Calculating Energy Absorbed by Eardrum from Intensity & Area

  • Thread starter Thread starter jmb07
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on calculating the energy absorbed by an eardrum when exposed to a sound wave with an intensity of 80.1 dB and an area of 0.7x10^-4 m² over a duration of 4 minutes. The initial calculation using the formula Power = Intensity × Area yields a power of 5.01x10^-6 W, leading to an energy absorption of 1.35 J. However, the correct energy absorbed is 1.72 J, indicating a need for proper conversion from decibels to power using the reference intensity of 10^-12 W/m².

PREREQUISITES
  • Understanding of sound intensity and its measurement in decibels (dB)
  • Knowledge of the formula for calculating power from intensity and area
  • Familiarity with energy calculations involving power and time
  • Basic grasp of logarithmic functions and their application in sound intensity
NEXT STEPS
  • Learn how to convert decibels to power using the formula 80.1 dB = 10 log10(P1/Po)
  • Study the reference intensity level for sound in air, specifically 10^-12 W/m²
  • Explore the relationship between intensity, area, and energy absorption in acoustics
  • Investigate the effects of sound intensity on human hearing and eardrum response
USEFUL FOR

Acoustics researchers, audio engineers, and students studying sound physics who are interested in the quantitative analysis of sound energy absorption by biological structures.

jmb07
Messages
27
Reaction score
0
A sound wave with intensity of 80.1 dB is incident on an eardrum of Area = 0.7x10^-4. How much energy is absorbed by the eardrum in 4 minutes?


I know this much... Power = (Intensity)(Area)..so...P=(80.1)(0.7x10^-4)

and Energy = (power)(time) so...E = (80.1)(0.7x10^-4)(240 sec) = 1.35


However, this is not the correct answer. Am i not doing the right conversions or what?? What looks wring here? (the correct answer is 1.72)

I know it probably has something to do with converting dB to something...im just not sure how..
 
Physics news on Phys.org
Isn't a decibel a log of the ratio of the sound to a reference level?

So ... 80.1 dB = 10log10(P1/Po)

Po is what? 10-12W/m2 ?
 

Similar threads

Calculating Energy Absorption by an Eardrum Exposed to Sound Waves
  • · Replies 4 ·
Replies
4
Views
2K
Intensity of sound wave and energy
  • · Replies 1 ·
Replies
1
Views
2K
Calculating Sound Energy & Intensity: A Scientist's Approach
  • · Replies 1 ·
Replies
1
Views
12K
How Much Energy Hits Each Eardrum from a 2500W Sound Source 20m Away?
  • · Replies 2 ·
Replies
2
Views
2K
Decibels, Waves, Kinetic Energy and Mosquitos
  • · Replies 2 ·
Replies
2
Views
20K
Intensity of Stars and the Inverse Square Law Problem
  • · Replies 10 ·
Replies
10
Views
3K
Calculating the power output from a persons mouth
  • · Replies 3 ·
Replies
3
Views
2K
Need help w/ problem: Sound intensity and energy
  • · Replies 1 ·
Replies
1
Views
5K
Calculating Energy Received by a Human Eardrum at Threshold of Hearing
  • · Replies 2 ·
Replies
2
Views
8K
Calculating Power at source from Intensity at distance R
  • · Replies 1 ·
Replies
1
Views
5K