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Refer to Interactive Solution \(17.817 .8\) at to review a method by which this problem can be solved. Two loudspeakers on a concert stage are vibrating in phase. A listener is \(50.5 \mathrm{~m}\) from the left speaker and \(26.0 \mathrm{~m}\) from the right one. The listener can respond to all frequencies from 20 to \(20000 \mathrm{~Hz}\), and the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What are the two lowest frequencies that can be heard loudly due to constructive interference?

Short Answer

Expert verified
The two lowest frequencies are 28 Hz and 42 Hz.

Step by step solution

01

Understand Constructive Interference

Constructive interference occurs when the path difference between two waves results in a phase difference that is a multiple of the wavelength. This results in louder sounds at the listener's location. Mathematically, the condition for constructive interference is \( \Delta x = n\lambda \) where \( n \) is an integer and \( \lambda \) is the wavelength of the sound.
02

Calculate Path Difference

Calculate the path difference between the two speakers: \( \Delta x = d_1 - d_2 = 50.5 \, \text{m} - 26.0 \, \text{m} = 24.5 \, \text{m} \). This difference will be used in finding the frequencies that cause constructive interference.
03

Relate Path Difference to Wavelength

Use the path difference equation \( \Delta x = n \lambda \) and relate it to the speed of sound and frequency. The wavelength \( \lambda \) is also given by \( \lambda = \frac{v}{f} \), where \( v = 343 \, \text{m/s} \) is the speed of sound and \( f \) is frequency.
04

Solve for Frequencies

Substitute \( \lambda = \frac{v}{f} \) into the path difference equation: \[ \Delta x = n \frac{v}{f} \] Simplifying gives \[ f = \frac{n v}{\Delta x} \]. With \( \Delta x = 24.5 \, \text{m} \), solve for \( f \) with \( n = 1 \) and \( n = 2 \):- For \( n = 1 \), \( f_1 = \frac{343}{24.5} \approx 14.0 \, \text{Hz} \).- For \( n = 2 \), \( f_2 = \frac{2 \times 343}{24.5} \approx 28.0 \, \text{Hz} \).
05

Validate the Frequencies

Given the valid listener's frequency range from 20 Hz to 20000 Hz, only frequencies within this range are valid. Therefore, \( f_1 = 14.0 \, \text{Hz} \) is below the audible range and not considered. The lowest audible frequency due to constructive interference is \( f_2 = 28.0 \, \text{Hz} \).

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Key Concepts

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

Sound Frequency
Sound frequency refers to the number of waves that pass a certain point per second. It is measured in hertz (Hz). When you listen to music or any sound, what you actually perceive are vibrations moving through the air. These vibrations can come fast or slow, which is what "frequency" measures:

Fast vibrations equal a high-frequency sound—it’s what we might define as a "high-pitched" sound, like a bird's tweet. Slow vibrations make a low-frequency sound, like a bass drum or a foghorn.

Sound frequency is significant because it affects not just pitch, but also the ability of sounds to interact. In physics, two sounds interacting constructively means they become louder due to the synchronization of their waves.

In the context of constructive interference, finding the right frequency enables sounds from two separate sources to align perfectly, resulting in louder sounds at the listener’s location.
Wavelength Calculation
The wavelength of a sound is the distance between two consecutive points that are in phase on the wave, like crest to crest or trough to trough. It’s usually denoted by the Greek letter lambda (\lambda).To compute the wavelength, you use the equation: \[\lambda = \frac{v}{f}\]where
  • \( v \)is the speed of sound, which is generally \( 343 \, \text{m/s} \)
  • \( f \)is the frequency of the sound.
Knowing the wavelength helps in understanding how sound travels through different mediums, affects constructive interference, and determines sound interactions.

For constructive interference, such as in the example from the exercise, the wavelength needs to fit the criteria based on the path difference. This adjustment allows the two sound waves from different sources to combine seamlessly, maximizing sound intensity.
Path Difference in Waves
Path difference in waves is the key concept when examining how sound waves interact. It’s simply the difference in distance that two waves travel to meet at a certain point. Sounds become louder through constructive interference when their path difference is a whole number of wavelengths.

Expressed as:\[ \Delta x = n\lambda \]this equation depicts scenarios where the sound waves are marching together, boosting the resultant sound’s amplitude. Here,
  • \( \Delta x \)is the path difference.
  • \( n \)is an integer (1, 2, 3, ...).
  • \( \lambda \)is the wavelength.
In the example given, the path difference is \( 24.5 \, \text{m} \), and we use it to determine which wave frequencies will maximize loudness. By recalculating for various\( n \), you determine the specific frequencies that lead to constructive interference, making sound louder and clearer to the listener.

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Most popular questions from this chapter

(a) When sound emerges from a loudspeaker, is the diffraction angle determined by the wavelength, the diameter of the speaker, or a combination of these two factors? (b) How is the wavelength of a sound related to its frequency? Explain your answers. The following two lists give diameters and sound frequencies for three loudspeakers. Pair each diameter with a frequency, so that the diffraction angle is the same for each of the speakers. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). Find the common diffraction angle. $$ \begin{array}{|l|c|} \hline \text { Diameter, } D & \text { Frequency, } f \\ \hline 0.050 \mathrm{~m} & 6.0 \mathrm{Khz} \\ \hline 0.10 \mathrm{~m} & 4.0 \mathrm{kHz} \\ \hline 0.15 \mathrm{~m} & 12.0 \mathrm{kHz} \\ \hline \end{array} $$

A string of length \(0.28 \mathrm{~m}\) is fixed at both ends. The string is plucked and a standing wave is set up that is vibrating at its second harmonic. The traveling waves that make up the standing wave have a speed of \(140 \mathrm{~m} / \mathrm{s}\). What is the frequency of vibration?

Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted outside on a pole. In winter, when the temperature is \(273 \mathrm{~K}\), the diffraction angle \(\theta\) has a value of \(15.0^{\circ} .\) What is the diffraction angle for the same sound on a summer day when the temperature is \(311 \mathrm{~K}\) ?

The \(\mathrm{E}\) string on an electric bass guitar has a length of \(0.628 \mathrm{~m}\) and, when producing the note E, vibrates at a fundamental frequency of \(41.2 \mathrm{~Hz}\). Players sometimes add to their instruments a device called a "D-tuner." This device allows the E string to be used to produce the note D, which has a fundamental frequency of \(36.7 \mathrm{~Hz}\). The D-tuner works by extending the length of the string, keeping all other factors the same. By how much does a D-tuner extend the length of the \(\mathrm{E}\) string?

Two speakers, one directly behind the other, are each generating a 245 -Hz sound wave. What is the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them? The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\).

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