Sound waves and eardrum problem

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SUMMARY

The discussion focuses on calculating the energy absorbed by an eardrum when exposed to a 50 dB sound wave with an area of 4.1 x 10-5 m2. The user attempted to use the formula I = 2π2f2A2pv but encountered errors in their calculations. The correct approach involves using the decibel formula I = 10 log(P/P0), where P0 = 10-12 W/m2. This method will yield the energy absorbed per second and the time required to accumulate 1.0 J of energy.

PREREQUISITES
  • Understanding of sound intensity and decibels
  • Familiarity with the formula I = 10 log(P/P0)
  • Basic knowledge of physics related to wave energy
  • Ability to manipulate scientific notation and units
NEXT STEPS
  • Calculate the energy absorbed by the eardrum using the correct intensity formula
  • Determine the time required to accumulate 1.0 J of energy at the calculated absorption rate
  • Explore the relationship between sound intensity and pressure using P = I * A
  • Investigate the effects of different sound levels on human hearing
USEFUL FOR

Students studying physics, audiologists, sound engineers, and anyone interested in the effects of sound waves on human anatomy.

thschica
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A 50 dB sound wave strikes an eardrum whose area is 4.1 * 10-5 m2.
(a) How much energy is absorbed by the eardrum per second?
W
(b) At this rate, how long would it take your eardrum to receive a total energy of 1.0 J?
years

I have tried to solve using I=2pi^2f^2A^2pv

and got f=280 ; v=343;p=1.29;I=1*10^-12 and A =1.28e-18 its wrong though
 
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thschica said:
A 50 dB sound wave strikes an eardrum whose area is 4.1 * 10-5 m2.
(a) How much energy is absorbed by the eardrum per second?
W
(b) At this rate, how long would it take your eardrum to receive a total energy of 1.0 J?
years

I have tried to solve using I=2pi^2f^2A^2pv

and got f=280 ; v=343;p=1.29;I=1*10^-12 and A =1.28e-18 its wrong though
Decibels are a little tricky to work with. I think all you need to know is that I = 10 log(P/P_0) where P is the power/area and P_0 = 10^{-12} W/m^2

AM
 

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