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Chapter 9: Problem 133

A closed organ pipe of length \(L\) is vibrating in its first overtone. There is a point \(Q\) inside the pipe at a distance \(7 L / 9\) from the open end. The ratio of pressure amplitude at \(Q\) to the maximum pressure amplitude in the pipe is (A) \(1: 2\) (B) \(2: 1\) (C) \(1: 1\) (D) \(2: 3\)

Short Answer

Expert verified
The ratio of the pressure amplitude at point Q to the maximum pressure amplitude in the pipe is (A) \(1: 2\).

Step by step solution

01

Understanding Closed Organ Pipe Vibrations

Closed organ pipe vibrates in its first overtone when only a single antinode is formed at the open end and a node is formed at the closed end. In a closed organ pipe, the fundamental frequency is defined by the relation: \(f_1=\frac{v}{4L}\), where \(v\) is the speed of sound in the air. The pressure amplitude is maximum at the node, which is at the closed end of the organ pipe, and minimum at the antinode, which is at the open end of the organ pipe.
02

Find the distance from the antinode to point Q

Given that point Q is at a distance of \(\frac{7L}{9}\) from the open end. Therefore, the distance from the antinode to point Q, denoted as \(x\), can be calculated as follows: \(x = L - \frac{7L}{9} = \frac{2L}{9}\)
03

Determine the pressure distribution function

Let's denote the maximum pressure amplitude in the organ pipe as \(P_m\). The pressure distribution function is given by: \(P(x) = P_m \cos(kx)\), where \(k = \frac{2\pi f}{v}\) is the wave number. To find the pressure amplitude at point Q, we will substitute the value of x in the above function. \(P_Q = P_m \cos(kx) = P_m \cos\left(k\frac{2L}{9}\right)\)
04

Write the equation for the first overtone

Since the organ pipe is vibrating in the first overtone, the wave number \(k\) can be substituted with the first overtone's wave number \(k_1\), and the equation for pressure amplitude at various points becomes: \(P_Q = P_m \cos\left(k_1\frac{2L}{9}\right)\)
05

Find the ratio of pressure amplitude

Now we need to find \(\frac{P_Q}{P_m}\). Hence, substituting the given values: \(\frac{P_Q}{P_m} = \frac{\cos\left(k_1\frac{2L}{9}\right)}{1}\) When the organ pipe vibrates in its first overtone, the relation between the wave number and length of the organ pipe is: \(k_1=\frac{3\pi}{2L}\) Substitute \(k_1\) in the equation: \(\frac{P_Q}{P_m} = \cos\left(\frac{3\pi}{2L} \cdot \frac{2L}{9}\right) = \cos\left(\frac{3\pi}{9}\right) = \cos(\frac{\pi}{3})\) Using the trigonometric property, we get: \(\frac{P_Q}{P_m} = \cos(\frac{\pi}{3}) = \frac{1}{2}\) Hence, the ratio of pressure amplitudes is \(\frac{1:2}\). Therefore, the correct answer is (A) \(1: 2\).

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