The frequency of first overtone of an open organ pipe is equal to fundamanetal frequency of a closed organ pipe. What will be the length of closed organ pipe if the length of open organ pipe is 16 cm ?
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Explanation

To solve the problem, we need to find the length of a closed organ pipe given that the frequency of the first overtone of an open organ pipe is equal to the fundamental frequency of the closed organ pipe. We know the length of the open organ pipe is 16 cm.

### Step-by-Step Solution:

1. **Understand the Frequencies**:
- The first overtone of an open organ pipe corresponds to the second harmonic. The fundamental frequency (first harmonic) of an open organ pipe is given by:
\[ f_1 = \frac{v}{2L_0} \]
- The frequency of the first overtone (second harmonic) is:
\[ f_2 = 2f_1 = 2 \left(\frac{v}{2L_0}\right) = \frac{v}{L_0} \]

2. **Fundamental Frequency of Closed Organ Pipe**:
- For a closed organ pipe, the fundamental frequency is given by:
\[ f_c = \frac{v}{4L_c} \]

3. **Set the Frequencies Equal**:
- According to the problem, the frequency of the first overtone of the open organ pipe is equal to the fundamental frequency of the closed organ pipe:
\[ f_2 = f_c \]
- Substituting the expressions for frequencies:
\[ \frac{v}{L_0} = \frac{v}{4L_c} \]

4. **Cancel the Velocity**:
- Since the speed of sound \( v \) is common in both equations, we can cancel it out (assuming \( v \neq 0 \)):
\[ \frac{1}{L_0} = \frac{1}{4L_c} \]

5. **Rearranging the Equation**:
- Cross-multiplying gives:
\[ 4L_c = L_0 \]

6. **Substituting the Length of Open Organ Pipe**:
- We know that \( L_0 = 16 \, \text{cm} \):
\[ 4L_c = 16 \, \text{cm} \]

7. **Solving for \( L_c \)**:
- Dividing both sides by 4:
\[ L_c = \frac{16 \, \text{cm}}{4} = 4 \, \text{cm} \]

### Final Answer:
The length of the closed organ pipe \( L_c \) is **4 cm**.

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