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 will follow these steps:

### Step 1: Understand the properties of closed and open organ pipes
- A closed organ pipe has one end closed and one end open. The fundamental frequency (first harmonic) of a closed organ pipe is given by:
\[ f_c = \frac{v}{4L} \]
where \( v \) is the speed of sound and \( L \) is the length of the pipe.

- An open organ pipe has both ends open. The first overtone (second harmonic) frequency of an open organ pipe is given by:
\[ f_o = \frac{v}{L'} \]
where \( L' \) is the length of the open pipe.

### Step 2: Set up the equations based on the problem statement
- Let the lengths of the two pieces after cutting the closed organ pipe be \( x \) and \( y \) such that \( x + y = 100 \) cm.
- According to the problem, the fundamental frequency of the closed pipe (length \( x \)) is equal to the first overtone frequency of the open pipe (length \( y \)):
\[ f_c = f_o \]

### Step 3: Write the equations for the frequencies
- For the closed organ pipe of length \( x \):
\[ f_c = \frac{320}{4x} = \frac{80}{x} \]

- For the open organ pipe of length \( y \):
\[ f_o = \frac{320}{y} \]

### Step 4: Set the frequencies equal to each other
- From the previous equations, we have:
\[ \frac{80}{x} = \frac{320}{y} \]

### Step 5: Cross-multiply and simplify
- Cross-multiplying gives:
\[ 80y = 320x \]
Simplifying this, we find:
\[ y = 4x \]

### Step 6: Substitute \( y \) in the length equation
- We know from the problem that:
\[ x + y = 100 \]
Substituting \( y = 4x \) into this equation gives:
\[ x + 4x = 100 \]
\[ 5x = 100 \]
\[ x = 20 \text{ cm} \]

### Step 7: Find \( y \)
- Now substituting \( x \) back to find \( y \):
\[ y = 4x = 4 \times 20 = 80 \text{ cm} \]

### Step 8: Calculate the fundamental frequency of the open pipe
- Now, we can find the fundamental frequency of the open pipe using its length \( y = 80 \) cm:
\[ f_o = \frac{320}{80} = 4 \text{ Hz} = 400 \text{ Hz} \]

### Final Result
- The lengths of the two pieces are:
- \( x = 20 \) cm (closed pipe)
- \( y = 80 \) cm (open pipe)
- The fundamental tone of the open pipe piece is:
- \( f_o = 400 \) Hz