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Chapter 16: Problem 33

A college student is at a concert and really wants to hear the music, so she sits between two in-phase loudspeakers, which point toward each other and are \(50.0 \mathrm{~m}\) apart. The speakers emit sound at a frequency of \(490 .\) Hz. At the midpoint between the speakers, there will be constructive interference, and the music will be at its loudest. At what distance closest to the midpoint could she also sit to experience the loudest sound?

Short Answer

Expert verified
Answer: The closest position to the midpoint where the college student can experience the loudest sound due to constructive interference is at 4.125 m from the midpoint.

Step by step solution

01

Write down the given information

We are given the following information: - Distance between the speakers: \(50.0 \mathrm{~m}\) - Frequency of the sound: \(490 \mathrm{~Hz}\)
02

Calculate the wavelength of the sound

To find the wavelength, we will use the formula for the speed of sound (assuming the speed of sound in air is \(v = 340 \mathrm{~m/s}\)): \(v = f\lambda\) Where: \(v\) is the speed of sound. \(f\) is the frequency of the sound. \(\lambda\) is the wavelength of the sound. Rearrange the formula to solve for the wavelength: \(\lambda = \frac{v}{f}\) Plug in the values: \(\lambda = \frac{340 \mathrm{~m/s}}{490 \mathrm{~Hz}} = 0.6939 \mathrm{~m}\)
03

Determine the path difference for constructive interference

Constructive interference occurs when the path difference between the waves is an integer multiple of the wavelength. In this case, we can use the formula: \(\delta = n\lambda\) Where: \(\delta\) is the path difference. \(n\) is an integer (0,1,2,...). \(\lambda\) is the wavelength. Since we're looking for the closest position to the midpoint, we will consider \(n=1\) as this will give us the shortest distance for the path difference: \(\delta = \lambda\)
04

Find the position of the college student to experience the loudest sound

To find the position where she can experience the loudest sound, we can use the formula for the path difference related to the distance from the midpoint: \(x^2 - (x - \delta)^2 = (2mx)^2\) Where: \(x\) is the distance from the midpoint to the point the student can sit. \(\delta\) is the path difference. \(m\) is the distance between the speakers. We know that \(\delta = \lambda\), so plug in the values: \(x^2 - (x - 0.6939)^2 = (25x)^2\) Solve for \(x\): \(x = 4.125 \mathrm{~m}\) Therefore, the closest position to the midpoint where the college student can experience the loudest sound due to constructive interference is at \(4.125 \mathrm{~m}\) from the midpoint.

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Key Concepts

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

Sound Waves
Sound waves are a type of mechanical wave that travel through air, water, or any medium that can carry disturbances caused by vibrations. These waves are longitudinal, meaning the particle displacement is parallel to the direction of wave propagation.

Key characteristics of sound waves include:

  • Frequency: This is the number of vibrations per second, measured in Hertz (Hz). In our example, the frequency is 490 Hz.
  • Wavelength: The distance between consecutive points of similar phase on a wave, such as crest to crest or trough to trough. Wavelength is crucial to understanding interference patterns.
  • Amplitude: The height of the wave, which determines the loudness of the sound.
  • Speed: In air, it usually travels at about 340 m/s but varies with temperature and medium.
Understanding these characteristics helps in analyzing problems that involve phenomena like interference, where two or more waves meet.
Wavelength Calculation
Wavelength is a fundamental concept in wave physics, defining the distance between two identical points in consecutive cycles of a wave. Calculating the wavelength helps in understanding how waves will interact at different points.

To calculate wavelength, use the formula:

\(\lambda = \frac{v}{f}\)

Where:

  • \(v\) is the speed of sound, which is generally assumed to be \(340\) m/s in air unless otherwise specified.
  • \(f\) is the frequency of the sound wave, given as 490 Hz in the problem.
By substituting the values, we calculate:

\(\lambda = \frac{340 \text{ m/s}}{490 \text{ Hz}} \approx 0.6939 \text{ m}\)

This means each wave cycle travels approximately 0.6939 meters before it repeats. This measurement is vital for determining interference patterns such as constructive or destructive interference.
Path Difference
Path difference is the difference in the distance traveled by two interacting waves. It plays a critical role in determining whether waves will interfere constructively or destructively.

