Problem 66 - Physics

Question

Hearing damage may occur when a person is exposed to a sound intensity level of \(90.0 \mathrm{~dB}\) (relative to the threshold of hearing) for a period of 9.0 hours. An eardrum has an area of \(2.0 \times 10^{-4} \mathrm{~m}^{2} .\) How much sound energy is incident on the eardrum during this time?

Step-by-Step Solution

Answer

The sound energy incident on the eardrum is 6.48 \(\times 10^{-3}\) joules.

Step 1: Understand Sound Intensity Level

The given sound intensity level is measured in decibels (dB), which is a logarithmic scale used to measure the intensity of sound relative to a reference level, typically the threshold of hearing, which is set at 0 dB. Here, we have a sound level of 90.0 dB.

Step 2: Convert Sound Intensity Level to Intensity

Sound intensity level (in dB) can be converted to intensity (I) using the formula: \(L = 10 \cdot \log_{10}\left(\frac{I}{I_0}\right)\), where \(L\) is the sound level in dB, \(I\) is the intensity in W/m², and \(I_0 = 10^{-12} \text{ W/m}^2\). Rearrange the formula to solve for \(I\): \(I = I_0 \cdot 10^{L/10}\). Substitute \(L = 90\) dB to find \(I\).

Step 3: Calculate Intensity

Using the formula from Step 2: \(I = 10^{-12} \times 10^{90/10} = 10^{-12} \times 10^{9} = 10^{-3}\) W/m². This gives us the intensity of the sound incident on the eardrum.

Step 4: Calculate Power on the Eardrum

Power is given by the formula: \(P = I \cdot A\), where \(I\) is the sound intensity and \(A\) is the area of the eardrum. Substitute \(I = 10^{-3} \text{ W/m}^2\) and \(A = 2.0 \times 10^{-4} \text{ m}^2\): \(P = 10^{-3} \times 2.0 \times 10^{-4} = 2.0 \times 10^{-7}\) W.

Step 5: Calculate Energy Incident on the Eardrum

Energy is given by the formula: \(E = P \cdot t\), where \(P\) is the power and \(t\) is the time in seconds. Convert the time from hours to seconds: \(9.0 \text{ hours} = 9.0 \times 3600 = 32400\) seconds. Now calculate the energy: \(E = 2.0 \times 10^{-7} \times 32400 = 6.48 \times 10^{-3}\) joules.

Key Concepts

Sound Intensity Level

Sound intensity level is a measure of how loud a sound is. It tells us about the pressure of sound waves as they move through the air. This is usually measured in decibels (dB). The decibel scale is used because it can handle the vast range of sound intensities humans can hear, from the faintest whisper to a jet engine. When you see a number like 90 dB, it tells you that the sound is quite loud, possibly loud enough to cause harm if you're exposed for too long. This scale is logarithmic, meaning each increase of 10 dB represents a tenfold increase in intensity. For example, a sound at 90 dB is ten times more intense than one at 80 dB.

Decibels (dB)

Decibels (dB) are used to measure sound intensity or loudness. The decibel scale is a convenient way to express large variations in sound intensity. The formula for calculating the sound level in decibels is:\[ L = 10 \cdot \log_{10}\left(\frac{I}{I_0}\right) \]Where:

• \(L\) is the sound level in decibels
• \(I\) is the intensity of the sound being measured
• \(I_0\) is the reference intensity, typically \(10^{-12} \text{ W/m}^2\) (the threshold of hearing for humans)

This formula shows the relationship between the actual sound intensity and its decibel level.

Eardrum Area

The human eardrum, or tympanic membrane, is a sensitive part of the ear that vibrates in response to sound waves. The area of the eardrum is important because it influences how much sound energy is received.In our example, the eardrum's area is \(2.0 \times 10^{-4} \text{ m}^2\). This might seem small, but it is adequate for transmitting sound waves to the inner ear, enabling us to hear. Understanding the area helps in calculating how much sound energy is incident upon the eardrum, which in turn can indicate how potentially damaging a sound might be to hearing.

Logarithmic Scale

A logarithmic scale is a way to express large ranges of values that differ greatly in scale. Instead of increasing linearly, each step on a logarithmic scale is multiplied by a constant, such as 10. This is useful for expressing decibels because sound intensity can vary over a huge range. In our context, the sound level is expressed in a logarithmic scale because the human ear perceives sound percentages more closely to this logarithmic style than a linear one. As a result, small changes in sound intensity can lead to a large difference in how loud it sounds to us.

Sound Intensity Conversion

Converting from sound intensity level (decibels) to sound intensity (W/m²) involves reversing the logarithmic calculation:\[ I = I_0 \times 10^{\frac{L}{10}} \]Here, \( I_0\) is the reference intensity \(10^{-12} \text{ W/m}^2\), and \(L\) is the sound level in decibels.By converting decibels to watts per square meter, one can calculate the precise physical power a sound wave exerts. This conversion is critical in scenarios such as calculating potential hearing damage or measuring environmental noise impact.