The fundamental frequency of an open organi pipe is 330 Hz. The first overtone of a closed organ pipe has the same frequency as the first overtone of the open pipe. How long is each pipe 2 Velocity of sound =330 m/s.

Answer

To solve the problem, we need to find the lengths of an open organ pipe and a closed organ pipe given their fundamental frequencies and the relationship between their overtones.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Fundamental frequency of the open organ pipe, \( f_0 = 330 \, \text{Hz} \)
- Velocity of sound, \( V = 330 \, \text{m/s} \)

2. **Determine the Length of the Open Organ Pipe:**
The fundamental frequency of an open organ pipe is given by the formula:
\[ f_0 = \frac{V}{2L_1} \]
Rearranging this to find \( L_1 \):
\[ L_1 = \frac{V}{2f_0} \]
Substituting the known values:
\[ L_1 = \frac{330 \, \text{m/s}}{2 \times 330 \, \text{Hz}} = \frac{330}{660} = 0.5 \, \text{m} \]

3. **Determine the First Overtone of the Open Organ Pipe:**
The first overtone of an open pipe is given by:
\[ f_1 = 2f_0 = 2 \times 330 \, \text{Hz} = 660 \, \text{Hz} \]

4. **Relate the First Overtone of the Closed Organ Pipe to the Open Pipe:**
The frequency of the first overtone of a closed organ pipe is given by:
\[ f' = \frac{3V}{4L_2} \]
Since the first overtone of the closed pipe has the same frequency as the first overtone of the open pipe:
\[ \frac{3V}{4L_2} = 660 \, \text{Hz} \]

5. **Rearranging to Find Length \( L_2 \):**
Rearranging the equation to find \( L_2 \):
\[ L_2 = \frac{3V}{4f'} \]
Substituting \( f' = 660 \, \text{Hz} \):
\[ L_2 = \frac{3 \times 330 \, \text{m/s}}{4 \times 660 \, \text{Hz}} = \frac{990}{2640} = 0.375 \, \text{m} \]

6. **Final Results:**
- Length of the open organ pipe, \( L_1 = 0.5 \, \text{m} \)
- Length of the closed organ pipe, \( L_2 = 0.375 \, \text{m} \)

### Summary:
- Length of the open organ pipe \( L_1 = 0.5 \, \text{m} \)
- Length of the closed organ pipe \( L_2 = 0.375 \, \text{m} \)

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