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Chapter 29: Problem 2832

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) \(1.0 \mathrm{~cm}\)

Short Answer

Expert verified
The end correction for this apparatus is \(1 \mathrm{~cm}\), which corresponds to option (D).

Step by step solution

01

Understand the resonance condition for a tube with one open end and one closed end

The formula for the resonance condition in a tube with one open end and one closed end is: \( L_n = (2n - 1)\frac{\lambda}{4} \) where: - \(L_n\) is the length of the air column for the nth resonance - \(n\) is the order of the resonance (1 for the first resonance, 2 for the second resonance, etc.) - \(\lambda\) is the wavelength of the sound wave
02

Find the wavelength from the given resonance lengths

We are given the lengths of the air column for the first and second resonances. We can treat these as two distinct resonances with an unknown end correction, denoted by \(e\): First resonance: \(L_1 = 10\text{ cm} + e\) Second resonance: \(L_2 = 32\text{ cm} + e\) Using the resonance condition formula for each resonance, we get: \(L_1 = \frac{\lambda}{4} \) \(L_2 = \frac{3\lambda}{4} \) Now we can substitute the expressions for \(L_1\) and \(L_2\) in terms of \(e\) into these equations: \(10\text{ cm} + e = \frac{\lambda}{4} \) \(32\text{ cm} + e = \frac{3\lambda}{4} \)
03

Solve for end correction e

We can now solve these two equations simultaneously to find the end correction \(e\). First, we will eliminate \(\lambda\) by multiplying the first equation by 3 and then subtracting it from the second equation: \((32\text{ cm} + e) - 3(10\text{ cm} + e) = \frac{3\lambda}{4} - \frac{3\lambda}{4}\) Solve for \(e\): \(32\text{ cm} - 30\text{ cm} = 2e \) \(2\text{ cm} = 2e \) Divide both sides by 2 to find \(e\): \(e = 1\text{ cm} \) So, the end correction for this apparatus is found to be 1 cm, which corresponds to option (D).

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Most popular questions from this chapter

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) \(10 \mathrm{~g}\)

Two vibrating strings of the same material but of lengths Land \(2 \mathrm{~L}\) have radii \(2 \mathrm{r}\) and \(\mathrm{r}\) respectively. They are stretched under the same tension. Both the string vibrate in their fundamental modes. The one of length \(\mathrm{L}\) with frequency \(\mathrm{U}_{1}\) and the other with frequency \(\mathrm{U}_{2}\). What is the ratio $\left(\mathrm{U}_{1} / \mathrm{U}_{2}\right)$ ? (A) 8 (B) 2 (C) 1 (D) 4

A closed organ pipe of length \(\mathrm{L}\) and an open organ pipe contain gases of densities \(\rho_{1}\) and \(\rho_{2}\) respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. What is the length of the open organ pipe? (A) \(\left.(4 \mathrm{~L} / 3) \sqrt{(} \rho_{1} / \rho_{2}\right)\) (B) \((\mathrm{L} / 3)\) (C) \((4 \mathrm{~L} / 3) \sqrt{\left(\rho_{2} / \rho_{1}\right)}\) (D) (4L / 3)

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) \(\mathrm{n}=3, \mathrm{f}_{2}=(5 / 4) \mathrm{f}_{1}\)

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
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