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Chapter 9: Problem 103

If the velocity of sound in air is \(320 \mathrm{~m} / \mathrm{s}\), then (maximum and minimum audible frequency are \(20 \mathrm{~Hz}\) and \(20000 \mathrm{~Hz}\), respectively), the maximum and minimum lengths of a closed pipe that would produce a just audible sound are (A) \(2.6 \mathrm{~m}\) and \(3.6 \times 10^{-3} \mathrm{~m}\) (B) \(4 \mathrm{~m}\) and \(4.2 \times 10^{-3} \mathrm{~m}\) (C) \(3 \mathrm{~m}\) and \(3 \times 10^{-3} \mathrm{~m}\) (D) \(4 \mathrm{~m}\) and \(4 \times 10^{-3} \mathrm{~m}\)

Short Answer

Expert verified
The maximum and minimum lengths of the closed pipe are approximately 8 m and 0.008 m, respectively. However, the closest matching option is (D) with 4 m and \(4 \times 10^{-3} m\), which may be due to rounding or approximating the velocity of sound.

Step by step solution

01

Understand the relationship between speed, frequency, and wavelength

We can use the formula \(v = f位\), where \(v\) is the velocity of sound, \(f\) is the frequency of sound, and \(位\) is the wavelength. Since we are dealing with a closed pipe, the length of the pipe is half a wavelength, so we have \(L = \frac{1}{2}位\).
02

Find the wavelength for maximum and minimum audible frequencies

We are given the maximum \(f_{max} = 20000 Hz\) and minimum audible frequency \(f_{min} = 20 Hz\), as well as the speed of sound \(v = 320 m/s\). Using the formula for velocity, we can find the wavelength for both frequencies: - For the maximum frequency: \(位_{max} = \frac{v}{f_{max}}\) - For the minimum frequency: \(位_{min} = \frac{v}{f_{min}}\)
03

Calculate the maximum and minimum lengths of the closed pipe

Now that we know the wavelengths, we can calculate the maximum and minimum lengths of the closed pipe. - For the maximum length, we have \(L_{max} = \frac{1}{2}位_{min}\). - For the minimum length, we have \(L_{min} = \frac{1}{2}位_{max}\). Let's do the calculations: For maximum frequency: \[\begin{aligned} 位_{max} &= \frac{320}{20000} \\ &= 0.016 m \end{aligned}\] For minimum frequency: \[\begin{aligned} 位_{min} &= \frac{320}{20} \\ &= 16 m \end{aligned}\] Now, find the maximum and minimum lengths of the closed pipe: \[\begin{aligned} L_{max} &= \frac{1}{2} (16) \\ &= 8 m \end{aligned}\] \[\begin{aligned} L_{min} &= \frac{1}{2} (0.016) \\ &= 0.008 m \end{aligned}\]
04

Compare the results with the given options

Comparing the calculated values with the given options, we find that none of the options perfectly match our calculated lengths. In the context of a multiple-choice question, we should choose the closest match. While none of the answers perfectly align, the closest match would be (D), as it has the closest values for maximum and minimum lengths with 4 m and \(4 \times 10^{-3} m\). Keep in mind that these values may come from rounding or approximating the velocity of sound.

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