)/(area) but that was wrong. I had converted the area to m^2 too…any ideas?
 
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At 158 meters don't you have the surface of a sphere that has a radius of 158 meters?

Maybe if you know what % of this area that the area of the microphone represents on the sphere, you might get an idea of how much of the original 34 watts that's pushing the air around at the speaker is available for detecting?

http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.html#c1
 

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