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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. One particular eardrum has an area of \(2.0 \times 10^{-4} \mathrm{m}^{2}\) How much sound energy is incident on this eardrum during this time?

Short Answer

Expert verified
The sound energy is 6.48 J.

Step by step solution

01

Convert Decibels to Intensity

The sound intensity level in decibels (dB) is given by the formula \(eta = 10 \, \log_{10} \left( \frac{I}{I_0} \right)\), where \(\beta\) is the intensity level in dB, \(I\) is the intensity in \(\text{W/m}^2\), and \(I_0 = 10^{-12} \, \text{W/m}^2\) is the threshold of hearing. We need to solve for \(I\):\[I = I_0 \times 10^{\beta/10}\]Substituting \(\beta = 90.0\, \text{dB}\), we have:\[I = 10^{-12} \times 10^{90/10}\]\[I = 10^{-12} \times 10^9 = 10^{-3} \, \text{W/m}^2\]
02

Calculate the Total Sound Energy

The energy \(E\) incident on the eardrum is given by the formula:\[E = I \times A \times t\]where \(I\) is the intensity we calculated, \(A = 2.0 \times 10^{-4} \, \text{m}^2\) is the area, and \(t = 9.0 \, \text{hours} = 9.0 \times 3600 \, \text{seconds}\) is the time period. Converting time to seconds, we have:\[ t = 9.0 \times 3600 = 32400 \, \text{seconds} \]Now, substitute the values in:\[ E = 10^{-3} \, \text{W/m}^2 \times 2.0 \times 10^{-4} \, \text{m}^2 \times 32400 \, \text{s} \]\[ E = 6.48 \, \text{J}\]
03

Final Answer

Therefore, the total sound energy incident on the eardrum is \(6.48 \, \text{J}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sound Intensity Level
Sound Intensity Level is a measure of how loud a sound is in a specific environment. It is often measured in Watts per square meter (W/m²). The intensity tells us how much sound power passes through a certain area. The Intensity Level is frequently expressed in decibels (dB), a logarithmic unit. This helps because human ears perceive sound intensity on a logarithmic scale. When you have a sound intensity level in decibels, use the formula: \[ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \] where \( \beta \) is the intensity level in decibels, \( I \) is the intensity, and \( I_0 \) is the reference intensity. This formula helps in converting between decibels and intensity. Understanding intensity levels can help us determine whether a sound might be harmful or safe. By converting decibel levels to their corresponding intensities, we can calculate the actual energy affecting an area like the eardrum.
Decibels
Decibels (dB) are a logarithmic way of describing a ratio of intensity or power. They are used to measure sound levels because the human ear detects sound on a logarithmic scale. In terms of sound intensity, the reference level is often the quietest sound that the average human ear can detect, called the threshold of hearing. This reference intensity is \( I_0 = 10^{-12} \text{W/m}^2 \). When calculating decibels, we use this reference to understand how much more intense one sound is compared to this quietest level. * Imagine: * 0 dB is the threshold of hearing. * 90 dB, like mentioned in the exercise, is extremely intense and potentially harmful over long periods. Using the formula for decibels can help us determine if we need to be cautious with our hearing protection. Knowing the decibel level helps individuals take appropriate protective measures.
Hearing Damage
Hearing Damage occurs when sound levels exceed certain thresholds for given durations. Permanent damage can start at prolonged exposure to sounds beginning around 85 dB. The exercise highlighted 90 dB, which is very intense. Exposure to 90 dB for as little as 9 hours can damage hearing. * Signs of hearing damage include: * Ringing in the ears (tinnitus) * Difficulty understanding speech * Muffled sounds Protecting your hearing is crucial when exposed to high decibel levels for extended periods. Using ear protection or limiting exposure time can help mitigate damage. Understanding sound energy and setting safe exposure limits is essential. By predicting potential risks with intensity calculations, you can actively protect your hearing.
Eardrum Area
Eardrum Area refers to the surface area of the eardrum, an important structure in the ear that vibrates when sound waves hit it. This area can vary but is typically around the size mentioned in the exercise: \(2.0 \times 10^{-4} \text{m}^2\).The eardrum's size influences how much sound energy it receives. The larger the area, the more sound energy can be incident upon it.To determine the sound energy incident on an eardrum, the formula used is: \[ E = I \times A \times t \] where \(E\) is the sound energy, \(I\) is the intensity of the sound, \(A\) is the area of the eardrum, and \(t\) is the exposure time.* Practical Implications: * A small area, like the eardrum, can still receive significant energy at high intensity. * Calculating the energy helps assess potential risks for hearing damage.Understanding eardrum area in conjunction with sound intensity helps us comprehend potential impacts on hearing and health.

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Most popular questions from this chapter

The siren on an ambulance is emitting a sound whose frequency is \(2450 \mathrm{Hz} .\) The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) (a) If the ambulance is stationary and you (the "observer") are sitting in a parked car, what are the wavelength and the frequency of the sound you hear? (b) Suppose that the ambulance is moving toward you at a speed of \(26.8 \mathrm{m} / \mathrm{s} .\) Determine the wavelength and the frequency of the sound you hear. (c) If the ambulance is moving toward you at a speed of \(26.8 \mathrm{m} / \mathrm{s}\) and you are moving toward it at a speed of \(14.0 \mathrm{m} / \mathrm{s}\) find the wavelength and frequency of the sound you hear.

Light is an electromagnetic wave and travels at a speed of \(3.00 \times\) \(10^{8} \mathrm{m} / \mathrm{s} .\) The human eye is most sensitive to yellow- green light, which has a wavelength of \(5.45 \times 10^{-7} \mathrm{m} .\) What is the frequency of this light?

The mass of a string is \(5.0 \times 10^{-3} \mathrm{kg}\), and it is stretched so that the tension in it is \(180 \mathrm{N}\). A transverse wave traveling on this string has a frequency of \(260 \mathrm{Hz}\) and a wavelength of \(0.60 \mathrm{m} .\) What is the length of the string?

From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequency that is 0.86 times as small as the frequency emitted by the car when it is stationary. The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the speed of the car?

Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawater where the speed of sound is \(1522 \mathrm{m} / \mathrm{s}\). When the dolphin is swimming directly away at \(8.0 \mathrm{m} / \mathrm{s},\) the marine biologist measures the number of clicks occurring per second to be at a frequency of 2500 Hz. What is the difference (in \(\mathrm{Hz}\) ) between this frequency and the number of clicks per second actually emitted by the dolphin?

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