Sound wave and energy can you help me?

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SUMMARY

The discussion centers on calculating the energy transferred to an eardrum from a sound wave with an intensity of 2x10^(-3) W/m^2. The user initially calculated the power using the formula P=IA, where A is the area of the eardrum, mistakenly using the radius instead of the diameter. The correct calculation yields a power of approximately 2.2610x10^(-7) W, leading to an energy transfer of 1.3572x10^(-5) J over one minute. The error was identified as a miscalculation in the area due to incorrect use of the diameter.

PREREQUISITES
  • Understanding of sound wave intensity and its measurement
  • Knowledge of the formula for power (P=IA)
  • Familiarity with the area calculation for a circle (A=πr^2)
  • Basic principles of energy transfer over time (E=Pt)
NEXT STEPS
  • Review the principles of sound wave intensity and its effects on human perception
  • Learn about the relationship between sound intensity and energy transfer
  • Study the mathematical derivation of power and energy formulas in physics
  • Explore common mistakes in area calculations and how to avoid them
USEFUL FOR

Students in physics, audio engineers, and anyone interested in the quantitative analysis of sound energy and its impact on human physiology.

kingwinner
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A sound wave with intensity 2x10^(-3) W/m^2 is perceived to be modestly loud. Your eardrum is 6.0 mm in diameter. How much energy will be transferred to your eardrum while listening to this sound for 1.0 min?

P=IA=(intensity)(area)
=2x10^(-3) * (pi (6x10^-3)^2)
=2.2610x10^-7 W

P=delta E/ delta t
=2.2610x10^-7 J/s * 60 s
=1.3572x10^-5 J

But this is not the right answer. I have no idea what I did wrong. Can someone please help me? Any help is greatly appreciated! Thanks!
 
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For one thing the diameter is 6mm. Not the radius. Change your pi*r^2 number.
 
Thanks for pointing it out! I rarely make this kind of silly mistake
 

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