The fundamental frequency of a closed organ pipe is 'n' . If its length is doubled then frequency will become (neglecting end correction)

Options

A. `(n)/(2)`
B. `(n)/(3)`
C. `n`
D. `2n`

Answer

To solve the problem, we need to understand the relationship between the length of a closed organ pipe and its fundamental frequency. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the fundamental frequency of a closed organ pipe
The fundamental frequency (f) of a closed organ pipe is given by the formula:
\[ f = \frac{V}{4L} \]
where:
- \( V \) is the speed of sound in air,
- \( L \) is the length of the pipe.

### Step 2: Identify the initial conditions
Let’s denote the initial length of the pipe as \( L \) and the initial fundamental frequency as \( n \). Therefore, we can write:
\[ n = \frac{V}{4L} \]

### Step 3: Change the length of the pipe
According to the problem, the length of the pipe is doubled. So the new length \( L' \) becomes:
\[ L' = 2L \]

### Step 4: Calculate the new frequency
Now, we need to find the new fundamental frequency \( f' \) with the new length:
\[ f' = \frac{V}{4L'} = \frac{V}{4(2L)} = \frac{V}{8L} \]

### Step 5: Relate the new frequency to the initial frequency
We can express the new frequency \( f' \) in terms of the initial frequency \( n \):
\[ f' = \frac{V}{8L} = \frac{1}{2} \cdot \frac{V}{4L} = \frac{1}{2}n \]

### Conclusion
Thus, if the length of the closed organ pipe is doubled, the new fundamental frequency becomes:
\[ f' = \frac{n}{2} \]

### Final Answer
The frequency will become \( \frac{n}{2} \).

---