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Chapter 10: Problem 1502

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 the nth harmonic. Choose the correct option. (A) \(n=3, f_{3}=(3 / 4) f\) (B) \(n=3, f_{2}=(5 / 4) f_{1}\) (C) \(\mathrm{n}=5, \mathrm{f}_{2}=(5 / 4) \mathrm{f}\) (D) \(\mathrm{n}=5, \mathrm{f}_{2}=(3 / 4) \mathrm{f}_{1}\)

Short Answer

Expert verified
(B) \(n=3, f_{2}=\frac{5}{4}f_{1}\)

Step by step solution

01

Calculate second harmonic frequency of the open pipe

The formula for the frequency of any harmonics in an open pipe is: \(f_n = \frac{nV}{2L}\) Where \(f_n\) is the frequency of the n-th harmonic, V is the speed of sound in air, and L is the length of the pipe. Given that the open pipe is in resonance with its second harmonic, \(f_1 = \frac{2V}{2L}\) and from this, we have \(f_1 = \frac{V}{L}\)
02

Calculate the harmonic frequency of the closed pipe

The formula for the frequency of any odd harmonics in a closed pipe is: \(f'_n = \frac{(2n-1)V}{4L}\) The closed end pipe resonates at its nth harmonic, so \(f_2 = \frac{(2n-1)V}{4L}\)
03

Solve for the relation between f1 and f2

Divide the equation of the closed pipe by the equation of the open pipe: \(\frac{f_2}{f_{1}} = \frac{(2n-1)V}{4L} \times \frac{L}{V}\) Simplifying the equation, we get \(\frac{f_2}{f_{1}} = \frac{2n-1}{4}\) We need to find the value of n for which the above equation holds true. Solution (contd.):
04

Check given options

Let's test the given options: (A) \(n=3\) and \(f_{3}=(3 / 4) f\) We are given that n should be equal to 3. However, the frequency is incorrect due to f3 being written instead of \(f_2\). Let's calculate the frequency if n = 3, \(\frac{f_2}{f_{1}} = \frac{2(3)-1}{4} = \frac{5}{4}\) Option (B) fits this condition, and thus the correct choice is: (B) \(n=3, f_{2}=(5 / 4) f_{1}\)

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