A closed organ pipe of length `l = 100 cm` is cut into two unequal pieces. The fundamental frequency of the new closed organ pipe place is found to be same as the frequency of first overtone of the open organ pipe place. Datermine the length of the two piece and the fundamental tone of the open pipe piece. Take velocity of sound `= 320 m//s`.

Answer

To solve the problem, we need to determine the lengths of the two pieces of the closed organ pipe and the fundamental frequency of the open organ pipe piece.

### Step-by-Step Solution:

1. **Identify the Lengths of the Pipes**:
Let the length of the closed organ pipe piece be \( L_C \) and the length of the open organ pipe piece be \( L_O \). Given that the total length of the original closed organ pipe is \( L = 100 \, \text{cm} \), we can write:
\[ L_C + L_O = 100 \, \text{cm} \]

2. **Fundamental Frequency of Closed Organ Pipe**:
The fundamental frequency \( f_C \) of a closed organ pipe is given by the formula:
\[ f_C = \frac{V}{4L_C} \]
where \( V \) is the velocity of sound (given as \( 320 \, \text{m/s} \)).

3. **First Overtone of Open Organ Pipe**:
The first overtone frequency \( f_O \) of an open organ pipe is given by:
\[ f_O = \frac{2V}{L_O} \]

4. **Setting the Frequencies Equal**:
According to the problem, the fundamental frequency of the closed organ pipe is equal to the first overtone frequency of the open organ pipe:
\[ \frac{V}{4L_C} = \frac{2V}{L_O} \]

5. **Canceling \( V \)**:
Since \( V \) is common in both equations, we can cancel it out (assuming \( V \neq 0 \)):
\[ \frac{1}{4L_C} = \frac{2}{L_O} \]

6. **Cross Multiplying**:
Cross multiplying gives us:
\[ L_O = 8L_C \]

7. **Substituting into the Length Equation**:
Now, substitute \( L_O \) in the total length equation:
\[ L_C + 8L_C = 100 \]
\[ 9L_C = 100 \]
\[ L_C = \frac{100}{9} \approx 11.11 \, \text{cm} \]

8. **Finding \( L_O \)**:
Now, substitute \( L_C \) back to find \( L_O \):
\[ L_O = 100 - L_C = 100 - \frac{100}{9} = \frac{800}{9} \approx 88.89 \, \text{cm} \]

9. **Calculating the Fundamental Frequency of the Open Pipe**:
The fundamental frequency \( f_O \) of the open organ pipe can be calculated using:
\[ f_O = \frac{V}{2L_O} \]
Substituting \( V = 320 \, \text{m/s} \) and \( L_O = \frac{800}{9} \, \text{cm} = \frac{800}{900} \, \text{m} = \frac{8}{9} \, \text{m} \):
\[ f_O = \frac{320}{2 \times \frac{8}{9}} = \frac{320 \times 9}{16} = 180 \, \text{Hz} \]

### Final Results:
- Length of the closed organ pipe piece \( L_C \approx 11.11 \, \text{cm} \)
- Length of the open organ pipe piece \( L_O \approx 88.89 \, \text{cm} \)
- Fundamental frequency of the open organ pipe \( f_O = 180 \, \text{Hz} \)