A closed organ pipe has length `L`. The air in it is vibrating in third overtone with maximum amplitude. The amplitude at distance`(L)/(7)` from closed of the pipe is

Options

A. `0`
B. `a`
C. `(a)/(2)`
D. Date insufficient

Explanation

To solve the problem step by step, we will analyze the behavior of the closed organ pipe vibrating in its third overtone and determine the amplitude at a specific point along the length of the pipe.

### Step 1: Understand the closed organ pipe and its harmonics
A closed organ pipe has one end closed and one end open. The harmonics for a closed pipe are given by the formula:
- Fundamental frequency (1st harmonic): \( \frac{1}{4} \lambda \)
- 1st overtone (2nd harmonic): \( \frac{3}{4} \lambda \)
- 2nd overtone (3rd harmonic): \( \frac{5}{4} \lambda \)
- 3rd overtone (4th harmonic): \( \frac{7}{4} \lambda \)

For the third overtone, the length of the pipe \( L \) is related to the wavelength \( \lambda \) by:
\[ L = \frac{7}{4} \lambda \]

### Step 2: Calculate the wavelength \( \lambda \)
From the equation above, we can express the wavelength \( \lambda \):
\[ \lambda = \frac{4L}{7} \]

### Step 3: Determine the wave number \( k \)
The wave number \( k \) is defined as:
\[ k = \frac{2\pi}{\lambda} \]
Substituting the value of \( \lambda \):
\[ k = \frac{2\pi}{\frac{4L}{7}} = \frac{7\pi}{2L} \]

### Step 4: Amplitude variation along the pipe
The amplitude of the wave at a distance \( x \) from the closed end can be expressed as:
\[ A(x) = A \sin(kx) \]
where \( A \) is the maximum amplitude.

### Step 5: Calculate the amplitude at \( x = \frac{L}{7} \)
Now we need to find the amplitude at a distance \( \frac{L}{7} \) from the closed end:
\[ A\left(\frac{L}{7}\right) = A \sin\left(k \cdot \frac{L}{7}\right) \]
Substituting the value of \( k \):
\[ A\left(\frac{L}{7}\right) = A \sin\left(\frac{7\pi}{2L} \cdot \frac{L}{7}\right) \]
This simplifies to:
\[ A\left(\frac{L}{7}\right) = A \sin\left(\frac{\pi}{2}\right) \]

### Step 6: Evaluate the sine function
Since \( \sin\left(\frac{\pi}{2}\right) = 1 \):
\[ A\left(\frac{L}{7}\right) = A \cdot 1 = A \]

### Final Answer
The amplitude at a distance \( \frac{L}{7} \) from the closed end of the pipe is \( A \).

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