Intensity of sound wave & energy passes into ear

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SUMMARY

The discussion centers on the calculation of sound intensity and its relationship to energy transfer into the ear. The formula used is I = P/(4∏r^2), where P represents power and r is the distance from the sound source. The calculated intensity was found to be 5.67 x 10^-8 W/m² using values of 6.3 x 10^-6 for power and 1.5 x 10^-4 for energy over time. The area of the ear canal is utilized in the calculation because it is at the same distance of 5 meters from the sound source, thereby incorporating the distance into the intensity assessment.

PREREQUISITES
  • Understanding of sound intensity and its calculation using I = P/(4∏r^2)
  • Familiarity with the concepts of power (P) and energy (E) over time (t)
  • Basic knowledge of wave energy and its propagation
  • Ability to perform scientific notation calculations
NEXT STEPS
  • Study the principles of sound wave propagation and intensity
  • Learn about the relationship between distance and sound intensity in different mediums
  • Explore the concept of power in acoustics and its measurement techniques
  • Investigate the effects of ear canal geometry on sound perception
USEFUL FOR

Students studying physics, particularly those focusing on acoustics, sound engineers, and anyone interested in the mechanics of sound wave energy transfer into the human ear.

coconut62
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Homework Statement



Please refer to the image. Question (a).

Homework Equations


I= P/(4∏r^2)
P=E/t


The Attempt at a Solution



After a few attempts, I found the way to get the answer:

(6.3 x 10^-6) x (1.5 x 10^-4) x 60 = 5.67 x 10^-8

But I don't understand why the area of the ear canal is used, instead of the area which the energy carried by the wave passes through. From what I understand, intensity depends on the distance from the source, so how come we don't need to take into account the distance of 5.0m?
 

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coconut62 said:

Homework Statement



Please refer to the image. Question (a).

Homework Equations


I= P/(4∏r^2)
P=E/t

The Attempt at a Solution



After a few attempts, I found the way to get the answer:

(6.3 x 10^-6) x (1.5 x 10^-4) x 60 = 5.67 x 10^-8

But I don't understand why the area of the ear canal is used, instead of the area which the energy carried by the wave passes through. From what I understand, intensity depends on the distance from the source, so how come we don't need to take into account the distance of 5.0m?
attachment.php?attachmentid=58367&d=1367331297.jpg


The sound intensity at 5 m from the sound source is given. The ear canal is also at 5 m from the source. Therefore the distance of 5 m is taken into account.
 
Thank you.
 

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