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When Gloria wears her hearing aid, the sound intensity level increases by 30.0 dB. By what factor does the sound intensity increase?

Short Answer

Expert verified
The sound intensity increases by a factor of 1000.

Step by step solution

01

Understanding Sound Intensity Level

Sound intensity levels in decibels (dB) are calculated using the formula: \[ L = 10 \log \left( \frac{I}{I_0} \right) \]where \(L\) is the sound level in dB, \(I\) is the sound intensity, and \(I_0\) is the reference intensity, typically \(10^{-12} \text{ W/m}^2\). An increase of 30 dB indicates a change in sound intensity.
02

Applying the Increase in Decibels

If the sound intensity level increases by 30 dB, this implies:\[ L' = L + 30 = 10 \log \left( \frac{I'}{I_0} \right) \]We want to find the factor \( \frac{I'}{I} \), where the original intensity level equation is:\[ L = 10 \log \left( \frac{I}{I_0} \right) \]
03

Relating Change in Intensity to dB

Having \( L' = L + 30 \), substitute and rearrange:\[ 10 \log \left( \frac{I'}{I_0} \right) = 10 \log \left( \frac{I}{I_0} \right) + 30 \]Simplifying gives:\[ \log \left( \frac{I'}{I_0} \right) - \log \left( \frac{I}{I_0} \right) = 3 \]
04

Solving for Intensity Factor

The properties of logarithms allow us to combine the logs:\[ \log \left( \frac{I'}{I} \right) = 3 \]To remove the logarithm, we exponentiate both sides:\[ \frac{I'}{I} = 10^3 = 1000 \]
05

Conclusion

The sound intensity increases by a factor of 1000.

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Key Concepts

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

Decibel Scale
Sound travels through the air as waves, and its intensity is an important factor in how we perceive sound. To quantify this intensity, especially for very weak or intense sounds, we use a logarithmic scale called the decibel scale. The decibel scale is a method of expressing sound intensity levels. It uses the logarithm of the intensity ratio to compress the wide range of sound intensities into a more manageable set of numbers. A key feature of the decibel scale is that it is not linear. Instead, it measures in ratios, and every 10 dB increase corresponds to a tenfold increase in intensity. For example:
  • 0 dB represents the threshold of hearing, which is the lowest sound level that the average person can hear.
  • 10 dB is ten times more intense than 0 dB.
  • 20 dB is one hundred times more intense than 0 dB.
Thus, when sound intensity increases by 30 dB, it doesn’t just increase by a factor of 30, but by a factor of 1000. This makes the decibel scale particularly useful for dealing with a wide range of sound intensities.
Logarithms in Physics
Logarithms are essential mathematical tools in physics for dealing with quantities that can vary over many orders of magnitude. In the case of sound intensity, the use of logarithms allows us to quantify levels of sound in a way that is more comparable to human perception.A logarithm is the power to which a base number must be raised to produce a given number. In the decibel scale, we often use a base of 10. The formula used to calculate the decibel level is:\[ L = 10 \log \left( \frac{I}{I_0} \right) \] where:
  • \(L\) is the sound level in decibels.
  • \(I\) is the intensity of the sound.
  • \(I_0\) is the reference intensity, usually taken as \(10^{-12} \text{ W/m}^2\).
This application of logarithms simplifies operations like multiplication and division into addition and subtraction, making calculations, especially involving large or small numbers, much more manageable. Understanding logarithms is essential for grasping how intensities change with decibel adjustments.
Sound Intensity Calculation
When analyzing how sound intensity changes, it's important to understand both the concept of intensity and the mathematics behind how it is measured. Sound intensity refers to the power per unit area carried by a wave. It is typically measured in watts per square meter (W/m") and indicates how much sound energy passes through a particular area.To calculate changes in sound intensity: 1. Start with the original intensity level formula:\[ L = 10 \log \left( \frac{I}{I_0} \right) \]2. For an increased decibel level, the new formula becomes:\[ L' = L + 30 \]3. Substitute into the expression and solve for the change in intensity.The relationship:\[ \log \left( \frac{I'}{I_0} \right) - \log \left( \frac{I}{I_0} \right) = 3 \]shows the mathematical properties of logarithms, where the difference in logarithms can be equated to multiplication. By solving:\[ \log \left( \frac{I'}{I} \right) = 3 \]we find:\[ \frac{I'}{I} = 10^3 = 1000 \], indicating that the sound intensity increases by a factor of 1000.This calculation offers a clear interpretation of what happens when the sound intensity level is increased by a given number of decibels, and highlights the practical use of the decibel scale in understanding real-world phenomena.

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Most popular questions from this chapter

A car is parked 20.0 m directly south of a railroad crossing. A train is approaching the crossing from the west, headed directly east at a speed of 55.0 m/s. The train sounds a short blast of its 289-Hz horn when it reaches a point 20.0 m west of the crossing. What frequency does the car’s driver hear when the horn blast reaches the car? The speed of sound in air is 343 m/s. (Hint: Assume that only the component of the train’s velocity that is directed toward the car affects the frequency heard by the driver.)

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at a location 22 \(\mathrm{m}\) away from the sound source is \(3.0 \times 10^{-4} \mathrm{W} / \mathrm{m}^{2} .\) What is the intensity at a spot that is 78 \(\mathrm{m}\) away?

A loudspeaker has a circular opening with a radius of 0.0950 \(\mathrm{m}\) . The electrical power needed to operate the speaker is 25.0 \(\mathrm{W}\) . The average sound intensity at the opening is 17.5 \(\mathrm{W} / \mathrm{m}^{2}\) What percentage of the electrical power is converted by the speaker into sound power?

A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1250 Hz. The bird-watcher, however, hears a frequency of 1290 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound?

To measure the acceleration due to gravity on a distant planet, an astronaut hangs a 0.055-kg ball from the end of a wire. The wire has a length of 0.95 \(\mathrm{m}\) and a linear density of \(1.2 \times 10^{-4} \mathrm{kg} / \mathrm{m}\) . Using electronic equipment, the astronaut measures the time for a transverse pulse to travel the length of the wire and obtains a value of 0.016 \(\mathrm{s}\) . The mass of the wire is negligible compared to the mass of the ball. Determine the acceleration due to gravity.

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