A Person Stands 3.50 M From One Speaker And 5.20 M From An Identical Speaker. If There Is Destructive Interference Where $n=1$, What Is The Frequency? F = [ ? ] Hz F = [?] \, \text{Hz} F = [ ?] Hz

Introduction

Destructive interference is a fundamental concept in physics that occurs when two or more waves overlap in a way that results in a decrease in amplitude. In the context of sound waves, destructive interference can be observed when two speakers are placed at specific distances from each other, creating a region where the sound waves cancel each other out. In this article, we will explore the concept of destructive interference in sound waves and use it to determine the frequency of a sound wave given the distances from two identical speakers.

The Concept of Destructive Interference

Destructive interference occurs when the peaks of two waves align with the troughs of another wave, resulting in a net decrease in amplitude. This can be represented mathematically using the equation:

$$y = A \sin(\omega t) + B \sin(\omega t + \phi)$$

where $y$ is the resulting wave, $A$ and $B$ are the amplitudes of the two waves, $\omega$ is the angular frequency, $t$ is time, and $\phi$ is the phase difference between the two waves.

For destructive interference to occur, the phase difference between the two waves must be $\pi$ radians, or 180 degrees. This can be represented mathematically as:

$$\phi = \pi$$

The Relationship Between Distance and Frequency

The distance between two speakers and the frequency of the sound wave are related through the equation:

$$\lambda = \frac{v}{f}$$

where $\lambda$ is the wavelength of the sound wave, $v$ is the speed of sound, and $f$ is the frequency of the sound wave.

The distance between the two speakers can be represented as:

$$d = \frac{n\lambda}{2}$$

where $d$ is the distance between the two speakers, $n$ is an integer representing the number of wavelengths, and $\lambda$ is the wavelength of the sound wave.

Solving for Frequency

Given the distances from the two speakers and the fact that $n=1$, we can use the equation:

$$d = \frac{n\lambda}{2}$$

to solve for the wavelength of the sound wave. We can then use the equation:

$$\lambda = \frac{v}{f}$$

to solve for the frequency of the sound wave.

Calculating the Wavelength

First, we need to calculate the wavelength of the sound wave. We are given that the distance from one speaker is 3.50 m and the distance from the other speaker is 5.20 m. Since $n=1$, we can use the equation:

$$d = \frac{n\lambda}{2}$$

to solve for the wavelength of the sound wave.

$$3.50 = \frac{1\lambda}{2}$$

$$\lambda = 7.00 \, \text{m}$$

Calculating the Frequency

Now that we have the wavelength of the sound wave, we can use the equation:

$$\lambda = \frac{v}{f}$$

to solve for the frequency of the sound wave. The speed of sound is approximately 343 m/s at room temperature and atmospheric pressure.

$$7.00 = \frac{343}{f}$$

$$f = \frac{343}{7.00}$$

$$f = 49.00 \, \text{Hz}$$

Conclusion

In this article, we used the concept of destructive interference in sound waves to determine the frequency of a sound wave given the distances from two identical speakers. We first calculated the wavelength of the sound wave using the equation:

$$d = \frac{n\lambda}{2}$$

and then used the equation:

$$\lambda = \frac{v}{f}$$

to solve for the frequency of the sound wave. The result was a frequency of 49.00 Hz.

References

• [1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of physics. John Wiley & Sons.
• [2] Serway, R. A., & Jewett, J. W. (2018). Physics for scientists and engineers. Cengage Learning.

Additional Resources

• [1] Khan Academy. (n.d.). Waves and Interference. Retrieved from https://www.khanacademy.org/science/physics/waves-and-interference
• [2] Physics Classroom. (n.d.). Interference of Waves. Retrieved from https://www.physicsclassroom.com/class/waves/Lesson-2/Interference-of-Waves
  Frequently Asked Questions (FAQs) About Destructive Interference in Sound Waves ====================================================================================

Q: What is destructive interference in sound waves?

A: Destructive interference in sound waves occurs when two or more sound waves overlap in a way that results in a decrease in amplitude. This can be observed when two speakers are placed at specific distances from each other, creating a region where the sound waves cancel each other out.

Q: What is the relationship between distance and frequency in destructive interference?

A: The distance between two speakers and the frequency of the sound wave are related through the equation:

$$\lambda = \frac{v}{f}$$

where $\lambda$ is the wavelength of the sound wave, $v$ is the speed of sound, and $f$ is the frequency of the sound wave.

Q: How do I calculate the wavelength of the sound wave in destructive interference?

A: To calculate the wavelength of the sound wave, you can use the equation:

$$d = \frac{n\lambda}{2}$$

where $d$ is the distance between the two speakers, $n$ is an integer representing the number of wavelengths, and $\lambda$ is the wavelength of the sound wave.

Q: What is the significance of the phase difference in destructive interference?

A: The phase difference between the two sound waves must be $\pi$ radians, or 180 degrees, for destructive interference to occur.

Q: Can destructive interference occur in other types of waves besides sound waves?

A: Yes, destructive interference can occur in other types of waves besides sound waves, such as light waves and water waves.

Q: What are some real-world applications of destructive interference in sound waves?

A: Some real-world applications of destructive interference in sound waves include:

• Soundproofing: Destructive interference can be used to create soundproofing materials that absorb sound waves.
• Acoustic design: Destructive interference can be used to design acoustic systems that minimize sound reflections and maximize sound quality.
• Music: Destructive interference can be used to create unique sound effects and musical instruments.

Q: How can I use destructive interference in sound waves to create a sound wave with a specific frequency?

A: To create a sound wave with a specific frequency using destructive interference, you can use the equation:

$$\lambda = \frac{v}{f}$$

to calculate the wavelength of the sound wave, and then use the equation:

$$d = \frac{n\lambda}{2}$$

to calculate the distance between the two speakers.

Q: What are some common mistakes to avoid when working with destructive interference in sound waves?

A: Some common mistakes to avoid when working with destructive interference in sound waves include:

• Not accounting for the phase difference between the two sound waves.
• Not using the correct equation to calculate the wavelength of the sound wave.
• Not considering the speed of sound in the environment.

Q: Where can I learn more about destructive interference in sound waves?

A: You can learn more about destructive interference in sound waves by:

• Reading books and articles on the subject.
• Watching online tutorials and videos.
• Taking online courses or attending workshops on the subject.
• Consulting with experts in the field.

Conclusion

Destructive interference in sound waves is a fascinating phenomenon that has many practical applications. By understanding the relationship between distance and frequency, and by using the correct equations to calculate the wavelength of the sound wave, you can create sound waves with specific frequencies and use them to create unique sound effects and musical instruments.