A closed pipe of length 10 cm has its fundamental frequency half that of the second overtone of an open pipe . The length of the open pipe .

Options

A. 10 cm
B. 20 cm
C. 30 cm
D. 40 cm

Answer

To solve the problem, we will follow these steps systematically:

### Step 1: Understand the fundamental frequency of a closed pipe
For a closed pipe, the fundamental frequency \( f_{c0} \) is given by the formula:
\[ f_{c0} = \frac{v}{4L} \]
where \( v \) is the speed of sound in air and \( L \) is the length of the pipe.

### Step 2: Substitute the length of the closed pipe
Given that the length of the closed pipe \( L = 10 \) cm = 0.1 m, we can substitute this value into the formula:
\[ f_{c0} = \frac{v}{4 \times 0.1} = \frac{v}{0.4} \]

### Step 3: Understand the second overtone of an open pipe
For an open pipe, the frequency of the nth overtone is given by:
\[ f_{n} = \frac{n v}{2L} \]
For the second overtone, \( n = 3 \):
\[ f_{o3} = \frac{3v}{2L} \]

### Step 4: Relate the frequencies
According to the problem, the fundamental frequency of the closed pipe is half that of the second overtone of the open pipe:
\[ f_{c0} = \frac{1}{2} f_{o3} \]
Substituting the expressions we derived:
\[ \frac{v}{0.4} = \frac{1}{2} \left( \frac{3v}{2L} \right) \]

### Step 5: Simplify the equation
Cancel \( v \) from both sides (assuming \( v \neq 0 \)):
\[ \frac{1}{0.4} = \frac{3}{4L} \]

### Step 6: Cross-multiply to solve for \( L \)
Cross-multiplying gives:
\[ 4L = 3 \times 0.4 \]
\[ 4L = 1.2 \]

### Step 7: Solve for \( L \)
Dividing both sides by 4:
\[ L = \frac{1.2}{4} = 0.3 \text{ m} = 30 \text{ cm} \]

### Conclusion
The length of the open pipe is **30 cm**.

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