A closed organ pipe and an open organ pipe of same length produce `4` beats when they are set into vibrations simultaneously. If the length of each of them were twice their initial lengths, the number of beats produced will be

Options

A. `2`
B. `4`
C. `1`
D. `8`

Explanation

To solve the problem, we need to analyze the frequencies of the closed and open organ pipes and how they change when the lengths are doubled.

### Step-by-step Solution:

1. **Identify the Frequencies of the Pipes:**
- For a closed organ pipe, the fundamental frequency \( f_1 \) is given by:
\[ f_1 = \frac{v}{4L} \]
- For an open organ pipe, the fundamental frequency \( f_2 \) is given by:
\[ f_2 = \frac{v}{2L} \]

2. **Calculate the Initial Beat Frequency:**
- The beat frequency \( f_b \) is the difference between the two frequencies:
\[ f_b = f_2 - f_1 = \frac{v}{2L} - \frac{v}{4L} \]
- To simplify, find a common denominator:
\[ f_b = \frac{2v}{4L} - \frac{v}{4L} = \frac{v}{4L} \]
- According to the problem, this beat frequency is given as 4 beats per second:
\[ \frac{v}{4L} = 4 \implies v = 16L \]

3. **Determine the New Frequencies When Length is Doubled:**
- If the length of each pipe is doubled, the new length \( L' \) is:
\[ L' = 2L \]
- The new frequency for the closed organ pipe \( f_1' \) becomes:
\[ f_1' = \frac{v}{4L'} = \frac{v}{4(2L)} = \frac{v}{8L} \]
- The new frequency for the open organ pipe \( f_2' \) becomes:
\[ f_2' = \frac{v}{2L'} = \frac{v}{2(2L)} = \frac{v}{4L} \]

4. **Calculate the New Beat Frequency:**
- The new beat frequency \( f_b' \) is:
\[ f_b' = f_2' - f_1' = \frac{v}{4L} - \frac{v}{8L} \]
- Again, find a common denominator:
\[ f_b' = \frac{2v}{8L} - \frac{v}{8L} = \frac{v}{8L} \]
- Since we previously found \( \frac{v}{4L} = 4 \):
\[ \frac{v}{8L} = \frac{1}{2} \cdot \frac{v}{4L} = \frac{1}{2} \cdot 4 = 2 \]

5. **Final Answer:**
- Therefore, the number of beats produced when the lengths are doubled is:
\[ \text{Number of beats} = 2 \]