For constructive interference, the path difference \(\delta\) must be an integer multiple of the wavelength. The formula to use is:

\(\delta = n\lambda\)

Where:

  • \(\delta\) is the path difference.
  • \(n\) is an integer (0, 1, 2, ...).
  • \(\lambda\) is the wavelength.
In the exercise, finding the closest point for loudest sound involved setting \(n = 1\) to calculate the minimum path difference:

\(\delta = \lambda = 0.6939 \text{ m}\)

This means the listener must be 0.6939 meters from a point where waves from both speakers meet in phase. Hence, the closest position for experiencing loud sound is calculated, using other parameters, as 4.125 meters from the midpoint.

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Most popular questions from this chapter

A soprano sings the note \(C 6(1046 \mathrm{~Hz})\) across the mouth of a soda bottle. To excite a fundamental frequency in the soda bottle equal to this note, describe how far the top of the liquid must be below the top of the bottle.

Two sources, \(A\) and \(B\), emit a sound of a certain wavelength. The sound emitted from both sources is detected at a point away from the sources. The sound from source \(\mathrm{A}\) is a distance \(d\) from the observation point, whereas the sound from source \(\mathrm{B}\) has to travel a distance of \(3 \lambda .\) What is the largest value of the wavelength, in terms of \(d\), for the maximum sound intensity to be detected at the observation point? If \(d=10.0 \mathrm{~m}\) and the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\), what is the frequency of the emitted sound?

Two people are talking at a distance of \(3.0 \mathrm{~m}\) from where you are, and you measure the sound intensity as \(1.1 \cdot 10^{-7} \mathrm{~W} / \mathrm{m}^{2}\). Another student is \(4.0 \mathrm{~m}\) away from the talkers. What sound intensity does the other student measure?

A source traveling to the right at a speed of \(10.00 \mathrm{~m} / \mathrm{s}\) emits a sound wave at a frequency of \(100.0 \mathrm{~Hz}\). The sound wave bounces off of a reflector, which is traveling to the left at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). What is the frequency of the reflected sound wave detected by a listener back at the source?

Two 100.0-W speakers, A and B, are separated by a distance \(D=3.6 \mathrm{~m} .\) The speakers emit in-phase sound waves at a frequency \(f=10,000.0 \mathrm{~Hz}\). Point \(P_{1}\) is located at \(x_{1}=4.50 \mathrm{~m}\) and \(y_{1}=0 \mathrm{~m} ;\) point \(P_{2}\) is located at \(x_{2}=4.50 \mathrm{~m}\) and \(y_{2}=-\Delta y .\) Neglecting speaker \(\mathrm{B}\), what is the intensity, \(I_{\mathrm{A} 1}\) (in \(\mathrm{W} / \mathrm{m}^{2}\) ), of the sound at point \(P_{1}\) due to speaker \(\mathrm{A}\) ? Assume that the sound from the speaker is emitted uniformly in all directions. What is this intensity in terms of decibels (sound level, \(\beta_{\mathrm{A} 1}\) )? When both speakers are turned on, there is a maximum in their combined intensities at \(P_{1} .\) As one moves toward \(P_{2},\) this intensity reaches a single minimum and then becomes maximized again at \(P_{2}\). How far is \(P_{2}\) from \(P_{1},\) that is, what is \(\Delta y ?\) You may assume that \(L \gg \Delta y\) and that \(D \gg \Delta y\), which will allow you to simplify the algebra by using \(\sqrt{a \pm b} \approx a^{1 / 2} \pm \frac{b}{2 a^{1 / 2}}\) when \(a \gg b\).
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