The fundamental frequency of a closed organ pipe is same as the first overtone frequency of an open pipe. If the length of open pipe is `50 cm`, the length of closed pipe is

Options

A. `25 cm`
B. `12.5 cm`
C. `100 cm`
D. `200 cm`

Explanation

To solve the problem, we need to find the length of the closed organ pipe (L1) given that its fundamental frequency is the same as the first overtone frequency of an open pipe, which has a length of 50 cm.

### Step-by-Step Solution:

1. **Understanding the Frequencies**:
- The fundamental frequency of a closed organ pipe is given by the formula:
\[ f_{closed} = \frac{V}{4L_1} \]
where \( L_1 \) is the length of the closed pipe and \( V \) is the speed of sound in air.
- The first overtone frequency of an open pipe is given by:
\[ f_{open} = \frac{V}{L_2} \]
where \( L_2 \) is the length of the open pipe.

2. **Given Information**:
- Length of the open pipe \( L_2 = 50 \, \text{cm} \).
- The fundamental frequency of the closed pipe is equal to the first overtone frequency of the open pipe:
\[ f_{closed} = f_{open} \]

3. **Setting Up the Equation**:
- From the equations for frequency, we have:
\[ \frac{V}{4L_1} = \frac{V}{L_2} \]

4. **Canceling the Speed of Sound**:
- Since \( V \) is the same for both pipes, we can cancel \( V \) from both sides:
\[ \frac{1}{4L_1} = \frac{1}{L_2} \]

5. **Substituting the Length of the Open Pipe**:
- Substitute \( L_2 = 50 \, \text{cm} \):
\[ \frac{1}{4L_1} = \frac{1}{50} \]

6. **Cross-Multiplying to Solve for \( L_1 \)**:
- Cross-multiplying gives:
\[ 50 = 4L_1 \]

7. **Solving for \( L_1 \)**:
- Dividing both sides by 4:
\[ L_1 = \frac{50}{4} = 12.5 \, \text{cm} \]

### Final Answer:
The length of the closed organ pipe is \( L_1 = 12.5 \, \text{cm} \).

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