The fundamental frequency of a closed organ pipe is sam eas the first overtone freuency of an open pipe. If the length of open pipe is 50cm, the length of closed pipe is

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understand the Frequencies**:
- For a closed organ pipe, the fundamental frequency (first harmonic) is given by:
\[ f_1 = \frac{V}{4L} \]
where \( V \) is the speed of sound in air and \( L \) is the length of the closed pipe.
- For an open pipe, the first overtone (second harmonic) frequency is given by:
\[ f_2 = \frac{2V}{L_o} \]
where \( L_o \) is the length of the open pipe.

2. **Set the Frequencies Equal**:
- According to the problem, the fundamental frequency of the closed pipe is equal to the first overtone frequency of the open pipe:
\[ f_1 = f_2 \]

3. **Substitute the Frequency Formulas**:
- From the equations above, we can write:
\[ \frac{V}{4L} = \frac{2V}{L_o} \]

4. **Cancel out \( V \)**:
- Since \( V \) is common on both sides, we can cancel it out:
\[ \frac{1}{4L} = \frac{2}{L_o} \]

5. **Cross Multiply**:
- Cross multiplying gives us:
\[ L_o = 8L \]

6. **Substitute the Length of the Open Pipe**:
- We know that \( L_o = 50 \) cm, so we can substitute this value:
\[ 50 = 8L \]

7. **Solve for \( L \)**:
- Rearranging gives us:
\[ L = \frac{50}{8} = 6.25 \text{ cm} \]

8. **Final Answer**:
- Therefore, the length of the closed organ pipe is:
\[ L = 6.25 \text{ cm} \]