The length of two open organ pipes are `l` and `(l+deltal)` respectively. Neglecting end correction, the frequency of beats between them will b approximately.

Options

A. `v/(2l)`
B. `v/(4l)`
C. `(vDeltal)/(2l^(2))`
D. `(vDeltal)/(l)`

Answer

To solve the problem of finding the frequency of beats between two open organ pipes of lengths \( l \) and \( l + \Delta l \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Frequency Formula for Open Organ Pipes**:
The frequency \( n \) of an open organ pipe is given by the formula:
\[ n = \frac{V}{2L} \]
where \( V \) is the speed of sound in air and \( L \) is the length of the pipe.

2. **Calculate the Frequencies of Both Pipes**:
- For the first pipe of length \( l \):
\[ n_1 = \frac{V}{2l} \]
- For the second pipe of length \( l + \Delta l \):
\[ n_2 = \frac{V}{2(l + \Delta l)} \]

3. **Determine the Beat Frequency**:
The beat frequency \( f_{beat} \) is the absolute difference between the two frequencies:
\[ f_{beat} = |n_1 - n_2| \]
Substituting the expressions for \( n_1 \) and \( n_2 \):
\[ f_{beat} = \left| \frac{V}{2l} - \frac{V}{2(l + \Delta l)} \right| \]

4. **Simplify the Expression**:
Factor out \( \frac{V}{2} \):
\[ f_{beat} = \frac{V}{2} \left| \frac{1}{l} - \frac{1}{l + \Delta l} \right| \]
Now, simplify the difference:
\[ \frac{1}{l} - \frac{1}{l + \Delta l} = \frac{(l + \Delta l) - l}{l(l + \Delta l)} = \frac{\Delta l}{l(l + \Delta l)} \]
Thus,
\[ f_{beat} = \frac{V}{2} \cdot \frac{\Delta l}{l(l + \Delta l)} \]

5. **Approximate for Small \( \Delta l \)**:
If \( \Delta l \) is small compared to \( l \), we can approximate \( l + \Delta l \) as \( l \):
\[ f_{beat} \approx \frac{V \Delta l}{2l^2} \]

### Final Result:
The frequency of beats between the two pipes is approximately:
\[ f_{beat} \approx \frac{V \Delta l}{2l^2} \]