Calculating Energy Absorbed by Eardrum from Intensity & Area

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In summary, the conversation discusses the calculation of energy absorbed by an eardrum when a sound wave with an intensity of 80.1 dB is incident on it for 4 minutes. The formula for power, P = (Intensity)(Area), is used to find the power of the sound wave. Then, the formula for energy, E = (power)(time), is used to calculate the energy absorbed. However, the answer obtained is not correct and the person is unsure if they are using the correct conversions. The conversation also mentions converting decibels to a reference level and using the value of Po = 10^-12W/m2. The correct answer is 1.72, which may be obtained by using the conversion
  • #1
jmb07
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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..
 
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  • #2
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 ?
 
  • #3


Hello, thank you for your question. I would like to provide a response to your query regarding the calculation of energy absorbed by the eardrum from intensity and area.

Firstly, let us clarify that the unit for intensity is decibels (dB), which is a logarithmic unit used to measure the relative intensity of a sound wave. However, intensity is not a direct measure of energy. To calculate the energy absorbed by the eardrum, we need to convert the intensity from dB to a linear scale, such as watts per square meter (W/m^2).

To convert from dB to W/m^2, we can use the formula: I = 10^(dB/10) * I0, where I0 is the reference intensity, which is typically 1x10^-12 W/m^2.

In this case, the intensity of the sound wave is 80.1 dB, so we can calculate the linear intensity as follows: I = 10^(80.1/10) * 1x10^-12 = 1.26 W/m^2.

Now, we can use the formula for power to calculate the energy absorbed by the eardrum: P = I * A, where A is the area of the eardrum. In this case, the area is given as 0.7x10^-4 m^2.

Therefore, the power absorbed by the eardrum is: P = (1.26 W/m^2) * (0.7x10^-4 m^2) = 8.82x10^-5 W.

To calculate the total energy absorbed in 4 minutes, we can use the formula: E = P * t, where t is the time in seconds. In this case, t = 4 minutes = 240 seconds.

Hence, the energy absorbed by the eardrum in 4 minutes is: E = (8.82x10^-5 W) * (240 seconds) = 0.0212 joules, which is equivalent to 1.72 in scientific notation.

In conclusion, the reason for the difference in the calculated energy and the given correct answer is due to not converting the intensity from dB to a linear scale. I hope this explanation helps to clarify any confusion. Thank you for your question and keep up the scientific curiosity!
 

FAQ: Calculating Energy Absorbed by Eardrum from Intensity & Area

How is energy absorbed by the eardrum calculated from intensity and area?

The energy absorbed by the eardrum can be calculated using the formula: energy absorbed = intensity x area. This formula takes into account the intensity of the sound wave (measured in watts per square meter) and the area of the eardrum that is exposed to the sound wave.

What units are used to measure intensity and area in this calculation?

The intensity of a sound wave is typically measured in watts per square meter (W/m²), while the area of the eardrum can be measured in square centimeters (cm²) or square meters (m²).

Is this calculation only applicable to the eardrum?

No, this calculation can also be used to determine the energy absorbed by any surface that is exposed to a sound wave. However, it is commonly used to measure the energy absorbed by the eardrum due to its importance in hearing and sound transmission.

How does the intensity and area affect the energy absorbed by the eardrum?

The intensity of a sound wave refers to the amount of energy that passes through a unit area per unit time. Therefore, the higher the intensity of the sound wave, the more energy will be absorbed by the eardrum. Similarly, a larger area of the eardrum exposed to the sound wave will result in more energy being absorbed.

Are there any other factors that can affect the energy absorbed by the eardrum?

Yes, there are other factors that can affect the energy absorbed by the eardrum, such as the duration of exposure to the sound wave, the frequency of the sound wave, and the sensitivity of the individual's ear. These variables can also impact the potential damage to the eardrum and overall hearing health.

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