Calculating Sound Intensity Levels for a Baby's Parents

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SUMMARY

The discussion focuses on calculating the difference in sound intensity levels experienced by a father and mother due to their respective distances from a baby's mouth. The father is 30 cm away, while the mother is 150 cm away. The solution involves using the formula for sound intensity levels, specifically B = 10 log (I/Io), and applying the area of a spherical shell, A = 4πr². The final calculated difference in sound intensity levels is approximately -13.98 dB, indicating that the mother perceives a significantly lower sound intensity than the father.

PREREQUISITES
  • Understanding of sound intensity and its calculation
  • Familiarity with logarithmic functions and their properties
  • Knowledge of spherical geometry and area calculations
  • Basic principles of wave propagation in physics
NEXT STEPS
  • Study the relationship between distance and sound intensity in acoustics
  • Learn about the properties of logarithms, particularly in the context of sound calculations
  • Explore the concept of sound intensity levels in different environments
  • Investigate the effects of distance on sound perception in real-world scenarios
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Students in physics, acoustics researchers, and parents interested in understanding sound intensity levels in relation to distance from sound sources.

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Homework Statement



A baby's mouth is 30 cm from her father's ear and 150 cm from her mother's ear. Compute the difference, in dB, between the sound intensity levels heard by the father and by the mother.

Homework Equations



Intensity = Power/ Area; power = 1/2 * mass per unit length*angular frequency^2*Area^2*v
Intensity = 1/2*Bulk modulus*Area^2*k*angular frequency
B = 10log (I/Io)

The Attempt at a Solution



i understand how to get the velocity (344m/s), but what about the mass per unit length, angular frequency and area

I don't know how to do this problem with such little info, unless I'm using the wrong formulas

ANY help would be much appreciated
 
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You're subtracting logs, what rule is there for subtracting logs?
 
u divide them, i understand that, however but where does the distance come into play? in the area?
 
Oh, use the first formula. You're dividing so the power should cancel, right?
 
yeah, the powers cancel out, but how does 30 cm and 150 cm relate to area

so far i have

answer =10 log (A * 1/A)
 
Well the sound is propagating like a spherical shell, A=4πr^2.
 
ah-ha

answer = 10 log (r^2/R^2) = 10 log ( 30^2 / 150^2) = -13.9794 dB

am i supposed to put 150^2/30^2 in the log, or is it fine the way it is?

thanks
 

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