The frequency of `2^(nd)` overtone of an open organ pipe is equal to frequency of first harmonic of a closed organ pipe. If the length of open pipe is 80 cm, then find the length of closed pipe.

Answer

To solve the problem, we need to relate the frequencies of the open and closed organ pipes using their respective formulas. Let's break it down step by step.

### Step 1: Understand the Harmonics
The second overtone of an open organ pipe corresponds to the third harmonic. The frequency of the nth harmonic for an open organ pipe is given by:
\[ f_n = \frac{nV}{2L} \]
where \( n \) is the harmonic number, \( V \) is the speed of sound, and \( L \) is the length of the pipe.

### Step 2: Write the Frequency for the Open Pipe
For the open organ pipe (where \( L_0 = 80 \, \text{cm} = 0.8 \, \text{m} \)):
\[ f_3 = \frac{3V}{2L_0} = \frac{3V}{2 \times 0.8} \]

### Step 3: Write the Frequency for the Closed Pipe
The frequency of the first harmonic for a closed organ pipe is given by:
\[ f_1 = \frac{V}{4L_C} \]
where \( L_C \) is the length of the closed pipe.

### Step 4: Set the Frequencies Equal
According to the problem, the frequency of the second overtone of the open pipe is equal to the first harmonic of the closed pipe:
\[ f_3 = f_1 \]
Substituting the expressions we derived:
\[ \frac{3V}{2L_0} = \frac{V}{4L_C} \]

### Step 5: Cancel Out the Speed of Sound
We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{3}{2L_0} = \frac{1}{4L_C} \]

### Step 6: Cross Multiply
Cross multiplying gives:
\[ 3 \cdot 4L_C = 2L_0 \]
\[ 12L_C = 2L_0 \]

### Step 7: Solve for \( L_C \)
Now, we can isolate \( L_C \):
\[ L_C = \frac{2L_0}{12} = \frac{L_0}{6} \]

### Step 8: Substitute the Value of \( L_0 \)
Substituting \( L_0 = 80 \, \text{cm} \):
\[ L_C = \frac{80}{6} \approx 13.33 \, \text{cm} \]

### Final Answer
The length of the closed organ pipe is approximately \( 13.33 \, \text{cm} \).

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