A closed organ pipe and an open organ pipe of same length produce 2beats when they are set into viberations simultaneously in their fundamental mode. The length of open organ pipe is now halved and of closed organ pipe is doubled, the nunber of beats produced wil be

Options

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

Answer

To solve the problem step by step, we will analyze the frequencies of both the closed and open organ pipes and how they change when their lengths are modified.

### Step 1: Understand the fundamental frequencies of the pipes
1. **Open Organ Pipe**: The fundamental frequency \( f_O \) is given by the formula:
\[ f_O = \frac{v}{2L} \]
where \( v \) is the speed of sound in air and \( L \) is the length of the open organ pipe.

2. **Closed Organ Pipe**: The fundamental frequency \( f_C \) is given by the formula:
\[ f_C = \frac{v}{4L} \]
where \( L \) is the length of the closed organ pipe.

### Step 2: Set up the equation for beats
According to the problem, the two pipes produce 2 beats when they vibrate simultaneously:
\[ |f_O - f_C| = 2 \]

### Step 3: Substitute the frequencies into the beat equation
Substituting the expressions for \( f_O \) and \( f_C \):
\[ \left| \frac{v}{2L} - \frac{v}{4L} \right| = 2 \]
Factoring out \( \frac{v}{4L} \):
\[ \left| \frac{v}{4L} \left(2 - 1\right) \right| = 2 \]
This simplifies to:
\[ \frac{v}{4L} = 2 \implies v = 8L \]

### Step 4: Modify the lengths of the pipes
1. The length of the open organ pipe is halved:
\[ L' = \frac{L}{2} \]

2. The length of the closed organ pipe is doubled:
\[ L'' = 2L \]

### Step 5: Calculate the new frequencies
1. **New Open Organ Pipe Frequency** \( f_O' \):
\[ f_O' = \frac{v}{2L'} = \frac{v}{2 \cdot \frac{L}{2}} = \frac{v}{L} \]

2. **New Closed Organ Pipe Frequency** \( f_C' \):
\[ f_C' = \frac{v}{4L''} = \frac{v}{4 \cdot 2L} = \frac{v}{8L} \]

### Step 6: Set up the new beat equation
Now we find the new number of beats produced:
\[ |f_O' - f_C'| = \left| \frac{v}{L} - \frac{v}{8L} \right| \]
Factoring out \( \frac{v}{L} \):
\[ = \left| \frac{v}{L} \left(1 - \frac{1}{8}\right) \right| = \left| \frac{v}{L} \cdot \frac{7}{8} \right| \]

### Step 7: Substitute the value of \( v \)
From Step 3, we know \( v = 8L \):
\[ = \left| \frac{8L}{L} \cdot \frac{7}{8} \right| = 7 \]

### Conclusion
The number of beats produced when the lengths of the pipes are modified is **7**.