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A closed organ pipe has length L. The ai...

A closed organ pipe has length `L`. The air in it is vibrating in third overtone with maximum amplitude a . The amplitude at distance`(L)/(7)` from closed of the pipe is a) 0 b) a c) a/2 d) data insufficient

A

`0`

B

`a`

C

`(a)/(2)`

D

Date insufficient

Text Solution

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The correct Answer is:
To solve the problem, we need to determine the amplitude of the sound wave at a distance \( \frac{L}{7} \) from the closed end of a closed organ pipe when it is vibrating in its third overtone. ### Step-by-Step Solution: 1. **Understanding the Closed Organ Pipe**: A closed organ pipe has one end closed and the other end open. The fundamental frequency and overtones can be determined based on the length of the pipe. 2. **Wavelength Calculation**: For a closed organ pipe, the relationship between the length \( L \) of the pipe and the wavelength \( \lambda \) is given by: \[ L = \frac{n \lambda}{4} \] where \( n \) is the harmonic number. For the third overtone, \( n = 7 \) (since the first overtone is the second harmonic, the second overtone is the third harmonic, and so on). Thus, we have: \[ L = \frac{7 \lambda}{4} \] From this, we can find the wavelength: \[ \lambda = \frac{4L}{7} \] 3. **Wave Number Calculation**: The wave number \( k \) is defined as: \[ k = \frac{2\pi}{\lambda} \] Substituting the value of \( \lambda \): \[ k = \frac{2\pi}{\frac{4L}{7}} = \frac{7\pi}{2L} \] 4. **Wave Equation**: The wave equation for a standing wave can be expressed as: \[ y(x) = A \sin(kx) \] where \( A \) is the maximum amplitude. For the third overtone, we substitute \( k \): \[ y(x) = A \sin\left(\frac{7\pi}{2L} x\right) \] 5. **Amplitude at \( \frac{L}{7} \)**: We need to find the amplitude at a distance \( x = \frac{L}{7} \): \[ y\left(\frac{L}{7}\right) = A \sin\left(\frac{7\pi}{2L} \cdot \frac{L}{7}\right) \] Simplifying this: \[ y\left(\frac{L}{7}\right) = A \sin\left(\frac{7\pi}{2L} \cdot \frac{L}{7}\right) = A \sin\left(\frac{\pi}{2}\right) = A \] 6. **Final Answer**: The amplitude at a distance \( \frac{L}{7} \) from the closed end of the pipe is \( A \). Thus, the correct answer is **(b) A**.

To solve the problem, we need to determine the amplitude of the sound wave at a distance \( \frac{L}{7} \) from the closed end of a closed organ pipe when it is vibrating in its third overtone. ### Step-by-Step Solution: 1. **Understanding the Closed Organ Pipe**: A closed organ pipe has one end closed and the other end open. The fundamental frequency and overtones can be determined based on the length of the pipe. 2. **Wavelength Calculation**: ...
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