The fundamental frequency of a closed organ pipe is equal to frequency of `2^(nd)` overtone of an open organ pipe. What will be the length of open pipe if length of closed pipe is 20 cm ?

Options

A. 40 cm
B. 80 cm
C. 120 cm
D. 100 cm

Explanation

To solve the problem, we need to find the length of an open organ pipe given that the fundamental frequency of a closed organ pipe is equal to the frequency of the second overtone of the open organ pipe. Let's break it down step by step.

### Step 1: Understand the Frequencies
For a closed organ pipe, the fundamental frequency (first harmonic) is given by the formula:
\[ f_c = \frac{V}{4L_c} \]
where \( V \) is the speed of sound in air and \( L_c \) is the length of the closed organ pipe.

For an open organ pipe, the frequency of the nth harmonic is given by:
\[ f_o = \frac{nV}{2L_o} \]
where \( L_o \) is the length of the open organ pipe and \( n \) is the harmonic number. The second overtone corresponds to the third harmonic (n=3).

### Step 2: Set the Frequencies Equal
According to the problem, the fundamental frequency of the closed pipe is equal to the frequency of the second overtone of the open pipe:
\[ f_c = f_o \]
This translates to:
\[ \frac{V}{4L_c} = \frac{3V}{2L_o} \]

### Step 3: Simplify the Equation
We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{1}{4L_c} = \frac{3}{2L_o} \]

### Step 4: Cross Multiply
Cross multiplying gives:
\[ 2L_o = 12L_c \]
or
\[ L_o = 6L_c \]

### Step 5: Substitute the Length of the Closed Pipe
We know that the length of the closed pipe \( L_c = 20 \) cm. Substituting this value in:
\[ L_o = 6 \times 20 \]
\[ L_o = 120 \text{ cm} \]

### Final Answer
The length of the open organ pipe is \( 120 \) cm.

---