A closed organ pipe and an open organ pipe of some length produce ` 2 beats ` when they are set up into vibration simultaneously in their fundamental mode . The length of the open organ pipe is now halved and of the closed organ pipe is doubled , the number of beats produced will be a) 7 b) 4 c) 8 d) 2

Options

A. 7
B. 4
C. 8
D. 2

Answer

To solve the problem, we need to determine the frequencies of the closed and open organ pipes before and after the changes in their lengths, and then calculate the number of beats produced.

### Step 1: Understand the initial conditions
We know that a closed organ pipe and an open organ pipe produce 2 beats when they vibrate simultaneously in their fundamental mode. The beat frequency is given by the difference in their frequencies.

### Step 2: Write the formulas for frequencies
For an open organ pipe, the frequency \( f_{\text{open}} \) is given by:
\[ f_{\text{open}} = \frac{v}{2L_{\text{open}}} \]
For a closed organ pipe, the frequency \( f_{\text{closed}} \) is given by:
\[ f_{\text{closed}} = \frac{v}{4L_{\text{closed}}} \]

### Step 3: Set up the initial equations
Let the initial lengths of the open and closed pipes be \( L_{\text{open}} \) and \( L_{\text{closed}} \) respectively. The condition for the beats is:
\[ f_{\text{open}} - f_{\text{closed}} = 2 \]
Substituting the frequency equations:
\[ \frac{v}{2L_{\text{open}}} - \frac{v}{4L_{\text{closed}}} = 2 \]

### Step 4: Simplify the equation
To simplify, we can find a common denominator:
\[ \frac{2v}{4L_{\text{open}}} - \frac{v}{4L_{\text{closed}}} = 2 \]
This leads to:
\[ \frac{2vL_{\text{closed}} - vL_{\text{open}}}{4L_{\text{open}}L_{\text{closed}}} = 2 \]
Multiplying both sides by \( 4L_{\text{open}}L_{\text{closed}} \):
\[ 2vL_{\text{closed}} - vL_{\text{open}} = 8L_{\text{open}}L_{\text{closed}} \]

### Step 5: Analyze the changes in lengths
Now, we change the lengths:
- The length of the open organ pipe is halved: \( L'_{\text{open}} = \frac{L_{\text{open}}}{2} \)
- The length of the closed organ pipe is doubled: \( L'_{\text{closed}} = 2L_{\text{closed}} \)

### Step 6: Calculate the new frequencies
The new frequency for the open pipe becomes:
\[ f'_{\text{open}} = \frac{v}{2L'_{\text{open}}} = \frac{v}{2 \cdot \frac{L_{\text{open}}}{2}} = \frac{v}{L_{\text{open}}} \]
The new frequency for the closed pipe becomes:
\[ f'_{\text{closed}} = \frac{v}{4L'_{\text{closed}}} = \frac{v}{4 \cdot 2L_{\text{closed}}} = \frac{v}{8L_{\text{closed}}} \]

### Step 7: Calculate the new beat frequency
Now we calculate the new beat frequency:
\[ \text{New beats} = f'_{\text{open}} - f'_{\text{closed}} = \frac{v}{L_{\text{open}}} - \frac{v}{8L_{\text{closed}}} \]
Finding a common denominator:
\[ \text{New beats} = \frac{8v}{8L_{\text{open}}} - \frac{v}{8L_{\text{closed}}} = \frac{8vL_{\text{closed}} - vL_{\text{open}}}{8L_{\text{open}}L_{\text{closed}}} \]

### Step 8: Substitute the earlier relation
From our earlier relation, we know:
\[ 2vL_{\text{closed}} - vL_{\text{open}} = 8L_{\text{open}}L_{\text{closed}} \]
Thus:
\[ 8vL_{\text{closed}} - vL_{\text{open}} = 4(2vL_{\text{closed}} - vL_{\text{open}}) = 4 \times 8L_{\text{open}}L_{\text{closed}} = 32L_{\text{open}}L_{\text{closed}} \]

### Step 9: Final calculation
Now we can substitute back into our beat frequency equation:
\[ \text{New beats} = \frac{32L_{\text{open}}L_{\text{closed}}}{8L_{\text{open}}L_{\text{closed}}} = 4 \]

### Conclusion
Thus, the number of beats produced after the changes in lengths is **4**.