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Chapter 11: Problem 44

While testing speakers for a concert. Tomás sets up two speakers to produce sound waves at the same frequency, which is between \(100 \mathrm{Hz}\) and $150 \mathrm{Hz}$. The two speakers vibrate in phase with one another. He notices that when he listens at certain locations, the sound is very soft (a minimum intensity compared to nearby points). One such point is \(25.8 \mathrm{m}\) from one speaker and \(37.1 \mathrm{m}\) from the other. What are the possible frequencies of the sound waves coming from the speakers? (The speed of sound in air is \(343 \mathrm{m} / \mathrm{s} .\) )

Short Answer

Expert verified
Answer: The possible frequency of the sound waves coming from the speakers is approximately 107.96 Hz.

Step by step solution

01

Write the given data

Here are the given data in the problem: - Distance from the listener to the first speaker: \(d_1 = 25.8 \mathrm{m}\) - Distance from the listener to the second speaker: \(d_2 = 37.1 \mathrm{m}\) - The frequency is between \(100 \mathrm{Hz}\) and \(150 \mathrm{Hz}\) - The speed of sound in air: \(v = 343 \mathrm{m}/\mathrm{s}\)
02

Find the path difference

The path difference is the difference in the distance traveled by the sound waves from each speaker to the listener. Calculate the path difference as follows: Path difference = \(|d_1 - d_2| = |25.8 - 37.1| = 11.3 \mathrm{m}\)
03

Destructive interference condition

For destructive interference, the condition is that the path difference must be equal to odd multiples of half-wavelengths, i.e., \((2n-1)\frac{\lambda}{2}\), where \(n\) is a positive integer, and \(\lambda\) is the wavelength of the sound wave. The relationship between wavelength, frequency, and speed of sound is given by \(\lambda = \frac{v}{f}\), where \(f\) is the frequency.
04

Write the destructive interference condition with frequency terms

Replace the wavelength in the destructive interference condition with the frequency terms using the relationship \(\lambda = \frac{v}{f}\). \((2n-1)\frac{\lambda}{2} = 11.3 \mathrm{m} \Rightarrow (2n-1) \frac{v}{2f} = 11.3 \mathrm{m}\)
05

Rearrange the equation for finding the frequency

Rearrange the equation from the previous step to solve for the frequency \(f\). \(f = \frac{(2n-1)v}{22.6}\)
06

Calculate the possible frequencies

Use the fact that the frequency is between \(100 \mathrm{Hz}\) and \(150 \mathrm{Hz}\). By trying integer values of \(n\), find the possible frequencies for destructive interference within this range. For \(n=1\): \(f = \frac{(2(1)-1)(343)}{22.6} \approx 107.96 \mathrm{Hz}\) For \(n=2\): \(f = \frac{(2(2)-1)(343)}{22.6} \approx 215.93 \mathrm{Hz}\) Since the second result is not within the given frequency range, the possible frequency of the sound waves coming from the speakers is approximately \(107.96 \mathrm{Hz}\).

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Most popular questions from this chapter

A cord of length \(1.5 \mathrm{m}\) is fixed at both ends. Its mass per unit length is \(1.2 \mathrm{g} / \mathrm{m}\) and the tension is \(12 \mathrm{N} .\) (a) What is the frequency of the fundamental oscillation? (b) What tension is required if the \(n=3\) mode has a frequency of \(0.50 \mathrm{kHz} ?\)
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