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Chapter 29: Problem 2824

A tube, closed at one end and containing air, produces, when excited, the fundamental note of frequency \(512 \mathrm{~Hz}\). If the tube is opened at both ends what is the fundamental frequency that can be excited (in \(\mathrm{Hz}\) )? (A) 256 (B) 1024 (C) 128 (D) 512

Short Answer

Expert verified
The fundamental frequency of the tube when opened at both ends is (B) 1024 Hz.

Step by step solution

01

Understand the fundamental frequency of a tube

The fundamental frequency of a tube depends on its length and whether it is open or closed at one or both ends. For a tube closed at one end, the fundamental frequency is determined by the following formula: \(f_1 = \frac{c}{4L}\), where \(f_1\) is the fundamental frequency, \(L\) is the length of the tube, and \(c\) is the speed of sound in air, which is approximately \(343 \frac{m}{s}\). For a tube opened at both ends, the fundamental frequency is given by: \(f_2 = \frac{c}{2L}\), where \(f_2\) is the new fundamental frequency.
02

Find the length of the tube using the closed tube formula

We are given the fundamental frequency of the tube when closed at one end (\(f_1\)) as \(512 Hz\). Therefore, we can find the length of the tube using the closed tube formula: \(L = \frac{c}{4f_1}\), Plugging in the given frequency and the speed of sound in air, we get: \(L = \frac{343}{4 \times 512} = \frac{343}{2048} m\)
03

Calculate the new fundamental frequency using the open tube formula

Now that we have the length of the tube, we can use the formula for the fundamental frequency of an open tube: \(f_2 = \frac{c}{2L}\), Plugging in the values of the speed of sound in air and the length of the tube, we get: \(f_2 = \frac{343}{2 \times \frac{343}{2048}} = 1024 Hz\)
04

Determine the correct answer

Comparing the calculated fundamental frequency (\(f_2 = 1024 Hz\)) to the given options: (A) 256 (B) 1024 (C) 128 (D) 512 The correct answer is (B) 1024 Hz.

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Most popular questions from this chapter

An air column in a pipe, which is closed at one end will be in resonance with a vibrating tuning fork of frequency \(264 \mathrm{~Hz}\), What is the length of the column if it is in \(\mathrm{cm} ?\) (speed of sound in \(\operatorname{air}=330 \mathrm{~m} / \mathrm{s}\) ) (A) \(62.50\) (B) \(15.62\) (C) 125 (D) \(93.75\)

A tuning fork of \(512 \mathrm{~Hz}\) is used to produce resonance in a resonance tube experiment. The level of water at first resonance is $30.7 \mathrm{~cm}\( and at second resonance is \)63.2 \mathrm{~cm}$. What is the error in calculating velocity of sound? Assume the speed of sound $330 \mathrm{~m} / \mathrm{s}$. (A) \(58(\mathrm{~cm} / \mathrm{s})\) (B) \(204.1(\mathrm{~cm} / \mathrm{s})\) (C) \(280(\mathrm{~cm} / \mathrm{s})\) (D) \(110(\mathrm{~cm} / \mathrm{s})\)

An open pipe is in resonance in \(2^{\text {nd }}\) harmonic with frequency \(\mathrm{f}_{1}\). Now one end of the tube is closed and frequency is increased to \(\mathrm{f}_{2}\) such that the resonance again occurs in nth harmonic. Choose the correct option. (A) \(\mathrm{n}=5, \mathrm{f}_{2}=(5 / 4) \mathrm{f}_{1}\) (B) \(\mathrm{n}=3, \mathrm{f}_{2}=(3 / 4) \mathrm{f}_{1}\) (C) \(\mathrm{n}=5, \mathrm{f}_{2}=(3 / 4) \mathrm{f}_{1}\) (D) \(\mathrm{n}=3, \mathrm{f}_{2}=(5 / 4) \mathrm{f}_{1}\)

In a resonance tube experiment, the first resonance is obtained for $10 \mathrm{~cm}\( of air column and the second for 32 \)\mathrm{cm}$. The end correction for this apparatus is equal to \(\ldots \ldots \ldots\) (A) \(1.9 \mathrm{~cm}\) (B) \(0.5 \mathrm{~cm}\) (C) \(2 \mathrm{~cm}\) (D) \(1.0 \mathrm{~cm}\)

A hollow pipe of length \(0.8 \mathrm{~m}\) is closed at one end. At its open end a \(0.5 \mathrm{~m}\) long uniform string is vibrating in its second harmonic and it resonates with the fundamentals frequency of the pipe if the tension in the wire is \(50 \mathrm{~N}\) and the speed of sound is $320 \mathrm{~ms}^{-1}$, what is the mass of the string. (A) \(20 \mathrm{~g}\) (B) \(5 \mathrm{~g}\) (C) \(40 \mathrm{~g}\) (D) \(10 \mathrm{~g}\)
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