The fundamental frequency of a closed organ pipe of length `20 cm` is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is

Options

A. 80cm
B. 100 cm
C. 120cm
D. 140 cm

Answer

To solve the problem step by step, we will use the formulas for the frequencies of closed and open organ pipes.

### Step 1: Identify the given values
We know:
- Length of the closed organ pipe (Lc) = 20 cm
- The fundamental frequency of the closed organ pipe is equal to the second overtone frequency of the open organ pipe.

### Step 2: Write the formula for the fundamental frequency of a closed organ pipe
The fundamental frequency (n1) of a closed organ pipe is given by the formula:
\[ n_1 = \frac{v}{4L_c} \]
where:
- \( v \) = speed of sound in air
- \( L_c \) = length of the closed organ pipe

### Step 3: Write the formula for the second overtone of an open organ pipe
The second overtone (n2) for an open organ pipe is given by:
\[ n_2 = \frac{3v}{2L_0} \]
where:
- \( L_0 \) = length of the open organ pipe

### Step 4: Set the two frequencies equal
Since the fundamental frequency of the closed organ pipe is equal to the second overtone of the open organ pipe, we can write:
\[ n_1 = n_2 \]
Thus,
\[ \frac{v}{4L_c} = \frac{3v}{2L_0} \]

### Step 5: Simplify the equation
We can cancel \( v \) from both sides (assuming \( v \neq 0 \)):
\[ \frac{1}{4L_c} = \frac{3}{2L_0} \]

### Step 6: Cross-multiply to solve for \( L_0 \)
Cross-multiplying gives us:
\[ 2L_0 = 12L_c \]
Now, substituting \( L_c = 20 \) cm:
\[ 2L_0 = 12 \times 20 \]
\[ 2L_0 = 240 \]

### Step 7: Solve for \( L_0 \)
Dividing both sides by 2:
\[ L_0 = \frac{240}{2} = 120 \text{ cm} \]

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

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