Calculating Sound Intensity Levels for a Baby's Parents

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Homework Help Overview

The problem involves calculating the difference in sound intensity levels experienced by a baby's father and mother based on their respective distances from the baby. The context is within the subject area of acoustics and sound intensity calculations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between distance and sound intensity, questioning how to incorporate distance into the calculations. There is exploration of the formulas related to intensity and sound propagation.

Discussion Status

Participants are actively engaging with the problem, with some offering insights into the mathematical relationships involved. There is a recognition of the need to clarify how distance affects area in the context of sound intensity calculations.

Contextual Notes

Some participants express concern about the adequacy of the information provided for solving the problem, particularly regarding the mass per unit length and angular frequency. The discussion reflects uncertainty about the application of formulas and the interpretation of distances in the context of sound intensity.

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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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