Select The Correct Answer.Sound Waves Can Be Ranked By Their Intensity, $I$, Given In The Formula Below, Where $r$ Is The Distance From The Source Of A Sound With A Power Output Of $P$:$[ P = 4 \pi I R^2

Introduction

Sound waves are a fundamental aspect of our daily lives, and understanding their properties is crucial in various fields, including physics, acoustics, and engineering. One of the key characteristics of sound waves is their intensity, which is a measure of the power per unit area. In this article, we will delve into the concept of sound wave intensity and explore the formula used to rank sound waves by their intensity.

What is Sound Wave Intensity?

Sound wave intensity is a measure of the power per unit area of a sound wave. It is typically denoted by the symbol $I$ and is expressed in units of watts per square meter (W/m²). The intensity of a sound wave is a critical factor in determining its perceived loudness. A higher intensity sound wave is perceived as louder than a lower intensity sound wave.

The Formula for Sound Wave Intensity

The formula for sound wave intensity is given by:

${ P = 4 \pi I r^2 }$

where:

• P$ is the power output of the sound source

• I$ is the intensity of the sound wave

• r$ is the distance from the source of the sound wave

Rearranging the Formula to Solve for Intensity

To solve for intensity, we need to rearrange the formula to isolate $I$. We can do this by dividing both sides of the equation by $4 \pi r^2$:

${ I = \frac{P}{4 \pi r^2} }$

Understanding the Variables

Now that we have the formula for intensity, let's take a closer look at the variables involved.

• Power Output ($P$): The power output of the sound source is a measure of the total energy emitted by the source per unit time. It is typically denoted by the symbol $P$ and is expressed in units of watts (W).
• Intensity ($I$): The intensity of the sound wave is a measure of the power per unit area. It is typically denoted by the symbol $I$ and is expressed in units of watts per square meter (W/m²).
• Distance ($r$): The distance from the source of the sound wave is a critical factor in determining the intensity of the sound wave. As the distance increases, the intensity of the sound wave decreases.

Example Problem

Let's consider an example problem to illustrate the use of the formula for intensity.

Suppose we have a sound source with a power output of 100 W, and we want to find the intensity of the sound wave at a distance of 10 m. Using the formula for intensity, we can plug in the values as follows:

${ I = \frac{P}{4 \pi r^2} }$

${ I = \frac{100}{4 \pi (10)^2} }$

${ I = \frac{100}{400 \pi} }$

${ I = \frac{1}{4 \pi} }$

${ I \approx 0.0796 \, \text{W/m}^2 }$

Conclusion

In conclusion, the formula for sound wave intensity is a critical tool in understanding the properties of sound waves. By rearranging the formula to solve for intensity, we can easily calculate the intensity of a sound wave given its power output and distance from the source. This knowledge is essential in various fields, including physics, acoustics, and engineering.

References

• [1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. John Wiley & Sons.
• [2] Young, H. D., & Freedman, R. A. (2012). University Physics. Addison-Wesley.

Further Reading

• [1] Acoustics: An Introduction to Its Principles and Applications. By Allan D. Pierce (2016)
• [2] The Physics of Sound. By Lawrence E. Kinsler, Austin R. Frey, Alan B. Coppens, and James V. Sanders (2000)

Glossary

• Intensity: A measure of the power per unit area of a sound wave.
• Power Output: A measure of the total energy emitted by a sound source per unit time.
• Distance: The distance from the source of a sound wave.

FAQs

• Q: What is the formula for sound wave intensity? A: The formula for sound wave intensity is given by: $P = 4 \pi I r^2$
• Q: How do I solve for intensity? A: To solve for intensity, you need to rearrange the formula to isolate $I$. You can do this by dividing both sides of the equation by $4 \pi r^2$.
• Q: What are the variables involved in the formula for intensity? A: The variables involved in the formula for intensity are power output ($P$), intensity ($I$), and distance ($r$).
  Sound Wave Intensity: A Q&A Guide =====================================

Introduction

In our previous article, we explored the concept of sound wave intensity and the formula used to rank sound waves by their intensity. In this article, we will delve deeper into the topic and provide a comprehensive Q&A guide to help you understand sound wave intensity better.

Q: What is sound wave intensity?

A: Sound wave intensity is a measure of the power per unit area of a sound wave. It is typically denoted by the symbol $I$ and is expressed in units of watts per square meter (W/m²).

Q: Why is sound wave intensity important?

A: Sound wave intensity is important because it determines the perceived loudness of a sound wave. A higher intensity sound wave is perceived as louder than a lower intensity sound wave.

Q: What is the formula for sound wave intensity?

A: The formula for sound wave intensity is given by:

${ P = 4 \pi I r^2 }$

where:

• P$ is the power output of the sound source

• I$ is the intensity of the sound wave

• r$ is the distance from the source of the sound wave

Q: How do I solve for intensity?

A: To solve for intensity, you need to rearrange the formula to isolate $I$. You can do this by dividing both sides of the equation by $4 \pi r^2$:

${ I = \frac{P}{4 \pi r^2} }$

Q: What are the variables involved in the formula for intensity?

A: The variables involved in the formula for intensity are power output ($P$), intensity ($I$), and distance ($r$).

Q: How does distance affect sound wave intensity?

A: As the distance from the source of the sound wave increases, the intensity of the sound wave decreases. This is because the power output of the sound source is spread out over a larger area.

Q: Can you provide an example problem to illustrate the use of the formula for intensity?

A: Suppose we have a sound source with a power output of 100 W, and we want to find the intensity of the sound wave at a distance of 10 m. Using the formula for intensity, we can plug in the values as follows:

${ I = \frac{P}{4 \pi r^2} }$

${ I = \frac{100}{4 \pi (10)^2} }$

${ I = \frac{100}{400 \pi} }$

${ I = \frac{1}{4 \pi} }$

${ I \approx 0.0796 \, \text{W/m}^2 }$

Q: What are some real-world applications of sound wave intensity?

A: Sound wave intensity has many real-world applications, including:

• Acoustics: Sound wave intensity is used to design and optimize acoustic systems, such as concert halls and recording studios.
• Noise reduction: Sound wave intensity is used to design and optimize noise reduction systems, such as soundproofing materials and noise-cancelling headphones.
• Medical applications: Sound wave intensity is used in medical applications, such as ultrasound imaging and lithotripsy.

Q: What are some common misconceptions about sound wave intensity?

A: Some common misconceptions about sound wave intensity include:

• Intensity is the same as loudness: While intensity is related to loudness, they are not the same thing. Loudness is a subjective measure of the perceived loudness of a sound wave, while intensity is an objective measure of the power per unit area of a sound wave.
• Intensity is only important for loud sounds: Intensity is important for all sounds, not just loud ones. Even quiet sounds have an intensity, and understanding intensity is essential for designing and optimizing acoustic systems.

Conclusion

In conclusion, sound wave intensity is a critical concept in understanding the properties of sound waves. By understanding the formula for intensity and the variables involved, you can design and optimize acoustic systems, reduce noise, and apply sound wave intensity in various medical and industrial applications.