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A pair of in-phase stereo speakers is placed side by side, \(0.85 \mathrm{m}\) apart. You stand directly in front of one of the speakers, \(1.1 \mathrm{m}\) from the speaker. What is the lowest frequency that will produce constructive interference at your location?

Short Answer

Expert verified
The lowest frequency for constructive interference is approximately 1183 Hz.

Step by step solution

01

Understand the Scenario

We have two stereo speakers that are 0.85 meters apart, and they are emitting sound waves that are in phase with each other. You are 1.1 meters away from one of the speakers, standing directly in front of it. We need to find the lowest frequency that causes constructive interference at your position.
02

Constructive Interference Condition

Constructive interference occurs when the path difference between the two sound waves (from both speakers) is a multiple of the wavelength, i.e., \(d_1 - d_2 = n\lambda\), where \(n\) is an integer, \(d_1\) and \(d_2\) are the distances from the sound source to the listener, and \(\lambda\) is the wavelength of the sound.
03

Determine Path Difference

Since you are directly in front of one speaker, the distance from the other speaker (which is not directly in front of you) needs to be calculated using the Pythagorean theorem. \(d_2 = \sqrt{(1.1)^2 + (0.85)^2} = \sqrt{1.21 + 0.7225} = \sqrt{1.9325}\approx 1.39\) meters.
04

Calculate the Path Difference

Calculate the path difference: \(d_1 - d_2 = 1.1 - 1.39 = -0.29\) meters.
05

Calculate Wavelength for Constructive Interference

Use the path difference equation for constructive interference and solve for the wavelength: \(-0.29 = n\lambda\). Since the speakers are in-phase, the smallest positive integer \(n\) that makes the path difference a positive wavelength is when \(-0.29 = -\lambda\), hence \(\lambda = 0.29\) meters.
06

Determine the Lowest Frequency

The speed of sound in air is approximately \(v = 343\ m/s\). The frequency is calculated using \(f = \frac{v}{\lambda}\). Substitute the values: \(f = \frac{343}{0.29} \approx 1182.76\) Hz.

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Key Concepts

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

Path Difference
When it comes to understanding wave interference, recognizing the concept of path difference is crucial. It refers to the difference in distance that two waves travel to reach a common point.
In this scenario with the stereo speakers, you stand in front of one, which means you'll hear two sound waves: one from the speaker directly in front of you and one from the additional speaker. Constructive interference occurs when the distance these waves travel from the speakers results in a path difference that is a multiple of their wavelength.
To calculate the path difference, you consider the distances from each speaker to your position. Use the Pythagorean theorem to determine the distance from the second speaker not directly in front of you. This calculation accounts for the placement of the speakers 0.85 meters apart. The distance, or path difference, helps determine where waves combine to enhance sound at points of constructive interference.
Wavelength Calculation
Wavelength is the distance between consecutive crests of a wave and is central to solving interference problems.
In the context of constructive interference, you need a situation where the path difference \(d_1 - d_2\) simplifies to an integer multiple of the wavelength \(\lambda\). This means aligning one full cycle of a wave from one speaker perfectly with another cycle from the other speaker.
For constructive interference in the given problem, the path difference calculated as -0.29 meters equals the wavelength when the smallest positive integer \(n\) makes the path difference positive and allows the wavelength condition \(n\lambda = -0.29\). It's through this relationship that we find \(\lambda = 0.29\) meters, ensuring that the sound waves enhance one another at the listener's position.
Sound Frequency
Sound frequency determines how often the wave cycles pass a given point in one second. Measured in hertz (Hz), this frequency will be directly affected by the speed of sound and the calculated wavelength.
In our speaker setup, with an approximated speed of sound in air \(v = 343\ m/s\), frequency \(f\) can be derived using the formula \(f = \frac{v}{\lambda}\). This equation connects how fast sound travels with how expansive each wave is, allowing us to calculate the number of cycles, i.e., frequency.
Given the calculated wavelength of 0.29 meters, substituting into the equation provides \(f = \frac{343}{0.29} \approx 1182.76\) Hz. This calculation gives a clear idea of the lowest frequency resulting in constructive interference at the listener's position where waves amplify instead of cancel out.

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Most popular questions from this chapter

The Bullet Train The Shinkansen, the Japanese "bullet" train, runs at high speed from Tokyo to Nagoya. Riding on the Shinkansen, you notice that the frequency of a crossing signal changes markedly as you pass the crossing. As you approach the crossing, the frequency you hear is \(f\); as you recede from the crossing the frequency you hear is \(2 f / 3 .\) What is the speed of the train?

Identical cellos are being tested. One is producing a fundamental frequency of \(130.9 \mathrm{Hz}\) on a string that is \(1.25 \mathrm{m}\) long and has a mass of 109 g. On the second cello the same string is fingered to reduce the length that can vibrate. If the beat frequency produced by these two strings is \(4.33 \mathrm{Hz}\), what is the vibrating length of the second string?

A friend in another city tells you that she has two organ pipes of different lengths, one open at both ends, the other open at one end only. In addition, she has determined that the beat frequency caused by the second-lowest frequency of each pipe is equal to the beat frequency caused by the third- lowest frequency of each pipe. Her challenge to you is to calculate the length of the organ pipe that is open at both ends, given that the length of the other pipe is \(1.00 \mathrm{m}\)

Predict/Explain Two strings are made of the same material and have equal tensions. String 1 is thick; string 2 is thin. (a) Is the speed of waves on string 1 greater than, less than, or equal to the speed of waves on string \(2 ?\) (b) Choose the best explanation from among the following: I. Since the strings are made of the same material, the wave speeds will also be the same. II. A thick string implies a large mass per length and a slow wave speed. III. A thick string has a greater force constant, and therefore a greater wave speed.

A particular jet engine produces a tone of \(495 \mathrm{Hz}\). Suppose that one jet is at rest on the tarmac while a second identical jet flies overhead at \(82.5 \%\) of the speed of sound. The pilot of each jet listens to the sound produced by the engine of the other jet. (a) Which pilot hears a greater Doppler shift? Explain. (b) Calculate the frequency heard by the pilot in the moving jet. (c) Calculate the frequency heard by the pilot in the stationary jet.

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